Essentials Of Modern Business Statistics With Microsoft Excel [PDF] [6hhaktaoiir0] (2023)

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Microsoft® Office Excel® Functions Function

Description

AVERAGE

Returns the arithmetic mean of a range its arguments.

BINOM.DIST

Returns the individual term binomial distribution probability.

CHISQ.DIST

Returns a probability from the chi-squared distribution.

CHISQ.DIST.RT

Returns the one-tailed probability of the chi-squared distribution.

CHISQ.INV

Returns the inverse of the left-tailed probability of the chi-squared distribution.

CHISQ.TEST

Returns the value from the chi-squared distribution for the statistic and the degrees of freedom.

CONFIDENCE.NORM

Returns the confidence interval for a population mean using the normal distribution.

CORREL

Returns the correlation coefficient between two data sets.

COUNT

Returns the number of cells in the range that contain numbers.

COUNTA

Returns the number of non-blank cells in the range.

COUNTIF

Returns the number of cells in a range that meet the specified criterion.

COVARIANCE.S

Returns the sample covariance.

EXPON.DIST

Returns a probability from the exponential distribution.

F.DIST.RT

Returns the right-tailed probability from the F distribution.

GEOMEAN

Returns the geometric mean of a range of cells.

HYPGEOM.DIST

Returns a probability from the hypergeometric distribution.

MAX

Returns the maximum value of the values in a range of cells.

MMEDIAN

Returns the median value of the values in a range of cells.

MIN

Returns the minimum value of the values in a range of cells.

MODE.SNGL

Returns the most-frequently occurring value in a range of cells.

NORM.S.DIST

Returns a probability from a standard normal distribution.

NORM.S.INV

Inverse of the standard normal distribution.

PERCENTILE.EXC

Returns the specified percentile of the values in a range of cells.

POISSON.DIST

Returns a probability from the poisson distribution.

POWER

Returns the result of a number raised to a power.

QUARTILE.EXC

Returns the specified quartile of the values in a range of cells.

RAND

Returns a real number from the uniform distribution between 0 and 1.

SQRT

Returns the positive square root of its argument.

STDEV.S

Returns the sample standard deviation of the values in a range of cells.

SUM

Returns the sum of the values in a range of cells.

SUMPRODUCT

Returns the sum of the products of the paired elements of the values in two ranges of cells.

T.DIST

Returns a left-tailed probability of the t distribution.

T.INV.2T

Returns the two-tailed inverse of the student's t-distribution.

VAR.S

Returns the sample variance of the values in a range of cells.

Essentials of Modern

Business Statistics

with Microsoft® Office Excel®

David R. Anderson

University of Cincinnati

Dennis J. Sweeney

University of Cincinnati

Thomas A. Williams Rochester Institute of Technology

Jeffrey D. Camm

Wake Forest University

James J. Cochran

University of Alabama

Australia Brazil Mexico Singapore United Kingdom United States

Essentials of Modern Business Statistics, Seventh Edition David R. Anderson, Dennis J. Sweeney, Thomas A. Williams, Jeffrey D. Camm, James J. Cochran Senior Vice President, General Manager, Social Science, Business and Humanities: Erin Joyner Product Director: Michael Schenk Product Team Manager: Joe Sabatino Sr. Product Manager: Aaron Arnsparger Content Developer: Anne Merrill Product Assistant: Renee Schnee Digital Content Designer: Brandon Foltz Sr. Marketing Manager: Nathan Anderson Sr. Content Project Manager: Colleen Farmer

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Printed in the United States of America Print Number: 01 Print Year: 2017

Brief Contents

Preface xvii

Chapter 1 Chapter 2

Data and Statistics 1 Descriptive Statistics: Tabular and Graphical Displays 35 Chapter 3 Descriptive Statistics: Numerical Measures 108 Chapter 4 Introduction to Probability 180 Chapter 5 Discrete Probability Distributions 228 Chapter 6 Continuous Probability Distributions 285 Chapter 7 Sampling and Sampling Distributions 317 Chapter 8 Interval Estimation 363 Chapter 9 Hypothesis Tests 405 Chapter 10 Inference About Means and Proportions with Two Populations 455 Chapter 11 Inferences About Population Variances 502 Chapter 12 Tests of Goodness of Fit, Independence, and Multiple Proportions 530 Chapter 13 Experimental design and Analysis of Variance 564 Chapter 14 Simple Linear Regression 620 Chapter 15 Multiple Regression 706 Appendix A References and Bibliography 762 Appendix B Tables 764 Appendix C Summation Notation 775 Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises (online) 777 Appendix E Microsoft Excel 2016 and Tools for Statistical Analysis 778 Index 786

Contents

Preface xvii

Chapter 1 Data and Statistics 1 Statistics in Practice: Bloomberg Businessweek 2 1.1 Applications in Business and Economics 3 Accounting 3 Finance 4 Marketing 4 Production 4 Economics 4 Information Systems 5 1.2 Data 5 Elements, Variables, and Observations 5 Scales of Measurement 7 Categorical and Quantitative Data 8 Cross-Sectional and Time Series Data 8 1.3 Data Sources 11 Existing Sources 11 Observational Study 12 Experiment 13 Time and Cost Issues 13 Data Acquisition Errors 13 1.4 Descriptive Statistics 14 1.5 Statistical Inference 16 1.6 Statistical Analysis Using Microsoft Excel 17 Data Sets and Excel Worksheets 18 Using Excel for Statistical Analysis 18 1.7 Analytics 21 1.8 Big Data and Data Mining 22 1.9 Ethical Guidelines for Statistical Practice 24 Summary 25 Glossary 26 Supplementary Exercises 27

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Chapter 2 Descriptive Statistics: Tabular and Graphical Displays 35

Statistics in Practice: Colgate-Palmolive Company 36 2.1 Summarizing Data for a Categorical Variable 37 Frequency Distribution 37 Relative Frequency and Percent Frequency Distributions 38 Using Excel to Construct a Frequency Distribution, a Relative Frequency Distribution, and a Percent Frequency Distribution 39 Bar Charts and Pie Charts 40 Using Excel to Construct a Bar Chart and a Pie Chart 42 2.2 Summarizing Data for a Quantitative Variable 48 Frequency Distribution 48 Relative Frequency and Percent Frequency Distributions 50 Using Excel to Construct a Frequency Distribution 50 Dot Plot 52 Histogram 53 Using Excel’s Recommended Charts Tool to Construct a Histogram 55 Cumulative Distributions 57 Stem-and-Leaf Display 58 2.3 Summarizing Data for Two Variables Using Tables 66 Crosstabulation 67 Using Excel’s PivotTable Tool to Construct a Crosstabulation 69 Simpson’s Paradox 71 2.4 Summarizing Data for Two Variables Using Graphical Displays 78 Scatter Diagram and Trendline 78 Using Excel to Construct a Scatter Diagram and a Trendline 80 Side-by-Side and Stacked Bar Charts 81 Using Excel’s Recommended Charts Tool to Construct Side-by-Side and Stacked Bar Charts 83 2.5 Data Visualization: Best Practices in Creating Effective Graphical Displays 88 Creating Effective Graphical Displays 88 Choosing the Type of Graphical Display 89 Data Dashboards 90 Data Visualization in Practice: Cincinnati Zoo and Botanical Garden 92 Summary 94 Glossary 95 Key Formulas 96 Supplementary Exercises 97 Case Problem 1 Pelican Stores 102 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Contents

vii

Case Problem 2 Motion Picture Industry 103 Case Problem 3 Queen City 104 Case Problem 4 Cut-Rate Machining, Inc. 104

Chapter 3 Descriptive Statistics: Numerical Measures 108 Statistics in Practice: Small Fry Design 109 3.1 Measures of Location 110 Mean 110 Median 112 Mode 113 Using Excel to Compute the Mean, Median, and Mode 114 Weighted Mean 115 Geometric Mean 116 Using Excel to Compute the Geometric Mean 118 Percentiles 119 Quartiles 120 Using Excel to Compute Percentiles and Quartiles 121 3.2 Measures of Variability 127 Range 128 Interquartile Range 128 Variance 128 Standard Deviation 130 Using Excel to Compute the Sample Variance and Sample Standard Deviation 131 Coefficient of Variation 132 Using Excel’s Descriptive Statistics Tool 132 3.3 Measures of Distribution Shape, Relative Location, and Detecting Outliers 137 Distribution Shape 137 z-Scores 137 Chebyshev’s Theorem 140 Empirical Rule 140 Detecting Outliers 141 3.4 Five-Number Summaries and Box Plots 144 Five-Number Summary 145 Box Plot 145 Using Excel to Construct a Box Plot 146 Comparative Analysis Using Box Plots 147 Using Excel to Construct a Comparative Analysis Using Box Plots 147 3.5 Measures of Association Between Two Variables 151 Covariance 152 Interpretation of the Covariance 153 Correlation Coefficient 156 Interpretation of the Correlation Coefficient 157 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Using Excel to Compute the Sample Covariance and Sample Correlation Coefficient 158 3.6 Data Dashboards: Adding Numerical Measures to Improve Effectiveness 161 Summary 164 Glossary 165 Key Formulas 166 Supplementary Exercises 168 Case Problem 1 Pelican Stores 173 Case Problem 2 Motion Picture Industry 174 Case Problem 3 Business Schools of Asia-Pacific 175 Case Problem 4 Heavenly Chocolates Website Transactions 177 Case Problem 5 African Elephant Populations 178

Chapter 4 Introduction to Probability 180 Statistics in Practice: National Aeronautics and Space Administration 181 4.1 Experiments, Counting Rules, and Assigning Probabilities 182 Counting Rules, Combinations, and Permutations 183 Assigning Probabilities 187 Probabilities for the KP&L Project 189 4.2 Events and Their Probabilities 192 4.3 Some Basic Relationships of Probability 196 Complement of an Event 196 Addition Law 197 4.4 Conditional Probability 203 Independent Events 206 Multiplication Law 206 4.5 Bayes’ Theorem 211 Tabular Approach 214 Summary 216 Glossary 217 Key Formulas 218 Supplementary Exercises 219 Case Problem 1 Hamilton County Judges 224 Case Problem 2 Rob’s Market 226

Chapter 5 Discrete Probability Distributions 228 Statistics in Practice: Citibank 229 5.1 Random Variables 230 Discrete Random Variables 230 Continuous Random Variables 230 5.2 Developing Discrete Probability Distributions 233 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Contents

5.3 Expected Value and Variance 238 Expected Value 238 Variance 238 Using Excel to Compute the Expected Value, Variance, and Standard Deviation 240 5.4 Bivariate Distributions, Covariance, and Financial Portfolios 244 A Bivariate Empirical Discrete Probability Distribution 244 Financial Applications 247 Summary 251 5.5 Binomial Probability Distribution 254 A Binomial Experiment 254 Martin Clothing Store Problem 256 Using Excel to Compute Binomial Probabilities 260 Expected Value and Variance for the Binomial Distribution 262 5.6 Poisson Probability Distribution 265 An Example Involving Time Intervals 266 An Example Involving Length or Distance Intervals 267 Using Excel to Compute Poisson Probabilities 267 5.7 Hypergeometric Probability Distribution 270 Using Excel to Compute Hypergeometric Probabilities 272 Summary 274 Glossary 275 Key Formulas 276 Supplementary Exercises 277 Case Problem 1 Go Bananas! 281 Case Problem 2 McNeil’s Auto Mall 282 Case Problem 3 Grievance Committee at Tuglar Corporation 283 Case Problem 4 Sagittarius Casino 283

Chapter 6 Continuous Probability Distributions 285 Statistics in Practice: Procter & Gamble 286 6.1 Uniform Probability Distribution 287 Area as a Measure of Probability 288 6.2 Normal Probability Distribution 291 Normal Curve 291 Standard Normal Probability Distribution 293 Computing Probabilities for Any Normal Probability Distribution 298 Grear Tire Company Problem 299 Using Excel to Compute Normal Probabilities 300 6.3 Exponential Probability Distribution 306 Computing Probabilities for the Exponential Distribution 307 Relationship Between the Poisson and Exponential Distributions 308 Using Excel to Compute Exponential Probabilities 308 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

x Contents

Summary 311 Glossary 311 Key Formulas 312 Supplementary Exercises 312 Case Problem 1 Specialty Toys 315 Case Problem 2 Gebhardt Electronics 316

Chapter 7 Sampling and Sampling Distributions 317 Statistics in Practice: Meadwestvaco Corporation 318 7.1 The Electronics Associates Sampling Problem 319 7.2 Selecting a Sample 320 Sampling from a Finite Population 320 Sampling from an Infinite Population 323 7.3 Point Estimation 327 Practical Advice 329 7.4 Introduction to Sampling Distributions 331 7.5 Sampling Distribution of x 334 Expected Value of x 334 Standard Deviation of x 335 Form of the Sampling Distribution of x 336 Sampling Distribution of x for the EAI Problem 338 Practical Value of the Sampling Distribution of x 338 Relationship Between the Sample Size and the Sampling Distribution of x 340 7.6 Sampling Distribution of p 344 Expected Value of p 344 Standard Deviation of p 345 Form of the Sampling Distribution of p 346 Practical Value of the Sampling Distribution of p 347 7.7 Other Sampling Methods 350 Stratified Random Sampling 350 Cluster Sampling 350 Systematic Sampling 351 Convenience Sampling 351 Judgment Sampling 352 7.8 Practical Advice: Big Data and Errors in Sampling 352 Sampling Error 353 Nonsampling Error 354 Implications of Big Data 355 Summary 357 Glossary 358 Key Formulas 359 Supplementary Exercises 359 Case Problem 1 Marion Dairies 362 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Contents

Chapter 8 Interval Estimation 363 Statistics in Practice: Food Lion 364 8.1 Population Mean: s Known 365 Margin of Error and the Interval Estimate 365 Using Excel 369 Practical Advice 370 8.2 Population Mean: s Unknown 372 Margin of Error and the Interval Estimate 374 Using Excel 376 Practical Advice 378 Using a Small Sample 378 Summary of Interval Estimation Procedures 379 8.3 Determining the Sample Size 383 8.4 Population Proportion 386 Using Excel 387 Determining the Sample Size 389 8.5 Practical Advice: Big Data and Interval Estimation 393 Big Data and the Precision of Confidence Intervals 393 Implications of Big Data 394 Summary 396 Glossary 396 Key Formulas 397 Supplementary Exercises 398 Case Problem 1 Young Professional Magazine 400 Case Problem 2 Gulf Real Estate Properties 402 Case Problem 3 Metropolitan Research, Inc. 403

Chapter 9 Hypothesis Tests 405 Statistics in Practice: John Morrell & Company 406 9.1 Developing Null and Alternative Hypotheses 407 The Alternative Hypothesis as a Research Hypothesis 407 The Null Hypothesis as an Assumption to Be Challenged 408 Summary of Forms for Null and Alternative Hypotheses 409 9.2 Type I and Type II Errors 411 9.3 Population Mean: s Known 413 One-Tailed Test 413 Two-Tailed Test 419 Using Excel 422 Summary and Practical Advice 424 Relationship Between Interval Estimation and Hypothesis Testing 425 9.4 Population Mean: s Unknown 429 One-Tailed Test 430 Two-Tailed Test 431 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

xii Contents

Using Excel 432 Summary and Practical Advice 434 9.5 Population Proportion 438 Using Excel 440 Summary 441 9.6 Practical Advice: Big Data and Hypothesis Testing 444 Big Data and p-Values 444 Implications of Big Data 445 Summary 447 Glossary 447 Key Formulas 448 Supplementary Exercises 448 Case Problem 1 Quality Associates, Inc. 451 Case Problem 2 Ethical Behavior of Business Students at Bayview University 452

Chapter 10 Inference About Means and Proportions with Two Populations 455

Statistics in Practice: U.S. Food and Drug Administration 456 10.1 Inferences About the Difference Between Two Population Means: s 1 and s 2 Known 457 Interval Estimation of m1 2 m2 457 Using Excel to Construct a Confidence Interval 459 Hypothesis Tests About m1 2 m2 461 Using Excel to Conduct a Hypothesis Test 463 Practical Advice 464 10.2 Inferences About the Difference Between Two Population Means: s 1 and s 2 Unknown 467 Interval Estimation of m1 2 m2 467 Using Excel to Construct a Confidence Interval 469 Hypothesis Tests About m1 2 m2 471 Using Excel to Conduct a Hypothesis Test 473 Practical Advice 475 10.3 Inferences About the Difference Between Two Population Means: Matched Samples 478 Using Excel to Conduct a Hypothesis Test 481 10.4 Inferences About the Difference Between Two Population Proportions 486 Interval Estimation of p1 2 p2 486 Using Excel to Construct a Confidence Interval 488 Hypothesis Tests About p1 2 p2 489 Using Excel to Conduct a Hypothesis Test 491 Summary 495 Glossary 496 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Contents

Key Formulas 496 Supplementary Exercises 497 Case Problem Par, Inc. 501

Chapter 11 Inferences About Population Variances 502 Statistics in Practice: U.S. Government Accountability Office 503 11.1 Inferences About a Population Variance 504 Interval Estimation 504 Using Excel to Construct a Confidence Interval 508 Hypothesis Testing 509 Using Excel to Conduct a Hypothesis Test 512 11.2 Inferences About Two Population Variances 516 Using Excel to Conduct a Hypothesis Test 520 Summary 525 Key Formulas 525 Supplementary Exercises 525 Case Problem 1 Air Force Training Program 527 Case Problem 2 Meticulous Drill & Reamer 528

Chapter 12 Tests of Goodness of Fit, Independence, and Multiple Proportions 530

tatistics in Practice: United Way 531 S 12.1 Goodness of Fit Test 532 Multinomial Probability Distribution 532 Using Excel to Conduct a Goodness of Fit Test 536 12.2 Test of Independence 538 Using Excel to Conduct a Test of Independence 542 12.3 Testing for Equality of Three or More Population Proportions 547 A Multiple Comparison Procedure 551 Using Excel to Conduct a Test of Multiple Proportions 553 Summary 557 Glossary 558 Key Formulas 558 Supplementary Exercises 558 Case Problem 1 A Bipartisan Agenda for Change 561 Case Problem 2 Fuentes Salty Snacks, Inc. 562 Case Problem 3 Fresno Board Games 563

Chapter 13 Experimental Design and Analysis of Variance 564

Statistics in Practice: Burke Marketing Services, Inc. 565 13.1 An Introduction to Experimental Design and Analysis of Variance 567 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

xiii

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Data Collection 567 Assumptions for Analysis of Variance 569 Analysis of Variance: A Conceptual Overview 569 13.2 Analysis of Variance and the Completely Randomized Design 572 Between-Treatments Estimate of Population Variance 573 Within-Treatments Estimate of Population Variance 574 Comparing the Variance Estimates: The F Test 574 ANOVA Table 576 Using Excel 578 Testing for the Equality of k Population Means: An Observational Study 579 13.3 Multiple Comparison Procedures 584 Fisher’s LSD 584 Type I Error Rates 587 13.4 Randomized Block Design 590 Air Traffic Controller Stress Test 591 ANOVA Procedure 592 Computations and Conclusions 593 Using Excel 595 13.5 Factorial Experiment 599 ANOVA Procedure 600 Computations and Conclusions 601 Using Excel 604 Summary 608 Glossary 609 Key Formulas 610 Supplementary Exercises 612 Case Problem 1 Wentworth Medical Center 616 Case Problem 2 Compensation for Sales Professionals 617 Case Problem 3 TourisTopia Travel 618

Chapter 14 Simple Linear Regression 620 Statistics in Practice: Alliance Data Systems 621 14.1 Simple Linear Regression Model 622 Regression Model and Regression Equation 622 Estimated Regression Equation 623 14.2 Least Squares Method 625 Using Excel to Construct a Scatter Diagram, Display the Estimated Regression Line, and Display the Estimated Regression Equation 629 14.3 Coefficient of Determination 637 Using Excel to Compute the Coefficient of Determination 641 Correlation Coefficient 641 14.4 Model Assumptions 645 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Contents

14.5 Testing for Significance 647 Estimate of s2 647 t Test 648 Confidence Interval for b1 650 F Test 651 Some Cautions About the Interpretation of Significance Tests 653 14.6 Using the Estimated Regression Equation for Estimation and Prediction 656 Interval Estimation 657 Confidence Interval for the Mean Value of y 658 Prediction Interval for an Individual Value of y 659 14.7 Excel’s Regression Tool 664 Using Excel’s Regression Tool for the Armand’s Pizza Parlors Example 664 Interpretation of Estimated Regression Equation Output 665 Interpretation of ANOVA Output 666 Interpretation of Regression Statistics Output 667 14.8 Residual Analysis: Validating Model Assumptions 670 Residual Plot Against x 671 ⁄ Residual Plot Against y 672 Standardized Residuals 674 Using Excel to Construct a Residual Plot 676 Normal Probability Plot 679 14.9 Outliers and Influential Observations 682 Detecting Outliers 682 Detecting Influential Observations 684 14.10 Practical Advice: Big Data and Hypothesis Testing in Simple Linear Regression 689 Summary 690 Glossary 690 Key Formulas 691 Supplementary Exercises 694 Case Problem 1 Measuring Stock Market Risk 698 Case Problem 2 U.S. Department of Transportation 699 Case Problem 3 Selecting a Point–and–Shoot Digital Camera 700 Case Problem 4 Finding the Best Car Value 701 Case Problem 5 Buckeye Creek Amusement Park 702 Appendix 14.1 Calculus-Based Derivation of Least Squares Formulas 703 Appendix 14.2 A Test for Significance Using Correlation 704

Chapter 15 Multiple Regression 706 Statistics in Practice: International Paper 707 15.1 Multiple Regression Model 708 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

xvi Contents

Regression Model and Regression Equation 708 Estimated Multiple Regression Equation 708 15.2 Least Squares Method 709 An Example: Butler Trucking Company 710 Using Excel’s Regression Tool to Develop the Estimated Multiple Regression Equation 713 Note on Interpretation of Coefficients 714 15.3 Multiple Coefficient of Determination 720 15.4 Model Assumptions 723 15.5 Testing for Significance 724 F Test 725 t Test 727 Multicollinearity 728 15.6 Using the Estimated Regression Equation for Estimation and Prediction 731 15.7 Categorical Independent Variables 734 An Example: Johnson Filtration, Inc. 734 Interpreting the Parameters 737 More Complex Categorical Variables 738 15.8 Residual Analysis 742 ⁄ Residual Plot Against y 743 ⁄ Standardized Residual Plot Against y 744 15.9 Practical Advice: Big Data and Hypothesis Testing in Multiple Regression 747 Summary 748 Glossary 748 Key Formulas 749 Supplementary Exercises 750 Case Problem 1 Consumer Research, Inc. 755 Case Problem 2 Predicting Winnings for NASCAR Drivers 756 Case Problem 3 Finding the Best Car Value 758

Appendix A References and Bibliography Appendix B Tables

762

764

Appendix C Summation Notation

775

Appendix D Self-Test Solutions and Answers to Even-Numbered Exercises (online) 777

Appendix E Microsoft Excel 2016 and Tools for Statistical Analysis Index

786

778

Preface

This text is the seventh edition of Essentials of Modern Business Statistics with Microsoft® Office Excel®. With this edition we welcome two eminent scholars to our author team: Jeffrey D. Camm of Wake Forest University and James J. Cochran of the University of Alabama. Both Jeff and Jim are accomplished teachers, researchers, and practitioners in the fields of statistics and business analytics. Jim is a fellow of the American Statistical Association. You can read more about their accomplishments in the About the Authors section that follows this preface. We believe that the addition of Jeff and Jim as our coauthors will both maintain and improve the effectiveness of Essentials of Modern Business Statistics with Microsoft Office Excel. The purpose of Essentials of Modern Business Statistics with Microsoft® Office Excel® is to give students, primarily those in the fields of business administration and economics, a conceptual introduction to the field of statistics and its many applications. The text is applications oriented and written with the needs of the nonmathematician in mind; the mathematical prerequisite is knowledge of algebra. Applications of data analysis and statistical methodology are an integral part of the organization and presentation of the text material. The discussion and development of each technique is presented in an applications setting, with the statistical results providing insights for decision making and solutions to applied problems. Although the book is applications oriented, we have taken care to provide sound methodological development and to use notation that is generally accepted for the topic being covered. Hence, students will find that this text provides good preparation for the study of more advanced statistical material. A bibliography to guide further study is included as an appendix.

Use of Microsoft Excel for Statistical Analysis Essentials of Modern Business Statistics with Microsoft® Office Excel® is first and foremost a statistics textbook that emphasizes statistical concepts and applications. But since most practical problems are too large to be solved using hand calculations, some type of statistical software package is required to solve these problems. There are several excellent statistical packages available today; however, because most students and potential employers value spreadsheet experience, many schools now use a spreadsheet package in their statistics courses. Microsoft Excel is the most widely used spreadsheet package in business as well as in colleges and universities. We have written Essentials of Modern Business Statistics with Microsoft® Office Excel® especially for statistics courses in which Microsoft Excel is used as the software package. Excel has been integrated within each of the chapters and plays an integral part in providing an application orientation. Although we assume that readers using this text are familiar with Excel basics such as selecting cells, entering formulas, copying, and so on, we do not assume that readers are familiar with Excel 2016 or Excel’s tools for statistical analysis. As a result, we have included Appendix E, which provides an introduction to Excel 2016 and tools for statistical analysis. Throughout the text the discussion of using Excel to perform a statistical procedure appears in a subsection immediately following the discussion of the statistical procedure. We believe that this style enables us to fully integrate the use of Excel throughout

xviii Preface

the text, but still maintain the primary emphasis on the statistical methodology being discussed. In each of these subsections, we provide a standard format for using Excel for statistical analysis. There are four primary tasks: Enter/Access Data, Enter Functions and Formulas, Apply Tools, and Editing Options. The Editing Options task is new with this edition. It primarily involves how to edit Excel output so that it is more suitable for presentations to users. We believe a consistent framework for applying Excel helps users to focus on the statistical methodology without getting bogged down in the details of using Excel. In presenting worksheet figures, we often use a nested approach in which the worksheet shown in the background of the figure displays the formulas and the worksheet shown in the foreground shows the values computed using the formulas. Different colors and shades of colors are used to differentiate worksheet cells containing data, highlight cells containing Excel functions and formulas, and highlight material printed by Excel as a result of using one or more data analysis tools.

Changes in the Seventh Edition We appreciate the acceptance and positive response to the previous editions of Essentials of Modern Business Statistics with Microsoft® Office Excel®. Accordingly, in making modifications for this new edition, we have maintained the presentation style and readability of those editions. The significant changes in the new edition are summarized here. Users of the previous edition will notice that the chapters offered and topics covered in this edition differ from previous editions. While the topical coverage of the first nine chapters remains the same, the organization and coverage in some of the later chapters have expanded. We have eliminated the coverage of the advanced topic of Time Series and Quality Control in favor of the expanded coverage in Chapters 10, 11, 12 and 13. Chapter10 now provides coverage of inferences of means and proportions for two populations, and chapter 11 is focused on inferences about population variances. Chapter 12 is a discussion of comparing multiple proportions, tests of independence and goodness of fit and chapter13 covers experimental design and ANOVA. We believe you will find the expanded coverage in these chapters useful in your classes. Coverage of regression is now in chapters14 and 15. These two chapters are revisions of the regression chapters from the 6th edition. In additions to these changes, we made the following revisions: ●

●

icrosoft Excel 2016. Step-by-step instructions and screen captures show how to M use the latest version of Excel to implement statistical procedures. Data and Statistics—Chapter 1. We have expanded our section on data mining to include a discussion of big data. We have added a new section on analytics. We have also placed greater emphasis on the distinction between observed and experimental data. Descriptive Statistics: Tabular and Graphical Displays—Chapter 2. Microsoft Excel now has the capability of creating box plots and comparative box plots. We have added to this chapter instruction on how to use this very useful new feature. Interval Estimation—Chapter 8. We have added a new section on the implications of big data (large data sets) on the interpretation of confidence intervals and importantly, the difference between statistical and practical significance. Hypothesis Tests—Chapter 9. Similar to our addition to Chapter 8, we have added a new section on the implications of big data (large data sets) on the interpretation of hypothesis tests and the difference between statistical and practical significance. Simple Linear Regression—Chapter 14. Similar to our addition to Chapter 8, we have added a new section on the implications of big data (large data sets) on

Preface

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xix

the interpretation of hypothesis tests in simple linear regression and the difference between statistical and practical significance. New Case Problems. We have added thirteen new case problems to this edition. The new case problems appear in the chapters on descriptive statistics and regression analysis. The case problems in the text provide students with the opportunity to analyze somewhat larger data sets and prepare managerial reports based on the results of their analysis. New Examples and Exercises Based on Real Data. We have added approximately 126 new examples and exercises based on real data and recently referenced sources of statistical information. Using data obtained from various data collection organizations, websites, and other sources such as The Wall Street Journal, USA Today, Fortune, and Barron’s, we have drawn upon actual studies to develop explanations and to create exercises that demonstrate many uses of statistics in business and economics. We believe the use of real data helps generate more student interest in the material and enables the student to learn about both the statistical methodology and its application. Updated and Improved End-of-Chapter Solutions and Solutions Manual. Partnering with accomplished instructor Dawn Bulriss at Maricopa Community Colleges, we took a deep audit of the solutions manual. Every question and solution was reviewed and reworked, as necessary. The solutions now contain additional detail: improved rounding instructions; expanded explanations with a studentfocus; and alternative answers using Excel and a statistical calculator. We believe this thorough review will enhance both the instructor and student learning experience in this digital age.

Features and Pedagogy Authors Anderson, Sweeney, Williams, Camm, and Cochran have continued many of the features that appeared in previous editions.

Methods Exercises and Applications Exercises The end-of-section exercises are split into two parts, Methods and Applications. The Methods exercises require students to use the formulas and make the necessary computations. The Applications exercises require students to use the chapter material in real-world situations. Thus, students first focus on the computational “nuts and bolts” and then move on to the subtleties of statistical application and interpretation.

Self-Test Exercises Certain exercises are identified as self-test exercises. Completely worked-out solutions for those exercises are provided in Appendix D in the Student Resources online. Students can attempt the self-test exercises and immediately check the solution to evaluate their understanding of the concepts presented in the chapter.

Margin Annotations and Notes and Comments Margin annotations that highlight key points and provide additional insights for the students are a key feature of this text. These annotations are designed to provide emphasis and enhance understanding of the terms and concepts being presented in the text. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

xx Preface

At the end of many sections, we provide Notes and Comments designed to give the student additional insights about the statistical methodology and its application. Notes and Comments include warnings about or limitations of the methodology, recommendations for application, brief descriptions of additional technical considerations, and other matters.

Data Files Accompany the Text Approximately 220 data files are available on the website that accompanies this text. The data sets are available in Excel 2016 format. DATAfile logos are used in the text to identify the data sets that are available on the website. Data sets for all case problems as well as data sets for larger exercises are included.

MindTap MindTap, featuring all new Excel Online integration powered by Microsoft, is a complete digital solution for the business statistics course. It has enhancements that take students from learning basic statistical concepts to actively engaging in critical thinking applications, while learning valuable software skills for their future careers. MindTap is a customizable digital course solution that includes an interactive eBook, autograded, algorithmic exercises from the textbook, Adaptive Test Prep, as well as interactive visualizations. All of these materials offer students better access to understand the materials within the course. For more information on MindTap, please contact your Cengage representative.

For Students Online resources are available to help the student work more efficiently. The resources can be accessed through www.cengagebrain.com.

For Instructors Instructor resources are available to adopters on the Instructor Companion Site, which can be found and accessed at www.cengage.com, including: ●

●

Solutions Manual: The Solutions Manual, prepared by the authors, includes solutions for all problems in the text. It is available online as well as print. Solutions to Case Problems: These are also prepared by the authors and contain solutions to all case problems presented in the text. PowerPoint Presentation Slides: The presentation slides contain a teaching outline that incorporates figures to complement instructor lectures. Test Bank: Cengage Learning Testing Powered by Cognero is a flexible, online system that allows you to: j author, edit, and manage test bank content from multiple Cengage Learning solutions, j create multiple test versions in an instant, and j deliver tests from your LMS, your classroom, or wherever you want. The Test Bank is also available in Microsoft Word.

Preface

xxi

Acknowledgments A special thanks goes to our associates from business and industry who supplied the Statistics in Practice features. We recognize them individually by a credit line in each of the articles. We are also indebted to our product manager, Aaron Arnsparger; our content developer, Anne Merrill; our content project manager, Colleen Farmer; our project manager at MPS Limited, Gaurav Prabhu; digital content designer, Brandon Foltz; and others at Cengage for their editorial counsel and support during the preparation of this text. We would like to acknowledge the work of our reviewers, who provided comments and suggestions of ways to continue to improve our text. Thanks to: James Bang, Virginia Military Institute Robert J. Banis, University of Missouri–St. Louis Timothy M. Bergquist, Northwest Christian College Gary Black, University of Southern Indiana William Bleuel, Pepperdine University Derrick Boone, Wake Forest University Lawrence J. Bos, Cornerstone University Dawn Bulriss, Maricopa Community Colleges Joseph Cavanaugh, Wright State University–Lake Campus Sheng-Kai Chang, Wayne State University Robert Christopherson, SUNY-Plattsburgh Michael Clark, University of Baltimore Robert D. Collins, Marquette University Ivona Contardo, Stellenbosch University Sean Eom, Southeast Missouri State University Samo Ghosh, Albright College Philip A. Gibbs, Washington & Lee University Daniel L. Gilbert, Tennessee Wesleyan College Michael Gorman, University of Dayton Erick Hofacker, University of Wisconsin, River Falls David Juriga, St. Louis Community College William Kasperski, Madonna University Kuldeep Kumar, Bond Business School Tenpao Lee, Niagara University Ying Liao, Meredith College Daniel Light, Northwest State College Ralph Maliszewski, Waynesburg University Saverio Manago, Salem State University Patricia A. Mullins, University of Wisconsin–Madison Jack Muryn, Cardinal Stritch University Anthony Narsing, Macon State College Robert M. Nauss, University of Missouri–St. Louis Elizabeth L. Rankin, Centenary College of Louisiana Surekha Rao, Indiana University, Northwest Jim Robison, Sonoma State University Farhad Saboori, Albright College Susan Sandblom, Scottsdale Community College Ahmad Saranjam, Bridgewater State University Jeff Sarbaum, University of North Carolina at Greensboro Robert Scott, Monmouth University Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

xxii Preface

Toni Somers, Wayne State University Jordan H. Stein, University of Arizona Bruce Thompson, Milwaukee School of Engineering Ahmad Vessal, California State University, Northridge Dave Vinson, Pellissippi State Daniel B. Widdis, Naval Postgraduate School Peter G. Wagner, University of Dayton Sheng-Ping Yang, Black Hills State University We would like to recognize the following individuals, who have helped us in the past and continue to influence our writing. Glen Archibald, University of Mississippi Darl Bien, University of Denver Thomas W. Bolland, Ohio University Mike Bourke, Houston Baptist University Peter Bryant, University of Colorado Terri L. Byczkowski, University of Cincinnati Robert Carver, Stonehill College Ying Chien, University of Scranton Robert Cochran, University of Wyoming Murray Côté, University of Florida David W. Cravens, Texas Christian University Eddine Dahel, Monterey Institute of International Studies Tom Dahlstrom, Eastern College Terry Dielman, Texas Christian University Joan Donohue, University of South Carolina Jianjun Du, University of Houston–Victoria Thomas J. Dudley, Pepperdine University Swarna Dutt, University of West Georgia Ronald Ehresman, Baldwin-Wallace College Mohammed A. El-Saidi, Ferris State University Robert Escudero, Pepperdine University Stacy Everly, Delaware County Community College Soheila Kahkashan Fardanesh, Towson University Nicholas Farnum, California State University–Fullerton Abe Feinberg, California State University, Northridge Michael Ford, Rochester Institute of Technology Phil Fry, Boise State University V. Daniel Guide, Duquesne University Paul Guy, California State University–Chico Charles Harrington, University of Southern Indiana Carl H. Hess, Marymount University Woodrow W. Hughes, Jr., Converse College Alan Humphrey, University of Rhode Island Ann Hussein, Philadelphia College of Textiles and Science Ben Isselhardt, Rochester Institute of Technology Jeffery Jarrett, University of Rhode Island Barry Kadets, Bryant College Homayoun Khamooshi, George Washington University Kenneth Klassen, California State University Northridge David Krueger, St. Cloud State University June Lapidus, Roosevelt University Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

xxiii

Preface

Martin S. Levy, University of Cincinnati Daniel M. Light, Northwest State College Ka-sing Man, Georgetown University Don Marx, University of Alaska, Anchorage Tom McCullough, University of California–Berkeley Timothy McDaniel, Buena Vista University Mario Miranda, The Ohio State University Barry J. Monk, Macon State College Mitchell Muesham, Sam Houston State University Richard O’Connell, Miami University of Ohio Alan Olinsky, Bryant College Lynne Pastor, Carnegie Mellon University Von Roderick Plessner, Northwest State University Robert D. Potter, University of Central Florida Tom Pray, Rochester Institute of Technology Harold Rahmlow, St. Joseph’s University Derrick Reagle, Fordham University Avuthu Rami Reddy, University of Wisconsin–Platteville Tom Ryan, Case Western Reserve University Ahmad Saranjam, Bridgewater State College Bill Seaver, University of Tennessee Alan Smith, Robert Morris College William Struning, Seton Hall University Ahmad Syamil, Arkansas State University David Tufte, University of New Orleans Jack Vaughn, University of Texas–El Paso Elizabeth Wark, Springfield College Ari Wijetunga, Morehead State University Nancy A. Williams, Loyola College in Maryland J. E. Willis, Louisiana State University Larry Woodward, University of Mary Hardin–Baylor Mustafa Yilmaz, Northeastern University David R. Anderson Dennis J. Sweeney Thomas A. Williams Jeffrey D. Camm James J. Cochran

CHAPTER

Data and Statistics CONTENTS STATISTICS IN PRACTICE: Bloomberg bUSINESSWEEK 1.1 APPLICATIONS IN BUSINESS AND ECONOMICS Accounting Finance Marketing Production Economics Information Systems 1.2 DATA Elements, Variables, and Observations Scales of Measurement Categorical and Quantitative Data Cross-Sectional and Time Series Data

1.3 DATA SOURCES Existing Sources Observational Study Experiment Time and Cost Issues Data Acquisition Errors 1.4 DESCRIPTIVE STATISTICS 1.5 STATISTICAL INFERENCE 1.6 Statistical Analysis Using Microsoft Excel Data Sets and Excel Worksheets Using Excel for Statistical Analysis 1.7 Analytics 1.8

Big Data and Data Mining

1.9 ETHICAL GUIDELINES FOR STATISTICAL PRACTICE

Chapter 1 Data and Statistics

STATISTICS in PRACTICE BLOOMBERG BUSINESSWEEK* With a global circulation of more than 1 million, Bloomberg Businessweek is one of the most widely read business magazines in the world. Bloomberg’s 1700 reporters in 145 service bureaus around the world enable Bloomberg Businessweek to deliver a variety of articles of interest to the global business and economic community. Along with feature articles on current topics, the magazine contains articles on international business, economic analysis, information processing, and science and technology. Information in the feature articles and the regular sections helps readers stay abreast of current developments and assess the impact of those developments on business and economic conditions. Most issues of Bloomberg Businessweek, formerly BusinessWeek, provide an in-depth report on a topic of current interest. Often, the in-depth reports contain statistical facts and summaries that help the reader understand the business and economic information. Examples of articles and reports include the impact of businesses moving important work to cloud computing, the crisis facing the U.S. Postal Service, and why the debt crisis is even worse than we think. In addition, Bloomberg Businessweek provides a variety of statistics about the state of the economy, including production indexes, stock prices, mutual funds, and interest rates. Bloomberg Businessweek also uses statistics and statistical information in managing its own business. For example, an annual survey of subscribers helps the company learn about subscriber demographics, reading habits, likely purchases, lifestyles, and so on. Bloomberg Businessweek managers use statistical summaries from the survey to provide better services to subscribers and advertisers. One recent North American subscriber

*The authors are indebted to Charlene Trentham, Research Manager, for providing this Statistics in Practice.

AP Photos/Weng lei - Imaginechina

NEW YORK, NEW YORK

Bloomberg Businessweek uses statistical facts and summaries in many of its articles. survey indicated that 90% of Bloomberg Businessweek subscribers use a personal computer at home and that 64% of Bloomberg Businessweek subscribers are involved with computer purchases at work. Such statistics alert Bloomberg Businessweek managers to subscriber interest in articles about new developments in computers. The results of the subscriber survey are also made available to potential advertisers. The high percentage of subscribers using personal computers at home and the high percentage of subscribers involved with computer purchases at work would be an incentive for a computer manufacturer to consider advertising in Bloomberg Businessweek. In this chapter, we discuss the types of data available for statistical analysis and describe how the data are obtained. We introduce descriptive statistics and statistical inference as ways of converting data into meaningful and easily interpreted statistical information.

Frequently, we see the following types of statements in newspapers and magazines: ●●

●●

ber Technologies Inc. is turning to the leveraged-loan market for the first time U to raise as much as $2 billion, a sign of the popular ride-sharing network’s hunger for cash as it expands around the world (The Wall Street Journal, June 14, 2016).

Against the U.S. dollar, the euro has lost nearly 30% of its value in the last year; the Australian dollar lost almost 20% (The Economist, April 25th–May 1st, 2015).

1.1 Applications in Business and Economics ●●

●●

W Group’s U.S. sales continue to slide, with total sales off by 13% from last JanuV ary, to 36,930 vehicles (Panorama, March 2014). A poll of 1320 corporate recruiters indicated that 68% of the recruiters ranked communication skills as one of the top five most important skills for new hires (Bloomberg Businessweek April 13–April 19, 2015). Green Mountain sold 18 billion coffee pods in two years (Harvard Business Review, January-February, 2016). Most homeowners spend between about $10,000 and roughly $27,000 converting a basement, depending on the size of the space, according to estimates from Home Advisor, a website that connects homeowners with prescreened service professionals (Consumer Reports, February 9, 2016). A full 88% of consumers say they buy private label, primarily because of price, according to Market Track (USA Today, May 17, 2016).

The numerical facts in the preceding statements—$2 billion, 30%, 20%, 13%, 36,930, 1320, 68%, 18 billion, $10,000, $27,000 and 88%—are called statistics. In this usage, the term statistics refers to numerical facts such as averages, medians, percentages, and maximums that help us understand a variety of business and economic situations. However, as you will see, the field, or subject, of statistics involves much more than numerical facts. In a broader sense, statistics is the art and science of collecting, analyzing, presenting, and interpreting data. Particularly in business and economics, the information provided by collecting, analyzing, presenting, and interpreting data gives managers and decision makers a better understanding of the business and economic environment and thus enables them to make more informed and better decisions. In this text, we emphasize the use of statistics for business and economic decision making. Chapter 1 begins with some illustrations of the applications of statistics in business and economics. In Section 1.2 we define the term data and introduce the concept of a data set. This section also introduces key terms such as variables and observations, discusses the difference between quantitative and categorical data, and illustrates the uses of cross- sectional and time series data. Section 1.3 discusses how data can be obtained from existing sources or through survey and experimental studies designed to obtain new data. The important role that the Internet now plays in obtaining data is also highlighted. The uses of data in developing descriptive statistics and in making statistical inferences are described in Sections 1.4 and 1.5. The last four sections of Chapter 1 provide the role of the computer in statistical analysis, an introduction to business analytics and the role statistics plays in it, an introduction to big data and data mining, and a discussion of ethical guidelines for statistical practice.

Applications in Business and Economics 1.1 In today’s global business and economic environment, anyone can access vast amounts of statistical information. The most successful managers and decision makers understand the information and know how to use it effectively. In this section, we provide examples that illustrate some of the uses of statistics in business and economics.

Accounting Public accounting firms use statistical sampling procedures when conducting audits for their clients. For instance, suppose an accounting firm wants to determine whether the amount of accounts receivable shown on a client’s balance sheet fairly represents the actual amount ofaccounts receivable. Usually the large number of individual accounts receivable makes reviewing and validating every account too time-consuming and expensive. As common Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Chapter 1 Data and Statistics

practice in such situations, the audit staff selects a subset of the accounts called a sample. After reviewing the accuracy of the sampled accounts, the auditors draw a conclusion as to whether the accounts receivable amount shown on the client’s balance sheet is acceptable.

Finance Financial analysts use a variety of statistical information to guide their investment recommendations. In the case of stocks, analysts review financial data such as price/ earnings ratios and dividend yields. By comparing the information for an individual stock with information about the stock market averages, an analyst can begin to draw a conclusion as to whether the stock is a good investment. For example, The Wall Street Journal (February 27, 2016) reported that the average dividend yield for the S&P 500 companies was 2.3%. Microsoft showed a dividend yield of 2.61%. In this case, the statistical information on dividend yield indicates a higher dividend yield for Microsoft than the average dividend yield for the S&P 500 companies. This and other information about Microsoft would help the analyst make an informed buy, sell, or hold recommendation for Microsoft stock.

Marketing Electronic scanners at retail checkout counters collect data for a variety of marketing research applications. For example, data suppliers such as ACNielsen and Information Resources, Inc. purchase point-of-sale scanner data from grocery stores, process the data, and then sell statistical summaries of the data to manufacturers. Manufacturers spend hundreds of thousands of dollars per product category to obtain this type of scanner data. Manufacturers also purchase data and statistical summaries on promotional activities such as special pricing and the use of in-store displays. Brand managers can review the scanner statistics and the promotional activity statistics to gain a better understanding of the relationship between promotional activities and sales. Such analyses often prove helpful in e stablishing future marketing strategies for the various products.

Production Today’s emphasis on quality makes quality control an important application of statistics in production. A variety of statistical quality control charts are used to monitor the output of a production process. In particular, an x-bar chart can be used to monitor the average output. Suppose, for example, that a machine fills containers with 12 ounces of a soft drink. Periodically, a production worker selects a sample of containers and computes the average number of ounces in the sample. This average, or x-bar value, is plotted on an x-bar chart. A plotted value above the chart’s upper control limit indicates overfilling, and a plotted value below the chart’s lower control limit indicates underfilling. The process is termed “in control” and allowed to continue as long as the plotted x-bar values fall between the chart’s upper and lower control limits. Properly interpreted, an x-bar chart can help determine when adjustments are necessary to correct a production process.

Economics Economists frequently provide forecasts about the future of the economy or some aspect of it. They use a variety of statistical information in making such forecasts. For instance, in forecasting inflation rates, economists use statistical information on such indicators as the Producer Price Index, the unemployment rate, and manufacturing capacity utilization. Often these statistical indicators are entered into computerized forecasting models that predict inflation rates. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

1.2 Data

Information Systems Information systems administrators are responsible for the day-to-day operation of an organization’s computer networks. A variety of statistical information helps administrators assess the performance of computer networks, including local area networks (LANs), wide area networks (WANs), network segments, intranets, and other data communication systems. Statistics such as the mean number of users on the system, the proportion of time any component of the system is down, and the proportion of bandwidth utilized at various times of the day are examples of statistical information that help the system administrator better understand and manage the computer network. Applications of statistics such as those described in this section are an integral part of this text. Such examples provide an overview of the breadth of statistical applications. To supplement these examples, practitioners in the fields of business and economics provided chapter-opening Statistics in Practice articles that introduce the material covered in each chapter. The Statistics in Practice applications show the importance of statistics in a wide variety of business and economic situations.

Data 1.2 Data are the facts and figures collected, analyzed, and summarized for presentation and interpretation. All the data collected in a particular study are referred to as the data set for the study. Table 1.1 shows a data set containing information for 60 nations that participate in the World Trade Organization (WTO). The WTO encourages the free flow of international trade and provides a forum for resolving trade disputes.

Elements, Variables, and Observations Elements are the entities on which data are collected. Each nation listed in Table 1.1 is an element with the nation or element name shown in the first column. With 60 nations, the data set contains 60 elements. A variable is a characteristic of interest for the elements. The data set in Table 1.1 includes the following five variables: ●●

●●

TO Status: The nation’s membership status in the World Trade Organization; W this can be either as a member or an observer. Per Capita GDP ($): The total market value ($) of all goods and services produced by the nation divided by the number of people in the nation; this is commonly used to compare economic productivity of the nations. Trade Deficit ($1000s): The difference between the total dollar value of the nation’s imports and the total dollar value of the nation’s exports. Fitch Rating: The nation’s sovereign credit rating as appraised by the Fitch Group1; the credit ratings range from a high of AAA to a low of F and can be modified by 1 or 2. Fitch Outlook: An indication of the direction the credit rating is likely to move over the upcoming two years; the outlook can be negative, stable, or positive.

Measurements collected on each variable for every element in a study provide the data. The set of measurements obtained for a particular element is called an observation. Referring to Table 1.1, we see that the first observation contains the following measurements: Member,

1 The Fitch Group is one of three nationally recognized statistical rating organizations designated by the U.S. Securities and Exchange Commission. The other two are Standard and Poor’s and Moody’s investor service.

Chapter 1 Data and Statistics

TABLE 1.1 DATA SET FOR 60 NATIONS IN THE WORLD TRADE ORGANIZATION

Nation

Nations

Data sets such as Nations are available on the companion site for this title.

WTO Status

Per Capita GDP ($)

Trade Deficit ($1000s)

Fitch Rating

Fitch Outlook

Armenia Member 5,400 2,673,359 BB2 Stable Australia Member 40,800 233,304,157 AAA Stable Austria Member 41,700 12,796,558 AAA Stable Azerbaijan Observer 5,400 216,747,320 BBB2 Positive Bahrain Member 27,300 3,102,665 BBB Stable Belgium Member 37,600 214,930,833 AA1 Negative Brazil Member 11,600 229,796,166 BBB Stable Bulgaria Member 13,500 4,049,237 BBB2 Positive Canada Member 40,300 21,611,380 AAA Stable Cape Verde Member 4,000 874,459 B1 Stable Chile Member 16,100 214,558,218 A1 Stable China Member 8,400 2156,705,311 A1 Stable Colombia Member 10,100 21,561,199 BBB2 Stable Costa Rica Member 11,500 5,807,509 BB1 Stable Croatia Member 18,300 8,108,103 BBB2 Negative Cyprus Member 29,100 6,623,337 BBB Negative Czech Republic Member 25,900 210,749,467 A1 Positive Denmark Member 40,200 215,057,343 AAA Stable Ecuador Member 8,300 1,993,819 B2 Stable Egypt Member 6,500 28,486,933 BB Negative El Salvador Member 7,600 5,019,363 BB Stable Estonia Member 20,200 802,234 A1 Stable France Member 35,000 118,841,542 AAA Stable Georgia Member 5,400 4,398,153 B1 Positive Germany Member 37,900 2213,367,685 AAA Stable Hungary Member 19,600 29,421,301 BBB2 Negative Iceland Member 38,000 2504,939 BB1 Stable Ireland Member 39,500 259,093,323 BBB1 Negative Israel Member 31,000 6,722,291 A Stable Italy Member 30,100 33,568,668 A1 Negative Japan Member 34,300 31,675,424 AA Negative Kazakhstan Observer 13,000 233,220,437 BBB Positive Kenya Member 1,700 9,174,198 B1 Stable Latvia Member 15,400 2,448,053 BBB2 Positive Lebanon Observer 15,600 13,715,550 B Stable Lithuania Member 18,700 3,359,641 BBB Positive Malaysia Member 15,600 239,420,064 A2 Stable Mexico Member 15,100 1,288,112 BBB Stable Peru Member 10,000 27,888,993 BBB Stable Philippines Member 4,100 15,667,209 BB1 Stable Poland Member 20,100 19,552,976 A2 Stable Portugal Member 23,200 21,060,508 BBB2 Negative South Korea Member 31,700 237,509,141 A1 Stable Romania Member 12,300 13,323,709 BBB2 Stable Russia Observer 16,700 2151,400,000 BBB Positive Rwanda Member 1,300 939,222 B Stable Serbia Observer 10,700 8,275,693 BB2 Stable Seychelles Observer 24,700 666,026 B Stable Singapore Member 59,900 227,110,421 AAA Stable Slovakia Member 23,400 22,110,626 A1 Stable

1.2 Data

Slovenia Member 29,100 2,310,617 AA2 Negative South Africa Member 11,000 3,321,801 BBB1 Stable Sweden Member 40,600 210,903,251 AAA Stable Switzerland Member 43,400 227,197,873 AAA Stable Thailand Member 9,700 2,049,669 BBB Stable Turkey Member 14,600 71,612,947 BB1 Positive UK Member 35,900 162,316,831 AAA Negative Uruguay Member 15,400 2,662,628 BB Positive USA Member 48,100 784,438,559 AAA Stable Zambia Member 1,600 21,805,198 B1 Stable

5400, 2,673,359, BB2, and Stable. The second observation contains the following measurements: Member, 40,800, 233,304,157, AAA, Stable, and so on. A data set with 60 elements contains 60 observations.

Scales of Measurement Data collection requires one of the following scales of measurement: nominal, ordinal, interval, or ratio. The scale of measurement determines the amount of information contained in the data and indicates the most appropriate data summarization and statistical analyses. When the data for a variable consist of labels or names used to identify an attribute of the element, the scale of measurement is considered a nominal scale. For example, referring to the data in Table 1.1, the scale of measurement for the WTO Status variable is nominal because the data “member” and “observer” are labels used to identify the status category for the nation. In cases where the scale of measurement is nominal, a numerical code as well as a nonnumerical label may be used. For example, to facilitate data collection and to prepare the data for entry into a computer database, we might use a numerical code for the WTO Status variable by letting 1 denote a member nation in the World Trade Organization and 2 denote an observer nation. The scale of measurement is nominal even though the data appear as numerical values. The scale of measurement for a variable is considered an ordinal scale if the data exhibit the properties of nominal data and in addition, the order or rank of the data is meaningful. For example, referring to the data in Table 1.1, the scale of measurement for the Fitch Rating is ordinal because the rating labels which range from AAA to F can be rank ordered from best credit rating AAA to poorest credit rating F. The rating letters provide the labels similar to nominal data, but in addition, the data can also be ranked or ordered based on the credit rating, which makes the measurement scale ordinal. Ordinal data can also be recorded by a numerical code, for example, your class rank in school. The scale of measurement for a variable is an interval scale if the data have all the properties of ordinal data and the interval between values is expressed in terms of a fixed unit of measure. Interval data are always numeric. College admission SAT scores are an example of interval-scaled data. For example, three students with SAT math scores of 620, 550, and 470 can be ranked or ordered in terms of best performance to poorest performance in math. In addition, the differences between the scores are meaningful. For instance, s tudent 1 scored 620 2 550 5 70 points more than student 2, while student 2 scored 550 2 470 5 80 points more than student 3. The scale of measurement for a variable is a ratio scale if the data have all the prop erties of interval data and the ratio of two values is meaningful. Variables such as distance, height, weight, and time use the ratio scale of measurement. This scale requires that Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Chapter 1 Data and Statistics

a zero value be included to indicate that nothing exists for the variable at the zero point. For example, consider the cost of an automobile. A zero value for the cost would indicate that the automobile has no cost and is free. In addition, if we compare the cost of $30,000 for one automobile to the cost of $15,000 for a second automobile, the ratio property shows that the first automobile is $30,000/$15,000 5 2 times, or twice, the cost of the second automobile.

Categorical and Quantitative Data

The statistical method appropriate for summarizing data depends upon whether the data are categorical or quantitative.

Data can be classified as either categorical or quantitative. Data that can be grouped by specific categories are referred to as categorical data. Categorical data use either the nominal or ordinal scale of measurement. Data that use numeric values to indicate how much or how many are referred to as quantitative data. Quantitative data are obtained using either the interval or ratio scale of measurement. A categorical variable is a variable with categorical data, and a quantitative variable is a variable with quantitative data. The statistical analysis appropriate for a particular variable depends upon whether the variable is categorical or quantitative. If the variable is categorical, the statistical analysis is limited. We can summarize categorical data by counting the number of observations in each category or by computing the proportion of the observations in each category. However, even when the categorical data are identified by a numerical code, arithmetic operations such as addition, subtraction, multiplication, and division do not provide meaningful results. Section 2.1 discusses ways of summarizing categorical data. Arithmetic operations provide meaningful results for quantitative variables. For example, quantitative data may be added and then divided by the number of observations to compute the average value. This average is usually meaningful and easily interpreted. In general, more alternatives for statistical analysis are possible when data are quantitative. Section 2.2 and Chapter 3 provide ways of summarizing quantitative data.

Cross-Sectional and Time Series Data For purposes of statistical analysis, distinguishing between cross-sectional data and time series data is important. Cross-sectional data are data collected at the same or approximately the same point in time. The data in Table 1.1 are cross-sectional because they describe the five variables for the 60 World Trade Organization nations at the same point in time. Time series data are data collected over several time periods. For example, the time series in Figure 1.1 shows the U.S. average price per gallon of conventional regular gasoline between 2010 and 2015. Note that gasoline prices peaked in May 2011. Between June 2014 and January 2015, the average price per gallon dropped dramatically. In August 2015, the average price per gallon was $2.52. Graphs of time series data are frequently found in business and economic publications. Such graphs help analysts understand what happened in the past, identify any trends over time, and project future values for the time series. The graphs of time series data can take on a variety of forms, as shown in Figure 1.2. With a little study, these graphs are usually easy to understand and interpret. For example, Panel (A) in Figure1.2 is a graph that shows the Dow Jones Industrial Average Index from 2005 to 2015. In September 2005, the popular stock market index was near 10,400. Over the next two years the index rose to almost 14,000 in October 2007. However, notice the sharp decline in the time series after the high in 2007. By March 2009, poor economic conditions had caused the Dow Jones Industrial Average Index to return to the 7000 level. This was a scary and discouraging period for investors. However, by late 2009, the index was showing a recovery by reaching 10,000 and rising to a high of over 18,000 in May 2015. By October 2015, the index had dropped substantially to just under 16,300. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

1.2 Data

FIGURE 1.1 U.S. AVERAGE PRICE PER GALLON FOR CONVENTIONAL

REGULAR GASOLINE $4.50

Average Price per Gallon

$4.00 $3.50 $3.00 $2.50 $2.00 $1.50 $1.00 $0.50 $0.00 Jan-10

Aug-10

Mar-11

Oct-11

May-12

Dec-12

Jul-13

Feb-14

Sep-14

Apr-15 Aug-15

Date Source: Energy Information Administration, U.S. Department of Energy, September 2015.

The graph in Panel (B) shows the net income of McDonald’s Inc. from 2007 to 2015. The declining economic conditions in 2008 and 2009 were actually beneficial to McDonald’s as the company’s net income rose to all-time highs. The growth in McDonald’s net income showed that the company was thriving during the economic downturn as people were cutting back on the more expensive sit-down restaurants and seeking less expensive alternatives offered by McDonald’s. McDonald’s net income continued to new all-time highs in2010 and 2011, remained at about 5.5 billion from 2011 to 2013, decreased substantially in 2014, and dropped again in 2015. Analysts suspect that the drop in net income was due to loss of customers to newer competition such as Chipotle. Panel (C) shows the time series for the occupancy rate of hotels in South Florida over a one-year period. The highest occupancy rates, 95% and 98%, occur during the months of February and March when the climate of South Florida is attractive to tourists. In fact, January to April of each year is typically the high-occupancy season for South Florida hotels. On the other hand, note the low occupancy rates during the months of August to October, with the lowest occupancy rate of 50% occurring in September. High temperatures and the hurricane season are the primary reasons for the drop in hotel occupancy during this period. NOTES AND COMMENTS 1. An observation is the set of measurements obtained for each element in a data set. Hence, the number of observations is always the same as the number of elements. The number of measurements obtained for each element equals the number of variables. Hence, the total number of data items can be determined by multiplying the number of observations by the number of variables.

2. Quantitative data may be discrete or continuous. Quantitative data that measure how many (e.g., number of calls received in 5 minutes) are discrete. Quantitative data that measure how much (e.g., weight or time) are continuous because no separation occurs between the possible data values.

Chapter 1 Data and Statistics

Dow Jones Industrial Average

FIGURE 1.2 A Variety of Graphs of Time Series Data 20,000 18,000 16,000 14,000 12,000 10,000 8,000 6,000 4,000 2,000 0 Sep-05 Sep-06 Sep-07 Sep-08 Sep-09 Sep-10 Sep-11 Sep-12 Sep-13 Sep-14 Sep-15

Date (A) Dow Jones Industrial Average

Net income ($ billions)

6 5 4 3 2 1 0

2007

2008

2009

2010

2011

2012

2013

2014

2015

Year (B) Net Income for McDonald’s Inc.

Percentage Occupied

100 80 60 40 20

ov D ec

Ju l A ug

n Ju

0 Month (C) Occupancy Rate of South Florida Hotels

1.3 Data Sources

Data Sources 1.3 Data can be obtained from existing sources, by conducting an observational study, or by conducting an experiment.

Existing Sources In some cases, data needed for a particular application already exist. Companies maintain a variety of databases about their employees, customers, and business operations. Data on employee salaries, ages, and years of experience can usually be obtained from internal personnel records. Other internal records contain data on sales, advertising expenditures, distri bution costs, inventory levels, and production quantities. Most companies also maintain detailed data about their customers. Table 1.2 shows some of the data commonly available from internal company records. Organizations that specialize in collecting and maintaining data make available substantial amounts of business and economic data. Companies access these external data sources through leasing arrangements or by purchase. Dun & Bradstreet, Bloomberg, and Dow Jones & Company are three firms that provide extensive business database services to clients. ACNielsen and Information Resources, Inc. built successful businesses collecting and processing data that they sell to advertisers and product manufacturers. Data are also available from a variety of industry associations and special interest organizations. The Travel Industry Association of America maintains travel-related information such as the number of tourists and travel expenditures by states. Such data would be of interest to firms and individuals in the travel industry. The Graduate Management Admission Council maintains data on test scores, student characteristics, and graduate management education programs. Most of the data from these types of sources are available to qualified users at a modest cost. The Internet is an important source of data and statistical information. Almost all companies maintain websites that provide general information about the company as well as data on sales, number of employees, number of products, product prices, and product specifications. In addition, a number of companies now specialize in making information available over the Internet. As a result, one can obtain access to stock quotes, meal prices at restaurants, salary data, and an almost infinite variety of information. Government agencies are another important source of existing data. For instance, the U.S. Department of Labor maintains considerable data on employment rates, wage TABLE 1.2 EXAMPLES OF DATA AVAILABLE FROM INTERNAL COMPANY RECORDS

Source

Some of the Data Typically Available

Employee records

Name, address, social security number, salary, number of vacation days, number of sick days, and bonus

Production records

Part or product number, quantity produced, direct labor cost, and materials cost

Inventory records

Part or product number, number of units on hand, reorder level, economic order quantity, and discount schedule

Sales records

Product number, sales volume, sales volume by region, and sales volume by customer type

Credit records

Customer name, address, phone number, credit limit, and accounts receivable balance

Customer profile

Age, gender, income level, household size, address, and preferences

Chapter 1 Data and Statistics

TABLE 1.3 EXAMPLES OF DATA AVAILABLE FROM SELECTED GOVERNMENT AGENCIES

Government Agency

Some of the Data Available

Census Bureau

Population data, number of households, and household income

Federal Reserve Board

Data on the money supply, installment credit, exchange rates, and discount rates

Office of Management and Budget

Data on revenue, expenditures, and debt of the federal government

Department of Commerce

Data on business activity, value of shipments by industry, level of profits by industry, and growing and declining industries

Bureau of Labor Statistics

Consumer spending, hourly earnings, unemployment rate, safety records, and international statistics

rates, size of the labor force, and union membership. Table 1.3 lists selected governmental agencies and some of the data they provide. Most government agencies that collect and process data also make the results available through a website. Figure 1.3 shows the homepage for the U.S. Bureau of Labor Statistics website.

Observational Study In an observational study we simply observe what is happening in a particular situation, record data on one or more variables of interest, and conduct a statistical analysis of the resulting data. For example, researchers might observe a randomly selected group of FIGURE 1.3 U.S. BUREAU of labor statistics HOMEPAGE

Courtesy of U.S. Bureau of Labor Statistics

1.3 Data Sources

Studies of smokers and nonsmokers are observational studies because researchers do not determine or control who will smoke and who will not smoke.

customers that enter a Walmart supercenter to collect data on variables such as the length of time the customer spends shopping, the gender of the customer, the amount spent, and so on. Statistical analysis of the data may help management determine how factors such as the length of time shopping and the gender of the customer affect the amount spent. As another example of an observational study, suppose that researchers were interested in investigating the relationship between the gender of the CEO for a Fortune 500 company and the performance of the company as measured by the return on equity (ROE). To obtain data, the researchers selected a sample of companies and recorded the gender of the CEO and the ROE for each company. Statistical analysis of the data can help determine the relationship between performance of the company and the gender of the CEO. This example is an observational study because the researchers had no control over the gender of the CEO or the ROE at each of the companies that were sampled. Surveys and public opinion polls are two other examples of commonly used observational studies. The data provided by these types of studies simply enable us to observe opinions of the respondents. For example, the New York State legislature commissioned a telephone survey in which residents were asked if they would support or oppose an increase in the state gasoline tax in order to provide funding for bridge and highway repairs. Statistical analysis of the survey results will assist the state legislature in determining if it should introduce a bill to increase gasoline taxes.

Experiment The largest experimental statistical study ever conducted is believed to be the 1954 Public Health Service experiment for the Salk polio vaccine. Nearly 2 million children in grades 1, 2, and 3 were selected from throughout the United States.

The key difference between an observational study and an experiment is that an experiment is conducted under controlled conditions. As a result, the data obtained from a welldesigned experiment can often provide more information as compared to the data obtained from existing sources or by conducting an observational study. For example, suppose a pharmaceutical company would like to learn about how a new drug it has developed affects blood pressure. To obtain data about how the new drug affects blood pressure, researchers selected a sample of individuals. Different groups of individuals are given different dosage levels of the new drug, and before and after data on blood pressure are collected for each group. Statistical analysis of the data can help determine how the new drug affects blood pressure. The types of experiments we deal with in statistics often begin with the identification of a particular variable of interest. Then one or more other variables are identified and controlled so that data can be obtained about how the other variables influence the primary variable of interest. In Chapter 13 we discuss statistical methods appropriate for analyzing the data from an experiment.

Time and Cost Issues Anyone wanting to use data and statistical analysis as aids to decision making must be aware of the time and cost required to obtain the data. The use of existing data sources is desirable when data must be obtained in a relatively short period of time. If important data are not readily available from an existing source, the additional time and cost involved in obtaining the data must be taken into account. In all cases, the decision maker should consider the contribution of the statistical analysis to the decision-making process. The cost of data acquisition and the subsequent statistical analysis should not exceed the savings generated by using the information to make a better decision.

Data Acquisition Errors Managers should always be aware of the possibility of data errors in statistical studies. Using erroneous data can be worse than not using any data at all. An error in data acquisition Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Chapter 1 Data and Statistics

o ccurs whenever the data value obtained is not equal to the true or actual value that would be obtained with a correct procedure. Such errors can occur in a number of ways. For example, an interviewer might make a recording error, such as a transposition in writing the age of a 24-year-old person as 42, or the person answering an interview question might misinterpret the question and provide an incorrect response. Experienced data analysts take great care in collecting and recording data to ensure that errors are not made. Special procedures can be used to check for internal consistency of the data. For instance, such procedures would indicate that the analyst should review the accuracy of data for a respondent shown to be 22 years of age but reporting 20 years of work experience. Data analysts also review data with unusually large and small values, called outliers, which are candidates for possible data errors. In Chapter 3 we present some of the methods statisticians use to identify outliers. Errors often occur during data acquisition. Blindly using any data that happen to be available or using data that were acquired with little care can result in misleading information and bad decisions. Thus, taking steps to acquire accurate data can help ensure reliable and valuable decision-making information.

Descriptive Statistics 1.4 Most of the statistical information in newspapers, magazines, company reports, and other publications consists of data that are summarized and presented in a form that is easy for the reader to understand. Such summaries of data, which may be tabular, graphical, or numerical, are referred to as descriptive statistics. Refer to the data set in Table 1.1 showing data for 60 nations that participate in the World Trade Organization. Methods of descriptive statistics can be used to summarize these data. For example, consider the variable Fitch Outlook, which indicates the direction the nation’s credit rating is likely to move over the next two years. The Fitch Outlook is recorded as being negative, stable, or positive. A tabular summary of the data showing the number of nations with each of the Fitch Outlook ratings is shown in Table1.4. A graphical summary of the same data, called a bar chart, is shown in Figure1.4. These types of summaries make the data easier to interpret. Referring to Table 1.4 and Figure1.4, we can see that the majority of Fitch Outlook credit ratings are stable, with 65% of the nations having this rating. Negative and positive outlook credit ratings are similar, with slightly more nations having a negative outlook (18.3%) than a positive outlook (16.7%). A graphical summary of the data for quantitative variable Per Capita GDP in Table1.1, called a histogram, is provided in Figure 1.5. Using the histogram, it is easy to see that P er Capita GDP for the 60 nations ranges from $0 to $60,000, with the highest concentration between $10,000 and $20,000. Only one nation had a Per Capita GDP exceeding $50,000. Table 1.4 FREQUENCIES AND PERCENT FREQUENCIES FOR THE FITCH CREDIT

RATING OUTLOOK OF 60 NATIONS Percent Fitch Outlook Frequency Frequency (%) Positive Stable Negative

10 16.7 39 65.0 11 18.3

1.4 Descriptive Statistics

FIGURE 1.4 BAR CHART FOR THE FITCH CREDIT RATING OUTLOOK FOR 60 NATIONS 70

Percent Frequency

60 50 40 30 20 10 0

Negative

Stable

Positive

Fitch Outlook

In addition to tabular and graphical displays, numerical descriptive statistics are used to summarize data. The most common numerical measure is the average, or mean. Using the data on Per Capita GDP for the 60 nations in Table 1.1, we can compute the average by adding Per Capita GDP for all 60 nations and dividing the total by 60. Doing so provides an average Per Capita GDP of $21,387. This average provides a measure of the central tendency, or central location of the data.

FIGURE 1.5 HISTOGRAM OF PER CAPITA GDP FOR 60 NATIONS 20 18 16 14 Frequency

12 10 8 6 4 2 0

0–9,999 10,000– 20,000– 30,000– 40,000– 50,000– 19,999 29,999 39,999 49,999 59,999 Per Capita GDP

Chapter 1 Data and Statistics

There is a great deal of interest in effective methods for developing and presenting descriptive statistics. Chapters 2 and 3 devote attention to the tabular, graphical, and numerical methods of descriptive statistics.

Statistical Inference 1.5 Many situations require information about a large group of elements (individuals, companies, voters, households, products, customers, and so on). But, because of time, cost, and other considerations, data can be collected from only a small portion of the group. The larger group of elements in a particular study is called the population, and the smaller group is called the sample. Formally, we use the following definitions.

POPULATION

A population is the set of all elements of interest in a particular study.

SAMPLE

A sample is a subset of the population.

The U.S. government conducts a census every 10 years. Market research firms conduct sample surveys every day.

The process of conducting a survey to collect data for the entire population is called a census. The process of conducting a survey to collect data for a sample is called a sample survey. As one of its major contributions, statistics uses data from a sample to make estimates and test hypotheses about the characteristics of a population through a process referred to as statistical inference. As an example of statistical inference, let us consider the study conducted by Norris Electronics. Norris manufactures a high-intensity lightbulb used in a variety of electrical products. In an attempt to increase the useful life of the lightbulb, the product design group developed a new lightbulb filament. In this case, the population is defined as all lightbulbs that could be produced with the new filament. To evaluate the advantages of the new filament, 200 bulbs with the new filament were manufactured and tested. Data collected from this sample showed the number of hours each lightbulb operated before filament burnout. See Table 1.5. Suppose Norris wants to use the sample data to make an inference about the average hours of useful life for the population of all lightbulbs that could be produced with the new filament. Adding the 200 values in Table 1.5 and dividing the total by 200 provides the sample average lifetime for the lightbulbs: 76 hours. We can use this sample result toestimate that the average lifetime for the lightbulbs in the population is 76 hours. F igure1.6 provides a graphical summary of the statistical inference process for Norris E lectronics. Whenever statisticians use a sample to estimate a population characteristic of interest, they usually provide a statement of the quality, or precision, associated with the estimate. For the Norris example, the statistician might state that the point estimate of the average lifetime for the population of new lightbulbs is 76 hours with a margin of error of 64 hours. Thus, an interval estimate of the average lifetime for all lightbulbs produced with the new filament is 72 hours to 80 hours. The statistician can also state how confident he or she is that the interval from 72 hours to 80 hours contains the population average.

1.6 Statistical Analysis Using Microsoft Excel

TABLE 1.5 HOURS UNTIL BURNOUT FOR A SAMPLE OF 200 LIGHTBULBS

FOR THE NORRIS ELECTRONICS EXAMPLE

Norris

107 73 68 97 76 79 94 59 98 57 54 65 71 70 84 88 62 61 79 98 66 62 79 86 68 74 61 82 65 98 62 116 65 88 64 79 78 79 77 86 74 85 73 80 68 78 89 72 58 69 92 78 88 77 103 88 63 68 88 81 75 90 62 89 71 71 74 70 74 70 65 81 75 62 94 71 85 84 83 63 81 62 79 83 93 61 65 62 92 65 83 70 70 81 77 72 84 67 59 58 78 66 66 94 77 63 66 75 68 76 90 78 71 101 78 43 59 67 61 71 96 75 64 76 72 77 74 65 82 86 66 86 96 89 81 71 85 99 59 92 68 72 77 60 87 84 75 77 51 45 85 67 87 80 84 93 69 76 89 75 83 68 72 67 92 89 82 96 77 102 74 91 76 83 66 68 61 73 72 76 73 77 79 94 63 59 62 71 81 65 73 63 63 89 82 64 85 92 64 73

FIGURE 1.6 THE PROCESS OF STATISTICAL INFERENCE FOR THE NORRIS

ELECTRONICS EXAMPLE

1.6

1. Population consists of all bulbs manufactured with the new filament. Average lifetime is unknown.

2. A sample of 200 bulbs is manufactured with the new filament.

4. The sample average is used to estimate the population average.

3. The sample data provide a sample average lifetime of 76 hours per bulb.

Statistical Analysis Using Microsoft Excel Because statistical analysis typically involves working with large amounts of data, computer software is frequently used to conduct the analysis. In this book we show how statistical analysis can be performed using Microsoft Excel.

Chapter 1 Data and Statistics

We want to emphasize that this book is about statistics; it is not a book about spreadsheets. Our focus is on showing the appropriate statistical procedures for collecting, analyzing, presenting, and interpreting data. Because Excel is widely available in business organizations, you can expect to put the knowledge gained here to use in the setting where you currently, or soon will, work. If, in the process of studying this material, you become more proficient with Excel, so much the better. We begin most sections with an application scenario in which a statistical procedure is useful. After showing what the statistical procedure is and how it is used, we turn to showing how to implement the procedure using Excel. Thus, you should gain an understanding of what the procedure is, the situation in which it is useful, and how to implement it using the capabilities of Excel.

Data Sets and Excel Worksheets

To hide rows 15 through 54 of the Excel worksheet, first select rows 15 through 54. Then, right-click and choose the Hide option. To redisplay rows 15 through 54, just select rows 14 through 55, right-click, and select the Unhide option.

Data sets are organized in Excel worksheets in much the same way as the data set for the 60 nations that participate in the World Trade Organization that appears in Table 1.1 is organized. Figure 1.7 shows an Excel worksheet for that data set. Note that row 1 and column A contain labels. Cells Bl:Fl contain the variable names; cells A2:A61 contain the observation names; and cells B2:F61 contain the data that were collected. A purple fill color is used to highlight the cells that contain the data. Displaying a worksheet with this many rows on a single page of a textbook is not practical. In such cases we will hide selected rows to conserve space. In the Excel worksheet shown in Figure 1.7 we have hidden rows15 through 54 (observations 14 through 53) to conserve space. The data are the focus of the statistical analysis. Except for the headings in row 1, each row of the worksheet corresponds to an observation and each column corresponds to a variable. For instance, row 2 of the worksheet contains the data for the first observation, Armenia; row 3 contains the data for the second observation, Australia; row 3 contains the data for the third observation, Austria; and so on. The names in column A provide a convenient way to refer to each of the 60 observations in the study. Note that column B of the worksheet contains the data for the variable WTO Status, column C contains the data for the Per Capita GDP ($), and so on. Suppose now that we want to use Excel to analyze the Norris Electronics data shown in Table 1.5. The data in Table 1.5 are organized into 10 columns with 20 data values in each column so that the data would fit nicely on a single page of the text. Even though the table has several columns, it shows data for only one variable (hours until burnout). In statistical worksheets it is customary to put all the data for each variable in a single column. Refer to the Excel worksheet shown in Figure 1.8. To make it easier to identify each observation in the data set, we entered the heading Observation into cell Al and the numbers 1–200 into cells A2:A201. The heading Hours Until Burnout has been entered into cell B1, and the data for the 200 observations have been entered into cells B2:B201. Note that rows 7 through 195 have been hidden to conserve space.

Using Excel for Statistical Analysis To separate the discussion of a statistical procedure from the discussion of using Excel to implement the procedure, the material that discusses the use of Excel will usually be set apart in sections with headings such as Using Excel to Construct a Bar Chart and a Pie

1.6 Statistical Analysis Using Microsoft Excel

Figure 1.7 Excel Worksheet for the 60 Nations that Participate

in the World Trade Organization

Note: Rows 15–54 are hidden.

Figure 1.8 Excel Worksheet for the Norris Electronics Data Set

Note: Rows 7–195 are hidden

Chapter 1 Data and Statistics

Chart, Using Excel to Construct a Frequency Distribution, and so on. In using Excel for statistical analysis, four tasks may be needed: Enter/Access Data; Enter Functions and Formulas; Apply Tools; and Editing Options. Enter/Access Data: Select cell locations for the data and enter the data along with appropriate labels; or open an existing Excel file such as one of the DATAfiles that accompany the text. Enter Functions and Formulas: Select cell locations, enter Excel functions and formulas, and provide descriptive labels to identify the results. Apply Tools: Use Excel’s tools for data analysis and presentation. Editing Options: Edit the results to better identify the output or to create a different type of presentation. For example, when using Excel’s chart tools, we can edit the chart that is created by adding, removing, or changing chart elements such as the title, legend, data labels, and so on. Our approach will be to describe how these tasks are performed each time we use Excel to implement a statistical procedure. It will always be necessary to enter data or open an existing Excel file. But, depending on the complexity of the statistical analysis, only one of the second or third tasks may be needed. To illustrate how the discussion of Excel will appear throughout the book, we will show how to use Excel’s AVERAGE function to compute the average lifetime for the 200 burnout times in Table 1.5. Refer to Figure 1.9 as we describe the tasks involved. The worksheet shown in the foreground of Figure 1.9 displays the data for the problem and shows the results of the analysis. It is called the value worksheet. The worksheet shown in the Figure 1.9 Computing the Average Lifetime of Lightbulbs for Norris Electronics

Using Excel’s Average Function

1.7 Analytics

background displays the Excel formula used to compute the average lifetime and is called the formula worksheet. A purple fill color is used to highlight the cells that contain the data in both worksheets. In addition, a green fill color is used to highlight the cells containing the functions and formulas in the formula worksheet and the corresponding results in the value worksheet. Enter/Access Data: Open the DATAfile named Norris. The data are in cells B2:B201 and labels are in column A and cell B1. Enter Functions and Formulas: Excel’s AVERAGE function can be used to compute the mean by entering the following formula into cell E2: 5AVERAGE(B2:201) Similarly, the formulas 5MEDIAN(B2:B201) and 5MODE.SNGL(B2:B201) are entered into cells E3 and E4, respectively, to compute the median and the mode. To identify the result, the label Average Lifetime is entered into cell D2. Note that for this illustration the Apply Tools and Editing Options tasks were not required. The value worksheet shows that the value computed using the AVERAGE function is 76 hours.

Analytics 1.7

We adopt the definition of analytics developed by the Institute for Operations Research and the Management Sciences (INFORMS).

Because of the dramatic increase in available data, more cost-effective data storage, faster computer processing, and recognition by managers that data can be extremely valuable for understanding customers and business operations, there has been a dramatic increase in data-driven decision making. The broad range of techniques that may be used to support data-driven decisions comprise what has become known as analytics. Analytics is the scientific process of transforming data into insight for making better decisions. Analytics is used for data-driven or fact-based decision making, which is often seen as more objective than alternative approaches to decision making. The tools of analytics can aid decision making by creating insights from data, improving our ability to more accurately forecast for planning, helping us quantify risk, and yielding better alternatives through analysis. Analytics can involve a variety of techniques from simple reports to the most advanced optimization techniques (algorithms for finding the best course of action). Analytics is now generally thought to comprise three broad categories of techniques. These categories are descriptive analytics, predictive analytics, and prescriptive analytics. Descriptive analytics encompasses the set of analytical techniques that describe what has happened in the past. Examples of these types of techniques are data queries, reports, descriptive statistics, data visualization, data dash boards, and basic what-if spreadsheet models. Predictive analytics consists of analytical techniques that use models constructed from past data to predict the future or to assess the impact of one variable on another. For example, past data on sales of a product may be used to construct a mathematical model that predicts future sales. Such a model can account for factors such as the growth trajectory and seasonality of the product’s sales based on past growth and seasonal patterns. Point-of-sale scanner data from retail outlets may be used by a packaged food manufacturer to help estimate the lift in unit sales associated with coupons or sales events. Survey data and past purchase behavior may be used to help predict the market share of a new product. Each of these is an example of predictive analytics. Linear regression, time series analysis, and forecasting models fall into the category of predictive analytics; these techniques are discussed later in this text. Simulation, which is the use of probability

Chapter 1 Data and Statistics

and statistical computer models to better understand risk, also falls under the category of predictive analytics. Prescriptive analytics differs greatly from descriptive or predictive analytics. What distinguishes prescriptive analytics is that prescriptive models yield a best course of action to take. That is, the output of a prescriptive model is a best decision. Hence, prescriptive analytics is the set of analytical techniques that yield a course of action. Optimization models, which generate solutions that maximize or minimize some objective subject to a set of constraints, fall into the category of prescriptive models. The airline industry’s use of revenue management is an example of a prescriptive model. The airline industry uses past purchasing data as inputs into a model that recommends the pricing strategy across all flights that will maximize revenue for the company. How does the study of statistics relate to analytics? Most of the techniques in descriptive and predictive analytics come from probability and statistics. These include descriptive statistics, data visualization, probability and probability distributions, samp ling, and predictive modeling, including regression analysis and time series forecasting. Each of these techniques is discussed in this text. The increased use of analytics for data-driven decision making makes it more important than ever for analysts and mana gers to understand statistics and data analysis. Companies are increasingly seeking data savvy managers who know how to use descriptive and predictive models to make data-driven decisions. At the beginning of this section, we mentioned the increased availability of data as one of the drivers of the interest in analytics. In the next section we discuss this explosion in available data and how it relates to the study of statistics.

Big Data and Data Mining 1.8 With the aid of magnetic card readers, bar code scanners, and point-of-sale terminals, most organizations obtain large amounts of data on a daily basis. And, even for a small local restaurant that uses touch screen monitors to enter orders and handle billing, the amount of data collected can be substantial. For large retail companies, the sheer volume of data collected is hard to conceptualize, and figuring out how to effectively use these data to improve profitability is a challenge. Mass retailers such as Walmart capture data on 20 to 30 million transactions every day, telecommunication companies such as France Telecom and AT&T generate over 300 million call records per day, and Visa processes 6800 payment transactions per second or approximately 600 million transactions per day. In addition to the sheer volume and speed with which companies now collect data, more complicated types of data are now available and are proving to be of great value to businesses. Text data are collected by monitoring what is being said about a company’s products or services on social media such as Twitter. Audio data are collected from service calls (on a service call, you will often hear “this call may be monitored for quality control”). Video data are collected by in-store video cameras to analyze shopping behavior. Analyzing information generated by these nontraditional sources is more complicated because of the complex process of transforming the information into data that can be analyzed. Larger and more complex data sets are now often referred to as big data. Although there does not seem to be a universally accepted definition of big data, many think if it as a set of data that cannot be managed, processed, or analyzed with commonly available software in a reasonable amount of time. Many data analysts define big data by referring to the three v’s of data: volume, velocity, and variety. Volume refers to the amount of available data (the typical unit of measure for data is now a terabyte, which is 1012 bytes); velocity refers to the speed at which data is collected and processed; and variety refers to the different data types. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

1.8 Big Data and Data Mining

Statistical methods play an important role in data mining, both in terms of discovering relationships in the data and predicting future outcomes. However, a thorough coverage of data mining and the use of statistics in data mining is outside the scope of this text.

The term data warehousing is used to refer to the process of capturing, storing, and maintaining the data. Computing power and data collection tools have reached the point where it is now feasible to store and retrieve extremely large quantities of data in seconds. Analysis of the data in the warehouse may result in decisions that will lead to new strategies and higher profits for the organization. For example, General Electric (GE) captures a large amount of data from sensors on its aircraft engines each time a plane takes off or lands. Capturing these data allows GE to offer an important service to its customers; GE monitors the engine performance and can alert its customer when service is needed or a problem is likely to occur. The subject of data mining deals with methods for developing useful decision-making information from large databases. Using a combination of procedures from statistics, mathematics, and computer science, analysts “mine the data” in the warehouse to convert it into useful information, hence the name data mining. Dr. Kurt Thearling, a leading practitioner in the field, defines data mining as “the automated extraction of predictive information from (large) databases.” The two key words in Dr. Thearling’s definition are “automated” and “predictive.” Data mining systems that are the most effective use automated procedures to extract information from the data using only the most general or even vague queries by the user. And data mining software automates the process of uncovering hidden predictive information that in the past required hands-on analysis. The major applications of data mining have been made by companies with a strong consumer focus, such as retail businesses, financial organizations, and communication companies. Data mining has been successfully used to help retailers such as Amazon and Barnes & Noble determine one or more related products that customers who have already purchased a specific product are also likely to purchase. Then, when a customer logs on to the company’s website and purchases a product, the website uses pop-ups to alert the customer about additional products that the customer is likely to purchase. In another application, data mining may be used to identify customers who are likely to spend more than $20 on a particular shopping trip. These customers may then be identified as the ones to receive special e-mail or regular mail discount offers to encourage them to make their next shopping trip before the discount termination date. Data mining is a technology that relies heavily on statistical methodology such as multiple regression, logistic regression, and correlation. But it takes a creative integration of all these methods and computer science technologies involving artificial intelligence and machine learning to make data mining effective. A substantial investment in time and money is required to implement commercial data mining software packages developed by firms such as Oracle, Teradata, and SAS. The statistical concepts introduced in this text will be helpful in understanding the statistical methodology used by data mining software packages and enable you to better understand the statistical information that is developed. Because statistical models play an important role in developing predictive models in data mining, many of the concerns that statisticians deal with in developing statistical models are also applicable. For instance, a concern in any statistical study involves the issue of model reliability. Finding a statistical model that works well for a particular sample of data does not necessarily mean that it can be reliably applied to other data. One of the common statistical approaches to evaluating model reliability is to divide the sample data set into two parts: a training data set and a test data set. If the model developed using the training data is able to accurately predict values in the test data, we say that the model is reliable. One advantage that data mining has over classical statistics is that the enormous amount of data available allows the data mining software to partition the data set so that a model developed for the training data set may be tested for reliability on other data. In this sense, the partitioning of the data set allows data mining to develop models and relationships and then quickly observe if they are repeatable and valid with new and different data. On the

Chapter 1 Data and Statistics

other hand, a warning for data mining applications is that with so much data available, there is a danger of overfitting the model to the point that misleading associations and cause/effect conclusions appear to exist. Careful interpretation of data mining results and additional testing will help avoid this pitfall.

Ethical Guidelines for Statistical Practice 1.9 Ethical behavior is something we should strive for in all that we do. Ethical issues arise in statistics because of the important role statistics plays in the collection, analysis, presentation, and interpretation of data. In a statistical study, unethical behavior can take a variety of forms including improper sampling, inappropriate analysis of the data, development of misleading graphs, use of inappropriate summary statistics, and/or a biased interpretation of the statistical results. As you begin to do your own statistical work, we encourage you to be fair, thorough, objective, and neutral as you collect data, conduct analyses, make oral presentations, and present written reports containing information developed. As a consumer of statistics, you should also be aware of the possibility of unethical statistical behavior by others. When you see statistics in newspapers, on television, on the Internet, and so on, it is a good idea to view the information with some skepticism, always being aware of the source as well as the purpose and objectivity of the statistics provided. The American Statistical Association, the nation’s leading professional organization for statistics and statisticians, developed the report “Ethical Guidelines for Statistical Practice”2 to help statistical practitioners make and communicate ethical decisions and assist students in learning how to perform statistical work responsibly. The report contains 67guidelines organized into eight topic areas: Professionalism; Responsibilities to Funders, Clients, and Employers; Responsibilities in Publications and Testimony; Responsibilities to Research Subjects; Responsibilities to Research Team Colleagues; Responsibilities to Other Statisticians or Statistical Practitioners; Responsibilities Regarding Allegations of Misconduct; and Responsibilities of Employers Including Organizations, Individuals, Attorneys, or Other Clients Employing Statistical Practitioners. One of the ethical guidelines in the professionalism area addresses the issue of running multiple tests until a desired result is obtained. Let us consider an example. In Section 1.5 we discussed a statistical study conducted by Norris Electronics involving a sample of 200 highintensity lightbulbs manufactured with a new filament. The average lifetime for the sample, 76 hours, provided an estimate of the average lifetime for all lightbulbs produced with the new filament. However, consider this. Because Norris selected a sample of bulbs, it is reasonable to assume that another sample would have provided a different average lifetime. Suppose Norris’s management had hoped the sample results would enable them to claim that the average lifetime for the new lightbulbs was 80 hours or more. Suppose further that Norris’s management decides to continue the study by manufacturing and testing repeated samples of 200 lightbulbs with the new filament until a sample mean of 80 hours or more is obtained. If the study is repeated enough times, a sample may eventually be obtained—by chance alone—that would provide the desired result and enable Norris to make such a claim. In this case, consumers would be misled into thinking the new product is better than it actually is. Clearly, this type of behavior is unethical and represents a gross misuse of statistics in practice. Several ethical guidelines in the responsibilities and publications and testimony area deal with issues involving the handling of data. For instance, a statistician must account for

American Statistical Association, “Ethical Guidelines for Statistical Practice,” 1999.

Summary

all data considered in a study and explain the sample(s) actually used. In the Norris Electronics study the average lifetime for the 200 bulbs in the original sample is 76 hours; this is considerably less than the 80 hours or more that management hoped to obtain. Suppose now that after reviewing the results showing a 76-hour average lifetime, Norris discards all the observations with 70 or fewer hours until burnout, allegedly because these bulbs contain imperfections caused by startup problems in the manufacturing process. After these lightbulbs are discarded, the average lifetime for the remaining lightbulbs in the sample turns out to be 82 hours. Would you be suspicious of Norris’s claim that the lifetime for their lightbulbs is 82 hours? If the Norris lightbulbs showing 70 or fewer hours until burnout were discarded simply to provide an average lifetime of 82 hours, there is no question that discarding the lightbulbs with 70 or fewer hours until burnout is unethical. But, even if the discarded lightbulbs contain imperfections due to startup problems in the manufacturing process—and, as a result, should not have been included in the analysis—the statistician who conducted the study must account for all the data that were considered and explain how the sample actually used was obtained. To do otherwise is potentially misleading and would constitute unethical behavior on the part of both the company and the statistician. A guideline in the shared values section of the American Statistical Association report states that statistical practitioners should avoid any tendency to slant statistical work toward predetermined outcomes. This type of unethical practice is often observed when unrepresentative samples are used to make claims. For instance, in many areas of the country smoking is not permitted in restaurants. Suppose, however, a lobbyist for the tobacco industry interviews people in restaurants where smoking is permitted in order to estimate the percentage of people who are in favor of allowing smoking in restaurants. The sample results show that 90% of the people interviewed are in favor of allowing smoking in restaurants. Based upon these sample results, the lobbyist claims that 90% of all people who eat in restaurants are in favor of permitting smoking in restaurants. In this case we would argue that sampling only persons eating in restaurants that allow smoking has biased the results. If only the final results of such a study are reported, readers unfamiliar with the details of the study (i.e., that the sample was collected only in restaurants allowing smoking) can be misled. The scope of the American Statistical Association’s report is broad and includes ethical guidelines that are appropriate not only for a statistician, but also for consumers of statistical information. We encourage you to read the report to obtain a better perspective of ethical issues as you continue your study of statistics and to gain the background for determining how to ensure that ethical standards are met when you start to use statistics in practice.

Summary Statistics is the art and science of collecting, analyzing, presenting, and interpreting data. Nearly every college student majoring in business or economics is required to take a course in statistics. We began the chapter by describing typical statistical applications for business and economics. Data consist of the facts and figures that are collected and analyzed. The four scales of measurement used to obtain data on a particular variable are nominal, ordinal, interval, and ratio. The scale of measurement for a variable is nominal when the data are labels or names used to identify an attribute of an element. The scale is ordinal if the data demonstrate the properties of nominal data and the order or rank of the data is meaningful. The scale is interval if the data demonstrate the properties of ordinal data and the interval b etween values is expressed in terms of a fixed unit of measure. Finally, the scale of measurement is ratio if the data show all the properties of interval data and the ratio of two values is meaningful. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Chapter 1 Data and Statistics

For purposes of statistical analysis, data can be classified as categorical or quantitative. Categorical data use labels or names to identify an attribute of each element. Categorical data use either the nominal or ordinal scale of measurement and may be nonnumeric or numeric. Quantitative data are numeric values that indicate how much or how many. Quantitative data use either the interval or ratio scale of measurement. Ordinary arithmetic operations are meaningful only if the data are quantitative. Therefore, statistical computations used for quantitative data are not always appropriate for categorical data. In Sections 1.4 and 1.5 we introduced the topics of descriptive statistics and statistical inference. Descriptive statistics are the tabular, graphical, and numerical methods used to summarize data. The process of statistical inference uses data obtained from a sample tomake estimates or test hypotheses about the characteristics of a population. The last four sections of the chapter provide information on the role of computers in statistical analysis, an introduction to the relatively new fields of analytics, data mining, and big data, and a summary of ethical guidelines for statistical practice.

Glossary AnalyticsThe scientific process of transforming data into insight for making better decisions. Big dataA set of data that cannot be managed, processed, or analyzed with commonly available software in a reasonable amount of time. Big data are characterized by great volume (a large amount of data), high velocity (fast collection and processing), or wide variety (could include nontraditional data such as video, audio, and text). Categorical dataLabels or names used to identify an attribute of each element. Categorical data use either the nominal or ordinal scale of measurement and may be nonnumeric or numeric. Categorical variableA variable with categorical data. CensusA survey to collect data on the entire population. Cross-sectional dataData collected at the same or approximately the same point in time. Data The facts and figures collected, analyzed, and summarized for presentation and interpretation. Data mining The process of using procedures from statistics and computer science to extract useful information from extremely large databases. Data setAll the data collected in a particular study. Descriptive analyticsAnalytical techniques that describe what has happened in the past. Descriptive statisticsTabular, graphical, and numerical summaries of data. ElementsThe entities on which data are collected. Interval scaleThe scale of measurement for a variable if the data demonstrate the properties of ordinal data and the interval between values is expressed in terms of a fixed unit of measure. Interval data are always numeric. Nominal scaleThe scale of measurement for a variable when the data are labels or names used to identify an attribute of an element. Nominal data may be nonnumeric or numeric. ObservationThe set of measurements obtained for a particular element. Ordinal scaleThe scale of measurement for a variable if the data exhibit the properties of nominal data and the order or rank of the data is meaningful. Ordinal data may be nonnumeric or numeric. PopulationThe set of all elements of interest in a particular study. Predictive analyticsAnalytical techniques that use models constructed from past data to predict the future or assess the impact of one variable on another. Prescriptive analyticsAnalytical techniques that yield a course of action. Quantitative dataNumeric values that indicate how much or how many of something. Quantitative data are obtained using either the interval or ratio scale of measurement. Quantitative variableA variable with quantitative data. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Supplementary Exercises

Ratio scaleThe scale of measurement for a variable if the data demonstrate all the properties of interval data and the ratio of two values is meaningful. Ratio data are always numeric. SampleA subset of the population. Sample surveyA survey to collect data on a sample. Statistical inferenceThe process of using data obtained from a sample to make estimates or test hypotheses about the characteristics of a population. StatisticsThe art and science of collecting, analyzing, presenting, and interpreting data. Time series dataData collected over several time periods. VariableA characteristic of interest for the elements.

Supplementary Exercises 1. D iscuss the differences between statistics as numerical facts and statistics as a discipline or field of study. 2. T ablet PC Comparison provides a wide variety of information about tablet computers. Their website enables consumers to easily compare different tablets using factors such as cost, type of operating system, display size, battery life, and CPU manufacturer. A sample of 10tablet computers is shown in Table 1.6 (Tablet PC Comparison website, February28, 2013). a. How many elements are in this data set? b. How many variables are in this data set? c. Which variables are categorical and which variables are quantitative? d. What type of measurement scale is used for each of the variables? 3. Refer to Table 1.6. a. What is the average cost for the tablets? b. Compare the average cost of tablets with a Windows operating system to the average cost of tablets with an Android operating system. c. What percentage of tablets use a CPU manufactured by TI OMAP? d. What percentage of tablets use an Android operating system? 4. T able 1.7 shows data for eight cordless telephones (Consumer Reports, November 2012). The Overall Score, a measure of the overall quality for the cordless telephone, ranges from 0 to 100. Voice Quality has possible ratings of poor, fair, good, very good, and excellent. Talk Time is the manufacturer’s claim of how long the handset can be used when it is fully charged. Table 1.6 Product Information for 10 Tablet Computers

Tablet Acer Iconia W510 Amazon Kindle Fire HD Apple iPad 4 HP Envy X2 Lenovo ThinkPad Tablet Microsoft Surface Pro Motorola Droid XYboard Samsung Ativ Smart PC Samsung Galaxy Tab Sony Tablet S

Cost ($)

Operating System

Display Size (inches)

Battery Life (hours)

CPU Manufacturer

599 299 499 860 668 899 530 590 525 360

Windows Android iOS Windows Windows Windows Android Windows Android Android

10.1 8.9 9.7 11.6 10.1 10.6 10.1 11.6 10.1 9.4

8.5 9 11 8 10.5 4 9 7 10 8

Intel TI OMAP Apple Intel Intel Intel TI OMAP Intel Nvidia Nvidia

Chapter 1 Data and Statistics

Table 1.7 DATA FOR EIGHT CORDLESS TELEPHONES

Brand

Model

AT&T AT&T Panasonic Panasonic Uniden Uniden Vtech Vtech

CL84100 TL92271 4773B 6592T D2997 D1788 DS6521 CS6649

Price ($)

Overall Score

Voice Quality

Handset on Base

Talk Time (Hours)

60 80 100 70 45 80 60 50

73 70 78 72 70 73 72 72

Excellent Very Good Very Good Very Good Very Good Very Good Excellent Very Good

Yes No Yes No No Yes No Yes

7 7 13 13 10 7 7 7

a. How many elements are in this data set? b. For the variables Price, Overall Score, Voice Quality, Handset on Base, and Talk Time, which variables are categorical and which variables are quantitative? c. What scale of measurement is used for each variable? 5. R efer to the data set in Table 1.7. a. What is the average price for the cordless telephones? b. What is the average talk time for the cordless telephones? c. What percentage of the cordless telephones have a voice quality of excellent? d. What percentage of the cordless telephones have a handset on the base? 6. J .D. Power and Associates surveys new automobile owners to learn about the quality of recently purchased vehicles. The following questions were asked in the J.D. Power Initial Quality Survey, May 2012. a. Did you purchase or lease the vehicle? b. What price did you pay? c. What is the overall attractiveness of your vehicle’s exterior? (Unacceptable, Average, Outstanding, or Truly Exceptional) d. What is your average number of miles per gallon? e. What is your overall rating of your new vehicle? (l- to 10-point scale with 1 Unacceptable and 10 Truly Exceptional) Comment on whether each question provides categorical or quantitative data. 7. T he Kroger Company is one of the largest grocery retailers in the United States, with over 2000 grocery stores across the country. Kroger uses an online customer opinion questionnaire to obtain performance data about its products and services and learn about what motivates its customers (Kroger website, April 2012). In the survey, Kroger customers were asked if they would be willing to pay more for products that had each of the following four characteristics. The four questions were as follows:

Would you pay more for

products that have a brand name? products that are environmentally friendly? products that are organic? products that have been recommended by others?

For each question, the customers had the option of responding Yes if they would pay more or No if they would not pay more. a. Are the data collected by Kroger in this example categorical or quantitative? b. What measurement scale is used? 8. T he Tennessean, an online newspaper located in Nashville, Tennessee, conducts a daily poll to obtain reader opinions on a variety of current issues. In a recent poll, 762 readers Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Supplementary Exercises

10.

11.

12.

responded to the following question: “If a constitutional amendment to ban a state income tax is placed on the ballot in Tennessee, would you want it to pass?” Possible responses were Yes, No, or Not Sure (The Tennessean website, February 15, 2013). a. What was the sample size for this poll? b. Are the data categorical or quantitative? c. Would it make more sense to use averages or percentages as a summary of the data for this question? d. Of the respondents, 67% said Yes, they would want it to pass. How many individuals provided this response? The Commerce Department reported receiving the following applications for the Malcolm Baldrige National Quality Award: 23 from large manufacturing firms, 18 from large service firms, and 30 from small businesses. a. Is type of business a categorical or quantitative variable? b. What percentage of the applications came from small businesses? The Bureau of Transportation Statistics Omnibus Household Survey is conducted annually and serves as an information source for the U.S. Department of Transportation. In one part of the survey the person being interviewed was asked to respond to the following statement: “Drivers of motor vehicles should be allowed to talk on a hand-held cell phone while driving.” Possible responses were strongly agree, somewhat agree, somewhat disagree, and strongly disagree. Forty-four respondents said that they strongly agree with this statement, 130 said that they somewhat agree, 165 said they somewhat disagree, and 741 said they strongly disagree with this statement. a. Do the responses for this statement provide categorical or quantitative data? b. Would it make more sense to use averages or percentages as a summary of the responses for this statement? c. What percentage of respondents strongly agree with allowing drivers of motor vehicles to talk on a hand-held cell phone while driving? d. Do the results indicate general support for or against allowing drivers of motor vehicles to talk on a hand-held cell phone while driving? In a Gallup telephone survey conducted on April 9–10, 2013, each person being interviewed was asked if they would vote for a law in their state that would increase the gas tax by up to 20 cents a gallon, with the new gas tax money going to improve roads and bridges and build more mass transportation in their state. Possible responses were vote for, vote against, and no opinion. Two hundred and ninety-five respondents said they would vote for the law, 672 said they would vote against the law, and 51 said they had no opinion (Gallup website, June 14, 2013). a. Do the responses for this question provide categorical or quantitative data? b. What was the sample size for this Gallup poll? c. What percentage of respondents would vote for a law increasing the gas tax? d. Do the results indicate general support for or against increasing the gas tax to improve roads and bridges and build more mass transportation? The Hawaii Visitors Bureau collects data on visitors to Hawaii. The following questions were among 16 asked in a questionnaire handed out to passengers during incoming airline flights. ●● This trip to Hawaii is my: 1st, 2nd, 3rd, 4th, etc. ●● The primary reason for this trip is: (10 categories, including vacation, convention, honeymoon) ●● Where I plan to stay: (11 categories, including hotel, apartment, relatives, camping) ●● Total days in Hawaii a. What is the population being studied? b. Is the use of a questionnaire a good way to reach the population of passengers on incoming airline flights? c. Comment on each of the four questions in terms of whether it will provide categorical or quantitative data.

Chapter 1 Data and Statistics

FIGURE 1.10 GOOGLE REVENUE 80 70 60 Revenue ($ billions)

50 40 30 20 10 0

2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 Year

13. Figure 1.10 provides a bar chart showing the annual revenue for Google from 2004 to 2014. (The Wall Street Journal, August 19, 2014). a. What is the variable of interest? b. Are the data categorical or quantitative? c. Are the data time series or cross-sectional? d. Comment on the trend in Google revenue over time. 14. T he following data show the number of rental cars in service for three rental car companies: Hertz, Avis, and Dollar, over a four-year period.

Company Hertz Dollar Avis

Year 1 327 167 204

Cars in Service (1000s) Year 2 Year 3 311 140 220

286 106 300

Year 4 290 108 270

a. C onstruct a time series graph for the years 1 to 4 showing the number of rental cars in service for each company. Show the time series for all three companies on the same graph. b. Comment on who appears to be the market share leader and how the market shares are changing over time. c. Construct a bar chart showing rental cars in service for Year 4. Is this chart based on cross-sectional or time series data? 15. T he U.S. Census Bureau tracks sales per month for various products and services through its Monthly Retail Trade Survey. Figure 1.11 shows monthly bookstore sales in millions of dollars for 2014. a. Are the data quantitative or categorical? b. Are the data cross-sectional or time series? c. Which month has the highest sales? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Supplementary Exercises

FIGURE 1.11 Monthly Bookstore Sales for 2014 1800 1600 1400

Sales (millions $)

1200 1000 800 600 400 200 0

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

Month

d. Which month has the second highest sales? e. Why do you think the answers to parts (c) and (d) might be the two highest months? Explain. 16. T he Energy Information Administration of the U.S. Department of Energy provided time series data for the U.S. average price per gallon of conventional regular gasoline between January 2007 and March 2012 (Energy Information Administration website, April 2012). Use the Internet to obtain the average price per gallon of conventional regular gasoline since March 2012. a. Extend the graph of the time series shown in Figure 1.1. b. What interpretations can you make about the average price per gallon of conventional regular gasoline since March 2012? c. Does the time series continue to show a summer increase in the average price per gallon? Explain. 17. A manager of a large corporation recommends a $10,000 raise be given to keep a valued subordinate from moving to another company. What internal and external sources of data might be used to decide whether such a salary increase is appropriate? 18. A random telephone survey of 1021 adults (aged 18 and older) was conducted by Opinion Research Corporation on behalf of CompleteTax, an online tax preparation and e-filing service. The survey results showed that 684 of those surveyed planned to file their taxes electronically. a. Develop a descriptive statistic that can be used to estimate the percentage of all taxpayers who file electronically. b. The survey reported that the most frequently used method for preparing the tax return was to hire an accountant or professional tax preparer. If 60% of the people surveyed had their tax return prepared this way, how many people used an accountant or professional tax preparer? c. Other methods that the person filing the return often used include manual preparation, use of an online tax service, and use of a software tax program. Would the data for the method for preparing the tax return be considered categorical or quantitative? 19. A Bloomberg Businessweek North American subscriber study collected data from a sample of 2861 subscribers. Fifty-nine percent of the respondents indicated an annual income of $75,000 or more, and 50% reported having an American Express creditcard. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Chapter 1 Data and Statistics

a. b. c. d. e.

hat is the population of interest in this study? W Is annual income a categorical or quantitative variable? Is ownership of an American Express card a categorical or quantitative variable? Does this study involve cross-sectional or time series data? Describe any statistical inferences Bloomberg Businessweek might make on the basis of the survey.

20. A survey of 131 investment managers in Barron’s Big Money poll revealed the following: ●● Forty-three percent of managers classified themselves as bullish or very bullish on the stock market. ●● The average expected return over the next 12 months for equities was 11.2%. ●● Twenty-one percent selected health care as the sector most likely to lead the market in the next 12months. ●● When asked to estimate how long it would take for technology and telecom stocks to resume sustainable growth, the managers’ average response was 2.5 years. a. Cite two descriptive statistics. b. Make an inference about the population of all investment managers concerning the average return expected on equities over the next 12 months. c. Make an inference about the length of time it will take for technology and telecom stocks to resume sustainable growth. 21. A seven-year medical research study reported that women whose mothers took the drug DES during pregnancy were twice as likely to develop tissue abnormalities that might lead to cancer as were women whose mothers did not take the drug. a. This study compared two populations. What were the p opulations? b. Do you suppose the data were obtained in a survey or an experiment? c. For the population of women whose mothers took the drug DES during pregnancy, a sample of 3980 women showed that 63 developed tissue abnormalities that might lead tocancer. Provide a descriptive statistic that could be used to estimate the number of women out of 1000 in this population who have tissue abnormalities. d. For the population of women whose mothers did not take the drug DES during pregnancy, what is the estimate of the number of women out of 1000 who would be expected to have tissue abnormalities? e. Medical studies often use a relatively large sample (in this case, 3980). Why? 22. A survey conducted by Better Homes and Gardens Real Estate LLC showed that one in five U.S. homeowners have either moved from their home or would like to move because their neighborhood or community isn’t ideal for their lifestyle (Better Homes and Gardens Real Estate website, September 26, 2013). The top lifestyle priorities of respondents when searching for their next home include ease of commuting by car, access to health and safety services, family-friendly neighborhood, availability of retail stores, access to cultural activities, public transportation access, and nightlife and restaurant access. Suppose a real estate agency in Denver, Colorado, hired you to conduct a similar study to determine the top lifestyle priorities for clients that currently have a home listed for sale with the agency or have hired the agency to help them locate a new home. a. What is the population for the survey you will be conducting? b. How would you collect the data for this study? 23. P ew Research Center is a nonpartisan polling organization that provides information about issues, attitudes, and trends shaping America. In a poll, Pew researchers found that 73% of teens aged 13–17 have a smartphone, 15% have a basic phone and 12% have no phone. The study also asked the respondents how they communicated with their closest friend. Of those with a smartphone, 58% responded texting, 17% social media and 10% phone calls. Of those with no smartphone, 25% responded texting, 29% social media and 21% phone calls. (Pew Research Center website, October 2015). Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Supplementary Exercises

a. O ne statistic (58%) concerned the use of texting to contact his/her closest friend, if the teen owns a smartphone. To what population is that applicable? b. Another statistic (25%) concerned the use of texting by those who do not own a smartphone. To what population is that applicable? c. Do you think the Pew researchers conducted a census or a sample survey to obtain their results? Why? 24. A sample of midterm grades for five students showed the following results: 72, 65, 82, 90, 76. Which of the following statements are correct, and which should be challenged as being too generalized? a. The average midterm grade for the sample of five students is 77. b. The average midterm grade for all students who took the exam is 77. c. An estimate of the average midterm grade for all students who took the exam is 77. d. More than half of the students who take this exam will score between 70 and 85. e. If five other students are included in the sample, their grades will be between 65 and 90. 25. T able 1.8 shows a data set containing information for 25 of the shadow stocks tracked by the American Association of Individual Investors. Shadow stocks are common stocks of smaller companies that are not closely followed by Wall Street analysts. The data set is also on the website that accompanies the text in the DATAfile named Shadow02.

TABLE 1.8 DATA SET FOR 25 SHADOW STOCKS

Gross Market Price/ Profit Ticker Cap Earnings Margin Company Exchange Symbol ($ millions) Ratio (%)

Shadow02

DeWolfe Companies North Coast Energy Hansen Natural Corp. MarineMax, Inc. Nanometrics Incorporated TeamStaff, Inc. Environmental Tectonics Measurement Specialties SEMCO Energy, Inc. Party City Corporation Embrex, Inc. Tech/Ops Sevcon, Inc. ARCADIS NV Qiao Xing Universal Tele. Energy West Incorporated Barnwell Industries, Inc. Innodata Corporation Medical Action Industries Instrumentarium Corp. Petroleum Development Drexler Technology Corp. Gerber Childrenswear Inc. Gaiam, Inc. Artesian Resources Corp. York Water Company

AMEX DWL OTC NCEB OTC HANS NYSE HZO OTC NANO OTC TSTF AMEX ETC AMEX MSS NYSE SEN OTC PCTY OTC EMBX AMEX TO OTC ARCAF OTC XING OTC EWST AMEX BRN OTC INOD OTC MDCI OTC INMRY OTC PETD OTC DRXR NYSE GCW OTC GAIA OTC ARTNA OTC YORW

36.4 52.5 41.1 111.5 228.6 92.1 51.1 101.8 193.4 97.2 136.5 23.2 173.4 64.3 29.1 27.3 66.1 137.1 240.9 95.9 233.6 126.9 295.5 62.8 92.2

8.4 6.2 14.6 7.2 38.0 33.5 35.8 26.8 18.7 15.9 18.9 20.7 8.8 22.1 9.7 7.4 11.0 26.9 3.6 6.1 45.6 7.9 68.2 20.5 22.9

36.7 59.3 44.8 23.8 53.3 4.1 35.9 37.6 23.6 36.4 59.5 35.7 9.6 30.8 16.3 73.4 29.6 30.6 52.1 19.4 53.6 25.8 60.7 45.5 74.2

Chapter 1 Data and Statistics

a. H ow many variables are in the data set? b. Which of the variables are categorical and which are quantitative? c. For the Exchange variable, show the frequency and the percent frequency for AMEX, NYSE, and OTC. Construct a bar graph similar to Figure 1.4 for the Exchange variable. d. Show the frequency distribution for the Gross Profit Margin using the five intervals: 0–14.9, 15–29.9, 30–44.9, 45–59.9, and 60–74.9. Construct a histogram similar to Figure 1.5. e. What is the average price/earnings ratio?

CHAPTER

Descriptive Statistics: Tabular and Graphical Displays CONTENTS STATISTICS IN PRACTICE: COLGATE-PALMOLIVE COMPANY 2.1

2.2

UMMARIZING data for a S CATEGORICAL Variable Frequency Distribution Relative Frequency and Percent Frequency Distributions Using Excel to Construct a Frequency Distribution, a Relative Frequency Distribution, and a Percent Frequency Distribution Bar Charts and Pie Charts Using Excel to Construct a Bar Chart and a Pie Chart UMMARIZING data for a S QUANTITATIVE Variable Frequency Distribution Relative Frequency and Percent Frequency Distributions Using Excel to Construct a Frequency Distribution Dot Plot Histogram Using Excel’s Recommended Charts Tool to Construct a Histogram Cumulative Distributions Stem-and-Leaf Display

2.3 Summarizing data for two variables using tables Crosstabulation Using Excel’s PivotTable Tool to Construct a Crosstabulation Simpson’s Paradox 2.4

ummarizing data for S two variables using graphical displays Scatter Diagram and Trendline Using Excel to Construct a Scatter Diagram and a Trendline Side-by-Side and Stacked Bar Charts Using Excel’s Recommended Charts Tool to Construct Sideby-Side and Stacked Bar Charts

2.5

Data visualizatIon: best practices in creating effective graphical displays Creating Effective Graphical Displays Choosing the Type of Graphical Display Data Dashboards Data Visualization in Practice: Cincinnati Zoo and Botanical Garden

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

STATISTICS in PRACTICE COLGATE-PALMOLIVE COMPANY*

*The authors are indebted to William R. Fowle, Manager of Quality Assurance, Colgate-Palmolive Company, for providing this Statistics in Practice.

The Colgate-Palmolive Company uses statistical summaries to help maintain the quality of its products.

these methods is to summarize data so that the data can be easily understood and interpreted.

Frequency Distribution of Density Data Density Frequency .29–.30 .31–.32 .33–.34 .35–.36 .37–.38 .39–.40 Total

30 75 32 9 3 1 150

H istogram of Density Data 75

Frequency

The Colgate-Palmolive Company started as a small soap and candle shop in New York City in 1806. Today, Colgate- Palmolive employs more than 38,000 people working in more than 200 countries and territories around the world. Although best known for its brand names of Colgate, Palmolive, and Fabuloso, the company also markets Irish Spring, Hill’s Science Diet, andAjax products. The Colgate-Palmolive Company uses statistics in its quality assurance program for home laundry detergent products. One concern is customer satisfaction with the quantity of detergent in a carton. Every carton in each size category is filled with the same amount of detergent by weight, but the volume of detergent is a ffected by the density of the detergent powder. For instance, if the powder density is on the heavy side, a smaller volume of detergent is needed to reach the carton’s specified weight. As a result, the carton may appear to be underfilled when opened by the consumer. To control the problem of heavy detergent powder, limits are placed on the acceptable range of powder density. Statistical samples are taken periodically, and the density of each powder sample is measured. Data summaries are then provided for operating personnel so that corrective action can be taken if necessary to keep the density within the desired quality specifications. A frequency distribution for the densities of 150 samples taken over a one-week period and a histogram are shown in the accompanying table and figure. Density levels above .40 are unacceptably high. The frequency distribution and histogram show that the operation is meeting its quality guidelines with all of the densities less than or equal to .40. Managers viewing these statistical summaries would be pleased with the quality of the detergent production process. In this chapter, you will learn about tabular and graphical methods of descriptive statistics such as frequency distributions, bar charts, histograms, stem-andleaf displays, crosstabulations, and others. The goal of

Kurt Brady/Alamy Stock Photo

NEW YORK, NEW YORK

Less than 1% of samples near the undesirable .40 level

.29– .31– .33– .35– .37– .39– .30 .32 .34 .36 .38 .40

Density

2.1 Summarizing Data for a Categorical Variable

As indicated in Chapter 1, data can be classified as either categorical or quantitative. Categorical data use labels or names to identify categories of like items, and quantitative data are numerical values that indicate how much or how many. This chapter introduces the use of tabular and graphical displays for summarizing both categorical and quantitative data. Tabular and graphical displays can be found in annual reports, newspaper articles, and research studies. Everyone is exposed to these types of presentations. Hence, it is important to understand how they are constructed and how they should be interpreted. We begin with a discussion of the use of tablular and graphical displays to summarize the data for a single variable. This is followed by a discussion of the use of tabular and graphical displays to summarize the data for two variables in a way that reveals the relationship between the two variables. Data visualization is a term often used to describe the use of graphical displays to summarize and present information about a data set. The last section of this chapter provides an introduction to data visualization and provides guidelines for creating effective graphical displays.

Summarizing Data 2.1 for a Categorical Variable Frequency Distribution We begin the discussion of how tabular and graphical displays can be used to summarize categorical data with the definition of a frequency distribution.

FREQUENCY DISTRIBUTION

A frequency distribution is a tabular summary of data showing the number (frequency) of observations in each of several nonoverlapping categories or classes.

Let us use the following example to demonstrate the construction and interpretation of a frequency distribution for categorical data. Coca-Cola, Diet Coke, Dr. Pepper, Pepsi, and Sprite are five popular soft drinks. Assume that the data in Table 2.1 show the soft drink selected in a sample of 50 soft drink purchases.

TABLE 2.1 DATA FROM A SAMPLE OF 50 SOFT DRINK PURCHASES

SoftDrink

Coca-Cola Diet Coke Pepsi Diet Coke Coca-Cola Coca-Cola Dr. Pepper Diet Coke Pepsi Pepsi

Coca-Cola Dr. Pepper Sprite Coca-Cola Diet Coke Coca-Cola Coca-Cola Sprite Coca-Cola Diet Coke

Coca-Cola Diet Coke Coca-Cola Sprite Pepsi Coca-Cola Coca-Cola Coca-Cola Pepsi Coca-Cola

Sprite Dr. Pepper Pepsi Diet Coke Pepsi Coca-Cola Coca-Cola Coca-Cola Pepsi Dr. Pepper

Coca-Cola Diet Coke Pepsi Pepsi Pepsi Pepsi Coca-Cola Dr. Pepper Pepsi Sprite

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

TABLE 2.2

To develop a frequency distribution for these data, we count the number of times each soft drink appears in Table 2.1. Coca-Cola appears 19 times, Diet Coke appears 8 times, Dr. Pepper appears 5 times, Pepsi appears 13 times, and Sprite appears 5 times. These counts are summarized in the frequency distribution in Table 2.2. This frequency distribution provides a summary of how the 50 soft drink purchases are distributed across the five soft drinks. This summary offers more insight than the original data shown in Table 2.1. Viewing the frequency distribution, we see that CocaCola is the leader, Pepsi is second, Diet Coke is third, and Sprite and Dr. Pepper are tied for fourth. The frequency distribution summarizes information about the popularity of the five soft drinks.

FREQUENCY

DISTRIBUTION OF SOFT DRINK PURCHASES Soft Drink Frequency Coca-Cola Diet Coke Dr. Pepper Pepsi Sprite Total

19 8 5 13 5 50

Relative Frequency and Percent Frequency Distributions A frequency distribution shows the number (frequency) of observations in each of several nonoverlapping classes. However, we are often interested in the proportion, or percentage, of observations in each class. The relative frequency of a class equals the fraction or proportion of observations belonging to a class. For a data set with n observations, the relative frequency of each class can be determined as follows:

Relative Frequency

Relative frequency of a class 5

frequency of the class n

(2.1)

The percent frequency of a class is the relative frequency multiplied by 100. A relative frequency distribution gives a tabular summary of data showing the relative frequency for each class. A percent frequency distribution summarizes the percent frequency of the data for each class. Table 2.3 shows a relative frequency distribution and a percent frequency distribution for the soft drink data. In Table 2.3 we see that the relative frequency for Coca-Cola is 19/50 5 .38, the relative frequency for Diet Coke is 8/50 5 .16, and so on. From the percent frequency distribution, we see that 38% of the purchases were Coca-Cola, 16% of the purchases were Diet Coke, and so on. We can also note that 38% 1 26% 1 16% 5 80% of the purchases were for the top three soft drinks.

TABLE 2.3 RELATIVE frequency AND PERCENT FREQUENCY DISTRIBUTIONS

OF SOFT DRINK PURCHASES Soft Drink

Relative Frequency

Percent Frequency

Coca-Cola .38 38 Diet Coke .16 16 Dr. Pepper .10 10 Pepsi .26 26 Sprite .10 10 Total

1.00

100

2.1 Summarizing Data for a Categorical Variable

Using Excel to Construct a Frequency Distribution, a Relative Frequency Distribution, and a Percent Frequency Distribution We can use Excel’s Recommended PivotTables tool to construct a frequency distribution for the sample of 50 soft drink purchases. Two tasks are involved: Enter/Access Data and Apply Tools. Enter/Access Data: Open the DATAfile named SoftDrink. The data are in cells A2:A51 and a label is in cell A1. Apply Tools: The following steps describe how to use Excel’s Recommended PivotTables tool to construct a frequency distribution for the sample of 50 soft drink purchases. Step 1. Select any cell in the data set (cells A1:A51) Step 2. Click Insert on the Ribbon Step 3. In the Tables group click Recommended PivotTables; a preview showing the frequency distribution appears Step 4. Click OK; the frequency distribution will appear in a new worksheet The worksheet in Figure 2.1 shows the frequency distribution for the 50 soft drink purchases created using these steps. Also shown is the PivotTable Fields dialog box, a key component of every PivotTable report. We will discuss the use of the PivotTable Fields dialog box later in the chapter. Figure 2.1 Frequency Distribution of Soft Drink Purchases Constructed

Using Excel’s Recommended PivotTables Tool

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Figure 2.2 Relative Frequency and Percent Frequency Distributions

of Soft Drink Purchases Constructed Using Excel’s Recommended PivotTables Tool

Editing Options: You can easily change the column headings in the frequency distribution output. For instance, to change the current heading in cell A3 (Row Labels) to “Soft Drink,” click in cell A3 and type “Soft Drink”; to change the current heading in cell B3 (Count of Brand Purchased) to “Frequency,” click in cell B3 and type “Frequency”; and to change the current heading in A9 (Grand Total) to “Total,” click in cell A9 and type “Total.” The foreground and background worksheets shown in Figure 2.2 contain the revised headings; in addition, the headings “Relative Frequency” and “Percent Frequency” were entered into cells C3 and D3. We will now show how to construct the relative frequency and percent frequency distributions. Enter Functions and Formulas: Refer to Figure 2.2 as we describe how to create the relative and percent frequency distributions for the soft drink purchases. The formula worksheet is in the background and the value worksheet in the foreground. To compute the relative frequency for Coca-Cola using equation (2.1), we entered the formula 5B4/$B$9 into cell C4; the result, 0.38, is the relative frequency for Coca-Cola. Copying cell C4 to cells C5:C8 computes the relative frequencies for each of the other soft drinks. To compute the percent frequency for Coca-Cola, we entered the formula 5 C4*100 into cell D4. The result, 38, indicates that 38% of the soft drink purchases were Coca-Cola. Copying cell D4 to cells D5:D8 computes the percent frequencies for each of the other soft drinks. To compute the total of the relative and percent frequencies we used Excel’s SUM function in cells C9 and D9.

Bar Charts and Pie Charts A bar chart is a graphical display for depicting categorical data summarized in a frequency, relative frequency, or percent frequency distribution. On one axis of the graph we specify the labels that are used for the classes (categories). A frequency, relative Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

2.1 Summarizing Data for a Categorical Variable

Frequency

FIGURE 2.3 BAR Chart OF SOFT DRINK PURCHASES 20 18 16 14 12 10 8 6 4 2 0

Coca-Cola

Diet Coke

Dr. Pepper

Pepsi

Sprite

Soft Drink

In quality control applications, bar charts are used to identify the most important causes of problems. When the bars are arranged in descending order of height from left to right with the most frequently occurring cause appearing first, the bar chart is called a Pareto diagram. This diagram is named for its founder, Vilfredo Pareto, an Italian economist.

frequency, or percent frequency scale can be used for the other axis of the chart. Then, using a bar of fixed width drawn above or next to each class label, we extend the length of the bar until we reach the frequency, relative frequency, or percent frequency of the class. For categorical data, the bars should be separated to emphasize the fact that each class is separate. Figure 2.3 shows a bar chart of the frequency distribution for the 50 soft drink purchases. Note how the graphical presentation shows Coca-Cola, Pepsi, and Diet Coke to be the most preferred brands. In Figure 2.3 the horizontal axis was used to specify the labels for the categories; thus, the bars of the chart appear vertically in the display. In Excel, this type of display is referred to as a column chart. We could also display the bars for the chart horizontally by using the vertical axis to display the labels; Excel refers to this type of display as a bar chart. The choice of whether to display the bars vertically or horizontally depends upon what you want the final chart to look like. Throughout the text we will refer to either type of display as a bar chart. The pie chart provides another graphical display for presenting relative frequency and percent frequency distributions for categorical data. To construct a pie chart, we first draw a circle to represent all the data. Then we use the relative frequencies to subdivide the circle into sectors, or parts, that correspond to the relative frequency for each class. For example, because a circle contains 360 degrees and Coca-Cola shows a relative frequency of .38, the sector of the pie chart labeled Coca-Cola consists of .38(360) 5 136.8 degrees. The sector of the pie chart labeled Diet Coke consists of .16(360) 5 57.6 degrees. Similar calculations for the other classes yield the pie chart in Figure 2.4. The numerical values shown for each sector can be frequencies, relative frequencies, or percent frequencies. Numerous options involving the use of colors, shading, legends, text font, and three- dimensional perspectives are available to enhance the visual appearance of bar and pie charts. When used carefully, such options can provide a more effective display. But this is not always the case. For instance, consider the three-dimensional pie chart for the soft drink data shown in Figure 2.5. Compare it to the simpler presentation shown in Figure 2.4. The three-dimensional perspective adds no new understanding. In fact, because you have to view the three-dimensional pie chart in Figure 2.5 at an angle rather than from straight overhead, it can be more difficult to visualize. The use of a legend in Figure 2.5 also forces your eyes to shift back and forth between the key and the chart. The simpler

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

FIGURE 2.4 PIE CHART OF SOFT DRINK PURCHASES

Coca-Cola 38% Pepsi 26%

Sprite 10% Dr. Pepper 10%

Diet Coke 16%

FIGURE 2.5 Three-dimensional pie CHART OF SOFT DRINK PURCHASES

Coca-Cola Pepsi Diet Coke Dr. Pepper Sprite

chart shown in Figure 2.4, which shows the percentages and classes directly on the pie, is more effective. In general, pie charts are not the best way to present percentages for comparison. Research has shown that people are much better at accurately judging differences in length rather than differences in angles (or slices). When making such comparisons, we recommend you use a bar chart similar to Figure 2.3. In Section 2.5 we provide additional guidelines for creating effective visual displays.

Using Excel to Construct a Bar Chart and a Pie Chart We can use Excel’s Recommended Charts tool to construct a bar chart and a pie chart for the sample of 50 soft drink purchases. Two tasks are involved: Enter/Access Data and Apply Tools. Enter/Access Data: Open the DATAfile named SoftDrink. The data are in cell A2:A51 and a label is in cell A1. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

2.1 Summarizing Data for a Categorical Variable

Figure 2.6 Bar Chart of Soft Drink Purchases Constructed Using Excel’s

Recommended Charts Tool

Apply Tools: The following steps describe how to use Excel’s Recommended Charts tool to construct a bar chart for the sample of 50 soft drink purchases. Step 1. Select any cell in the data set (cells A1:A51) Step 2. Click Insert on the Ribbon Step 3. In the Charts group click Recommended Charts; a preview showing the bar chart appears Step 4. Click OK; the bar chart will appear in a new worksheet Excel refers to the bar chart in Figure 2.6 as a Clustered Column chart.

The worksheet in Figure 2.6 shows the bar chart for the 50 soft drink purchases created using these steps. Also shown are the frequency distribution and PivotTable Fields dialog box that were created by Excel in order to construct the bar chart. Thus, using Excel’s Recommended Charts tool you can construct a bar chart and a frequency distribution at the same time. Editing Options: You can easily edit the bar chart to display a different chart title and add axis titles. For instance, suppose you would like to use “Bar Chart of Soft Drink Purchases” as the chart title and insert “Soft Drink” for the horizontal axis title and “Frequency” for the vertical axis title. Step 1. Click the Chart Title and replace it with Bar Chart of Soft Drink Purchases Step 2. Click the Chart Elements button (located next to the top right corner of the chart)

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Step 3. When the list of chart elements appears: Click Axis Titles (creates placeholders for the axis titles) Step 4. Click the Horizontal (Category) Axis Title and replace it with Soft Drink Step 5. Click the Vertical (Value) Axis Title and replace it with Frequency The edited bar chart is shown in Figure 2.7. Creating a Pie Chart: To display a pie chart, select the bar chart (by clicking anywhere in the chart) to display three tabs (Analyze, Design, and Format) located on the Ribbon under the heading PivotChart Tools. Click the Design Tab and choose the Change Chart Type option to display the Change Chart Type dialog box. Click the Pie option and then OK to display a pie chart of the soft drink purchases.

Figure 2.7 Edited Bar Chart of Soft Drink Purchases Constructed Using

Excel’s Recommended Charts Tool

Exercises

Methods 1. T he response to a question has three alternatives: A, B, and C. A sample of 120 responses provides 60 A, 24 B, and 36 C. Show the frequency and relative frequency distributions. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

2.1 Summarizing Data for a Categorical Variable

2. A partial relative frequency distribution is given.

Class

Relative Frequency

A .22 B .18 C .40 D

a. b. c. d.

What is the relative frequency of class D? The total sample size is 200. What is the frequency of class D? Show the frequency distribution. Show the percent frequency distribution.

3. A questionnaire provides 58 Yes, 42 No, and 20 No-Opinion answers. a. In the construction of a pie chart, how many degrees would be in the section of the pie showing the Yes answers? b. How many degrees would be in the section of the pie showing the No answers? c. Construct a pie chart. d. Construct a bar chart.

Applications

Syndicated

4. For the 2010–2011 viewing season, the top five syndicated programs were Wheel of Fortune (WoF), Two and Half Men (THM), Jeopardy (Jep), Judge Judy (JJ), and the Oprah Winfrey Show (OWS) (Nielsen Media Research website, April 16, 2012). Data indicating the preferred shows for a sample of 50 viewers follow. WoF Jep JJ Jep THM THM WoF OWS Jep THM Jep OWS WoF WoF WoF WoF THM OWS THM WoF THM JJ JJ Jep THM OWS OWS JJ JJ Jep JJ WoF THM WoF WoF THM THM WoF JJ JJ Jep THM WoF Jep Jep WoF THM OWS OWS Jep a. b. c. d.

re these data categorical or quantitative? A Provide frequency and percent frequency distributions. Construct a bar chart and a pie chart. On the basis of the sample, which television show has the largest viewing audience? Which one is second?

5. In alphabetical order, the six most common last names in the United States are Brown, Johnson, Jones, Miller, Smith, and Williams (The World Almanac, 2012). Assume that a sample of 50 individuals with one of these last names provided the following data.

2012Names

Brown Williams Williams Williams Brown Smith Jones Smith Johnson Smith Miller Smith Brown Williams Johnson Johnson Smith Smith Johnson Brown Williams Miller Johnson Williams Johnson Williams Johnson Jones Smith Brown Johnson Smith Smith Brown Jones Jones Jones Smith Smith Miller Miller Jones Williams Miller Smith Jones Johnson Brown Johnson Miller

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Summarize the data by constructing the following: a. Relative and percent frequency distributions b. A bar chart c. A pie chart d. Based on these data, what are the three most common last names? 6. N ielsen Media Research provided the list of the 25 top-rated single shows in television history. The following data show the television network that produced each of these 25 top-rated shows.

Networks

CBS CBS NBC FOX CBS CBS NBC NBC NBC ABC ABC NBC ABC ABC NBC CBS NBC CBS ABC NBC NBC CBS CBS ABC CBS a. C onstruct a frequency distribution, percent frequency distribution, and bar chart for the data. b. Which network or networks have done the best in terms of presenting top-rated television shows? Compare the performance of ABC, CBS, and NBC. 7. T he Canmark Research Center Airport Customer Satisfaction Survey uses an online questionnaire to provide airlines and airports with customer satisfaction ratings for all aspects of the customers’ flight experience (airportsurvey website, July 2012). After completing a flight, customers receive an e-mail asking them to go to the website and rate a variety of factors, including the reservation process, the check-in process, luggage policy, cleanliness of gate area, service by flight attendants, food/beverage selection, on-time arrival, and so on. A five-point scale, with Excellent (E), Very Good (V), Good (G), Fair (F), and Poor (P), is used to record customer ratings. Assume that passengers on a Delta Airlines flight from Myrtle Beach, South Carolina to Atlanta, Georgia provided the following ratings for the question “Please rate the airline based on your overall experience with this flight.” The sample ratings are shown below.

AirSurvey

E E G V V E V V V E E G V E E V E E E V V V V F V E V E G E G E V E V E V V V V E E V V E P E V P V a. U se a percent frequency distribution and a bar chart to summarize these data. What do these summaries indicate about the overall customer satisfaction with the Delta flight? b. The online survey questionnaire enabled respondents to explain any aspect of the flight that failed to meet expectations. Would this be helpful information to a manager looking for ways to improve the overall customer satisfaction on Delta flights? Explain. 8. Data for a sample of 55 members of the Baseball Hall of Fame in Cooperstown, New York are shown here. Each observation indicates the primary position played by the Hall of Famers: pitcher (P), catcher (H), first base (1), second base (2), third base (3), shortstop (S), left field (L), center field (C), and right field (R).

BaseballHall

L P C H 2 P R 1 S S 1 L P R P P P P R C S L R P C C P P R P 2 3 P H L P 1 C P P P S 1 L R R 1 2 H S 3 H 2 L P a. Construct frequency and relative frequency distributions to summarize the data. b. What position provides the most Hall of Famers?

2.1 Summarizing Data for a Categorical Variable

c. What position provides the fewest Hall of Famers? d. What outfield position (L, C, or R) provides the most Hall of Famers? e. Compare infielders (1, 2, 3, and S) to outfielders (L, C, and R). 9. N early 1.8 million bachelor’s degrees and over 750,000 master’s degrees are awarded annually by U.S. postsecondary institutions (National Center for Education Statistics website, November 2014). The Department of Education tracks the field of study for these graduates in the following categories: Business (B), Computer Sciences and Engineering (CSE), Education (E), Humanities (H), Natural Sciences and Mathematics (NSM), Social and Behavioral Sciences (SBS), and Other (O). Consider the following samples of 100 graduates: Bachelor’s Degree Field of Study

Majors

SBS O H SBS B O SBS NSM H E

H B CSE SBS H H NSM NSM B B

H B CSE B SBS SBS NSM SBS B O

H O O H O H CSE O O B

E O CSE NSM B CSE H H O B

B H B B B CSE H H O B

O B H B O B E B NSM O

SBS O O O O E E SBS H O

NSM SBS O SBS B CSE SBS SBS E O

CSE O SBS SBS O SBS CSE NSM B O

B CSE E SBS H B B E B E

E H CSE E B E NSM E O O

B B NSM E B CSE H H B SBS

H E O B CSE O E O E B

E E B O SBS E B O CSE B

B O E E B O E O O O

Master’s Degree Field of Study O O O H SBS CSE B B CSE E

a. b. c. d. e.

HotelRatings

O E B H B B O E O O

B SBS B B B E E B O SBS

O B O E CSE CSE O O H E

Provide a percent frequency distribution of field of study for each degree. Construct a bar chart for field of study for each degree. What is the lowest percentage field of study for each degree? What is the highest percentage field of study for each degree? Which field of study has the largest increase in percentage from bachelor’s to masters’?

10. V irtualTourist provides ratings for hotels throughout the world. Ratings provided by 649 guests at the Sheraton Anaheim Hotel, located near the Disneyland Resort in Anaheim, California, can be found in the DATAfile named HotelRatings (VirtualTourist website, February 25, 2013). Possible responses were Excellent, Very Good, Average, Poor, and Terrible. a. Construct a frequency distribution. b. Construct a percent frequency distribution. c. Construct a bar chart for the percent frequency distribution. d. Comment on how guests rate their stay at the Sheraton Anaheim Hotel. e. Results for 1679 guests who stayed at Disney’s Grand Californian provided the following frequency distribution.

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Rating

Frequency

Excellent 807 Very Good 521 Average 200 Poor 107 Terrible 44

Compare the ratings for Disney’s Grand Californian with the results obtained for the Sheraton Anaheim Hotel.

Summarizing Data for a Quantitative Variable 2.2 Frequency Distribution TABLE 2.4

EAR-END AUDIT Y TIMES (IN DAYS) 12 14 19 18 15 15 18 17 20 27 22 23 22 21 33 28 14 18 16 13

Audit

As defined in Section 2.1, a frequency distribution is a tabular summary of data showing the number (frequency) of observations in each of several nonoverlapping categories or classes. This definition holds for quantitative as well as categorical data. However, with quantitative data we must be more careful in defining the nonoverlapping classes to be used in the frequency distribution. For example, consider the quantitative data in Table 2.4. These data show the time in days required to complete year-end audits for a sample of 20 clients of Sanderson and Clifford, a small public accounting firm. The three steps necessary to define the classes for a frequency distribution with quantitative data are as follows: 1. Determine the number of nonoverlapping classes. 2. Determine the width of each class. 3. Determine the class limits. Let us demonstrate these steps by developing a frequency distribution for the audit time data in Table 2.4. Number of classes Classes are formed by specifying ranges that will be used to group

the data. As a general guideline, we recommend using between 5 and 20 classes. For a small number of data items, as few as 5 or 6 classes may be used to summarize the data. For a larger number of data items, a larger number of classes is usually required. The goal is to use enough classes to show the variation in the data, but not so many classes that some contain only a few data items. Because the number of data items in Table 2.4 is relatively small (n 5 20), we chose to develop a frequency distribution with five classes. Making the classes the same width reduces the chance of inappropriate interpretations by the user.

Width of the classes The second step in constructing a frequency distribution for quan-

titative data is to choose a width for the classes. As a general guideline, we recommend that the width be the same for each class. Thus the choices of the number of classes and the width of classes are not independent decisions. A larger number of classes means a smaller class width, and vice versa. To determine an approximate class width, we begin by identifying the largest and smallest data values. Then, with the desired number of classes specified, we can use the following expression to determine the approximate class width.

Approximate class width 5

Largest data value 2 Smallest data value Number of classes

(2.2)

2.2 Summarizing Data for a Quantitative Variable

No single frequency distribution is best for a data set. Different people may construct different, but equally acceptable, frequency distributions. The goal is to reveal the natural grouping and variation in the data.

The approximate class width given by equation (2.2) can be rounded to a more convenient value based on the preference of the person developing the frequency distribution. For example, an approximate class width of 9.28 might be rounded to 10 simply because 10 is a more convenient class width to use in presenting a frequency distribution. For the data involving the year-end audit times, the largest data value is 33 and the smallest data value is 12. Because we decided to summarize the data with five classes, using equation (2.2) provides an approximate class width of (33 2 12)/5 5 4.2. We therefore decided to round up and use a class width of five days in the frequency distribution. In practice, the number of classes and the appropriate class width are determined by trial and error. Once a possible number of classes is chosen, equation (2.2) is used to find the approximate class width. The process can be repeated for a different number of classes. Ultimately, the analyst uses judgment to determine the combination of the number of classes and class width that provides the best frequency distribution for summarizing the data. For the audit time data in Table 2.4, after deciding to use five classes, each with a width of five days, the next task is to specify the class limits for each of the classes. Class limits Class limits must be chosen so that each data item belongs to one and only

TABLE 2.5

FREQUENCY DISTRIBUTION FOR THE AUDIT TIME DATA Audit Time (days) Frequency 10 –14 4 15–19 8 20 –24 5 25–29 2 30 –34 1 Total 20

one class. The lower class limit identifies the smallest possible data value assigned to the class. The upper class limit identifies the largest possible data value assigned to the class. In developing frequency distributions for categorical data, we did not need to specify class limits because each data item naturally fell into a separate class. But with quantitative data, such as the audit times in Table 2.4, class limits are necessary to determine where each data value belongs. Using the audit time data in Table 2.4, we selected 10 days as the lower class limit and 14 days as the upper class limit for the first class. This class is denoted 10–14 in Table 2.5. The smallest data value, 12, is included in the 10–14 class. We then selected 15 days as the lower class limit and 19 days as the upper class limit of the next class. We continued defining the lower and upper class limits to obtain a total of five classes: 10–14, 15–19, 20–24, 25–29, and 30–34. The largest data value, 33, is included in the 30–34 class. The difference between the lower class limits of adjacent classes is the class width. Using the first two lower class limits of 10 and 15, we see that the class width is 15 2 10 5 5. With the number of classes, class width, and class limits determined, a frequency distribution can be obtained by counting the number of data values belonging to each class. For example, the data in Table 2.4 show that four values—12, 14, 14, and 13— belong to the 10 –14 class. Thus, the frequency for the 10 –14 class is 4. Continuing this counting process for the 15–19, 20 –24, 25–29, and 30 –34 classes provides the frequency distribution in Table 2.5. Using this frequency distribution, we can observe the following: 1. The most frequently occurring audit times are in the class of 15–19 days. Eight of the 20 audit times belong to this class. 2. Only one audit required 30 or more days. Other conclusions are possible, depending on the interests of the person viewing the frequency distribution. The value of a frequency distribution is that it provides insights about the data that are not easily obtained by viewing the data in their original unorganized form. Class midpoint In some applications, we want to know the midpoints of the classes in a frequency distribution for quantitative data. The class midpoint is the value halfway between the lower and upper class limits. For the audit time data, the five class midpoints are 12, 17, 22, 27, and 32.

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Relative Frequency and Percent Frequency Distributions We define the relative frequency and percent frequency distributions for quantitative data in the same manner as for categorical data. First, recall that the relative frequency is the proportion of the observations belonging to a class. With n observations,

Relative frequency of class 5

frequency of the class n

The percent frequency of a class is the relative frequency multiplied by 100. Based on the class frequencies in Table 2.5 and with n 5 20, Table 2.6 shows the relative frequency distribution and percent frequency distribution for the audit time data. Note that .40 of the audits, or 40%, required from 15 to 19 days. Only .05 of the audits, or 5%, required 30 or more days. Again, additional interpretations and insights can be obtained by using Table 2.6.

Using Excel to Construct a Frequency Distribution We can use Excel’s PivotTable tool to construct a frequency distribution for the audit time data. Two tasks are involved: Enter/Access Data and Apply Tools. Enter/Access Data: Open the DATAfile named Audit. The data are in cells A2:A21 and a label is in cell A1. Apply Tools: The following steps describe how to use Excel’s PivotTable tool to construct a frequency distribution for the audit time data. When using Excel’s PivotTable tool, each column of data is referred to as a field. Thus, for the audit time example, the data appearing in cells A2:A21 and the label in cell A1 are referred to as the Audit Time field. Step 1. Select any cell in the data set (cells A1:A21) Step 2. Click Insert on the Ribbon Step 3. In the Tables group click PivotTable Step 4. When the Create PivotTable dialog box appears: Click OK; a PivotTable and PivotTable Fields dialog box will appear in a new worksheet Step 5. In the PivotTable Fields dialog box: Drag Audit Time to the Rows area Drag Audit Time to the Values area Step 6. Click on Sum of Audit Time in the Values area Step 7. Click Value Field Settings from the list of options that appears TABLE 2.6 RELATIVE frequency AND PERCENT FREQUENCY DISTRIBUTIONS FOR

THE AUDIT TIME DATA

Audit Time (days)

Relative Frequency

Percent Frequency

10 –14 .20 20 15 –19 .40 40 20 –24 .25 25 25 –29 .10 10 30 –34 .05 5 Total 1.00 100

2.2 Summarizing Data for a Quantitative Variable

Step 8. When the Value Field Settings dialog box appears: Under Summarize value field by, choose Count Click OK Figure 2.8 shows the resulting PivotTable Fields Dialog and the corresponding PivotTable. To construct the frequency distribution shown in Table 2.5, we must group the rows containing the audit times. The following steps accomplish this. Step 1. Right-click cell A4 in the PivotTable or any other cell containing an audit time. Step 2. Choose Group from the list of options that appears Step 3. When the Grouping dialog box appears: Enter 10 in the Starting at box Enter 34 in the Ending at box Enter 5 in the By box Click OK

The same Excel procedures we followed in the previous section can now be used to develop relative and percent frequency distributions if desired.

Figure 2.9 shows the completed PivotTable Fields dialog box and the corresponding PivotTable. We see that with the exception of the column headings, the PivotTable provides the same information as the frequency distribution shown in Table 2.5. Editing Options: You can easily change the labels in the PivotTable to match the labels in Table 2.5. For instance, to change the current heading in cell A3 (Row Labels) to “Audit Time (days),” click in cell A3 and type “Audit Time (days)”; to change the current heading in cell B3 (Count of Audit Time) to “Frequency,” click in cell B3 and type

Figure 2.8 PivotTable Fields Dialog Box and Initial PivotTable Used

to Construct a Frequency Distribution for the Audit Time Data

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Figure 2.9 Frequency Distribution for the Audit Time Data

Constructed Using Excel’s PivotTable Tool

“Frequency”; and to change the current heading in A9 (Grand Total) to “Total,” click in cell A9 and type “Total.” In smaller data sets or when there are a large number of classes, some classes may have no data values. In such cases, Excel’s PivotTable tool will remove these classes when constructing the frequency distribution. For presentation, however, we recommend editing the results to show all the classes, including those with no data values. The following steps show how this can be done. Exercise 21 is an example that has some classes with no data values.

Step 1. Right click on any cell in the Row Labels column of the PivotTable Step 2. Click Field Settings Step 3. When the Field Settings dialog box appears: Click the Layout and Print tab Choose Show items with no data Click OK

Dot Plot One of the simplest graphical summaries of data is a dot plot. A horizontal axis shows the range for the data. Each data value is represented by a dot placed above the axis. Figure 2.10 is the dot plot for the audit time data in Table 2.4. The three dots located above 18 on the horizontal axis indicate that an audit time of 18 days occurred three Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

2.2 Summarizing Data for a Quantitative Variable

FIGURE 2.10 DOT PLOT FOR THE AUDIT TIME DATA

Audit Time (days)

times. Dot plots show the details of the data and are useful for comparing the distribution of the data for two or more variables.

Histogram A common graphical display of quantitative data is a histogram. This graphical display can be prepared for data previously summarized in either a frequency, relative frequency, or percent frequency distribution. A histogram is constructed by placing the variable of interest on the horizontal axis and the frequency, relative frequency, or percent frequency on the vertical axis. The frequency, relative frequency, or percent frequency of each class is shown by drawing a rectangle whose base is determined by the class limits on the horizontal axis and whose height is the corresponding frequency, relative frequency, or percent frequency. Figure 2.11 is a histogram for the audit time data. Note that the class with the greatest frequency is shown by the rectangle appearing above the class of 15–19 days. The height of the rectangle shows that the frequency of this class is 8. A histogram for the relative or percent frequency distribution of these data would look the same as the histogram in Figure 2.11 with the exception that the vertical axis would be labeled with relative or percent frequency values. As Figure 2.11 shows, the adjacent rectangles of a histogram touch one another. Unlike a bar chart, a histogram contains no natural separation between the rectangles of adjacent classes. This format is the usual convention for histograms. Because the classes for the audit time data are stated as 10–14, 15–19, 20–24, 25–29, and 30–34, one-unit spaces of 14 to 15, 19 to 20, 24 to 25, and 29 to 30 would seem to be needed between the classes. These spaces are eliminated when constructing a histogram. Eliminating the spaces between classes in a histogram for the audit time data helps show that all values between the lower limit of the first class and the upper limit of the last class are p ossible. FIGURE 2.11 HISTOGRAM FOR THE AUDIT TIME DATA

8 7

Frequency

6 5 4 3 2 1 10–14

15–19

20–24

25–29

30–34

Audit Time (days)

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

One of the most important uses of a histogram is to provide information about the shape, or form, of a distribution. Figure 2.12 contains four histograms constructed from relative frequency distributions. Panel A shows the histogram for a set of data moderately skewed to the left. A histogram is said to be skewed to the left if its tail extends farther to the left. This histogram is typical for exam scores, with no scores above 100%, most of the scores above 70%, and only a few really low scores. Panel B shows the histogram for a set of data moderately skewed to the right. A histogram is said to be skewed to the right if its tail extends farther to the right. An example of this type of histogram would be for data such as housing prices; a few expensive houses create the skewness in the right tail. Panel C shows a symmetric histogram. In a symmetric histogram, the left tail mirrors the shape of the right tail. Histograms for data found in applications are never perfectly symmetric, but the histogram for many applications may be roughly symmetric. Data for SAT scores, heights and weights of people, and so on lead to histograms that are roughly symmetric. Panel D shows a histogram highly skewed to the right. This histogram was constructed from data on the dollar amount of customer purchases over one day at a women’s apparel store. Data from applications in business and economics often lead to histograms that are skewed to the right. For instance, data on housing prices, salaries, purchase amounts, and so on often result in histograms skewed to the right.

FIGURE 2.12 HISTOGRAMS SHOWING DIFFERING LEVELS OF SKEWNESS Panel B: Moderately Skewed Right

Panel A: Moderately Skewed Left 0.35

0.35

0.3

0.25

0.2

0.15

0.1

0.05

Panel D: Highly Skewed Right

Panel C: Symmetric 0.3 0.25

0.4 0.35 0.3

0.2 0.15 0.1

0.25 0.2 0.15 0.1

0.05 0

2.2 Summarizing Data for a Quantitative Variable

Using Excel’s Recommended Charts Tool to Construct a Histogram In Figure 2.9 we showed the results of using Excel’s PivotTable tool to construct a frequency distribution for the audit time data. We will use these results to illustrate how Excel’s Recommended Charts tool can be used to construct a histogram for depicting quantitative data summarized in a frequency distribution. Refer to Figure 2.13 as we describe the steps involved. Apply Tools: The following steps describe how to use Excel’s Recommended Charts tool to construct a histogram for the audit time data. Step 1. Select any cell in the PivotTable report (cells A3:B9) Step 2. Click Insert on the Ribbon Step 3. In the Charts group click Recommended Charts; a preview showing the recommended chart appears Step 4. Click OK The worksheet in Figure 2.13 shows the chart for the audit time data created using these steps. With the exception of the gaps separating the bars, this resembles the histogram for the audit time data shown in Figure 2.11. We can easily edit this chart to remove the gaps between the bars and enter more descriptive axis labels and a chart heading.

FIGURE 2.13 Initial Chart Used to Construct a Histogram for the Audit

Time Data

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Editing Options: In addition to removing the gaps between the bars, suppose you would like to use “Histogram for Audit Time Data” as the chart title and insert “Audit Time (days)” for the horizontal axis title and “Frequency” for the vertical axis title. Step 1. Right-click any bar in the chart and choose Format Data Series from the list of options that appears Step 2. When the Format Data Series dialog box appears: Go to the Series Options section Set the Gap Width to 0 Click the Close button at the top right of the dialog box Step 3. Click the Chart Title and replace it with Histogram for Audit Time Data (located next to the top right corner of Step 4. Click the Chart Elements button the chart) Step 5. When the list of chart elements appears: Click Axis Titles (creates placeholders for the axis titles) Click Legend to remove the check in the Legend box Step 6. Click the Horizontal (Category) Axis Title and replace it with Audit Time (days) Step 7. Click the Vertical (Value) Axis Title and replace it with Frequency The edited histogram for the audit time is shown in Figure 2.14.

ool

Figure 2.14 Histogram for the Audit Time Data Created Using Excel’s

Recommended Charts T

2.2 Summarizing Data for a Quantitative Variable

In smaller data sets or when there are a large number of classes, some classes may have no data values. In such cases, Excel’s PivotTable tool will automatically remove these classes when constructing the frequency distribution, and the histogram created using Excel’s Recommended Charts tool will not show the classes with no data values. For presentation, however, we recommend editing the results to show all the classes, including those with no data values. The following steps show how this can be done by editing the PivotTable tool output before constructing the histogram. Step 1. Right click on any cell in the Row Labels column of the PivotTable Step 2. Click Field Settings Step 3. When the Field Settings dialog box appears: Click the Layout & Print tab Choose Show items with no data Click OK Exercise 21 is an example that has some classes with no data values.

After performing steps 1–3, the PivotTable will display all classes, including those with no data values. Using the Recommended Charts tool will then create a bar chart that displays all classes.

Cumulative Distributions A variation of the frequency distribution that provides another tabular summary of quantitative data is the cumulative frequency distribution. The cumulative frequency distribution uses the number of classes, class widths, and class limits developed for the frequency distribution. However, rather than showing the frequency of each class, the cumulative frequency distribution shows the number of data items with values less than or equal to the upper class limit of each class. The first two columns of Table 2.7 provide the cumulative frequency distribution for the audit time data. To understand how the cumulative frequencies are determined, consider the class with the description “less than or equal to 24.” The cumulative frequency for this class is simply the sum of the frequencies for all classes with data values less than or equal to 24. For the frequency distribution in Table 2.5, the sum of the frequencies for classes 10–14, 15–19, and 20–24 indicates that 4 1 8 1 5 5 17 data values are less than or equal to 24. Hence, the cumulative frequency for this class is 17. In addition, the cumulative frequency d istribution in Table 2.7 shows that four audits were completed in 14 days or less and 19 audits were completed in 29 days or less. As a final point, we note that a cumulative relative frequency distribution shows the proportion of data items, and a cumulative percent frequency distribution shows the percentage of data items with values less than or equal to the upper limit of each class. TABLE 2.7 CUMULATIVE FREQUENCY, CUMULATIVE RELATIVE FREQUENCY,

AND CUMULATIVE PERCENT FREQUENCY DISTRIBUTIONS FOR THE AUDIT TIME DATA Audit Time (days)

Cumulative Frequency

Cumulative Relative Frequency

Cumulative Percent Frequency

Less than or equal to 14 4 .20 20 Less than or equal to 19 12 .60 60 Less than or equal to 24 17 .85 85 Less than or equal to 29 19 .95 95 Less than or equal to 34 20 1.00 100

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

The cumulative relative frequency distribution can be computed either by summing the relative frequencies in the relative frequency distribution or by dividing the cumulative frequencies by the total number of items. Using the latter approach, we found the cumulative relative frequencies in column 3 of Table 2.7 by dividing the cumulative frequencies in column 2 by the total number of items (n 5 20). The cumulative percent f requencies were again computed by multiplying the relative frequencies by 100. The cumulative relative and percent frequency distributions show that .85 of the audits, or 85%, were completed in 24 days or less, .95 of the audits, or 95%, were completed in 29 days or less, and so on.

Stem-and-Leaf Display A stem-and-leaf display is a graphical display used to show simultaneously the rank order and shape of a distribution of data. To illustrate the use of a stem-and-leaf display, consider the data in Table 2.8. These data result from a 150-question aptitude test given to 50 individuals recently interviewed for a position at Haskens Manufacturing. The data indicate the number of questions answered correctly. To develop a stem-and-leaf display, we first arrange the leading digits of each data value to the left of a vertical line. To the right of the vertical line, we record the last digit for each data value. Based on the top row of data in Table 2.8 (112, 72, 69, 97, and 107), the first five entries in constructing a stem-and-leaf display would be as follows: 6 9 7 2 8 9 7 10 7 11 2 12 13 14 For example, the data value 112 shows the leading digits 11 to the left of the line and the last digit 2 to the right of the line. Similarly, the data value 72 shows the leading digit 7 to the left of the line and last digit 2 to the right of the line. Continuing to place the last digit of each data value on the line corresponding to its leading digit(s) provides the following: TABLE 2.8 NUMBER OF QUESTIONS ANSWERED CORRECTLY

ON AN APTITUDE TEST

ApTest

112 72 69 97 107 73 92 76 86 73 126 128 118 127 124 82 104 132 134 83 92 108 96 100 92 115 76 91 102 81 95 141 81 80 106 84 119 113 98 75 68 98 115 106 95 100 85 94 106 119

2.2 Summarizing Data for a Quantitative Variable

6 9 8 7 2 3 6 3 6 5 8 6 2 3 1 1 0 4 5 9 7 2 2 6 2 1 5 8 8 5 4 10 7 4 8 0 2 6 6 0 6 11 2 8 5 9 3 5 9 12 6 8 7 4 13 2 4 14 1 With this organization of the data, sorting the digits on each line into rank order is simple. Doing so provides the stem-and-leaf display shown here. 6 8 9 7 2 3 3 5 6 6 8 9 10 11 12 13 14

0 1 1 2 3 4 5 6 1 2 2 2 4 5 5 6 7 8 8 0 0 2 4 6 6 6 7 8 2 3 5 5 8 9 9 4 6 7 8 2 4 1

The numbers to the left of the vertical line (6, 7, 8, 9, 10, 11, 12, 13, and 14) form the stem, and each digit to the right of the vertical line is a leaf. For example, consider the first row with a stem value of 6 and leaves of 8 and 9. 6 8 9 This row indicates that two data values have a first digit of six. The leaves show that the data values are 68 and 69. Similarly, the second row 7 2 3 3 5 6 6 indicates that six data values have a first digit of seven. The leaves show that the data values are 72, 73, 73, 75, 76, and 76. To focus on the shape indicated by the stem-and-leaf display, let us use a rectangle to contain the leaves of each stem. Doing so, we obtain the following: 6 8 9 7 2 3 3 5 6 6 8 0 1 1 2 3 4 5 6 9 1 2 2 2 4 5 5 6 7 8 8 10 0 0 2 4 6 6 6 7 8 11 2 3 5 5 8 9 9 12 4 6 7 8 13 2 4 14 1 Rotating this page counterclockwise onto its side provides a picture of the data that is simi lar to a histogram with classes of 60–69, 70–79, 80–89, and so on. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Although the stem-and-leaf display may appear to offer the same information as a histogram, it has two primary advantages. 1. The stem-and-leaf display is easier to construct by hand. 2. Within a class interval, the stem-and-leaf display provides more information than the histogram because the stem-and-leaf shows the actual data.

In a stretched stem-and-leaf display, whenever a stem value is stated twice, the first value corresponds to leaf values of 0–4, and the second value corresponds to leaf values of 5–9.

Just as a frequency distribution or histogram has no absolute number of classes, neither does a stem-and-leaf display have an absolute number of rows or stems. If we believe that our original stem-and-leaf display condensed the data too much, we can easily stretch the display by using two or more stems for each leading digit. For example, to use two stems for each leading digit, we would place all data values ending in 0, 1, 2, 3, and 4 in one row and all values ending in 5, 6, 7, 8, and 9 in a second row. The following stretched stem-and-leaf display illustrates this approach. 6 8 9 7 2 3 3 7 5 6 6 8 0 1 1 2 3 4 8 5 6 9 1 2 2 2 4 9 5 5 6 7 8 8 10 0 0 2 4 10 6 6 6 7 8 11 2 3 11 5 5 8 9 9 12 4 12 6 7 8 13 2 4 13 14 1 Note that values 72, 73, and 73 have leaves in the 0–4 range and are shown with the first stem value of 7. The values 75, 76, and 76 have leaves in the 5–9 range and are shown with the second stem value of 7. This stretched stem-and-leaf display is similar to a frequency distribution with intervals of 65–69, 70–74, 75–79, and so on. The preceding example showed a stem-and-leaf display for data with as many as three digits. Stem-and-leaf displays for data with more than three digits are possible. For example, consider the following data on the number of hamburgers sold by a fast-food restaurant for each of 15 weeks. 1565 1852 1644 1766 1888 1912 2044 1812 1790 1679 2008 1852 1967 1954 1733 A stem-and-leaf display of these data follows. Leaf unit 5 10 15 6 16 4 7 17 3 6 9 18 1 5 5 8 19 1 5 6 20 0 4

2.2 Summarizing Data for a Quantitative Variable

A single digit is used to define each leaf in a stem-and-leaf display. The leaf unit indicates how to multiply the stem-andleaf numbers in order to approximate the original data. Leaf units may be 100, 10, 1, 0.1, and so on.

Note that a single digit is used to define each leaf and that only the first three digits of each data value have been used to construct the display. At the top of the display we have specified Leaf unit 5 10. To illustrate how to interpret the values in the display, consider the first stem, 15, and its associated leaf, 6. Combining these numbers, we obtain 156. To reconstruct an approximation of the original data value, we must multiply this number by 10, the value of the leaf unit. Thus, 156 3 10 5 1560 is an approximation of the original data value used to construct the stem-and-leaf display. Although it is not possible to reconstruct the exact data value from this stem-and-leaf display, the convention of using a single digit for each leaf enables stem-and-leaf displays to be constructed for data having a large number of digits. For stem-and-leaf displays where the leaf unit is not shown, the leaf unit is assumed to equal 1.

NOTES AND COMMENTS 1. A bar chart and a histogram are essentially the same thing; both are graphical presentations of the data in a frequency distribution. A histogram is just a bar chart with no separation between bars. For some discrete quantitative data, a separation between bars is also appropriate. Consider, for example, the number of classes in which a college student is enrolled. The data may only assume integer values. Intermediate values such as 1.5, 2.73, and so on are not possible. With continuous quantitative data, however, such as the audit times in Table 2.4, a separation between bars is not appropriate. 2. The appropriate values for the class limits with quantitative data depend on the level of accuracy of the data. For instance, with the audit time data of Table 2.4 the limits used were integer values. If the data were rounded to the nearest tenth of a day (e.g., 12.3, 14.4, and so on), then the limits would be stated in tenths of days. For instance, the first class would be 10.0–14.9. If the data were recorded to the nearest hundredth

of a day (e.g., 12.34, 14.45, and so on), the limits would be stated in hundredths of days. For instance, the first class would be 10.00–14.99. 3. An open-end class requires only a lower class limit or an upper class limit. For example, in the audit time data of Table 2.4, suppose two of the audits had taken 58 and 65 days. Rather than continue with the classes of width 5 with classes 35–39, 40–44, 45–49, and so on, we could simplify the frequency distribution to show an open-end class of “35 or more.” This class would have a frequency of 2. Most often the open-end class appears at the upper end of the distribution. Sometimes an open-end class appears at the lower end of the distribution, and occasionally such classes appear at both ends. 4. The last entry in a cumulative frequency distribution always equals the total number of observations. The last entry in a cumulative relative frequency distribution always equals 1.00 and the last entry in a cumulative percent frequency distribution always equals 100.

Exercises

Methods 11. Consider the following data.

Frequency

14 21 23 21 16 19 22 25 16 16 24 24 25 19 16 19 18 19 21 12 16 17 18 23 25 20 23 16 20 19 24 26 15 22 24 20 22 24 22 20 a. Develop a frequency distribution using classes of 12–14, 15–17, 18–20, 21–23, and 24–26. b. Develop a relative frequency distribution and a percent frequency distribution using the classes in part (a).

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

12. Consider the following frequency distribution.

Class Frequency 10–19 10 20–29 14 30–39 17 40–49 7 50–59 2

Construct a cumulative frequency distribution and a cumulative relative frequency distribution. 13. Construct a histogram for the data in exercise 12. 14. Consider the following data. 8.9 6.8

10.2 9.5

11.5 11.5

7.8 11.2

10.0 14.9

12.2 7.5

13.5 10.0

14.1 6.0

10.0 15.8

12.2 11.5

a. C onstruct a dot plot. b. Construct a frequency distribution. c. Construct a percent frequency distribution. 15. Construct a stem-and-leaf display for the following data. 11.3 9.3

9.6 8.1

10.4 7.7

7.5 7.5

8.3 8.4

10.5 6.3

10.0 8.8

16. Construct a stem-and-leaf display for the following data. Use a leaf unit of 10. 1161 1221

1206 1378

1478 1623

1300 1426

1604 1557

1725 1730

1361 1706

1422 1689

Applications 17. A doctor’s office staff studied the waiting times for patients who arrive at the office with a request for emergency service. The following data with waiting times in minutes were collected over a one-month period.

2 5 10 12 4 4 5 17 11 8 9 8 12 21 6 8 7 13 18 3 Use classes of 0–4, 5–9, and so on in the following: a. Show the frequency distribution. b. Show the relative frequency distribution. c. Show the cumulative frequency distribution. d. Show the cumulative relative frequency distribution. e. What proportion of patients needing emergency service wait 9 minutes or less? 18. CBSSports.com developed the Total Player Ratings system to rate players in the National Basketball Association (NBA) based upon various offensive and defensive statistics. The following data show the average number of points scored per game (PPG) for 50 players with the highest ratings for a portion of the 2012–2013 NBA season (CBSSports.com website, February 25, 2013).

NBAPlayerPts

27.0 21.1 23.3 15.7 17.0

28.8 19.2 16.4 17.2 17.3

26.4 21.2 18.9 18.2 17.5

27.1 15.5 16.5 17.5 14.0

22.9 17.2 17.0 13.6 16.9

28.4 16.7 11.7 16.3 16.3

19.2 17.6 15.7 16.2 15.1

21.0 18.5 18.0 13.6 12.3

20.8 18.3 17.7 17.1 18.7

17.6 18.3 14.6 16.7 14.6

2.2 Summarizing Data for a Quantitative Variable

Use classes starting at 10 and ending at 30 in increments of 2 for PPG in the following. a. Show the frequency distribution. b. Show the relative frequency distribution. c. Show the cumulative percent frequency distribution. d. Develop a histogram for the average number of points scored per game. e. Do the data appear to be skewed? Explain. f. What percentage of the players averaged at least 20 points per game? 19. Based on the tons handled in a year, the ports listed below are the 25 busiest ports in the United States (The 2013 World Almanac).

Tons Handled (Millions)

Port

Ports

Baltimore Baton Rouge Beaumont Corpus Christi Duluth-Superior Houston Huntington Lake Charles Long Beach Los Angeles Mobile New Orleans New York

39.6 55.5 77.0 73.7 36.6 227.1 61.5 54.6 75.4 62.4 55.7 72.4 139.2

Port Norfolk Harbor Pascagoula Philadelphia Pittsburgh Plaquemines Port Arthur Savannah South Louisiana St. Louis Tampa Texas City Valdez

Tons Handled (Millions) 41.6 37.3 34.0 33.8 55.8 30.2 34.7 236.3 30.8 34.2 56.6 31.9

a. What is the largest number of tons handled? What is the smallest number of tons handled? b. Using a class width of 25, develop a frequency distribution of the data starting with 25–49.9, 50–74.9, 75–99.9, and so on. c. Prepare a histogram. Interpret the histogram. 20. T he London School of Economics and the Harvard Business School conducted a study of how chief executive officers (CEOs) spend their day. The study found that CEOs spend on a verage about 18 hours per week in meetings, not including conference calls, business meals, and public events (The Wall Street Journal, February 14, 2012). Shown below is the time spent per week in meetings (hours) for a sample of 25 CEOs.

CEOTime

14 15 18 23 15 19 20 13 15 23 23 21 15 20 21 16 15 18 18 19 19 22 23 21 12 a. What is the lowest amount of time spent per week on meetings? The highest? b. Use a class width of two hours to prepare a frequency distribution and a percent frequency distribution for the data. c. Prepare a histogram and comment on the shape of the distribution. 21. Quantcast.com provides the number of people from the United States who visit a given website. The list below shows the number of U.S. visitors for the 50 most highly visited websites (Quantcast.com website, December 2015).

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Website

WebVisitors

about.com adobe.com amazon.com answers.com aol.com ask.com bing.com blogger.com blogspot.com bustle.com buzzfeed.com craigslist.org curiositystream.com ebay.com facebook.com footlocker.ca foxnews.com go.com godaddy.com google.com linkedin.com live.com lowermybills.com microsoft.com monster.com

Number of Visitors in a Month 31 23 72 67 38 32 87 31 27 23 68 26 40 58 129 23 26 23 32 220 39 49 30 67 40

Website

Number of Visitors in a Month

mozilla.org msn.com nytimes.com paypal.com pinterest.com quizlet.com sascdn.com smartyads.com southwest.com stackexchange.com tumblr.com twitter.com urbandictionary.com urbanoutfitters.com videoamp.com walmart.com wayfair.com weather.com whitepages.com wikia.com wikipedia.org wordpress.com yahoo.com yelp.com youtube.com

24 110 23 38 58 23 33 24 28 25 25 198 28 39 89 25 24 40 26 41 46 35 43 73 222

Summarize the data by constructing the following: a. A frequency distribution (classes in millions: 20–29.999, 30–30.999, and so on). b. A relative frequency distribution. c. A cumulative distribution. d. A cumulative relative frequency distribution. e. Show a histogram. Comment on the shape of the distribution. f. What is the website with the most U.S. visitors? How many people from the U.S. visited the site? 22. E ntrepreneur magazine ranks franchises using performance measures such as growth rate, number of locations, startup costs, and financial stability. The number of locations for the top 20 U.S. franchises follow (The World Almanac, 2012).

Franchise

No. U.S. Franchise Locations Franchise Hampton Inn 1,864 Jan-Pro Franchising Intl. Inc. ampm 3,183 Hardee’s McDonald’s 32,805 Pizza Hut Inc. 7-Eleven Inc. 37,496 Kumon Math & Reading Centers Supercuts 2,130 Dunkin’ Donuts Days Inn 1,877 KFC Corp. Vanguard Cleaning 2,155 Jazzercise Inc. Systems Anytime Fitness Servpro 1,572 Matco Tools Subway 34,871 Stratus Building Solutions Denny’s Inc. 1,668

No. U.S. Locations 12,394 1,901 13,281 25,199 9,947 16,224 7,683 1,618 1,431 5,018

2.2 Summarizing Data for a Quantitative Variable

Use classes 0–4999, 5000–9999, 10,000–14,999, and so forth to answer the following q uestions. a. Construct a frequency distribution and a percent frequency distribution of the number of U.S. locations for these top-ranked franchises. b. Construct a histogram of these data. c. Comment on the shape of the distribution. 23. T he following data show the year-to-date percent change (YTD % Change) for 30 stockmarket indexes from around the word (The Wall Street Journal, August 26, 2013).

MarketIndexes

Country

Index

Australia Belgium Brazil Canada Chile China Eurozone France Germany Hong Kong India Israel Italy Japan Mexico Netherlands Singapore South Korea Spain Sweden Switzerland Taiwan U.K. U.S. U.S. U.S. U.S. U.S. World World

S&P/ASX200 Bel-20 São Paulo Bovespa S&P/TSX Comp Santiago IPSA Shanghai Composite EURO Stoxx CAC 40 DAX Hang Seng S&P BSE Sensex Tel Aviv FTSE MIB Nikkei IPC All-Share AEX Straits Times Kospi IBEX 35 SX All Share Swiss Market Weighted FTSE 100 S&P 500 DJIA Dow Jones Utility Nasdaq 100 Nasdaq Composite DJ Global ex U.S. DJ Global Index

YTD % Change 10.2 12.6 214.4 2.6 216.3 29.3 10.0 11.8 10.6 23.5 24.7 1.3 6.6 31.4 26.4 9.3 22.5 26.4 6.4 13.8 17.4 2.3 10.1 16.6 14.5 6.6 17.4 21.1 4.2 9.9

a. What index has the largest positive YTD % Change? b. Using a class width of 5 beginning with 220 and going to 40, develop a frequency distribution for the data. c. Prepare a histogram. Interpret the histogram, including a discussion of the general shape of the histogram. d. Use The Wall Street Journal or another media source to find the current percent changes for these stock market indexes in the current year. Which index has had the largest percent increase? Which index has had the smallest percent decrease? Prepare a summary of the data.

EngineeringSalary

24. T he DATAfile EngineeringSalary contains the median starting salary and median mid-career salary (measured 10 years after graduation) for graduates from 19 engineering schools (The Wall Street Journal website, November 2014). Develop a stem-and-leaf display for both the median starting salary and the median mid-career salary. Comment on any differences you observe. 25. E ach year America.EDU ranks the best paying college degrees in America. The following data show the median starting salary, the mid-career salary, and the percentage increase from starting salary to mid-career salary for the 20 college degrees with the highest mid-career salary (America.EDU website, August 29, 2013).

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Degree

BestPayingDegrees

Aerospace engineering Applied mathematics Biomedical engineering Chemical engineering Civil engineering Computer engineering Computer science Construction management Economics Electrical engineering Finance Government Information systems Management info. systems Mathematics Nuclear engineering Petroleum engineering Physics Software engineering Statistics

Starting Salary

Mid-Career Salary

% Increase

59,400 56,400 54,800 64,800 53,500 61,200 56,200 50,400 48,800 60,800 47,500 41,500 49,300 50,900 46,400 63,900 93,000 50,700 56,700 50,000

108,000 101,000 101,000 108,000 93,400 87,700 97,700 87,000 97,800 104,000 91,500 88,300 87,100 90,300 88,300 104,000 157,000 99,600 91,300 93,400

82 79 84 67 75 43 74 73 100 71 93 113 77 77 90 63 69 96 61 87

a. Using a class width of 10, construct a histogram for the percentage increase in the starting salary. b. Comment on the shape of the distribution. c. Develop a stem-and-leaf display for the percentage increase in the starting salary. d. What are the primary advantages of the stem-and-leaf display as compared to the histogram? 26. T he Flying Pig Half-Marathon (13.1 miles) had 10,897 finishers. The following data show the ages for a sample of 40half-marathoners.

Marathon

49 33 40 37 56 44 46 57 55 32 50 52 43 64 40 46 24 30 37 43 31 43 50 36 61 27 44 35 31 43 52 43 66 31 50 72 26 59 21 47 a. Construct a stretched stem-and-leaf display. b. What age group had the largest number of runners? c. What age occurred most frequently?

Summarizing Data for Two Variables 2.3 Using Tables Thus far in this chapter, we have focused on using tabular and graphical displays to summarize the data for a single categorical or quantitative variable. Often a manager or decision maker needs to summarize the data for two variables in order to reveal the relationship—if any—between the variables. In this section, we show how to construct a tabular summary of the data for two variables. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

2.3 Summarizing Data for Two Variables Using Tables

TABLE 2.9 QUALITY RATING AND MEAL PRICE DAta FOR 300 LOS ANGELES

RESTAURANTS

Restaurant

Quality Rating

1 2 3 4 5 6 7 8 9 10 ? ? ?

Meal Price ($)

Good Very Good Good Excellent Very Good Good Very Good Very Good Very Good Good ? ? ?

18 22 28 38 33 28 19 11 23 13 ? ? ?

Crosstabulation

Grouping the data for a quantitative variable enables us to treat the quantitative variable as if it were a categorical variable when creating a crosstabulation.

A crosstabulation is a tabular summary of data for two variables. Although both variables can be either categorical or quantitative, crosstabulations in which one variable is categorical and the other variable is quantitative are just as common. We will illustrate this latter case byconsidering the following application based on data from Zagat’s Restaurant Review. Datashowing the quality rating and the typical meal price were collected for a sample of 300restaurants in the Los Angeles area. Table 2.9 shows the data for the first 10 restaurants. Quality rating is a categorical variable with rating categories of Good, Very Good, and Excellent. Meal Price is a quantitative variable that ranges from $10 to $49. A crosstabulation of the data for this application is shown in Table 2.10. The labels shown in the margins of the table define the categories (classes) for the two variables. In the left margin, the row labels (Good, Very Good, and Excellent) correspond to the three rating categories for the quality rating variable. In the top margin, the column labels ($10–19, $20–29, $30–39, and $40–49) show that the Meal Price data have been grouped into four classes. Because each restaurant in the sample provides a quality rating and a meal price, each restaurant is associated with a cell appearing in one of the rows and one of the columns of the crosstabulation.For example, Table 2.9 shows restaurant 5 as having a Very Good quality rating and a Meal Price of $33. This restaurant belongs to the cell in row 2 and column 3 of the crosstabulation shown in Table 2.10. In constructing a crosstabulation, we simply count the number of restaurants that belong to each of the cells.

TABLE 2.10 CROSSTABULATION OF QUALITY RATING AND MEAL PRICE Data

FOR 300 LOS ANGELES RESTAURANTS Meal Price Quality Rating $10 –19 $20 –29 $30 –39 $40 – 49 Total Good 42 40 2 0 84 Very Good 34 64 46 6 150 Excellent 2 14 28 22 66 Total

78 118 76 28 300

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Although four classes of the Meal Price variable were used to construct the crosstabulation shown in Table 2.10, the crosstabulation of quality rating and meal price could have been developed using fewer or more classes for the meal price variable. The issues involved in deciding how to group the data for a quantitative variable in a crosstabulation are similar to the issues involved in deciding the number of classes to use when constructing a frequency distribution for a quantitative variable. For this application, four classes of meal price was considered a reasonable number of classes to reveal any relationship between quality rating and meal price. In reviewing Table 2.10, we see that the greatest number of restaurants in the sample (64) have a very good rating and a meal price in the $20–29 range. Only two restaurants have an excellent rating and a meal price in the $10–19 range. Similar interpretations of the other frequencies can be made. In addition, note that the right and bottom margins of the crosstabulation provide the frequency distributions for quality rating and meal price separately. From the frequency distribution in the right margin, we see that data on quality ratings show 84 restaurants with a good quality rating, 150 restaurants with a very good quality rating, and 66 restaurants with an excellent quality rating. Similarly, the bottom margin shows the frequency distribution for the meal price variable. Dividing the totals in the right margin of the crosstabulation by the total for that column provides a relative and percent frequency distribution for the quality rating v ariable.

Quality Rating Good Very Good Excellent

Relative Frequency Percent Frequency .28 28 .50 50 .22 22

Total

1.00 100

From the percent frequency distribution we see that 28% of the restaurants were rated good, 50% were rated very good, and 22% were rated excellent. Dividing the totals in the bottom row of the crosstabulation by the total for that row provides a relative and percent frequency distribution for the meal price variable.

Meal Price $10 –19 $20 –29 $30 –39 $40 – 49

Relative Frequency Percent Frequency .26 26 .39 39 .25 25 .09 9

Total 1.00 100

Note that the sum of the values in the relative frequency column does not add exactly to 1.00 and the sum of the values in the percent frequency distribution does not add exactly to 100; the reason is that the values being summed are rounded. From the percent frequency distribution we see that 26% of the meal prices are in the lowest price class ($10–19), 39% are in the next higher class, and so on. The frequency and relative frequency distributions constructed from the margins of a crosstabulation provide information about each of the variables individually, but they do not shed any light on the relationship between the variables. The primary value of a crosstabulation lies in the insight it offers about the relationship between the variables. A review of Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

2.3 Summarizing Data for Two Variables Using Tables

TABLE 2.11 ROW PERCENTAGES FOR EACH QUALITY RATING CATEGORY

Meal Price Quality Rating $10–19 $20–29 $30–39 $40–49 Total Good 50.0 47.6 2.4 0.0 100 Very Good 22.7 42.7 30.6 4.0 100 Excellent 3.0 21.2 42.4 33.4 100

the crosstabulation in Table 2.10 reveals that restaurants with higher meal prices received higher quality ratings than restaurants with lower meal prices. Converting the entries in a crosstabulation into row percentages or column percentages can provide more insight into the relationship between the two variables. For row percentages, the results of dividing each frequency in Table 2.10 by its corresponding row total are shown in Table 2.11. Each row of Table 2.11 is a percent frequency distribution of meal price for one of the quality rating categories. Of the restaurants with the lowest quality rating (good), we see that the greatest percentages are for the less expensive restaurants (50% have $10–19 meal prices and 47.6% have $20–29 meal prices). Of the restaurants with the highest quality rating (excellent), we see that the greatest p ercentages are for the more expensive restaurants (42.4% have $30–39 meal prices and 33.4% have $40–49 meal prices). Thus, we continue to see that restaurants with higher meal prices received higher quality ratings. Crosstabulations are widely used to investigate the relationship between two variables. In practice, the final reports for many statistical studies include a large number of cross tabulations. In the Los Angeles restaurant survey, the crosstabulation is based on one categorical variable (Quality Rating) and one quantitative variable (Meal Price). Crosstabulations can also be developed when both variables are categorical and when both variables are quantitative. When quantitative variables are used, however, we must first create classes for the values of the variable. For instance, in the restaurant example we grouped the meal prices into four classes ($10–19, $20–29, $30–39, and $40–49).

Using Excel’s PivotTable Tool to Construct a Crosstabulation Excel’s PivotTable tool can be used to summarize the data for two or more variables simultaneously. We will illustrate the use of Excel’s PivotTable tool by showing how to develop a crosstabulation of quality ratings and meal prices for the sample of 300 restaurants located in the Los Angeles area. Enter/Access Data: Open the DATAfile named Restaurant. The data are in cells B2:C301 and labels are in column A and cells B1:C1. Apply Tools: Each of the three columns in the Restaurant data set [labeled Restaurant, Quality Rating, and Meal Price ($)] is considered a field by Excel. Fields may be chosen to represent rows, columns, or values in the PivotTable. The following steps describe how to use Excel’s PivotTable tool to construct a crosstabulation of quality ratings and meal prices. Step 1. Select cell A1 or any cell in the data set Step 2. Click Insert on the Ribbon Step 3. In the Tables group click PivotTable Step 4. When the Create PivotTable dialog box appears: Click OK; a PivotTable and PivotTable Fields dialog box will appear in a new worksheet Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Step 5. In the PivotTable Fields dialog box: Drag Quality Rating to the Rows area Drag Meal Price to the Columns area Drag Restaurant to the Values area Step 6. Click on Sum of Restaurant in the Values area Step 7. Click Value Field Settings from the list of options that appears Step 8. When the Value Field Settings dialog box appears: Under Summarize value field by, choose Count Click OK Figure 2.15 shows the PivotTable Fields dialog box and the corresponding PivotTable created following the above steps. For readability, columns H:AC have been hidden. Editing Options: To complete the PivotTable we need to group the rows containing the meal prices and place the rows for quality rating in the proper order. The following steps accomplish this. Step 1. Right-click cell B4 in the PivotTable or any other cell containing meal prices. Step 2. Choose Group from the list of options that appears Step 3. When the Grouping dialog box appears: Enter 10 in the Starting at box Enter 49 in the Ending at box Enter 10 in the By box Click OK Step 4. Right-click on Excellent in cell A5 Step 5. Choose Move and click Move “Excellent” to End The final PivotTable is shown in Figure 2.16. Note that it provides the same information as the crosstabulation shown in Table 2.10.

FIGURE 2.15 Initial PivotTable Fields Dialog Box and PivotTable For

the Restaurant Data

2.3 Summarizing Data for Two Variables Using Tables

Simpson’s Paradox The data in two or more crosstabulations are often combined or aggregated to produce a summary crosstabulation showing how two variables are related. In such cases, conclusions drawn from two or more separate crosstabulations can be reversed when the data are aggregated into a single crosstabulation. The reversal of conclusions based on aggregate and unaggregated data is called Simpson’s paradox. To provide an illustration of Simpson’s paradox we consider an example involving the analysis of verdicts for two judges in two different courts. Judges Ron Luckett and Dennis Kendall presided over cases in Common Pleas Court and Municipal Court during the past three years. Some of the verdicts they rendered were appealed. In most of these cases the appeals court upheld the original verdicts, but in some

FIGURE 2.16 Final PivotTable for the Restaurant Data

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

cases those verdicts were reversed. For each judge a crosstabulation was developed based upon two variables: Verdict (upheld or reversed) and Type of Court (Common Pleas and Municipal). Suppose that the two crosstabulations were then combined by aggregating the type of court data. The resulting aggregated crosstabulation contains two variables: Verdict (upheld or reversed) and Judge (Luckett or Kendall). This crosstabulation shows the number of appeals in which the verdict was upheld and the number in which the verdict was reversed for both judges. The following crosstabulation shows these results along with the column percentages in parentheses next to each value.

Judge Verdict Luckett Kendall Total Upheld 129 (86%) 110 (88%) 239 Reversed 21 (14%) 15 (12%) 36 Total (%) 150 (100%) 125 (100%) 275

A review of the column percentages shows that 86% of the verdicts were upheld for Judge Luckett, whereas 88% of the verdicts were upheld for Judge Kendall. From this aggregated crosstabulation, we conclude that Judge Kendall is doing the better job because a greater percentage of Judge Kendall’s verdicts are being upheld. The following unaggregated crosstabulations show the cases tried by Judge Luckett and Judge Kendall in each court; column percentages are shown in parentheses next to each value.

Judge Luckett

Judge Kendall

Common Municipal Common Municipal Verdict Pleas Court Total Verdict Pleas Court Total Upheld 29 (91%) 100 (85%) 129 Upheld 90 (90%) 20 (80%) 110 Reversed 3 (9%) 18 (15%) 21 Reversed 10 (10%) 5 (20%) 15 Total (%)

32 (100%)

118 (100%)

150

Total (%)

100 (100%)

25 (100%)

125

From the crosstabulation and column percentages for Judge Luckett, we see that the verdicts were upheld in 91% of the Common Pleas Court cases and in 85% of the Municipal Court cases. From the crosstabulation and column percentages for Judge Kendall, we see that the verdicts were upheld in 90% of the Common Pleas Court cases and in 80% of the Municipal Court cases. Thus, when we unaggregate the data, we see that Judge Luckett has a better record because a greater percentage of Judge Luckett’s verdicts are being upheld in both courts. This result contradicts the conclusion we reached with the aggregated data crosstabulation that showed Judge Kendall had the better record. This reversal of conclusions based on aggregated and unaggregated data illustrates Simpson’s paradox. The original crosstabulation was obtained by aggregating the data in the separate crosstabulations for the two courts. Note that for both judges the percentage of appeals that resulted in reversals was much higher in Municipal Court than in Common Pleas Court. Because Judge Luckett tried a much higher percentage of his cases in Municipal Court, the aggregated data favored Judge Kendall. When we look at the crosstabulations for the two courts separately, however, Judge Luckett shows the better record. Thus, for the original crosstabulation, we see that the type of court is a hidden variable that cannot be ignored when evaluating the records of the two judges. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

2.3 Summarizing Data for Two Variables Using Tables

Because of the possibility of Simpson’s paradox, realize that the conclusion or interpretation may be reversed depending upon whether you are viewing unaggregated or aggregate crosstabulation data. Before drawing a conclusion, you may want to investigate whether the aggregate or unaggregate form of the crosstabulation provides the better insight and conclusion. Especially when the crosstabulation involves aggregated data, you should investigate whether a hidden variable could affect the results such that separate or unaggregated crosstabulations provide a different and possibly better insight and conclusion.

Exercises

Methods 27. T he following data are for 30 observations involving two categorical variables, x and y. The categories for x are A, B, and C; the categories for y are 1 and 2.

Crosstab

Observation x y Observation x y 1 A 1 16 B 2 2 B 1 17 C 1 3 B 1 18 B 1 4 C 2 19 C 1 5 B 1 20 B 1 6 C 2 21 C 2 7 B 1 22 B 1 8 C 2 23 C 2 9 A 1 24 A 1 10 B 1 25 B 1 11 A 1 26 C 2 12 B 1 27 C 2 13 C 2 28 A 1 14 C 2 29 B 1 15 C 2 30 B 2

a. Develop a crosstabulation for the data, with x as the row variable and y as the column variable. b. Compute the row percentages. c. Compute the column percentages. d. What is the relationship, if any, between x and y? 28. The following observations are for two quantitative variables, x and y.

Crosstab2

Observation 1 2 3 4 5 6 7 8 9 10

x y Observation x y 28 72 11 13 98 17 99 12 84 21 52 58 13 59 32 79 34 14 17 81 37 60 15 70 34 71 22 16 47 64 37 77 17 35 68 27 85 18 62 67 64 45 19 30 39 53 47 20 43 28

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

a. Develop a crosstabulation for the data, with x as the row variable and y as the column variable. For x use classes of 10–29, 30–49, and so on; for y use classes of 40–59, 60–79, and so on. b. Compute the row percentages. c. Compute the column percentages. d. What is the relationship, if any, between x and y?

Applications 29. The Daytona 500 is a 500-mile automobile race held annually at the Daytona International Speedway in Daytona Beach, Florida. The following crosstabulation shows the automobile make by average speed of the 25 winners from 1988 to 2012 (The 2013 World Almanac).

Average Speed in Miles per Hour Make

130–139.9

140–149.9

150–159.9

160–169.9

Buick Chevrolet Dodge Ford

1 3 2

5 2 1

Total

170–179.9

Total

1 16 2 6

a. Compute the row percentages. b. What percentage of winners driving a Chevrolet won with an average speed of at least 150 miles per hour? c. Compute the column percentages. d. What percentage of winning average speeds 160–169.9 miles per hour were Chevrolets? 30. The following crosstabulation shows the average speed of the 25 winners by year of the Daytona 500 automobile race (The 2013 World Almanac).

Year Average Speed

1988–1992

130–139.9 140–149.9 150–159.9 160–169.9 170–179.9

1 2

Total

1993–1997

1998–2002

2 3

1 1 2 1

2003–2007

2008–2012

Total

2 2 1

3 1 1

6 8 6 4 1

a. Calculate the row percentages. b. What is the apparent relationship between average winning speed and year? What might be the cause of this apparent relationship?

2.3 Summarizing Data for Two Variables Using Tables

31. R ecently, management at Oak Tree Golf Course received a few complaints about the condition of the greens. Several players complained that the greens are too fast. Rather than react to the comments of just a few, the Golf Association conducted a survey of 100 male and 100 female golfers. The survey results are summarized here.

Male Golfers

Greens Condition

Handicap

Too Fast

Female Golfers

Fine

Handicap

Under 15 10 40 15 or more 25 25

Greens Condition Too Fast

Fine

Under 15 1 9 15 or more 39 51

a. C ombine these two crosstabulations into one with Male and Female as the row labels and Too Fast and Fine as the column labels. Which group shows the highest percentage saying that the greens are too fast? b. Refer to the initial crosstabulations. For those players with low handicaps (better players), which group (male or female) shows the highest percentage saying the greens are too fast? c. Refer to the initial crosstabulations. For those players with higher handicaps, which group (male or female) shows the highest percentage saying the greens are too fast? d. What conclusions can you draw about the preferences of men and women concerning the speed of the greens? Are the conclusions you draw from part (a) as compared with parts (b) and (c) consistent? Explain any apparent inconsistencies. 32. The following crosstabulation shows the number of households (1000s) in each of the four regions of the United States and the number of households at each income level (U.S. Census Bureau website, August 2013). Income Level of Household

Region Northeast Midwest South West Total

Under $15,000

$15,000 to $24,999

2,733 3,273 6,235 3,086 15,327

Number of $100,000 Households and over (1000s)

$25,000 to $34,999

$35,000 to $49,999

$50,000 to $74,999

$75,000 to $99,999

2,244 3,326 5,657 2,796

2,264 3,056 5,038 2,644

2,807 3,767 6,476 3,557

3,699 5,044 7,730 4,804

2,486 3,183 4,813 3,066

5,246 4,742 7,660 6,104

21,479 26,391 43,609 26,057

14,023

13,002

16,607

21,277

13,548

23,752

117,536

a. Compute the row percentages and identify the percent frequency distributions of income for households in each region. b. What percentage of households in the West region have an income level of $50,000 or more? What percentage of households in the South region have an income level of $50,000 or more? c. Construct percent frequency histograms for each region of households. Do any relationships between regions and income level appear to be evident in your findings?

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

d. Compute the column percentages. What information do the column percentages provide? e. What percent of households with a household income of $100,000 and over are from the South region? What percentage of households from the South region have a household income of $100,000 and over? Why are these two percentages different?

33. Each year Forbes ranks the world’s most valuable brands. A portion of the data for 82 of the brands in the 2013 Forbes list is shown in Table 2.12 (Forbes website, February 2014). The data set includes the following variables: Brand: The name of the brand. Industry: The type of industry associated with the brand, labeled Automotive & Luxury, Consumer Packaged Goods, Financial Services, Other, Technology. Brand Value ($ billions): A measure of the brand’s value in billions of dollars developed by Forbes based on a variety of financial information about the brand. 1-Yr Value Change (%): The percentage change in the value of the brand over the previous year. Brand Revenue ($ billions): The total revenue in billions of dollars for the brand. a. Prepare a crosstabulation of the data on Industry (rows) and Brand Value ($ billions). Use classes of 0–10, 10–20, 20–30, 30–40, 40–50, and 50–60 for Brand Value ($ billions). b. Prepare a frequency distribution for the data on Industry. c. Prepare a frequency distribution for the data on Brand Value ($ billions). d. How has the crosstabulation helped in preparing the frequency distributions in parts (b) and (c)? e. What conclusions can you draw about the type of industry and the brand value? 34. Refer to Table 2.12. a. Prepare a crosstabulation of the data on Industry (rows) and Brand Revenue ($ billions). Use class intervals of 25 starting at 0 for Brand Revenue ($ billions). b. Prepare a frequency distribution for the data on Brand Revenue ($ billions).

Table 2.12 Data for 82 of the Most Valuable Brands

BrandValue

Brand

Industry

Accenture Adidas Allianz Amazon.com ? ? ? Heinz Hermès ? ? ? Wells Fargo Zara

Other Other Financial Services Technology ? ? ? Consumer Packaged Goods Automotive & Luxury ? ? ? Financial Services Other

Brand Value ($ billions) 9.7 8.4 6.9 14.7 ? ? ? 5.6 9.3 ? ? ? 9 9.4

1-Yr Value Change (%) 10 23 5 44 ? ? ? 2 20 ? ? ? 214 11

Brand Revenue ($ billions) 30.4 14.5 130.8 60.6 ? ? ? 4.4 4.5 ? ? ? 91.2 13.5

Source: Data from Forbes, 2014.

2.3 Summarizing Data for Two Variables Using Tables

c. What conclusions can you draw about the type of industry and the brand revenue? d. Prepare a crosstabulation of the data on Industry (rows) and the 1-Yr Value Change (%). Use class intervals of 20 starting at 260 for 1-Yr Value Change (%). e. Prepare a frequency distribution for the data on 1-Yr Value Change (%). f. What conclusions can you draw about the type of industry and the 1-year change in value? 35. T he U.S. Department of Energy’s Fuel Economy Guide provides fuel efficiency data for cars and trucks (Fuel Economy website, September 8, 2012). A portion of the data for 149 compact, midsize, and large cars is shown in Table 2.13. The data set contains the following variables:

Size: Compact, Midsize, and Large

Displacement: Engine size in liters Cylinders: Number of cylinders in the engine Drive: All wheel (A), front wheel (F), and rear wheel (R) Fuel Type: Premium (P) or regular (R) fuel

City MPG: Fuel efficiency rating for city driving in terms of miles per gallon

Hwy MPG: Fuel efficiency rating for highway driving in terms of miles per g allon

The complete data set is contained in the DATAfile named FuelData2012. a. Prepare a crosstabulation of the data on Size (rows) and Hwy MPG (columns). Use classes of 15–19, 20–24, 25–29, 30–34, 35–39, and 40–44 for Hwy MPG. b. Comment on the relationship beween Size and Hwy MPG. c. Prepare a crosstabulation of the data on Drive (rows) and City MPG (columns). Use classes of 10–14, 15–19, 20–24, 25–29, 30–34, and 35–39, and 40–44 for City MPG. d. Comment on the relationship between Drive and City MPG. e. Prepare a crosstabulation of the data on Fuel Type (rows) and City MPG (columns). Use classes of 10–14, 15–19, 20–24, 25–29, 30–34, 35–39, and 40–44 for City MPG. f. Comment on the relationship between Fuel Type and City MPG.

TABLE 2.13 Fuel efficiency data

FuelData2012

Car

Size

Displacement Cylinders Drive Fuel Type City MPG Hwy MPG

1 Compact 2 Compact 3 Compact ? ? ? ? ? ? 94 Midsize 95 Midsize ? ? ? ? ? ? 148 Large 149 Large

2.0 2.0 2.0 ? ? ? 3.5 2.5 ? ? ? 6.7 6.7

4 F 4 A 4 A ? ? ? ? ? ? 6 A 4 F ? ? ? ? ? ? 12 R 12 R

P P P ? ? ? R R ? ? ? P P

22 21 21 ? ? ? 17 23 ? ? ? 11 11

30 29 31 ? ? ? 25 33 ? ? ? 18 18

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Summarizing Data for Two Variables 2.4 Using Graphical Displays In the previous section we showed how a crosstabulation can be used to summarize the data for two variables and help reveal the relationship between the variables. In most cases, a graphical display is more useful for recognizing patterns and trends in the data. In this section, we introduce a variety of graphical displays for exploring the relationships between two variables. Displaying data in creative ways can lead to powerful insights and allow us to make “common-sense inferences” based on our ability to visually compare, contrast, and recognize patterns. We begin with a discussion of scatter diagrams and trendlines.

Scatter Diagram and Trendline A scatter diagram is a graphical display of the relationship between two quantitative variables, and a trendline is a line that provides an approximation of the relationship. As an illustration, consider the advertising/sales relationship for a stereo and sound equipment store in San Francisco. On 10 occasions during the past three months, the store used weekend television commercials to promote sales at its stores. The managers want to investigate whether a relationship exists between the number of commercials shown and sales at the store during the following week. Sample data for the 10 weeks with sales in hundreds of dollars are shown in Table 2.14. Figure 2.17 shows the scatter diagram and the trendline1 for the data in Table 2.14. The number of commercials (x) is shown on the horizontal axis and the sales (y) are shown on the vertical axis. For week 1, x 5 2 and y 5 50. A point with those coordinates is plotted on the scatter diagram. Similar points are plotted for the other nine weeks. Note that during two of the weeks one commercial was shown, during two of the weeks two commercials were shown, and so on. The scatter diagram in Figure 2.17 indicates a positive relationship between the number of commercials and sales. Higher sales are associated with a higher number of c ommercials. The relationship is not perfect in that all points are not on a straight line. However, the general pattern of the points and the trendline suggest that the overall relationship is positive. Some general scatter diagram patterns and the types of relationships they suggest are shown in Figure 2.18. The top left panel depicts a positive relationship similar to the one for TABLE 2.14 SAMPLE DATA FOR THE STEREO AND SOUND EQUIPMENT STORE Number of Commercials Stereo

Week

1 2 3 4 5 6 7 8 9 10

2 5 1 3 4 1 5 3 4 2

Sales ($100s) y 50 57 41 54 54 38 63 48 59 46

The equation of the trendline is y 5 36.15 1 4.95x. The slope of the trendline is 4.95 and the y-intercept (the point where the trendline intersects the y-axis) is 36.15. We will discuss in detail the interpretation of the slope and y-intercept for a linear trendline in Chapter 14 when we study simple linear regression.

2.4 Summarizing Data for Two Variables Using Graphical Displays

FIGURE 2.17 SCATTER DIAGRAM AND TRENDLINE FOR THE STEREO AND SOUND

EQUIPMENT STORE

Sales ($100s)

60 55 50 45 40 35

2 3 Number of Commercials

FIGURE 2.18 TYPES OF RELATIONSHIPS DEPICTED BY SCATTER DIAGRAMS y

Positive Relationship

No Apparent Relationship

Negative Relationship

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

the number of commercials and sales example. In the top right panel, the scatter diagram shows no apparent relationship between the variables. The bottom panel depicts a negative relationship where y tends to decrease as x increases.

Using Excel to Construct a Scatter Diagram and a Trendline We can use Excel to construct a scatter diagram and a trendline for the stereo and sound equipment store data. Enter/Access Data: Open the DATAfile named Stereo. The data are in cells B2:C11 and labels are in column A and cells B1:C1. Apply Tools: The following steps describe how to use Excel to construct a scatter diagram from the data in the worksheet. Step 1. Select cells B1:C11 Step 2. Click the Insert tab on the Ribbon Step 3. In the Charts group, click Insert Scatter (X,Y) or Bubble Chart Step 4. When the list of scatter diagram subtypes appears: Click Scatter (the chart in the upper left corner) The worksheet in Figure 2.19 shows the scatter diagram produced using these steps. Editing Options: You can easily edit the scatter diagram to display a different chart title, add axis titles, and display the trendline. For instance, suppose you would like to use “Scatter Diagram for the Stereo and Sound Equipment Store” as the chart title and insert “Number of Commercials” for the horizontal axis title and “Sales ($100s)” for the vertical axis title. Step 1. Click the Chart Title and replace it with Scatter Diagram for the Stereo and Sound Equipment Store Step 2. Click the Chart Elements button (located next to the top right corner of the chart) FIGURE 2.19 Initial Scatter Diagram for the Stereo and Sound Equipment Store

Using Excel’s Recommended Charts Tool

2.4 Summarizing Data for Two Variables Using Graphical Displays

Step 3. When the list of chart elements appears: Click Axis Titles (creates placeholders for the axis titles) Click Gridlines (to deselect the Gridlines option) Click Trendline Step 4. Click the Horizontal (Value) Axis Title and replace it with Number of Commercials Step 5. Click the Vertical (Value) Axis Title and replace it with Sales ($100s) Step 6. To change the trendline from a dashed line to a solid line, right-click on the trendline and choose the Format Trendline option Step 7. When the Format Trendline dialog box appears: Select the Fill & Line option In the Dash type box, select Solid Close the Format Trendline dialog box The edited scatter diagram and trendline are shown in Figure 2.20.

Side-by-Side and Stacked Bar Charts In Section 2.1 we said that a bar chart is a graphical display for depicting categorical data summarized in a frequency, relative frequency, or percent frequency distribution. Side-byside bar charts and stacked bar charts are extensions of basic bar charts that are used to display and compare two variables. By displaying two variables on the same chart, we may better understand the relationship between the variables. A side-by-side bar chart is a graphical display for depicting multiple bar charts on the same display. To illustrate the construction of a side-by-side chart, recall the application involving the quality rating and meal price data for a sample of 300 restaurants located in the Los Angeles area. Quality rating is a categorical variable with rating categories of Good, Very Good, and Excellent. Meal Price is a quantitative variable that ranges from $10 to $49. The crosstabulation displayed in Table 2.10 shows that the data for meal price were

FIGURE 2.20 Edited Scatter Diagram and Trendline for the Stereo and Sound

Equipment Store Using Excel’s Recommended Charts Tool

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

groupedinto four classes: $10–19, $20–29, $30–39, and $40–49. We will use these classes to construct a side-by-side bar chart. Figure 2.21 shows a side-by-side chart for the restaurant data. The color of each bar indicates the quality rating (blue 5 good, red 5 very good, and green 5 excellent). Each bar is constructed by extending the bar to the point on the vertical axis that represents the frequency with which that quality rating occurred for each of the meal price categories. Placing each meal price category’s quality rating frequency adjacent to one another allows us to quickly determine how a particular meal price category is rated. We see that the lowest meal price category ($10–$19) received mostly good and very good ratings, but very few excellent ratings. The highest price category ($40–49), however, shows a much different result. This meal price category received mostly excellent ratings, some very good ratings, but no good ratings. Figure 2.21 also provides a good sense of the relationship between meal price and quality rating. Notice that as the price increases (left to right), the height of the blue bars decreases and the height of the green bars generally increases. This indicates that as price increases, the quality rating tends to be better. The very good rating, as expected, tends to be more prominent in the middle price categories as indicated by the dominance of the red bars in the middle of the chart. Stacked bar charts are another way to display and compare two variables on the same display. A stacked bar chart is a bar chart in which each bar is broken into rectangular segments of a different color showing the relative frequency of each class in a manner similar to a pie chart. To illustrate a stacked bar chart we will use the quality rating and meal price data summarized in the crosstabulation shown in Table 2.10. We can convert the frequency data in Table 2.10 into column percentages by dividing each element in a particular column by the total for that column. For instance, 42 of the 78 restaurants with a meal price in the $10–19 range had a good quality rating. In other words, (42/78)100 or 53.8% of the 78 restaurants had a good rating. Table 2.15 shows the column percentages for each meal price category. Using the data in Table 2.15 we constructed the stacked bar chart shown in Figure 2.22. Because the stacked bar chart is based on percentages, Figure 2.22 shows even more clearly than Figure 2.21 the relationship between the variables. As we move from the low price category ($10–19) to the high price category ($40–49), the length of the blue bars decreases and the length of the green bars increases. FIGURE 2.21 SIDE-BY-SIDE BAR CHART FOR THE QUALITY AND MEAL PRICE DATA 70 60 50 Frequency

Good

Very Good

Excellent

20 10 0

10–19

20–29 30 –39 Meal Price ($)

40 –49

2.4 Summarizing Data for Two Variables Using Graphical Displays

TABLE 2.15 Column Percentages for Each Meal Price Category

Quality Rating $10–19 Good 53.8% Very Good 43.6 Excellent 2.6 Total

Meal Price $20–29 $30–39 $40– 49 33.9% 54.2 11.9

2.6% 60.5 36.8

100.0% 100.0%

0.0% 21.4 78.6

100.0% 100.0%

FIGURE 2.22 STACKED BAR CHART FOR the QUALITY rating AND MEAL

PRICE DATA 100% 90% 80% 70% 60% 50% 40% 30% 20% 10% 0%

Excellent Very Good Good

10 –19

20–29 30 –39 Meal Price ($)

40 –49

Using Excel’s Recommended Charts Tool to Construct Side-by-Side and Stacked Bar Charts In Figure 2.16 we showed the results of using Excel’s PivotTable tool to construct a frequency distribution for the sample of 300 restaurants in the Los Angeles area. We will use these results to illustrate how Excel’s Recommended Charts tool can be used to construct side-by-side and stacked bar charts for the restaurant data using the PivotTable output. Apply Tools: The following steps describe how to use Excel’s Recommended Charts tool to construct a side-by-side bar chart for the restaurant data using the PivotTable tool output shown in Figure 2.16. Step 1. Select any cell in the PivotTable report (cells A3:F8) Step 2. Click Insert on the Ribbon. Step 3. In the Charts group click Recommended Charts; a preview showing a bar chart with quality rating on the horizontal axis appears Step 4. Click OK

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Step 5. Click Design on the Ribbon (located below the Pivotchart Tools heading) Step 6. In the Data group click Switch Row/Column; a side-by-side bar chart with meal price on the horizontal axis appears Excel refers to the bar chart in Figure 2.23 as a Clustered Column chart.

The worksheet in Figure 2.23 shows the side-by-side chart for the restaurant data created using these steps. Editing Options: We can easily edit the side-by-side bar chart to enter a chart heading and axis labels. Suppose you would like to use “Side-by-Side Bar Chart” as the chart title, insert “Meal Price ($)” for the horizontal axis title, and insert “Frequency” for the vertical axis title. (located next to the top right corner of Step 1. Click the Chart Elements button the chart) Step 2. When the list of chart elements appears: Click Chart title (creates placeholder for the chart title) Click Axis Titles (creates placeholder for the axis titles) Step 3. Click the Chart Title and replace it with Side-by-Side Bar Chart Step 4. Click the Horizontal (Category) Axis Title and replace it with Meal Price ($) Step 5. Click the Vertical (Value) Axis Title and replace it with Frequency The edited side-by-side chart is shown in Figure 2.24.

FIGURE 2.23 Side-By-Side Chart for the Restaurant Data Constructed Using Excel’s

Recommended Charts Tool

2.4 Summarizing Data for Two Variables Using Graphical Displays

FIGURE 2.24 Edited Side-By-Side Chart for the Restaurant Data Constructed Using

Excel’s Recommended Charts Tool

You can easily change the side-by-side bar chart to a stacked bar chart using the following steps. Step 6. In the Type group click Change Chart Type Step 7. When the Change Chart Type dialog box appears: Select the Stacked Columns option Click OK Once you have created a side-by-side bar chart or a stacked bar chart, you can easily switch back and forth between the two chart types by reapplying steps 6 and 7.

NOTES AND COMMENTS 1. A time series is a sequence of observations on a variable measured at successive points in time or over successive periods of time. A scatter diagram in which the value of time is shown on the horizontal axis and the time series values are shown on the vertical axis is referred to in time series analysis as a time series plot.

2. A stacked bar chart can also be used to display frequencies rather than percentage frequencies. In this case, the different color segments of each bar represent the contribution to the total for that bar, rather than the percentage contribution.

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Exercises

Methods 36. The following 20 observations are for two quantitative variables, x and y.

Scatter

Observation x y Observation x y 1 222 22 11 237 48 2 233 49 12 34 229 3 2 8 13 9 218 4 29 216 14 233 31 5 213 10 15 20 216 6 21 228 16 23 14 7 213 27 17 215 18 8 223 35 18 12 17 9 14 25 19 220 211 10 3 23 20 27 222

a. Develop a scatter diagram for the relationship between x and y. b. What is the relationship, if any, between x and y? 37. C onsider the following data on two categorical variables. The first variable, x, can take on values A, B, C, or D. The second variable, y, can take on values I or II. The following table gives the frequency with which each combination occurs.

y I II

A B C D

143 857 200 800 321 679 420 580

a. Construct a side-by-side bar chart with x on the horizontal axis. b. Comment on the relationship between x and y. 38. T he following crosstabulation summarizes the data for two categorical variables, x and y. The variable x can take on values Low, Medium, or High and the variable y can take on values Yes or No.

Low Medium High Total

y Yes No Total 20 10 30 15 35 50 20 5 25 55 50 105

a. Compute the row percentages. b. Construct a stacked percent frequency bar chart with x on the horizontal axis. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

2.4 Summarizing Data for Two Variables Using Graphical Displays

Applications 39. A study on driving speed (miles per hour) and fuel efficiency (miles per gallon) for m idsize automobiles resulted in the following data:

MPG

Driving Speed

30 50 40 55 30 25 60 25 50 55

Fuel Efficiency

28 25 25 23 30 32 21 35 26 25

a. C onstruct a scatter diagram with driving speed on the horizontal axis and fuel effi‑ ciency on the vertical axis. b. Comment on any apparent relationship between these two variables.

Snow

40. T he DATAfile Snow contains average annual snowfall (inches) for 51 major U.S. cities over 30 years. For example, the average low temperature for Columbus, Ohio is 44 degrees and the average annual snowfall is 27.5 inches. a. Construct a scatter diagram with the average annual low temperature on the horizontal axis and the average annual snowfall on the vertical axis. b. Does there appear to be any relationship between these two variables? c. Based on the scatter diagram, comment on any data points that seem unusual. 41. P eople often wait until middle age to worry about having a healthy heart. However, recent studies have shown that earlier monitoring of risk factors such as blood pressure can be very beneficial (The Wall Street Journal, January 10, 2012). Having higher than normal blood pressure, a condition known as hypertension, is a major risk factor for heart d isease. Suppose a large sample of male and female individuals of various ages was selected and that each individual’s blood pressure was measured to determine if they have hypertension. For the sample data, the following table shows the percentage of individuals with hypertension.

Hypertension

Age Male Female 20–34 11.00% 9.00% 35–44 24.00% 19.00% 45–54 39.00% 37.00% 55–64 57.00% 56.00% 65–74 62.00% 64.00% 751 73.30% 79.00%

a. Develop a side-by-side bar chart with age on the horizontal axis, the percentage of individuals with hypertension on the vertical axis, and side-by-side bars based on gender. b. What does the display you developed in part (a), indicate about hypertension and age? c. Comment on differences by gender. 42. S martphones are mobile phones with Internet, photo, music, and video capability. The following survey results show smartphone ownership by age.

Smartphones

Age Category Smartphone (%) Other Cell Phone (%) No Cell Phone (%) 18–24 49 46 5 25–34 58 35 7 35–44 44 45 11 45–54 28 58 14 55–64 22 59 19 651 11 45 44

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

a. C onstruct a stacked bar chart to display the above survey data on type of mobile phone ownership. Use age category as the variable on the horizontal axis. b. Comment on the relationship between age and smartphone ownership. c. How would you expect the results of this survey to be different if conducted in 2021? 43. T he Northwest regional manager of an outdoor equipment retailer conducted a study to determine how managers at three store locations are using their time. A summary of the results is shown in the following table.

ManagerTime

Percentage of Manager’s Work Week Spent on

Store Location Meetings Reports Customers Idle Bend 18 11 52 19 Portland 52 11 24 13 Seattle 32 17 37 14

a. Create a stacked bar chart with store location on the horizontal axis and percentage of

time spent on each task on the vertical axis. b. C reate a side-by-side bar chart with store location on the horizontal axis and side-byside bars of the percentage of time spent on each task. c. Which type of bar chart (stacked or side-by-side) do you prefer for these data? Why?

Data Visualization: Best Practices in Creating 2.5 Effective Graphical Displays Data visualization is a term used to describe the use of graphical displays to summarize and present information about a data set. The goal of data visualization is to communicate as effectively and clearly as possible the key information about the data. In this section, we provide guidelines for creating an effective graphical display, discuss how to select an appropriate type of display given the purpose of the study, illustrate the use of data dashboards, and show how the Cincinnati Zoo and Botanical Garden uses data visualization techniques to improve decision making.

Creating Effective Graphical Displays The data presented in Table 2.16 show the forecasted or planned value of sales ($1000s) and the actual value of sales ($1000s) by sales region in the United States for Gustin Chemical for the past year. Note that there are two quantitative variables (planned sales and actual sales) and one categorical variable (sales region). Suppose we would like to develop a graphical display that would enable management of Gustin Chemical to visualize how each sales region did relative to planned sales and simultaneously enable management to visualize sales performance across regions. Figure 2.25 shows a side-by-side bar chart of the planned versus actual sales data. Note how this bar chart makes it very easy to compare the planned versus actual sales in a region, as well as across regions. This graphical display is simple, contains a title, is well labeled, and uses distinct colors to represent the two types of sales. Note also that the scale of the vertical axis begins at zero. The four sales regions are separated by space so that it is clear that they are distinct, whereas the planned versus actual sales values are side by side for easy comparison within each region. The side-by-side bar chart in Figure 2.25 makes it easy to see that the Southwest region is the lowest in both planned and actual sales and that the Northwest region slightly exceeded its planned sales. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

2.5 Data Visualization: Best Practices in Creating Effective Graphical Displays

TABLE 2.16 PLANNED AND ACTUAL SALES BY SALES REGION ($1000s) Sales Region

Planned Sales ($1000s)

Actual Sales ($1000s)

Northeast 540 Northwest 420 Southeast 575 Southwest 360

447 447 556 341

Creating an effective graphical display is as much art as it is science. By following the general guidelines listed below you can increase the likelihood that your display will effectively convey the key information in the data. ●● ●●

●● ●● ●●

Give the display a clear and concise title. Keep the display simple. Do not use three dimensions when two dimensions are sufficient. Clearly label each axis and provide the units of measure. If color is used to distinguish categories, make sure the colors are distinct. If multiple colors or line types are used, use a legend to define how they are used and place the legend close to the representation of the data.

Choosing the Type of Graphical Display In this chapter we discussed a variety of graphical displays, including bar charts, pie charts, dot plots, histograms, stem-and-leaf plots, scatter diagrams, side-by-side bar charts, and stacked bar charts. Each of these types of displays was developed for a specific purpose. In order to provide guidelines for choosing the appropriate type of graphical display, we now provide a summary of the types of graphical displays categorized by their purpose. We note that some types of graphical displays may be used effectively for multiple purposes. Displays Used to Show the Distribution of Data ●●

Bar Chart—Used to show the frequency distribution and relative frequency distribution for categorical data

FIGURE 2.25 SIDE-BY-SIDE BAR CHART FOR PLANNED VERSUS ACTUAL SALES U.S.PlannedVersusActualSales 700 600 Sales ($1000)

500 400 300

Planned Actual

200 100 0

Northeast Northwest Southeast Southwest Region

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays ●●

●●

ie Chart—Used to show the relative frequency and percent frequency for categoriP cal data Dot Plot—Used to show the distribution for quantitative data over the entire range of the data Histogram—Used to show the frequency distribution for quantitative data over a set of class intervals Stem-and-Leaf Display—Used to show both the rank order and shape of the distribution for quantitative data

Displays Used to Make Comparisons ●● ●●

Side-by-Side Bar Chart—Used to compare two variables Stacked Bar Charts—Used to compare the relative frequency or percent frequency of two categorical variables

Displays Used to Show Relationships ●●

●●

catter Diagram—Used to show the relationship between two quantitative variS ables Trendline—Used to approximate the relationship of data in a scatter diagram

Data Dashboards Data dashboards are also referred to as digital dashboards.

One of the most widely used data visualization tools is a data dashboard. If you drive a car, you are already familiar with the concept of a data dashboard. In an automobile, the car’s dashboard contains gauges and other visual displays that provide the key information that is important when operating the vehicle. For example, the gauges used to display the car’s speed, fuel level, engine temperature, and oil level are critical to ensure safe and efficient operation of the automobile. In some vehicles, this information is even displayed visually on the windshield to provide an even more effective display for the driver. Data dashboards play a similar role for managerial decision making. A data dashboard is a set of visual displays that organizes and presents information that is used to monitor the performance of a company or organization in a manner that is easy to read, understand, and interpret. Just as a car’s speed, fuel level, engine temperature, and oil level are important information to monitor in a car, every business has key performance indicators (KPIs)2 that need to be monitored to assess how a company is performing. Examples of KPIs are inventory on hand, daily sales, percentage of on-time deliveries, and sales revenue per quarter. A data dashboard should provide timely summary information (potentially from various sources) on KPIs that is important to the user, and it should do so in a manner that informs rather than overwhelms its user. To illustrate the use of a data dashboard in decision making, we will discuss an application involving the Grogan Oil Company. Grogan has offices located in three cities in Texas: Austin (its headquarters), Houston, and Dallas. Grogan’s Information Technology (IT) call center, located in the Austin office, handles calls from employees regarding computer-related problems involving software, Internet, and e-mail issues. For example, if a Grogan employee in Dallas has a computer software problem, the employee can call the IT call center for assistance. The data dashboard shown in Figure 2.26 was developed to monitor the performance of the call center. This data dashboard combines several displays to monitor the call center’s KPIs. The data presented are for the current shift, which started at 8:00 a.m. The stacked bar chart in the upper left-hand corner shows the

Key performance indicators are sometimes referred to as Key Performance Metrics (KPMs).

2.5 Data Visualization: Best Practices in Creating Effective Graphical Displays

FIGURE 2.26 GROGAN OIL INFORMATION TECHNOLOGY CALL CENTER DATA

DASHBOARD Grogan Oil

IT Call Center

Shift 1

19–Sep–12

12:44:00 PM

Call Volume

Time Breakdown This Shift Idle 14%

Software

Internet

E-mail

10 5 8:00

9:00

10:00 Hour

11:00

12:00

W59

E-mail Austin 0

100

200 Minutes

300

400

10 15 Number of Calls

321

31–32

30–31

29–30

28–29

27–28

26–27

25–26

22–23

21–22

20–21

19–20

18–19

17–18

16–17

15–16

14–15

13–14

12–13

11–12

10–11

8–9

9–10

7–8

6–7

5–6

4–5

3–4

2–3

1–2

Time to Resolve a Case

Frequency

E-mail

Dallas

T57

14 12 10 8 6 4 2 0

Software Internet

Internet

Internet 18%

Houston

Software

W24 Case Number

E-mail 22%

Call Volume by Office

Unresolved Cases Beyond 15 Minutes

24–25

Software 46%

23–24

Number of Calls

Minutes

call volume for each type of problem (software, Internet, or e-mail) over time. This chart shows that call volume is heavier during the first few hours of the shift, calls concerning e-mail issues appear to decrease over time, and volume of calls regarding software issues are highest at midmorning. The pie chart in the upper right-hand corner of the dashboard shows the percentage of time that call center employees spent on each type of problem or not working on a call (idle). Both of these charts are helpful in determining optimal staffing levels. For instance, knowing the call mix and how stressed the system is—as measured by percentage of idle time—can help the IT manager make sure there are enough call center employees available with the right level of expertise. The side-by-side bar chart below the pie chart shows the call volume by type of problem for each of Grogan’s offices. This allows the IT manager to quickly identify if there is a particular type of problem by location. For example, it appears that the office in Austin is reporting a relatively high number of issues with e-mail. If the source of the problem can be identified quickly, then the problem for many might be resolved quickly. Also, note that a relatively high number of software problems are coming from the Dallas office. The higher call volume in this case was simply Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

due to the fact that the Dallas office is currently installing new software, and this has resulted in more calls to the IT call center. Because the IT manager was alerted to this by the Dallas office last week, the IT manager knew there would be an increase in calls coming from the Dallas office and was able to increase staffing levels to handle the expected increase in calls. For each unresolved case that was received more than 15 minutes ago, the bar chart shown in the middle left-hand side of the data dashboard displays the length of time that each of these cases has been unresolved. This chart enables Grogan to quickly monitor the key problem cases and decide whether additional resources may be needed to resolve them. The worst case, T57, has been unresolved for over 300 minutes and is actually left over from the previous shift. Finally, the histogram at the bottom shows the distribution of the time to r esolve the problem for all resolved cases for the current shift. The Grogan Oil data dashboard illustrates the use of a dashboard at the operational level. The data dashboard is updated in real time and used for operational decisions such as staffing levels. Data dashboards may also be used at the tactical and strategic levels of management. For example, a logistics manager might monitor KPIs for on-time performance and cost for its third-party carriers. This could assist in tactical decisions such as transportation mode and carrier selection. At the highest level, a more strategic dashboard would allow upper management to quickly assess the financial health of the company by monitoring more aggregate financial, service level, and capacity utilization information. The guidelines for good data visualization discussed previously apply to the individual charts in a data dashboard, as well as to the entire dashboard. In addition to those guidelines, it is important to minimize the need for screen scrolling, avoid unnecessary use of color or three-dimensional displays, and use borders between charts to improve readability. As with individual charts, simpler is almost always better.

Data Visualization in Practice: Cincinnati Zoo and Botanical Garden3 The Cincinnati Zoo and Botanical Garden, located in Cincinnati, Ohio, is the second oldest zoo in the world. In order to improve decision making by becoming more data-driven, management decided they needed to link together the different facets of their business and provide nontechnical managers and executives with an intuitive way to better understand their data. A complicating factor is that when the zoo is busy, managers are expected to be on the grounds interacting with guests, checking on operations, and anticipating issues as they arise or before they become an issue. Therefore, being able to monitor what is happening on a real-time basis was a key factor in deciding what to do. Zoo management concluded that a data visualization strategy was needed to address the problem. Because of its ease of use, real-time updating capability, and iPad compatibility, the Cincinnati Zoo decided to implement its data visualization strategy using IBM’s Cognos advanced data visualization software. Using this software, the Cincinnati Zoo developed the data dashboard shown in Figure 2.27 to enable zoo management to track the following key performance indicators: ●● ●●

Item Analysis (sales volumes and sales dollars by location within the zoo) Geoanalytics (using maps and displays of where the day’s visitors are spending their time at the zoo)

The authors are indebted to John Lucas of the Cincinnati Zoo and Botanical Garden for providing this application.

2.5 Data Visualization: Best Practices in Creating Effective Graphical Displays

FIGURE 2.27 Data Dashboard For The Cincinnati Zoo

Food YTD

Zoo Cafe Ice Cream 27%

LaRosa’s 12% Skyline Safari 2%

Zoo Cafe 27% Skyline Entry 4%

Safari Grill 12%

5.00

Revenue Amount $151,367.00 ($981.00) ($153,329.00) ($305,677.00) ($458,025.00) ($610,373.00)

Food Watering Hole LaRosa’s Skyline Safari UDF Safari Grill Safari Ice Cream Skyline Entry Cantina Funnel Cake Zoo Cafe Zoo Cafe Ice Cream Dippin Dots Cart 1 Dippin Dots Cart 2 Outpost

Food and Retail Food - Non-Memb Food - Member Retail - Non-Mem Retail - Member

4.00 3.00 Per Cap

Watering Hole 9%

2.00 1.00

YTD Revenue Amount

0.00

2011 Q1

2011 Q2 YTD

Revenue Amount Zoo Shop Actual Attendance 4.00

5.00

6.00

7.50

10.00

12.50

7.00

3.00

5.00

15.00

2.50

17.50

8.00

2.00 1.00 0.00

9.00

Food

Adult Admission

166,286

160,115

–6,171

–3.71%

Membership Admission

440,758

422,060

–18,698

–4.24%

47,500

41,963

–5,537

–11.66%

0.00

Food and Retail

20.00

Retail

86,658

–9,126

16,241

18,050

Comp Admission

27,118

29,387

2,269

8.37%

School Admissions

86,512

75,283

–11,229

–12.98%

1,809

Giraffe Sitting 16*

–9.53%

Group Sales - Events

Hoody Double Play Zo Wolf Sitting 16* Pro

Subtotal (included) Product Name

11.14%

Event Admission

11,180

11,891

711

6.36%

Education Program Admissions

25,562

20,538

–5,024

–19.65%

2,514

2,418

2,518,75%

Total (Access Code Groups) 917,037 868,459

–48,578

2,482.84%

Online Tickets

Sweat Rhino/Tree Ki

YTD Change YTD Growth

95,784

Group Sales Tickets

10.00

Prior YTD YTD

Child Admission

Jacket Fleece Men’s

4D Shop

2011 Q3

2011/09/01 2011/09/04 MTD $36.99

$1,305.73

$1,342.72

$79.98

$1,223.68

$1,303.66

$284.81

$659.56

$944.37

$46.99

$789.42

$836.41

$224.85

$524.65

$749.50

$673.62 $4,503.04 $5,176.66 $11,279.85 $31,520.16 $42,800.01

Bottled Pop

$99.86

$107.02

$206.88

Sweat Rhino Behind B

$29.99

$111.97

$141.96

$107.27

Jacket Fleece Ladies Giraffe Sitting 16*

$89.94

$9.99

$90.93

$100.92

Subtotal (included)

$154.83

$507.13

$661.96

Product Name

$842.89 $2,600.49 $3,443.38

Family Shop Giraffe Sitting 16* Wolf Sitting 16* Pro

$224.85

$149.90

$374.75

$104.93

$119.92

$224.85

$119.96

Sweat Rhino Behind B Giraffe Mini Flopsie

Top

●● ●● ●● ●●

$104.93

$14.99

Backpack Logo Fold

$12.59

$94.00

$106.59

Penguin Emp Mini Flo

$8.09

$96.69

$104.78

Subtotal (included)

$350.46

$580.47

$930.93

Page up

Customer Spending Cashier Sales Performance Sales and Attendance Data Versus Weather Patterns Performance of the Zoo’s Loyalty Rewards Program

An iPad mobile application was also developed to enable the zoo’s managers to be out on the grounds and still see and anticipate what is occurring on a real-time basis. The Cincinnati Zoo’s iPad data dashboard, shown in Figure 2.28, provides managers with access to the following information: ●● ●● ●●

eal-time attendance data, including what types of guests are coming to the zoo R Real-time analysis showing which items are selling the fastest inside the zoo Real-time geographical representation of where the zoo’s visitors live

Having access to the data shown in Figures 2.27 and 2.28 allows the zoo managers to make better decisions on staffing levels within the zoo, which items to stock based upon weather and other conditions, and how to better target its advertising based on geodemographics. The impact that data visualization has had on the zoo has been substantial. Within the first year of use, the system has been directly responsible for revenue growth of over $500,000, increased visitation to the zoo, enhanced customer service, and reduced marketing costs. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

FIGURE 2.28 The Cincinnati Zoo iPad Data Dashboard

NOTES AND COMMENTS 1. A variety of software is available for data visualization. Among the more popular packages are Cognos, JMP, Spotfire, and Tableau. 2. Radar charts and bubble charts are two other commonly used charts for displaying relationships among multiple variables. However, many experts in data visualization recommend against using these charts because they can be overcomplicated. Instead, the use of simpler displays such as bar charts and scatter diagrams is recommended. 3. A very powerful tool for visualizing geographic data is a Geographic Information System (GIS).

A GIS uses color, symbols, and text on a map to help you understand how variables are distributed geographically. For example, a company interested in trying to locate a new distribution center might wish to better understand how the demand for its product varies throughout the United States. A GIS can be used to map the demand, with red regions indicating high demand, blue lower demand, and no color indicating regions where the product is not sold. Locations closer to red high-demand regions might be good candidate sites for further consideration.

Summary A set of data, even if modest in size, is often difficult to interpret directly in the form in which it is gathered. Tabular and graphical displays can be used to summarize and present data so that patterns are revealed and the data are more easily interpreted. Frequency distributions, relative frequency distributions, percent frequency distributions, bar charts, and pie charts were presented as tabular and graphical displays for summarizing the data for a single categorical variable. Frequency distributions, relative frequency distributions, percent Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Glossary

FIGURE 2.29 TABULAR AND GRAPHICAL displays FOR SUMMARIZING DATA

Data

Quantitative Data

Categorical Data

Tabular Displays

Frequency Distribution Relative Frequency Distribution Percent Frequency Distribution Crosstabulation

Graphical Displays

Bar Chart Pie Chart Side-by-Side Bar Chart Stacked Bar Chart

Tabular Displays

Frequency Distribution Relative Frequency Distribution Percent Frequency Distribution Cumulative Frequency Distribution Cumulative Relative Frequency Distribution

Graphical Displays

Dot Plot Histogram Stem-and-Leaf Display Scatter Diagram

Cumulative Percent Frequency Distribution Crosstabulation

frequency distributions, histograms, cumulative frequency distributions, cumulative relative frequency distributions, cumulative percent frequency distributions, and stem-and-leaf displays were presented as ways of summarizing the data for a single quantitative variable. A crosstabulation was presented as a tabular display for summarizing the data for two variables, and a scatter diagram was introduced as a graphical display for summarizing the data for two quantitative variables. We also showed that side-by-side bar charts and stacked bar charts are just extensions of basic bar charts that can be used to display and compare two categorical variables. Guidelines for creating effective graphical displays and how to choose the most appropriate type of display were discussed. Data dashboards were introduced to illustrate how a set of visual displays can be developed that organizes and presents information that is used to monitor a company’s performance in a manner that is easy to read, understand, and interpret. Figure 2.29 provides a summary of the tabular and graphical methods presented in this chapter.

Glossary Bar chartA graphical device for depicting categorical data that have been summarized in a frequency, relative frequency, or percent frequency distribution. Categorical dataLabels or names used to identify categories of like items. Class midpointThe value halfway between the lower and upper class limits. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

CrosstabulationA tabular summary of data for two variables. The classes for one variable are represented by the rows; the classes for the other variable are represented by the columns. Cumulative frequency distributionA tabular summary of quantitative data showing the number of data values that are less than or equal to the upper class limit of each class. Cumulative percent frequency distribution A tabular summary of quantitative data showing the percentage of data values that are less than or equal to the upper class limit of each class. Cumulative relative frequency distribution A tabular summary of quantitative data showing the fraction or proportion of data values that are less than or equal to the upper class limit of each class. Data dashboardA set of visual displays that organizes and presents information that is used to monitor the performance of a company or organization in a manner that is easy to read, understand, and interpret. Data visualizationA term used to describe the use of graphical displays to summarize and present information about a data set. Dot plotA graphical device that summarizes data by the number of dots above each data value on the horizontal axis. Frequency distributionA tabular summary of data showing the number (frequency) of observations in each of several nonoverlapping categories or classes. HistogramA graphical display of a frequency distribution, relative frequency distribution, or percent frequency distribution of quantitative data constructed by placing the class intervals on the horizontal axis and the frequencies, relative frequencies, or percent frequencies on the vertical axis. Percent frequency distribution A tabular summary of data showing the percentage of observations in each of several nonoverlapping classes. Pie chartA graphical device for presenting data summaries based on subdivision of a circle into sectors that correspond to the relative frequency for each class. Quantitative dataNumerical values that indicate how much or how many. Relative frequency distributionA tabular summary of data showing the fraction or proportion of observations in each of several nonoverlapping categories or classes. Scatter diagramA graphical display of the relationship between two quantitative variables. One variable is shown on the horizontal axis and the other variable is shown on the vertical axis. Side-by-side bar chartA graphical display for depicting multiple bar charts on the same display. Simpson’s paradoxConclusions drawn from two or more separate crosstabulations that can be reversed when the data are aggregated into a single crosstabulation. Stacked bar chartA bar chart in which each bar is broken into rectangular segments of a different color showing the relative frequency of each class in a manner similar to a pie chart. Stem-and-leaf displayA graphical display used to show simultaneously the rank order and shape of a distribution of data. TrendlineA line that provides an approximation of the relationship between two variables.

Key Formulas Relative Frequency

frequency of the class n

(2.1)

Supplementary Exercises

Approximate Class Width

Largest data value 2 Smallest data value Number of classes

(2.2)

Supplementary Exercises 44. A pproximately 1.5 million high school students take the SAT each year and nearly 80% of colleges and universities without open admissions policies use SAT scores in making admission decisions.The current version of the SAT includes three parts: reading comprehension, mathematics, and writing. A perfect c ombined score for all three parts is 2400. A sample of SAT scores for the combined three-part SAT is as follows.

NewSAT

1665 1525 1355 1645 1780 1275 2135 1280 1060 1585 1650 1560 1150 1485 1990 1590 1880 1420 1755 1375 1475 1680 1440 1260 1730 1490 1560 940 1390 1175 a. S how a frequency distribution and histogram. Begin with the first class starting at 800 and use a class width of 200. b. Comment on the shape of the distribution. c. What other observations can be made about the SAT scores based on the tabular and graphical summaries?

MedianHousehold

45. The DATAfile MedianHousehold contains the median household income for a family with two earners for each of the fifty states (American Community Survey, 2013). a. Construct a frequency and a percent frequency distribution of median household income. Begin the first class at 65.0 and use a class width of 5. b. Construct a histogram. c. Comment on the shape of the distribution. d. Which state has the highest median income for two-earner households? e. Which state has the lowest median income for two-earner households? 46. Data showing the population by state in millions of people follow (The World Almanac, 2012).

2012Population

State Population State Population State Population Alabama 4.8 Louisiana 4.5 Ohio 11.5 Alaska 0.7 Maine 1.3 Oklahoma 3.8 Arizona 6.4 Maryland 5.8 Oregon 4.3 Arkansas 2.9 Massachusetts 6.5 Pennsylvania 12.7 California 37.3 Michigan 9.9 Rhode Island 1.0 Colorado 5.0 Minnesota 5.3 South Carolina 4.6 Connecticut 3.6 Mississippi 3.0 South Dakota 0.8 Delaware 0.9 Missouri 6.0 Tennessee 6.3 Florida 18.8 Montana 0.9 Texas 25.1 Georgia 9.7 Nebraska 1.8 Utah 2.8 Hawaii 1.4 Nevada 2.7 Vermont 0.6 Idaho 1.6 New Hampshire 1.3 Virginia 8.0 Illinois 12.8 New Jersey 8.8 Washington 6.7 Indiana 6.5 New Mexico 2.0 West Virginia 1.9 Iowa 3.0 New York 19.4 Wisconsin 5.7 Kansas 2.9 North Carolina 9.5 Wyoming 0.6 Kentucky 4.3 North Dakota 0.7

Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

a. Develop a frequency distribution, a percent frequency distribution, and a histogram. Use a class width of 2.5 million. b. Does there appear to be any skewness in the distribution? Explain. c. What observations can you make about the population of the 50 states? 47. A startup company’s ability to gain funding is a key to success. The funds raised (in millions of dollars) by 50 startup companies follow (The Wall Street Journal, March 10, 2011). 81 80 69 73 192 91 47 154 48 54

StartUps

61 51 119 50 18 272 24 72 118 112

103 130 81 110 54 58 57 38 40 129

166 77 60 21 49 54 78 131 49 156

168 78 20 60 63 40 78 52 55 31

a. Construct a stem-and-leaf display. b. Comment on the display.

BBB

48. C onsumer complaints are frequently reported to the Better Business Bureau. In 2011, the industries with the most complaints to the Better Business Bureau were banks, cable and satellite television companies, collection agencies, cellular phone providers, and new car dealerships (USA Today, April 16, 2012). The results for a sample of 200 complaints are contained in the DATAfile named BBB. a. Show the frequency and percent frequency of complaints by industry. b. Construct a bar chart of the percent frequency distribution. c. Which industry had the highest number of complaints? d. Comment on the percentage frequency distribution for complaints. 49. The term Beta refers to a measure of a stock’s price volatility relative to the stock market as a whole. A Beta of 1 means the stock’s price moves exactly with the market. A Beta of 1.6 means the stock’s price would increase by 1.6% for an increase of 1% in the stock market. A larger Beta means the stock price is more volatile. The Betas for the stocks of the companies that make up the Dow Jones Industrial Average are shown in Table 2.17 (Yahoo Finance, November 2014).

Table 2.17 Betas for Dow Jones Industrial Average Companies

DJIABeta

Company

Beta

Company

Beta

American Express Company The Boeing Company Caterpillar Inc. Cisco Systems, Inc. Chevron Corporation E. I. du Pont de Nemours and Company The Walt Disney Company General Electric Company The Goldman Sachs Group, Inc. The Home Depot, Inc. International Business Machines Corporation Intel Corporation Johnson & Johnson JPMorgan Chase & Co. The Coca-Cola Company

1.24 0.99 1.2 1.36 1.11 1.36 0.97 1.19 1.79 1.22 0.92 0.9 0.84 1.84 0.68

McDonald's Corp. 3M Company Merck & Co. Inc. Microsoft Corporation Nike, Inc. Pfizer Inc. The Procter & Gamble Company AT&T, Inc. The Travelers Companies, Inc. UnitedHealth Group Incorporated United Technologies Corporation Visa Inc. Verizon Communications Inc. Walmart Stores Inc. Exxon Mobil Corporation

0.62 1.23 0.56 0.69 0.47 0.72 0.73 0.18 0.86 0.88 1.22 0.82 0.04 0.26 1.1

Supplementary Exercises

a. b. c. d.

Construct a frequency distribution and percent frequency distribution. Construct a histogram. Comment on the shape of the distribution. Which stock has the highest beta? Which has the lowest beta?

50. The U.S. Census Bureau serves as the leading source of quantitative data about the nation’s people and economy. The following crosstabulation shows the number of households (1000s) and the household income by the level of education for heads of household having received a high school degree or more education (U.S. Census Bureau website, 2013).

Household Income Level of Education High school graduate Bachelor’s degree Master’s degree Doctoral degree

Total

Under $25,000

$25,000 to $49,999

$50,000 to $99,999

9,880 2,484 685 79 13,128

9,970 4,164 1,205 160 15,499

9,441 7,666 3,019 422 20,548

$100,000 and Over

Total

3,482 7,817 4,094 1,076 16,469

32,773 22,131 9,003 1,737 65,644

a. Construct a percent frequency distribution for the level of education variable. What percentage of heads of households have a master’s or doctoral degree? b. Construct a percent frequency distribution for the household income variable. What percentage of households have an income of $50,000 or more? c. Convert the entries in the crosstabulation into column percentages. Compare the level of education of households with a household income of under $25,000 to the level of education of households with a household income of $100,000 or more. Comment on any other items of interest when reviewing the crosstabulation showing column percentages. 51. Western University has only one women’s softball scholarship remaining for the coming year. The final two players that Western is considering are Allison Fealey and Emily Janson. The coaching staff has concluded that the speed and defensive skills are virtually identical for the two players, and that the final decision will be based on which player has the best batting average. Crosstabulations of each player’s batting performance in their junior and senior years of high school are as follows:

Allison Fealey

Outcome Junior Senior Hit 15 75 No Hit 25 175 Total At-Bats 40 250

Emily Janson Outcome Junior Senior Hit 70 35 No Hit 130 85 Total At Bats 200 120

A player’s batting average is computed by dividing the number of hits a player has by the total number of at-bats. Batting averages are represented as a decimal number with three places after the decimal. a. Calculate the batting average for each player in her junior year. Then calculate the batting average of each player in her senior year. Using this analysis, which player should be awarded the scholarship? Explain. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

b. C ombine or aggregate the data for the junior and senior years into one crosstabulation as follows: Outcome Hit No Hit

Player Fealey Janson

Total At-Bats

Calculate each player’s batting average for the combined two years. Using this analysis, which player should be awarded the scholarship? Explain. c. Are the recommendations you made in parts (a) and (b) consistent? Explain any apparent inconsistencies.

FortuneBest100

52. F ortune magazine publishes an annual survey of the 100 best companies to work for. The data in the DATAfile named FortuneBest100 shows the rank, company name, the size of the company, and the percentage job growth for full-time employees for 98 of the Fortune 100 companies for which percentage job growth data were available (Fortune magazine website, February 25, 2013). The column labeled Rank shows the rank of the company in the Fortune 100 list; the column labeled Size indicates whether the company is a small company (less than 2500 employees), a midsized company (2500 to 10,000 employees), or a large company (more than 10,000 employees); and the column labeled Growth Rate (%) shows the percentage growth rate for full-time employees. a. Construct a crosstabulation with Job Growth (%) as the row variable and Size as the column variable. Use classes starting at 210 and ending at 70 in increments of 10 for Growth Rate (%). b. Show the frequency distribution for Job Growth (%) and the frequency distribution for Size. c. Using the crosstabulation constructed in part (a), develop a crosstabulation showing column percentages. d. Using the crosstabulation constructed in part (a), develop a crosstabulation showing row percentages. e. Comment on the relationship between the percentage job growth for full-time employees and the size of the company. 53. T able 2.18 shows a portion of the data for a sample of 103 private colleges and universities. The complete data set is contained in the DATAfile named Colleges. The data include the name of the college or university, the year the institution was founded, the tuition and fees (not including room and board) for the most recent academic year, and the percentage of full time, first-time bachelor’s degree–seeking undergraduate students who obtain their degree in six years or less (The World Almanac, 2012)

TABLE 2.18 DATA FOR A SAMPLE OF private colleges and universities

Colleges

School American University Baylor University Belmont University ? ? ? Wofford College Xavier University Yale University

Year Founded 1893 1845 1951 ? ? ? 1854 1831 1701

Tuition & Fees $36,697 $29,754 $23,680 ? ? ? $31,710 $29,970 $38,300

% Graduate 79.00 70.00 68.00 ? ? ? 82.00 79.00 98.00

101

Supplementary Exercises

a. C onstruct a crosstabulation with Year Founded as the row variable and Tuition & Fees as the column variable. Use classes starting with 1600 and ending with 2000 in increments of 50 for Year Founded. For Tuition & Fees, use classes starting with 1 and ending 45,000 in increments of 5000. b. Compute the row percentages for the crosstabulation in part (a). c. What relationship, if any, do you notice between Year Founded and Tuition & Fees? 54. R efer to the data set in Table 2.18. a. Construct a crosstabulation with Year Founded as the row variable and % Graduate as the column variable. Use classes starting with 1600 and ending with 2000 in increments of 50 for Year Founded. For % Graduate, use classes starting with 35% and ending with 100% in increments of 5%. b. Compute the row percentages for your crosstabulation in part (a). c. Comment on any relationship between the variables. 55. R efer to the data set in Table 2.18. a. Construct a scatter diagram to show the relationship between Year Founded and Tuition & Fees. b. Comment on any relationship between the variables. 56. R efer to the data set in Table 2.18. a. Prepare a scatter diagram to show the relationship between Tuition & Fees and % Graduate. b. Comment on any relationship between the variables. 57. G oogle has changed its strategy with regard to how much it invests in advertising, and which media it uses to do so. The following table shows Google’s marketing budget in millions of dollars for 2008 and 2011 (The Wall Street Journal, March 27, 2012).

2008 2011 Internet 26.0 123.3 Newspaper, etc. 4.0 20.7 Television 0.0 69.3

a. C onstruct a side-by-side bar chart with year as the variable on the horizontal axis. Comment on any trend in the display. b. Convert the above table to percentage allocation for each year. Construct a stacked bar chart with year as the variable on the horizontal axis. c. Is the display in part (a) or part (b) more insightful? Explain. 58. A zoo has categorized its visitors into three categories: member, school, and general. The member category refers to visitors who pay an annual fee to support the zoo. Members receive certain benefits such as discounts on merchandise and trips planned by the zoo. The school category includes faculty and students from day care and elementary and secondary schools; these visitors generally receive a discounted rate. The general category includes all other visitors. The zoo has been concerned about a recent drop in attendance. To help better understand attendance and membership, a zoo staff member has collected the following data:

Zoo

Attendance

Visitor Category 2008 2009 2010 2011 General 153,713 158,704 163,433 169,106 Member 115,523 104,795 98,437 81,217 School 82,885 79,876 81,970 81,290 Total 352,121 343,375 343,840 331,613

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Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

a. C onstruct a bar chart of total attendance over time. Comment on any trend in the data. b. Construct a side-by-side bar chart showing attendance by visitor category with year as the variable on the horizontal axis. c. Comment on what is happening to zoo attendance based on the charts from parts (a) and (b).

Case Problem 1

Pelican Stores Pelican Stores, a division of National Clothing, is a chain of women’s apparel stores operating throughout the country. The chain recently ran a promotion in which discount coupons were sent to customers of other National Clothing stores. Data collected for a sample of 100 in-store credit card transactions at Pelican Stores during one day while the promotion was running are contained in the DATAfile named PelicanStores. Table 2.19 shows a portion of the data set. The Proprietary Card method of payment refers to charges made using a National Clothing charge card. Customers who made a purchase using a discount coupon are referred to as promotional customers, and customers who made a purchase but did not use a discount coupon are referred to as regular customers. Because the promotional coupons were not sent to regular Pelican Stores customers, management considers the sales made to people presenting the promotional coupons as sales it would not otherwise make. Of course, Pelican also hopes that the promotional customers will continue to shop at its stores. Most of the variables shown in Table 2.19 are self-explanatory, but two of the variables require some clarification. Items The total number of items purchased Net Sales The total amount ($) charged to the credit card

Pelican’s management would like to use this sample data to learn about its customer base and to evaluate the promotion involving discount coupons.

Managerial Report Use the tabular and graphical methods of descriptive statistics to help management develop a customer profile and to evaluate the promotional campaign. At a minimum, your report should include the following: 1. Percent frequency distribution for key variables. 2. A bar chart or pie chart showing the number of customer purchases attributable to the method of payment. TABLE 2.19 DATA FOR A SAMPLE OF 100 CREDIT CARD PURCHASES AT PELICAN STORES

Type of Customer Customer Items Net Sales

PelicanStores

1 2 3 4 5 ? ? ? 96 97 98 99 100

Regular 1 Promotional 1 Regular 1 Promotional 5 Regular 2 ? ? ? ? ? ? Regular 1 Promotional 9 Promotional 10 Promotional 2 Promotional 1

Method of Marital Payment Gender Status Age

39.50 Discover Male 102.40 Proprietary Card Female 22.50 Proprietary Card Female 100.40 Proprietary Card Female 54.00 MasterCard Female ? ? ? ? ? ? ? ? ? 39.50 MasterCard Female 253.00 Proprietary Card Female 287.59 Proprietary Card Female 47.60 Proprietary Card Female 28.44 Proprietary Card Female

Married Married Married Married Married ? ? ? Married Married Married Married Married

32 36 32 28 34

? ? ?

44 30 52 30 44

103

Case Problem 2 Motion Picture Industry

3. A crosstabulation of type of customer (regular or promotional) versus net sales. Comment on any similarities or differences present. 4. A scatter diagram to explore the relationship between net sales and customer age.

Case Problem 2

Motion Picture Industry The motion picture industry is a competitive business. More than 50 studios produce a total of 300 to 400 new motion pictures each year, and the financial success of each motion picture varies considerably. The opening weekend gross sales ($ millions), the total gross sales ($ millions), the number of theaters the movie was shown in, and the number of weeks the motion picture was in release are common variables used to measure the success of a motion picture. Data collected for the top 100 motion pictures produced in 2011 are contained in the DATAfile named 2011Movies (Box Office Mojo, March 17, 2012). Table 2.20 shows the data for the first 10 motion pictures in this file.

Managerial Report Use the tabular and graphical methods of descriptive statistics to learn how these variables contribute to the success of a motion picture. Include the following in your report. 1. Tabular and graphical summaries for each of the four variables along with a discus sion of what each summary tells us about the motion picture industry. 2. A scatter diagram to explore the relationship between Total Gross Sales and Opening Weekend Gross Sales. Discuss. 3. A scatter diagram to explore the relationship between Total Gross Sales and Number of Theaters. Discuss. 4. A scatter diagram to explore the relationship between Total Gross Sales and Number of Weeks in Release. Discuss. TABLE 2.20 Performance Data for 10 Motion Pictures Motion Picture

2011Movies

Harry Potter and the Deathly Hallows Part 2 Transformers: Dark of the Moon The Twilight Saga: Breaking Dawn Part 1 The Hangover Part II Pirates of the Caribbean: On Stranger Tides Fast Five Mission: Impossible— Ghost Protocol Cars 2 Sherlock Holmes: A Game of Shadows Thor

Opening Gross Sales ($ millions)

Total Gross Sales Number of ($ millions) Theaters

Weeks in Release

169.19 381.01 4375 19

97.85 352.39 4088 15 138.12 281.29 4066 14

85.95 254.46 3675 16 90.15 241.07 4164 19

86.20 12.79

209.84 208.55

3793 3555

15 13

66.14 191.45 4115 25 39.64 186.59 3703 13 65.72 181.03 3963 16

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Chapter 2 Descriptive Statistics: Tabular and Graphical Displays

Case Problem 3

Queen City Cincinnati, Ohio, also known as the Queen City, has a population of approximately 298,000 and is the third largest city in the state of Ohio. The Cincinnati metropolitan area has a population of about 2.2 million. The city is governed by a mayor and a ninemember city council. The city manager, who is responsible for the day-to-day operation of the city, reports to the mayor and city council. The city manager recently created the Office of Performance and Data Analytics with the goal of improving the efficiency of city operations. One of the first tasks of this new office is to review the previous year’s expenditures. The file QueenCity contains data on the previous year’s expenditures, including the following: Department The number of the department incurring the expenditure Department Description The name of the department incurring the description Category The category of the expenditure Fund The fund to which the expenditure was charged Expenditure The dollar amount of the expense

QueenCity

Table 2.21 shows the first four entries of the 5427 expenditures for the year. The city manager would like to use this data to better understand how the city’s budget is being spent.

Managerial Report Use tabular and graphical methods of descriptive statistics to help the city manager get a better understanding of how the city is spending its funding. Your report should include the following: 1. Tables and/or graphical displays that show the amount of expenditures by category and percentage of total expenditures by category. 2. A table that shows the amount of expenditures by department and the percentage of total expenditures by department. Combine any department with less than 1% into a category named “Other.” 3. A table that shows the amount of expenditures by fund and the percentage of total expenditures by fund. Combine any fund with less than 1% into a category named “Other.”

Case Problem 4

Cut-Rate Machining, Inc. Jon Weideman, first shift foreman for Cut-Rate Machining, Inc., is attempting to decide on a vendor from whom to purchase a drilling machine. He narrows his alternatives to four vendors: The Hole-Maker, Inc. (HM); Shafts & Slips, Inc. (SS); Judge’s Jigs (JJ); and

TABLE 2.21 ANNUAL EXPENDITURES FOR QUEEN CITY (FIRST FOUR ENTRIES)

Department

Department Description

See Also

North park san jose apartments Idea

E-mail

10 5 8:00

9:00

10:00 Hour

11:00

12:00

W59 Case Number

E-mail Austin 0

100

200 Minutes

300

400

10 15 Number of Calls

321

31–32

30–31

29–30

28–29

27–28

26–27

25–26

22–23

21–22

20–21

19–20

18–19

17–18

16–17

15–16

14–15

13–14

12–13

11–12

10–11

8–9

9–10

7–8

6–7

5–6

4–5

3–4

2–3

1–2

Time to Resolve a Case

Frequency

E-mail

Dallas

T57

14 12 10 8 6 4 2 0

Software Internet

Internet

Internet 18%

Houston

Software

W24

E-mail 22%

Call Volume by Office

Unresolved Cases Beyond 15 Minutes

24–25

Software 46%

23–24

Number of Calls

Idle 14%

Software

Minutes

●●

he pie chart in the upper right corner of the dashboard shows the percentage of time T that call center employees spent on each type of problem or not working on a call (idle). For each unresolved case that extended beyond 15 minutes, the bar chart shown in the middle left portion of the dashboard shows the length of time that each of these cases has been unresolved. The bar chart in the middle right portion of the dashboard shows the call volume by office (Houston, Dallas, Austin) for each type of problem. The histogram at the bottom of the dashboard shows the distribution of the time to resolve a case for all resolved cases for the current shift.

In order to gain additional insight into the performance of the call center, Grogan’s IT manager has decided to expand the current dashboard by adding box plots for the time required to resolve calls received for each type of problem (e-mail, Internet, and software). In addition, a graph showing the time to resolve individual cases has been added in the lower left portion of the dashboard. Finally, the IT manager added a display of summary statistics for each type of problem and summary statistics for each of the first few hours of the shift. The updated dashboard is shown in Figure 3.21. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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3.6 Data Dashboards: Adding Numerical Measures to Improve Effectiveness

FIGURE 3.21 UPDATED GROGAN OIL INfORMATION TECHNOLOGY CALL CENTER DATA

DASHBOARD Grogan Oil

IT Call Center

Shift 1

19–Sep–12

12:44:00 PM

Call Volume

Time Breakdown This Shift

Internet

15 5 8:00

9:00

10:00 Hour

11:00

12:00

Call Volume by Office

Unresolved Cases Beyond 15 Minutes W59 Case Number

Internet

Austin 0

100

200 Minutes

300

400

10 15 Number of Calls

321

31–32

30–31

29–30

28–29

27–28

26–27

25–26

22–23

21–22

20–21

19–20

18–19

17–18

16–17

15–16

14–15

13–14

12–13

11–12

10–11

8–9

9–10

7–8

6–7

5–6

4–5

3–4

2–3

1–2

Time to Resolve a Case

Frequency

E-mail

Dallas

E-mail

T57

14 12 10 8 6 4 2 0

Software

Houston

Software

W24

Internet 18%

Software 46%

24–25

E-mail 22%

E-mail

23–24

Number of Calls

Idle 14%

Software

Minutes

8:00

Time to Resolve Cases 9:00 10:00

Box Plot of Time to Resolve by Type of Case

11:00 12:00

25 20

Minutes

10 5

5 0

0 1

6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 Call E-mail Software Internet

Internet

E-mail

Software

Type of Case

Summary Statistics - Resolved Cases

Type of Case E-mail Internet Software

Cases 34 19 23

Mean 4.6 5.4 5.2

Median 2.0 3.0 4.0

Std. Dev 5.6 4.9 4.2

Hour 8:00 9:00 10:00 11:00 12:00

Cases 22 19 19 9 6

Mean 3.5 5.8 5.3 6.9 4.8

Median 2.0 3.0 4.0 6.0 3.5

Std. Dev 3.7 6.6 4.8 5.1 3.9

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Chapter 3 Descriptive Statistics: Numerical Measures

Drilling down refers to functionality in interactive data dashboards that allows the user to access information and analyses at an increasingly detailed level.

The IT call center has set a target performance level or benchmark of 10 minutes for the mean time to resolve a case. Furthermore, the center has decided it is undesirable for the time to resolve a case to exceed 15 minutes. To reflect these benchmarks, a black horizontal line at the mean target value of 10 minutes and a red horizontal line at the maximum acceptable level of 15 minutes have been added to both the graph showing the time to resolve cases and the box plots of the time required to resolve calls received for each type of problem. The summary statistics in the dashboard in Figure 3.21 show that the mean time to resolve an e-mail case is 4.6 minutes, the mean time to resolve an Internet case is 5.4 minutes, and the mean time to resolve a software case is 5.2 minutes. Thus, the mean time to resolve each type of case is better than the target mean (10 minutes). Reviewing the box plots, we see that the box associated with the e-mail cases is larger than the boxes associated with the other two types of cases. The summary statistics also show that the standard deviation of the time to resolve e-mail cases is larger than the standard deviations of the times to resolve the other types of cases. This leads us to take a closer look at the e-mail cases in the two new graphs. The box plot for the e-mail cases has a whisker that extends beyond 15 minutes and an outlier well beyond 15 minutes. The graph of the time to resolve individual cases (in the lower left position of the dashboard) shows that this is because of two calls on e-mail cases during the 9:00 hour that took longer than the target maximum time (15 minutes) to resolve. This analysis may lead the IT call center manager to further investigate why resolution times are more variable for e-mail cases than for Internet or software cases. Based on this analysis, the IT manager may also decide to investigate the circumstances that led to inordinately long resolution times for the two e-mail cases that took longer than 15 minutes to resolve. The graph of the time to resolve individual cases also shows that most calls received during the first hour of the shift were resolved relatively quickly; the graph also shows that the time to resolve cases increased gradually throughout the morning. This could be due to a tendency for complex problems to arise later in the shift or possibly to the backlog of calls that accumulates over time. Although the summary statistics suggest that cases submitted during the 9:00 hour take the longest to resolve, the graph of time to resolve individual cases shows that two time-consuming e-mail cases and one time-consuming software case were reported during that hour, and this may explain why the mean time to resolve cases during the 9:00 hour is larger than during any other hour of the shift. Overall, reported cases have generally been resolved in 15 minutes or less during this shift. Dashboards such as the Grogan Oil data dashboard are often interactive. For instance, when a manager uses a mouse or a touch screen monitor to position the cursor over the display or point to something on the display, additional information, such as the time to resolve the problem, the time the call was received, and the individual and/or the location that reported the problem, may appear. Clicking on the individual item may also take the user to a new level of analysis at the individual case level.

Summary In this chapter we introduced several descriptive statistics that can be used to summarize the location, variability, and shape of a data distribution. Unlike the tabular and graphical displays introduced in Chapter 2, the measures introduced in this chapter summarize the data in terms of numerical values. When the numerical values obtained are for a sample, they are called sample statistics. When the numerical values obtained are for a population,

165

Glossary

they are called population parameters. Some of the notation used for sample statistics and population parameters follow. In statistical inference, a sample statistic is referred to as a point estimator of the population parameter.

Mean Variance Standard deviation Covariance Correlation

Sample Statistic

Population Parameter

x s2 s sxy rxy

2 xy xy

As measures of location, we defined the mean, median, mode, weighted mean, geometric mean, percentiles, and quartiles. Next, we presented the range, interquartile range, variance, standard deviation, and coefficient of variation as measures of variability or dispersion. Our primary measure of the shape of a data distribution was the skewness. Negative values of skewness indicate a data distribution skewed to the left, and positive values of skewness indicate a data distribution skewed to the right. We then described how the mean and standard deviation could be used, applying Chebyshev’s theorem and the empirical rule, to provide more information about the distribution of data and to identify outliers. In Section 3.4 we showed how to develop a five-number summary and a box plot to p rovide simultaneous information about the location, variability, and shape of the distribution. In Section 3.5 we introduced covariance and the correlation coefficient as m easures of association between two variables. In the final section, we showed how adding numerical measures can improve the effectiveness of data dashboards. The descriptive statistics we discussed can be developed using statistical software packages and spreadsheets.

Glossary Box plotA graphical summary of data based on a five-number summary. Chebyshev’s theoremA theorem that can be used to make statements about the proportion of data values that must be within a specified number of standard deviations of the mean. Coefficient of variationA measure of relative variability computed by dividing the standard deviation by the mean and multiplying by 100. Correlation coefficientA measure of linear association between two variables that takes on values between 21 and 11. Values near 11 indicate a strong positive linear relationship; values near 21 indicate a strong negative linear relationship; and values near zero indicate the lack of a linear relationship. CovarianceA measure of linear association between two variables. Positive values indicate a positive relationship; negative values indicate a negative relationship. Empirical rule A rule that can be used to compute the percentage of data values that mustbe within one, two, and three standard deviations of the mean for data that exhibit a bell-shaped distribution. Five-number summaryA technique that uses five numbers to summarize the data: smallest value, first quartile, median, third quartile, and largest value. Geometric meanA measure of location that is calculated by finding the nth root of the product of n values. Interquartile range (IQR)A measure of variability, defined as the difference between the third and first quartiles. MeanA measure of central location computed by summing the data values and dividing by the number of observations. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Chapter 3 Descriptive Statistics: Numerical Measures

MedianA measure of central location provided by the value in the middle when the data are arranged in ascending order. ModeA measure of location, defined as the value that occurs with greatest frequency. OutlierAn unusually small or unusually large data value. Pearson product moment correlation coefficient A measure of the linear relationship between two variables. PercentileA value that provides information about how the data are spread over the interval from the smallest to the largest value. Point estimatorA sample statistic, such as x, s 2, and s, used to estimate the corresponding population parameter. Population parameterA numerical value used as a summary measure for a population (e.g., the population mean, , the population variance, 2, and the population standard deviation, ). pth percentileFor a data set containing n observations, the pth percentile divides the data into two parts: Approximately p% of the observation are less than the pth percentile and approximately (100 2 p)% of the observations are greater than the pth percentile. QuartilesThe 25th, 50th, and 75th percentiles, referred to as the first quartile, the second quartile (median), and third quartile, respectively. The quartiles can be used to divide a data set into four parts, with each part containing approximately 25% of the data. RangeA measure of variability, defined to be the largest value minus the smallest value. Sample statisticA numerical value used as a summary measure for a sample (e.g., the sample mean, x, the sample variance, s 2, and the sample standard deviation, s). SkewnessA measure of the shape of a data distribution. Data skewed to the left result in negative skewness; a symmetric data distribution results in zero skewness; and data skewed to the right result in positive skewness. Standard deviationA measure of variability computed by taking the positive square root of the variance. VarianceA measure of variability based on the squared deviations of the data values about the mean. Weighted meanThe mean obtained by assigning each observation a weight that reflects its importance. z-scoreA value computed by dividing the deviation about the mean (xi 2 x) by the standard deviation s. A z-score is referred to as a standardized value and denotes the number of standard deviations xi is from the mean.

Key Formulas Sample Mean

oxi

(3.1)

Population Mean

(3.2)

Weighted Mean

owi xi owi

(3.3)

Geometric Mean

n xg 5 Ï sx1dsx2d Á sxnd 5 [sx1dsx2d Á sxnd]1/n

(3.4)

167

Key Formulas

Location of the pth Percentile

p sn 1 1d 100

(3.5)

IQR 5 Q3 2 Q1

(3.6)

Lp 5

Interquartile Range Population Variance

2 5

o(xi 2 )2

s2 5

o(xi 2 x)2

(3.7)

(3.8)

Sample Variance

n21

Standard Deviation

Sample standard deviation 5 s 5 Ïs2

Population standard deviation 5 5 Ï2

(3.9) (3.10)

Coefficient of Variation

1StandardMeandeviation 3 1002%

(3.11)

z-Score

zi 5

xi 2 x s

(3.12)

Sample Covariance

sxy 5

o(xi 2 x)( yi 2 y) n21

(3.13)

Population Covariance

o(xi 2 x)( yi 2 y) xy 5 N

(3.14)

Pearson Product Moment Correlation Coefficient: Sample Data

rxy 5

sxy sx sy

(3.15)

Pearson Product Moment Correlation Coefficient: Population Data

xy 5

xy x y

(3.16)

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Chapter 3 Descriptive Statistics: Numerical Measures

Supplementary Exercises 62. T he average number of times Americans dine out in a week fell from 4.0 in 2008 to 3.8 in 2012 (Zagat.com, April 1, 2012). The number of times a sample of 20 families dined out last week provides the following data. 6 1 5 3 7 3 0 3 1 3 4 1 2 4 1 0 5 6 3 1 a. b. c. d. e.

Coaches

Compute the mean and median. Compute the first and third quartiles. Compute the range and interquartile range. Compute the variance and standard deviation. The skewness measure for these data is 0.34. Comment on the shape of this distribution. Is it the shape you would expect? Why or why not? f. Do the data contain outliers? 63. USA Today reports that NCAA colleges and universities are paying higher salaries to a newly recruited football coach compared to what they paid their previous football coach. (USA Today, February 12, 2013). The annual base salaries for the previous head football coach and the new head football coach at 23 schools are given in the DATAfile Coaches. a. Determine the median annual salary for a previous head football coach and a new head football coach. b. Compute the range for salaries for both previous and new head football coaches. c. Compute the standard deviation for salaries for both previous and new head football coaches. d. Based on your answers to (a) to (c), comment on any differences between the annual base salary a school pays a new head football coach compared to what it paid its previous head football coach. 64. T he average waiting time for a patient at an El Paso physician’s office is just over 29 minutes, well above the national average of 21 minutes. In fact, El Paso has the longest physician’s office waiting times in the United States (El Paso Times, January 8, 2012). In order to address the issue of long patient wait times, some physician’s offices are using wait tracking systems to notify patients of expected wait times. Patients can adjust their arrival times based on this information and spend less time in waiting rooms. The following data show wait times (minutes) for a sample of patients at offices that do not have an office tracking system and wait times for a sample of patients at offices with an office tracking system.

Without Wait Tracking System WaitTracking

With Wait Tracking System

24 31 67 11 17 14 20 18 31 12 44 37 12 9 23 13 16 12 37 15

169

Supplementary Exercises

a. W hat are the mean and median patient wait times for offices with a wait tracking system? What are the mean and median patient wait times for offices without a wait tracking system? b. What are the variance and standard deviation of patient wait times for offices with a wait tracking system? What are the variance and standard deviation of patient wait times for visits to offices without a wait tracking system? c. Do offices with a wait tracking system have shorter patient wait times than offices without a wait tracking system? Explain. d. Considering only offices without a wait tracking system, what is the z-score for the tenth patient in the sample? e. Considering only offices with a wait tracking system, what is the z-score for the sixth patient in the sample? How does this z-score compare with the z-score you calculated for part (d)? f. Based on z-scores, do the data for offices without a wait tracking system contain any outliers? Based on z-scores, do the data for offices with a wait tracking system contain any outliers?

Sleep

65. U.S. companies lose $63.2 billion per year from workers with insomnia. Workers lose an average of 7.8 days of productivity per year due to lack of sleep (Wall Street Journal, January 23, 2013). The following data show the number of hours of sleep attained during a recent night for a sample of 20 workers. 6 8

5 10 5 6 9 9 5 9 5 7 8 6 9 8 9 6 10 8

a. What is the mean number of hours of sleep for this sample? b. What is the variance? Standard deviation?

Smartphone

66. A study of smartphone users shows that 68% of smartphone use occurs at home and a user spends an average of 410 minutes per month using a smartphone to interact with other people (Harvard Business Review, January–February 2013). Consider the following data indicating the number of minutes in a month spent interacting with others via a smartphone for a sample of 50 smartphone users. 353 458 404 394 416 437 430 369 448 430 431 469 446 387 445 354 468 422 402 360 444 424 441 357 435 461 407 470 413 351 464 374 417 460 352 445 387 468 368 430 384 367 436 390 464 405 372 401 388 367 a. W hat is the mean number of minutes spent interacting with others for this sample? How does it compare to the mean reported in the study? b. What is the standard deviation for this sample? c. Are there any outliers in this sample? 67. P ublic transportation and the automobile are two methods an employee can use to get to work each day. Samples of travel times recorded for each method are shown. Times are in m inutes.

Transportation

Public Transportation: 28 Automobile: 29

29 31

32 33

37 32

33 34

25 30

29 31

32 32

41 35

34 33

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Chapter 3 Descriptive Statistics: Numerical Measures

a. C ompute the sample mean time to get to work for each method. b. Compute the sample standard deviation for each method. c. On the basis of your results from parts (a) and (b), which method of transportation should be preferred? Explain. d. Develop a box plot for each method. Does a comparison of the box plots support your conclusion in part (c)? 68. In 2007 the New York Times reported that the median annual household income in the United States was $55,500 (New York Times website, August 21, 2013). Answer the following questions based on the following sample of 14 household incomes for 2013 ($1000s). 49.4 52.2

52.4 64.5

53.4 51.6

51.3 46.5

52.1 52.9

48.7 52.5

52.1 51.2

a. What is the median household income for the sample data for 2013? b. Based on the sample data, estimate the percentage change in the median household income from 2007 to 2013. c. Compute the first and third quartiles. d. Provide a five-number summary. e. Using the z-score approach, do the data contain any outliers? Does the approach that uses the values of the first and third quartiles and the interquartile range to detect outliers provide the same results?

FoodIndustry

69. T he data contained in the DATAfile named FoodIndustry show the company/chain name, the average sales per store ($1000s), and the food segment industry for 47 restaurant chains (Quick Service Restaurant Magazine website, August 2013). a. What was the mean U.S. sales per store for the 47 restaurant chains? b. What are the first and third quartiles? What is your interpretation of the quartiles? c. Show a box plot for the level of sales and discuss if there are any outliers in terms of sales that would skew the results. d. Develop a frequency distribution showing the average sales per store for each segment. Comment on the results obtained. 70. T ravel 1 Leisure magazine presented its annual list of the 500 best hotels in the world. The magazine provides a rating for each hotel along with a brief description that includes the size of the hotel, amenities, and the cost per night for a double room. A sample of 12 of the top-rated hotels in the United States f ollows.

Travel

Hotel

Location

Boulders Resort & Spa Disney’s Wilderness Lodge Four Seasons Hotel Beverly Hills Four Seasons Hotel Hay-Adams Inn on Biltmore Estate Loews Ventana Canyon Resort Mauna Lani Bay Hotel Montage Laguna Beach Sofitel Water Tower St. Regis Monarch Beach The Broadmoor

Phoenix, AZ Orlando, FL Los Angeles, CA Boston, MA Washington, DC Asheville, NC Phoenix, AZ Island of Hawaii Laguna Beach, CA Chicago, IL Dana Point, CA Colorado Springs, CO

RoomsCost/Night($) 220 727 285 273 145 213 398 343 250 414 400 700

499 340 585 495 495 279 279 455 595 367 675 420

a. W hat is the mean number of rooms? b. What is the mean cost per night for a double room? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Supplementary Exercises

c. D evelop a scatter diagram with the number of rooms on the horizontal axis and the cost per night on the vertical axis. Does there appear to be a relationship between the number of rooms and the cost per night? Discuss. d. What is the sample correlation coefficient? What does it tell you about the relationship between the number of rooms and the cost per night for a double room? Does this appear reasonable? Discuss. 71. T he 32 teams in the National Football League (NFL) are worth, on average, $1.17 billion, 5% more than last year. The following data show the annual revenue ($ millions) and the estimated team value ($ millions) for the 32 NFL teams (Forbes website, February 28, 2014).

Team

NFLTeamValue

Arizona Cardinals Atlanta Falcons Baltimore Ravens Buffalo Bills Carolina Panthers Chicago Bears Cincinnati Bengals Cleveland Browns Dallas Cowboys Denver Broncos Detroit Lions Green Bay Packers Houston Texans Indianapolis Colts Jacksonville Jaguars Kansas City Chiefs Miami Dolphins Minnesota Vikings New England Patriots New Orleans Saints New York Giants New York Jets Oakland Raiders Philadelphia Eagles Pittsburgh Steelers San Diego Chargers San Francisco 49ers Seattle Seahawks St. Louis Rams Tampa Bay Buccaneers Tennessee Titans Washington Redskins

Revenue ($ millions)

Current Value ($ millions)

253 252 292 256 271 298 250 264 539 283 248 282 320 276 260 245 268 234 408 276 338 321 229 306 266 250 255 270 239 267 270 381

961 933 1227 870 1057 1252 924 1005 2300 1161 900 1183 1450 1200 840 1009 1074 1007 1800 1004 1550 1380 825 1314 1118 949 1224 1081 875 1067 1055 1700

a. D evelop a scatter diagram with Revenue on the horizontal axis and Value on the vertical axis. Does there appear that there is any relationship between the two variables? b. What is the sample correlation coefficient? What can you say about the strength of the relationship between Revenue and Value? 72. D oes a major league baseball team’s record during spring training indicate how the team will play during the regular season? Over the last six years, the correlation coefficient between a team’s winning percentage in spring training and its winning percentage in the regular season is .18. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Chapter 3 Descriptive Statistics: Numerical Measures

Team SpringTraining

Baltimore Orioles Boston Red Sox Chicago White Sox Cleveland Indians Detroit Tigers Kansas City Royals Los Angeles Angels

Spring Regular Training Season Team .407 .429 .417 .569 .569 .533 .724

.422 .586 .546 .500 .457 .463 .617

Minnesota Twins New York Yankees Oakland A’s Seattle Mariners Tampa Bay Rays Texas Rangers Toronto Blue Jays

Spring Regular Training Season .500 .577 .692 .500 .731 .643 .448

.540 .549 .466 .377 .599 .488 .531

Shown are the winning percentages for the 14 American League teams during a previous season. a. What is the correlation coefficient between the spring training and the regular season winning percentages? b. What is your conclusion about a team’s record during spring training indicating how the team will play during the regular season? What are some of the reasons why this occurs? Discuss. 73. T he days to maturity for a sample of five money market funds are shown here. The dollar amounts invested in the funds are provided. Use the weighted mean to determine the mean number of days to maturity for dollars invested in these five money market funds.

Days to Dollar Value Maturity ($millions)

20 20 12 30 7 10 5 15 6 10

74. A utomobiles traveling on a road with a posted speed limit of 55 miles per hour are checked for speed by a state police radar system. Following is a frequency distribution of speeds. a. What is the mean speed of the automobiles traveling on this road? b. Compute the variance and the standard deviation.

Speed (miles per hour)

Frequency

45–49 10 50–54 40 55–59 150 60–64 175 65–69 75 70–74 15 75–79 10 Total 475

173

Case Problem 1 Pelican Stores

75. T he Panama Railroad Company was established in 1850 to construct a railroad across the isthmus that would allow fast and easy access between the Atlantic and Pacific Oceans. The following table (The Big Ditch, Mauer and Yu, 2011) provides annual returns for Panama Railroad stock from 1853 through 1880.

PanamaRailroad

Year

Return on Panama Railroad Company Stock (%)

1853 21 1854 29 1855 19 1856 2 1857 3 1858 36 1859 21 1860 16 1861 25 1862 43 1863 44 1864 48 1865 7 1866 11 1867 23 1868 20 1869 211 1870 251 1871 242 1872 39 1873 42 1874 12 1875 26 1876 9 1877 26 1878 25 1879 31 1880 30

a. C reate a graph of the annual returns on the stock. The New York Stock Exchange earned an annual average return of 8.4% from 1853 through 1880. Can you tell from the graph if the Panama Railroad Company stock outperformed the New York Stock Exchange? b. Calculate the mean annual return on Panama Railroad Company stock from 1853 through 1880. Did the stock outperform the New York Stock Exchange over the same period?

Case Problem 1 Pelican Stores Pelican Stores, a division of National Clothing, is a chain of women’s apparel stores o perating throughout the country. The chain recently ran a promotion in which discount coupons were sent to customers of other National Clothing stores. Data collected for a sample of 100 in-store credit card transactions at Pelican Stores during one day while the promotion was Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Chapter 3 Descriptive Statistics: Numerical Measures

TABLE 3.9 SAMPLE OF 100 CREDIT CARD PURCHASES AT PELICAN STORES

PelicanStores

Customer

Type of Customer Items Net Sales

1 2 3 4 5 6 7 8 9 10 . . . 96 97 98 99 100

Regular 1 Promotional 1 Regular 1 Promotional 5 Regular 2 Regular 1 Promotional 2 Regular 1 Promotional 2 Regular 1 . . . . . . Regular 1 Promotional 9 Promotional 10 Promotional 2 Promotional 1

39.50 102.40 22.50 100.40 54.00 44.50 78.00 22.50 56.52 44.50 . . . 39.50 253.00 287.59 47.60 28.44

Method of Payment Gender Discover Proprietary Card Proprietary Card Proprietary Card MasterCard MasterCard Proprietary Card Visa Proprietary Card Proprietary Card . . . MasterCard Proprietary Card Proprietary Card Proprietary Card Proprietary Card

Male Female Female Female Female Female Female Female Female Female . . . Female Female Female Female Female

Marital Status

Age

Married Married Married Married Married Married Married Married Married Married . . . Married Married Married Married Married

32 36 32 28 34 44 30 40 46 36 . . . 44 30 52 30 44

running are contained in the DATAfile named PelicanStores. Table 3.9 shows a portion of the data set. The proprietary card method of payment refers to charges made using a N ational Clothing charge card. Customers who made a purchase using a discount coupon are referred to as promotional customers and customers who made a purchase but did not use a discount coupon are referred to as regular customers. Because the promotional coupons were not sent to regular Pelican Stores customers, management considers the sales made to people presenting the promotional coupons as sales it would not otherwise make. Of course, Pelican also hopes that the promotional customers will continue to shop at its stores. Most of the variables shown in Table 3.9 are self-explanatory, but two of the variables require some clarification. Items The total number of items purchased Net Sales The total amount ($) charged to the credit card Pelican’s management would like to use this sample data to learn about its customer base and to evaluate the promotion involving discount coupons.

Managerial Report Use the methods of descriptive statistics presented in this chapter to summarize the data and comment on your findings. At a minimum, your report should include the following: 1. Descriptive statistics on net sales and descriptive statistics on net sales by various classifications of customers. 2. Descriptive statistics concerning the relationship between age and net sales.

Case Problem 2 Motion Picture Industry The motion picture industry is a competitive business. More than 50 studios produce several hundred new motion pictures each year, and the financial success of the motion pictures varies considerably. The opening weekend gross sales, the total gross sales, the number of Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

175

Case Problem 3 Business Schools of Asia-Pacific

TABLE 3.10 Performance Data for 10 Motion Pictures

Motion Picture Harry Potter and the Deathly Hallows Part 2 Transformers: Dark of the Moon The Twilight Saga: Breaking Dawn Part 1 The Hangover Part II Pirates of the Caribbean: On Stranger Tides Fast Five Mission: Impossible—Ghost Protocol Cars 2 Sherlock Holmes: A Game of Shadows Thor

2011Movies

Opening Gross Sales ($millions)

Total Gross Sales ($millions)

Number of Theaters

Weeks in Release

169.19 381.01 4375 19 97.85 352.39 4088 15 138.12 281.29 4066 14 85.95 254.46 3675 16 90.15 241.07 4164 19 86.20 209.84 3793 15 12.79 208.55 3555 13 66.14 191.45 4115 25 39.64 186.59 3703 13 65.72 181.03 3963 16

theaters the movie was shown in, and the number of weeks the motion picture was in r elease are common variables used to measure the success of a motion picture. Data on the top 100 grossing motion pictures released in 2011 (Box Office Mojo website, March 17, 2012) are contained in the DATAfile named 2011Movies. Table 3.10 shows the data for the first 10 motion pictures in this file. Note that some movies, such as War Horse, were released late in 2011 and continued to run in 2012.

Managerial Report Use the numerical methods of descriptive statistics presented in this chapter to learn how these variables contribute to the success of a motion picture. Include the following in your report: 1. Descriptive statistics for each of the four variables along with a discussion of what the descriptive statistics tell us about the motion picture industry. 2. What motion pictures, if any, should be considered high-performance outliers? E xplain. 3. Descriptive statistics showing the relationship between total gross sales and each of the other variables. Discuss.

Case Problem 3 Business Schools of Asia-Pacific Asian

The pursuit of a higher education degree in business is now international. A survey shows that more and more Asians choose the master of business administration (MBA) degree route to corporate success. As a result, the number of applicants for MBA courses at Asia-Pacific schools continues to increase. Across the region, thousands of Asians show an increasing willingness to temporarily shelve their careers and spend two years in pursuit of a theoretical business qualification. Courses in these schools are notoriously tough and include economics, banking, marketing, behavioral sciences, labor relations, decision making, strategic thinking, business law, and more. The data set in Table 3.11 shows some of the characteristics of the leading Asia-Pacific business schools.

200 5 24,420 29,600 28 University of New South Wales (Sydney) 228 4 19,993 32,582 29 Indian Institute of Management (Ahmedabad) 392 5 4,300 4,300 22 Chinese University of Hong Kong 90 5 11,140 11,140 29 International University of Japan (Niigata) 126 4 33,060 33,060 28 Asian Institute of Management (Manila) 389 5 7,562 9,000 25 Indian Institute of Management (Bangalore) 5 380 3,935 16,000 23 6 6,146 7,170 29 National University of Singapore 147 8 Indian Institute of Management (Calcutta) 463 2,880 16,000 23 2 20,300 20,300 30 Australian National University (Canberra) 42 5 8,500 8,500 32 Nanyang Technological University (Singapore) 50 University of Queensland (Brisbane) 138 17 16,000 22,800 32 2 11,513 11,513 26 Hong Kong University of Science and Technology 60 8 17,172 19,778 34 Macquarie Graduate School of Management (Sydney) 12 7 17,355 17,355 25 Chulalongkorn University (Bangkok) 200 Monash Mt. Eliza Business School (Melbourne) 350 13 16,200 22,500 30 Asian Institute of Management (Bangkok) 300 10 18,200 18,200 29 University of Adelaide 20 19 16,426 23,100 30 Massey University (Palmerston North, New Zealand) 30 15 13,106 21,625 37 Royal Melbourne Institute of Technology Business Graduate School 7 13,880 17,765 32 30 Jamnalal Bajaj Institute of Management Studies (Mumbai) 240 9 1,000 1,000 24 Curtin Institute of Technology (Perth) 98 15 9,475 19,097 29 Lahore University of Management Sciences 70 14 11,250 26,300 23 5 2,260 2,260 32 Universiti Sains Malaysia (Penang) 30 De La Salle University (Manila) 44 17 3,300 3,600 28

Melbourne Business School Yes Yes No Yes Yes Yes Yes Yes No Yes Yes No Yes No Yes Yes No No No No No Yes No No Yes

47 28 0 10 60 50 1 51 0 80 20 26 37 27 6 30 90 10 35 30 0 43 2.5 15 3.5

No No No Yes No

Yes

No No No No Yes No No Yes No Yes No No No No No Yes Yes No Yes

Yes Yes No Yes Yes

Yes

Yes Yes No No No Yes No Yes No Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes

7,000 55,000 7,500 16,000 13,100

48,900

71,400 65,200 7,100 31,000 87,000 22,800 7,500 43,300 7,400 46,600 49,300 49,600 34,000 60,100 17,600 52,500 25,000 66,000 41,400

Students Local Foreign Starting Full-Time per Tuition Tuition English Work Salary Business School Enrollment Faculty ($) ($) Age %Foreign GMAT Test Experience ($)

TABLE 3.11 DATA FOR 25 ASIA-PACIFIC BUSINESS SCHOOLS

177

Case Problem 4 Heavenly Chocolates Website Transactions

Managerial Report Use the methods of descriptive statistics to summarize the data in Table 3.11. Discuss your findings. 1. Include a summary for each variable in the data set. Make comments and interpretations based on maximums and minimums, as well as the appropriate means and proportions. What new insights do these descriptive statistics provide concerning Asia-Pacific business schools? 2. Summarize the data to compare the following: a. Any difference between local and foreign tuition costs. b. Any difference between mean starting salaries for schools requiring and not requiring work experience. c. Any difference between starting salaries for schools requiring and not requiring English tests. 3. Do starting salaries appear to be related to tuition? 4. Present any additional graphical and numerical summaries that will be beneficial in communicating the data in Table 3.11 to others.

Case Problem 4 Heavenly Chocolates Website Transactions Heavenly Chocolates manufactures and sells quality chocolate products at its plant and retail store located in Saratoga Springs, New York. Two years ago the company developed a website and began selling its products over the Internet. Website sales have exceeded the company’s expectations, and mangement is now considering stragegies to increase sales even further. To learn more about the website customers, a sample of 50 Heavenly Chocolate transactions was selected from the previous month’s sales. Data showing the day of the week each transaction was made, the type of browser the customer used, the time spent on the website, the number of website pages viewed, and the amount spent by each of the 50 customers are contained in the DATAfile named Shoppers. A portion of the data is shown in Table 3.12. TABLE 3.12 A SAMPLE OF 50 HEAVENLY CHOCOLATES WEBSITE

TRANSACTIONS ustomer C Day Browser Time (min)

Shoppers

1 2 3 4 5 6 7 . . . . 48 49 50

Mon Internet Explorer Wed Other Mon Internet Explorer Tue Firefox Wed Internet Explorer Sat Firefox Sun Internet Explorer . . . . . . . . Fri Internet Explorer Mon Other Fri Internet Explorer

12.0 19.5 8.5 11.4 11.3 10.5 11.4 . . . . 9.7 7.3 13.4

Pages Amount Viewed Spent ($) 4 6 4 2 4 6 2 . . . . 5 6 3

54.52 94.90 26.68 44.73 66.27 67.80 36.04 . . . . 103.15 52.15 98.75

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Chapter 3 Descriptive Statistics: Numerical Measures

Heavenly Chocolates would like to use the sample data to determine if online shoppers who spend more time and view more pages also spend more money during their visit to the website. The company would also like to investigate the effect that the day of the week and the type of browser have on sales.

Managerial Report Use the methods of descriptive statistics to learn about the customers who visit the Heavenly Chocolates website. Include the following in your report. 1. Graphical and numerical summaries for the length of time the shopper spends on the website, the number of pages viewed, and the mean amount spent per transaction. Discuss what you learn about Heavenly Chocolates’ online shoppers from these numerical summaries. 2. Summarize the frequency, the total dollars spent, and the mean amount spent per transaction for each day of week. What observations can you make about Heavenly Chocolates’ business based on the day of the week? Discuss. 3. Summarize the frequency, the total dollars spent, and the mean amount spent per transaction for each type of browser. What observations can you make about Heavenly Chocolates’ business based on the type of browser? Discuss. 4. Develop a scatter diagram and compute the sample correlation coefficient to explore the relationship between the time spent on the website and the dollar amount spent. Use the horizontal axis for the time spent on the website. Discuss. 5. Develop a scatter diagram and compute the sample correlation coefficient to explore the relationship between the the number of website pages viewed and the amount spent. Use the horizontal axis for the number of website pages viewed. Discuss. 6. Develop a scatter diagram and compute the sample correlation coefficient to explore the relationship between the time spent on the website and the number of pages viewed. Use the horizontal axis to represent the number of pages viewed. Discuss.

Case Problem 5 African Elephant Populations Although millions of elephants once roamed across Africa, by the mid-1980s elephant populations in African nations had been devastated by poaching. Elephants are important to African ecosystems. In tropical forests, elephants create clearings in the canopy that encourage new tree growth. In savannas, elephants reduce bush cover to create an environment that is favorable to browsing and grazing animals. In addition, the seeds of many plant species depend on passing through an elephant’s digestive tract before germination. The status of the elephant now varies greatly across the continent; in some nations, strong measures have been taken to effectively protect elephant populations, while in other nations the elephant populations remain in danger due to poaching for meat and ivory, loss of habitat, and conflict with humans. Table 3.13 shows elephant populations for several African nations in 1979, 1989, and 2007 (Lemieux and Clarke, “The International Ban on Ivory Sales and Its Effects on Elephant Poaching in Africa,” British Journal of Criminology, 49(4), 2009). The David Sheldrick Wildlife Trust was established in 1977 to honor the memory of naturalist David Leslie William Sheldrick, who founded Warden of Tsavo East National Park in Kenya and headed the Planning Unit of the Wildlife Conservation and Management Department in that country. Management of the Sheldrick Trust would like to know what these data indicate about elephant populations in various African countries since 1979.

179

Case Problem 5 African Elephant Populations

TABLE 3.13 ELEPHANT POPULATIONS FOR SEVERAL AFRICAN NATIONS IN 1979,

1989, AND 2007 Country 1979

AfricanElephants

Angola Botswana Cameroon Cen. African Rep. Chad Congo Dem. Rep. of Congo Gabon Kenya Mozambique Somalia Sudan Tanzania Zambia Zimbabwe

12,400 20,000 16,200 63,000 15,000 10,800 377,700 13,400 65,000 54,800 24,300 134,000 316,300 150,000 30,000

Elephant population 1989

2007

12,400 51,000 21,200 19,000 3,100 70,000 85,000 76,000 19,000 18,600 6,000 4,000 80,000 41,000 43,000

2,530 175,487 15,387 3,334 6,435 22,102 23,714 70,637 31,636 26,088 70 300 167,003 29,231 99,107

Managerial Report Use methods of descriptive statistics to summarize the data and comment on changes in elephant populations in African nations since 1979. At a minimum your report should include the following. 1. The mean annual change in elephant population for each country in the 10 years from 1979 to 1989, and a discussion of which countries saw the largest changes in elephant population over this 10-year period. 2. The mean annual change in elephant population for each country from 1989 to 2007, and a discussion of which countries saw the largest changes in elephant population over this 18-year period. 3. A comparison of your results from parts 1 and 2, and a discussion of the conclusions you can draw from this comparison.

CHAPTER

Introduction to Probability CONTENTS STATISTICS IN PRACTICE: NATIONAL AERONAUTICS AND SPACE ADMINISTRATION 4.1 EXPERIMENTS, COUNTING RULES, AND ASSIGNING PROBABILITIES Counting Rules, Combinations, and Permutations Assigning Probabilities Probabilities for the KP&L Project

4.3 SOME BASIC RELATIONSHIPS OF PROBABILITY Complement of an Event Addition Law 4.4 CONDITIONAL PROBABILITY Independent Events Multiplication Law 4.5 BAYES’ THEOREM Tabular Approach

4.2 EVENTS AND THEIR PROBABILITIES

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Statistics in Practice

STATISTICS in PRACTICE National Aeronautics and Space Administration* The National Aeronautics and Space Administration ( NASA) is the agency of the United States government that is responsible for the U.S. civilian space program and aeronautics and aerospace research. NASA is best known for its manned space exploration; its mission statement is to “pioneer the future in space exploration, scientific discovery and aeronautics research.” NASA, with its 18,800 employees, is currently working on the design of a new Space Launch System that will take the astronauts farther into space than ever before and provide the cornerstone for future human space exploration. Although NASA’s primary mission is space exploration, their expertise has been called upon to assist countries and organizations throughout the world. In one such situation, the San José copper and gold mine in Copiapó, Chile, caved in, trapping 33 men more than 2000 feet underground. While it was important to bring the men safely to the surface as quickly as possible, it was imperative that the rescue effort be carefully designed and implemented to save as many miners as possible. The Chilean government asked NASA to provide assistance in developing a rescue method. In response, NASA sent a four-person team consisting of an engineer, two physicians, and a psychologist with expertise in vehicle design and issues of long-term confinement. The probability of success and failure of various rescue methods was prominent in the thoughts of everyone involved. Since there were no historical data available that applied to this unique rescue situation, NASA scientists developed subjective probability estimates for the success *The authors are indebted to Dr. Michael Duncan and Clinton Cragg at NASA for providing this Statistics in Practice.

REUTERS/Hugo Infante/Government of Chile/Handout

Washington, DC

NASA scientists based probabilities on similar circumstances experienced during space flights. and failure of various rescue methods based on similar circumstances experienced by astronauts returning from short- and long-term space missions. The probability estimates provided by NASA guided officials in the selection of a rescue method and provided insight as to how the miners would survive the ascent in a rescue cage. The rescue method designed by the Chilean officials in consultation with the NASA team resulted in the construction of a 13-foot-long, 924-pound steel rescue capsule that would be used to bring up the miners one at a time. All miners were rescued, with the last miner emerging 68 days after the cave-in occurred. In this chapter you will learn about probability as well as how to compute and interpret probabilities for a variety of situations. In addition to subjective probabilities, you will learn about classical and relative frequency methods for assigning probabilities. The basic relationships of probability, conditional probability, and Bayes’ theorem will be covered.

Managers often base their decisions on an analysis of uncertainties such as the following:

Some of the earliest work on probability originated in a series of letters between Pierre de Fermat and Blaise Pascal in the 1650s.

1. 2. 3. 4.

hat are the chances that sales will decrease if we increase prices? W What is the likelihood a new assembly method will increase productivity? How likely is it that the project will be finished on time? What is the chance that a new investment will be profitable?

Probability is a numerical measure of the likelihood that an event will occur. Thus, probabilities can be used as measures of the degree of uncertainty associated with the

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FIGURE 4.1 PROBABILITY AS A NUMERICAL MEASURE OF THE LIKELIHOOD

OF AN EVENT OCCURRING Increasing Likelihood of Occurrence 0

1.0

Probability: The occurrence of the event is just as likely as it is unlikely.

fourevents previously listed. If probabilities are available, we can determine the likelihood of each event occurring. Probability values are always assigned on a scale from 0 to 1. A probability near zero indicates an event is unlikely to occur; a probability near 1 indicates an event is almost certain to occur. Other probabilities between 0 and 1 represent degrees of likelihood that an event will occur. For example, if we consider the event “rain tomorrow,” we understand that when the weather report indicates “a near-zero probability of rain,” it means almost no chance of rain. However, if a .90 probability of rain is reported, we know that rain is likely to occur. A .50 probability indicates that rain is just as likely to occur as not. Figure4.1 depicts the view of probability as a numerical measure of the likelihood of an event occurring.

Experiments, Counting Rules, 4.1 and Assigning Probabilities In discussing probability, we define an experiment as a process that generates well-defined outcomes. On any single repetition of an experiment, one and only one of the possible experimental outcomes will occur. Several examples of experiments and their associated outcomes follow.

Experiment

Experimental Outcomes

Toss a coin Select a part for inspection Conduct a sales call Roll a die Play a football game

Head, tail Defective, nondefective Purchase, no purchase 1, 2, 3, 4, 5, 6 Win, lose, tie

By specifying all possible experimental outcomes, we identify the sample space for an experiment.

Sample Space

The sample space for an experiment is the set of all experimental outcomes.

4.1 Experiments, Counting Rules, and Assigning Probabilities

Experimental outcomes are also called sample points.

183

An experimental outcome is also called a sample point to identify it as an element of the sample space. Consider the first experiment in the preceding table—tossing a coin. The upward face of the coin—a head or a tail—determines the experimental outcomes (sample points). If we let S denote the sample space, we can use the following notation to describe the sample space.

S 5 {Head, Tail}

The sample space for the second experiment in the table—selecting a part for inspection— can be described as follows:

S 5 {Defective, Nondefective}

Both of these experiments have two experimental outcomes (sample points). However, suppose we consider the fourth experiment listed in the table—rolling a die. The possible experimental outcomes, defined as the number of dots appearing on the upward face of the die, are the six points in the sample space for this experiment.

S 5 {1, 2, 3, 4, 5, 6}

Counting Rules, Combinations, and Permutations Being able to identify and count the experimental outcomes is a necessary step in assigning probabilities. We now discuss three useful counting rules. Multiple-step experiments The first counting rule applies to multiple-step experiments.

Consider the experiment of tossing two coins. Let the experimental outcomes be defined in terms of the pattern of heads and tails appearing on the upward faces of the two coins. How many experimental outcomes are possible for this experiment? The experiment of tossing two coins can be thought of as a two-step experiment in which step 1 is the tossing of the first coin and step 2 is the tossing of the second coin. If we use H to denote a head and T to denote a tail, (H, H ) indicates the experimental outcome with a head on the first coin and a head on the second coin. Continuing this notation, we can describe the sample space (S ) for this coin-tossing experiment as follows:

S 5 {(H, H ), (H, T ), (T, H ), (T, T )}

Thus, we see that four experimental outcomes are possible. In this case, we can easily list all the experimental outcomes. The counting rule for multiple-step experiments makes it possible to determine the number of experimental outcomes without listing them.

Counting Rule for Multiple-Step Experiments

If an experiment can be described as a sequence of k steps with n1 possible outcomes on the first step, n2 possible outcomes on the second step, and so on, then the total number of experimental outcomes is given by (n1) (n2) . . . (nk).

Viewing the experiment of tossing two coins as a sequence of first tossing one coin (n1 5 2) and then tossing the other coin (n2 5 2), we can see from the counting rule that (2)(2) 5 4 distinct experimental outcomes are possible. As shown, they are S 5 {(H, H ), Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Chapter 4 Introduction to Probability

Without the tree diagram, one might think only three experimental outcomes are possible for two tosses of a coin: 0 heads, 1 head, and 2 heads.

(H, T ), (T, H ), (T, T )}. The number of experimental outcomes in an experiment involving tossing six coins is (2)(2)(2)(2)(2)(2) 5 64. A tree diagram is a graphical representation that helps in visualizing a multiple-step experiment. Figure 4.2 shows a tree diagram for the experiment of tossing two coins. The sequence of steps moves from left to right through the tree. Step 1 corresponds to tossing the first coin, and step 2 corresponds to tossing the second coin. For each step, the two possible outcomes are head or tail. Note that for each possible outcome at step 1 two branches correspond to the two possible outcomes at step 2. Each of the points on the right end of the tree corresponds to an experimental outcome. Each path through the tree from the leftmost node to one of the nodes at the right side of the tree corresponds to a unique sequence of outcomes. Let us now see how the counting rule for multiple-step experiments can be used in the analysis of a capacity expansion project for the Kentucky Power & Light Company (KP&L). KP&L is starting a project designed to increase the generating capacity of one of its plants in northern Kentucky. The project is divided into two sequential stages or steps: stage 1 (design) and stage 2 (construction). Even though each stage will be scheduled and controlled as closely as possible, management cannot predict beforehand the exact time required to complete each stage of the project. An analysis of similar construction projects revealed possible completion times for the design stage of 2, 3, or 4 months and possible completion times for the construction stage of 6, 7, or 8 months. In addition, because of the critical need for additional electrical power, management set a goal of 10 months for the completion of the entire project. Because this project has three possible completion times for the design stage (step1) and three possible completion times for the construction stage (step 2), the counting rule for multiple-step experiments can be applied here to determine a total of (3)(3) 5 9 experi mental outcomes. To describe the experimental outcomes, we use a two-number notation; for instance, (2, 6) indicates that the design stage is completed in 2 months and the construction stage is completed in 6 months. This experimental outcome results in a total of 2 1 6 5 8 months to complete the entire project. Table 4.1 summarizes the nine experimental outcomes for the KP&L problem. The tree diagram in Figure 4.3 shows how the nine outcomes (sample points) occur.

FIGURE 4.2 TREE DIAGRAM FOR THE EXPERIMENT OF TOSSING TWO COINS

Step 1 First Coin

Step 2 Second Coin Head

(H, H )

Tail

d Hea

Tai

Experimental Outcome (Sample Point)

(H, T )

Head

(T, H )

Tail

(T, T )

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TABLE 4.1 EXPERIMENTAL OUTCOMES (SAMPLE POINTS) FOR THE KP&L PROJECT

Completion Time (months) Stage 1 Stage 2 Design Construction

2 2 2 3 3 3 4 4 4

Notation for Experimental Outcome

Total Project Completion Time (months)

(2, 6) (2, 7) (2, 8) (3, 6) (3, 7) (3, 8) (4, 6) (4, 7) (4, 8)

8 9 10 9 10 11 10 11 12

6 7 8 6 7 8 6 7 8

The counting rule and tree diagram help the project manager identify the experimental outcomes and determine the possible project completion times. From the information in Figure 4.3, we see that the project will be completed in 8 to 12 months, with six of the nine experimental outcomes providing the desired completion time of 10 months or less. FIGURE 4.3 TREE DIAGRAM FOR THE KP&L PROJECT

Step 1 Design

Step 2 Construction

7 mo.

Experimental Outcome (Sample Point)

Total Project Completion Time

(2, 6)

8 months

(2, 7)

9 months

(2, 8)

10 months

(3, 6)

9 months

(3, 7)

10 months

(3, 8)

11 months

(4, 6)

10 months

(4, 7)

11 months

(4, 8)

12 months

2m o.

8m o.

6m 3 mo.

7 mo.

o. 4m

7 mo.

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Even though identifying the experimental outcomes may be helpful, we need to consider how probability values can be assigned to the experimental outcomes before making an assessment of the probability that the project will be completed within the desired 10 months. Combinations A second useful counting rule allows one to count the number of experi

mental outcomes when the experiment involves selecting n objects from a set of N objects. It is called the counting rule for combinations.

Counting Rule for Combinations

The number of combinations of N objects taken n at a time is CnN 5

(4.1)

N! 5 NsN 2 1dsN 2 2d Á s2ds1d n! 5 nsn 2 1dsn 2 2d Á s2ds1d

where and, by definition,

In sampling from a finite population of size N, the counting rule for combinations is used to find the number of different samples of size n that can be selected.

N N! 5 n n!sN 2 nd!

0! 5 1

The notation ! means factorial; for example, 5 factorial is 5! 5 (5)(4)(3)(2)(1) 5 120. As an illustration of the counting rule for combinations, consider a quality control procedure in which an inspector randomly selects two of five parts to test for defects. In a group of five parts, how many combinations of two parts can be selected? The counting rule in equation (4.1) shows that with N 5 5 and n 5 2, we have

C25 5

5 10 1522 5 2!s55!2 2d! 5 s2ds1ds3ds2ds1d 5 120 12 s5ds4ds3ds2ds1d

Thus, 10 outcomes are possible for the experiment of randomly selecting two parts from a group of five. If we label the five parts as A, B, C, D, and E, the 10 combinations or experi mental outcomes can be identified as AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE. As another example, consider that the Florida lottery system uses the random selec tion of 6 integers from a group of 53 to determine the weekly winner. The counting rule for combinations, equation (4.1), can be used to determine the number of ways 6 different integers can be selected from a group of 53. The counting rule for combinations shows that the chance of winning the lottery is very unlikely.

53! 5 15362 5 6!s5353!2 6d! 5 6!47!

s53ds52ds51ds50ds49ds48d 5 22,957,480 s6ds5ds4ds3ds2ds1d

The counting rule for combinations tells us that almost 23 million experimental outcomes are possible in the lottery drawing. An individual who buys a lottery ticket has 1 chance in 22,957,480 of winning. Permutations A third counting rule that is sometimes useful is the counting rule for ermutations. It allows one to compute the number of experimental outcomes when n p objects are to be selected from a set of N objects where the order of selection is important. The same n objects selected in a different order are considered a different e xperimental outcome.

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4.1 Experiments, Counting Rules, and Assigning Probabilities

Counting Rule for Permutations

The number of permutations of N objects taken n at a time is given by

1Nn2 5 sN N!2 nd!

PnN 5 n!

(4.2)

The counting rule for permutations closely relates to the one for combinations; how ever, an experiment results in more permutations than combinations for the same number of objects because every selection of n objects can be ordered in n! different ways. As an example, consider again the quality control process in which an inspector selects two of five parts to inspect for defects. How many permutations may be selected? The counting rule in equation (4.2) shows that with N 5 5 and n 5 2, we have P25 5

5! 5! s5ds4ds3ds2ds1d 120 5 5 5 5 20 s5 2 2d! 3! s3ds2ds1d 6

Thus, 20 outcomes are possible for the experiment of randomly selecting two parts from a group of five when the order of selection must be taken into account. If we label the parts A, B, C, D, and E, the 20 permutations are AB, BA, AC, CA, AD, DA, AE, EA, BC, CB, BD, DB, BE, EB, CD, DC, CE, EC, DE, and ED.

Assigning Probabilities Now let us see how probabilities can be assigned to experimental outcomes. The three approaches most frequently used are the classical, relative frequency, and subjective meth ods. Regardless of the method used, two basic requirements for assigning probabilities must be met.

Basic Requirements for Assigning Probabilities

1. T he probability assigned to each experimental outcome must be between 0 and 1, inclusively. If we let Ei denote the ith experimental outcome and P(Ei ) its probability, then this requirement can be written as

0 # PsEid # 1 for all i

(4.3)

2. T he sum of the probabilities for all the experimental outcomes must equal 1.0. For n experimental outcomes, this requirement can be written as

PsE1d 1 PsE2d 1 Á 1 PsEn d 5 1

(4.4)

The classical method of assigning probabilities is appropriate when all the experi mental outcomes are equally likely. If n experimental outcomes are possible, a probability of 1/n is assigned to each experimental outcome. When using this approach, the two basic requirements for assigning probabilities are automatically satisfied. For an example, consider the experiment of tossing a fair coin; the two experimen tal outcomes—head and tail—are equally likely. Because one of the two equally likely Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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outcomes is a head, the probability of observing a head is 1/2, or .50. Similarly, the probability of observing a tail is also 1/2, or .50. As another example, consider the experiment of rolling a die. It would seem reasonable to conclude that the six possible outcomes are equally likely, and hence each outcome is assigned a probability of 1/6. If P(1) denotes the probability that one dot appears on the upward face of the die, then P(1) 5 1/6. Similarly, P(2) 5 1/6, P(3) 5 1/6, P(4) 5 1/6, P(5) 5 1/6, and P(6) 5 1/6. Note that these probabilities satisfy the two basic requirements of equations (4.3) and (4.4) because each of the probabilities is greater than or equal to zero and they sum to 1.0. The relative frequency method of assigning probabilities is appropriate when data are available to estimate the proportion of the time the experimental outcome will occur if the experiment is repeated a large number of times. As an example, consider a study of waiting times in the X-ray department for a local hospital. A clerk recorded the number of patients waiting for service at 9:00 a.m. on 20 successive days and obtained the following results. Number Waiting

Number of Days Outcome Occurred

0 1 2 3 4

2 5 6 4 3

Total20

These data show that on 2 of the 20 days, zero patients were waiting for service; on 5 of the days, one patient was waiting for service; and so on. Using the relative frequency method, we would assign a probability of 2/20 5 .10 to the experimental outcome of zero patients waiting for service, 5/20 5 .25 to the experimental outcome of one patient waiting, 6/20 5 .30 to two patients waiting, 4/20 5 .20 to three patients waiting, and 3/20 5 .15 to four patients waiting. As with the classical method, using the relative frequency method automatically satisfies the two basic requirements of equations (4.3) and (4.4). The subjective method of assigning probabilities is most appropriate when one cannot realistically assume that the experimental outcomes are equally likely and when little relevant data are available. When the subjective method is used to assign probabilities to the experimental outcomes, we may use any information available, such as our experience or intuition. After considering all available information, a probability value that expresses our degree of belief (on a scale from 0 to 1) that the experimental outcome will occur is specified. Because subjective probability expresses a person’s degree of belief, it is personal. Using the subjective method, different people can be expected to assign different proba bilities to the same experimental outcome. The subjective method requires extra care to ensure that the two basic requirements of equations (4.3) and (4.4) are satisfied. Regardless of a person’s degree of belief, the proba bility value assigned to each experimental outcome must be between 0 and 1, inclusive, and the sum of all the probabilities for the experimental outcomes must equal 1.0. Consider the case in which Tom and Judy Elsbernd make an offer to purchase a house. Two outcomes are possible:

E1 5 their offer is accepted E2 5 their offer is rejected

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Bayes’ theorem (see Section 4.5) provides a means for combining subjectively determined prior probabilities with probabilities obtained by other means to obtain revised, or posterior, probabilities.

Judy believes that the probability their offer will be accepted is .8; thus, Judy would set P(E1) 5 .8 and P(E2) 5 .2. Tom, however, believes that the probability that their offer will be accepted is .6; hence, Tom would set P(E1) 5 .6 and P(E2) 5 .4. Note that Tom’s proba bility estimate for E1 reflects a greater pessimism that their offer will be accepted. Both Judy and Tom assigned probabilities that satisfy the two basic requirements. The fact that their probability estimates are different emphasizes the personal nature of the subjective method. Even in business situations where either the classical or the relative frequency approach can be applied, managers may want to provide subjective probability estimates. In such cases, the best probability estimates often are obtained by combining the estimates from the classical or relative frequency approach with subjective probability estimates.

Probabilities for the KP&L Project To perform further analysis on the KP&L project, we must develop probabilities for each of the nine experimental outcomes listed in Table 4.1. On the basis of experience and judgment, management concluded that the experimental outcomes were not equally likely. Hence, the classical method of assigning probabilities could not be used. Management then decided to conduct a study of the completion times for similar projects undertaken by KP&L over the past three years. The results of a study of 40 similar projects are summarized in Table 4.2. After reviewing the results of the study, management decided to employ the relative frequency method of assigning probabilities. Management could have provided subjective probability estimates but felt that the current project was quite similar to the 40 previous projects. Thus, the relative frequency method was judged best. In using the data in Table 4.2 to compute probabilities, we note that outcome (2, 6)— stage 1 completed in 2 months and stage 2 completed in 6 months—occurred six times in the 40 projects. We can use the relative frequency method to assign a probability of 6/40 5 .15 to this outcome. Similarly, outcome (2, 7) also occurred in six of the 40 projects, providing a 6/40 5 .15 probability. Continuing in this manner, we obtain the probability assignments for the sample points of the KP&L project shown in Table 4.3. Note that P(2, 6) represents the probability of the sample point (2, 6), P(2, 7) represents the probability of the sample point (2, 7), and so on. TABLE 4.2 COMPLETION RESULTS FOR 40 KP&L PROJECTS

Completion Time (months) Stage 1 Stage 2 Design Construction Sample Point

2 2 2 3 3 3 4 4 4

6 7 8 6 7 8 6 7 8

(2, 6) (2, 7) (2, 8) (3, 6) (3, 7) (3, 8) (4, 6) (4, 7) (4, 8)

Number of Past Projects Having These Completion Times 6 6 2 4 8 2 2 4 6 Total40

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TABLE 4.3 PROBABILITY ASSIGNMENTS FOR THE KP&L PROJECT BASED

ON THE RELATIVE FREQUENCY METHOD Sample Point

Project Completion Time

(2, 6) 8 months (2, 7) 9 months (2, 8) 10 months (3, 6) 9 months (3, 7) 10 months (3, 8) 11 months (4, 6) 10 months (4, 7) 11 months (4, 8) 12 months

Probability of Sample Point P(2, 6) 5 6/40 5 .15 P(2, 7) 5 6/40 5 .15 P(2, 8) 5 2/40 5 .05 P(3, 6) 5 4/40 5 .10 P(3, 7) 5 8/40 5 .20 P(3, 8) 5 2/40 5 .05 P(4, 6) 5 2/40 5 .05 P(4, 7) 5 4/40 5 .10 P(4, 8) 5 6/40 5 .15 Total 1.00

NOTES AND COMMENTS 1. In statistics, the notion of an experiment differs somewhat from the notion of an experiment in the physical sciences. In the physical sciences, researchers usually conduct an experiment in a laboratory or a controlled environment in order to learn about cause and effect. In statistical experiments, probability determines outcomes. Even though the experiment is repeated in exactly the

same way, an entirely different outcome may occur. Because of this influence of probability on the outcome, the experimentsof statistics are sometimes called random experiments. 2. When drawing a random sample without replace ment from a population of size N, the counting rule for combinations is used to find the number of different samples of size n that can be s elected.

Exercises

Methods 1. A n experiment has three steps with three outcomes possible for the first step, two outcomes possible for the second step, and four outcomes possible for the third step. How many experimental outcomes exist for the entire experiment? 2. H ow many ways can three items be selected from a group of six items? Use the letters A, B, C, D, E, and F to identify the items, and list each of the different combinations of three items. 3. H ow many permutations of three items can be selected from a group of six? Use the letters A, B, C, D, E, and F to identify the items, and list each of the permutations of items B, D, and F. 4. C onsider the experiment of tossing a coin three times. a. Develop a tree diagram for the experiment. b. List the experimental outcomes. c. What is the probability for each experimental outcome? 5. S uppose an experiment has five equally likely outcomes: E1, E2, E3, E4, E5. Assign prob abilities to each outcome and show that the requirements in equations (4.3) and (4.4) are satisfied. What method did you use? 6. A n experiment with three outcomes has been repeated 50 times, and it was learned that E1 occurred 20 times, E2 occurred 13 times, and E3 occurred 17 times. Assign probabilities to the outcomes. What method did you use? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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4.1 Experiments, Counting Rules, and Assigning Probabilities

7. A decision maker subjectively assigned the following probabilities to the four outcomes of an experiment: P(E1) 5 .10, P(E2) 5 .15, P(E3) 5 .40, and P(E4) 5 .20. Are these proba bility assignments valid? Explain.

Applications 8. In the city of Milford, applications for zoning changes go through a two-step process: a review by the planning commission and a final decision by the city council. At step 1 the planning commission reviews the zoning change request and makes a positive or negative recommendation concerning the change. At step 2 the city council reviews the planning commission’s recommendation and then votes to approve or to disapprove the zoning change. Suppose the developer of an apartment complex submits an application for a zoning change. Consider the application process as an experiment. a. How many sample points are there for this experiment? List the sample points. b. Construct a tree diagram for the experiment. 9. Simple random sampling uses a sample of size n from a population of size N to obtain data that can be used to make inferences about the characteristics of a population. Suppose that, from a population of 50 bank accounts, we want to take a random sample of four accounts in order to learn about the population. How many different random samples of four accounts are possible? 10. The following table shows the percentage of on-time arrivals, the number of mishandled baggage reports per 1000 passengers, and the number of customer complaints per 1000 passengers for 10 airlines (Forbes website, February 2014).

Airline Virgin America JetBlue AirTran Airways Delta Air Lines Alaska Airlines Frontier Airlines Southwest Airlines US Airways American Airlines United Airlines

On-Time Arrivals (%) 83.5 79.1 87.1 86.5 87.5 77.9 83.1 85.9 76.9 77.4

Mishandled Baggage per 1000 Passengers 0.87 1.88 1.58 2.10 2.93 2.22 3.08 2.14 2.92 3.87

Customer Complaints per 1000 Passengers 1.50 0.79 0.91 0.73 0.51 1.05 0.25 1.74 1.80 4.24

a. I f you randomly choose a Delta Air Lines flight, what is the probability that this individual flight has an on-time arrival? b. If you randomly choose one of the 10 airlines for a follow-up study on airline quality ratings, what is the probability that you will choose an airline with less than two mishandled baggage reports per 1000 passengers? c. If you randomly choose 1 of the 10 airlines for a follow-up study on airline quality ratings, what is the probability that you will choose an airline with more than one customer complaint per 1000 passengers? d. What is the probability that a randomly selected AirTran Airways flight will not arrive on time? 11. A Gallup Poll of U.S. adults indicated that Kentucky is the state with the highest percentage of smokers (Gallup.com, December 2015). Consider the following example data from Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Chapter 4 Introduction to Probability

State

Smoker

Non-Smoker

47 Kentucky Indiana 32 39 Ohio Total: 118

176 134 182 492

the Tri-State region, an area that comprises northern Kentucky, southeastern Indiana, and southwestern Ohio. a. Use the data to compute the probability that an adult in the Tri-State region smokes. b. What is the probability of an adult in each state of the Tri-State region being a smoker? Which state in the Tri-State region has the lowest probability of an adult being a smoker? 12. T he Powerball lottery is played twice each week in 31 states, the District of Columbia, and the Virgin Islands. To play Powerball, a participant must purchase a $2 ticket, select five numbers from the digits 1 through 59, and then select a Powerball number from the digits1 through 35. To determine the winning numbers for each game, lottery officials draw five white balls out a drum of 59 white balls numbered 1 through 59 and one red ball out of a drum of 35 red balls numbered 1 through 35. To win the Powerball jackpot, a participant’s numbers must match the numbers on the five white balls in any order and must also match the number on the red Powerball. The numbers 5–16–22–23–29 with a Powerball number of 6 provided the record jackpot of $580 million ( Powerball website, November29, 2012). a. How many Powerball lottery outcomes are possible? (Hint: Consider this a two-step experiment. Select the five white ball numbers and then select the one red Powerball number.) b. What is the probability that a $2 lottery ticket wins the Powerball lottery? 13. A company that manufactures toothpaste is studying five different package designs. Assuming that one design is just as likely to be selected by a consumer as any other design, what selection probability would you assign to each of the package designs? In an actual experiment, 100 consumers were asked to pick the design they preferred. The following data were obtained. Do the data confirm the belief that one design is just as likely to be selected as another? Explain.

Design

Number of Times Preferred

1 5 2 15 3 30 4 40 5 10

Events and Their Probabilities 4.2 In the introduction to this chapter we used the term event much as it would be used in everyday language. Then, in Section 4.1 we introduced the concept of an experiment and its associated experimental outcomes or sample points. Sample points and events provide the foundation for the study of probability. As a result, we must now introduce the formal definition of an event as it relates to sample points. Doing so will provide the basis for determining the probability of an event. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

4.2 Events and Their Probabilities

193

Event

An event is a collection of sample points.

For an example, let us return to the KP&L project and assume that the project manager is interested in the event that the entire project can be completed in 10 months or less. Referring to Table 4.3, we see that six sample points—(2, 6), (2, 7), (2, 8), (3, 6), (3, 7), and (4, 6)—provide a project completion time of 10 months or less. Let C denote the event that the project is completed in 10 months or less; we write

C 5 {(2, 6), (2, 7), (2, 8), (3, 6), (3, 7), (4, 6)}

Event C is said to occur if any one of these six sample points appears as the experimental outcome. Other events that might be of interest to KP&L management include the following.

L 5 The event that the project is completed in less than 10 months M 5 The event that the project is completed in more than 10 months

Using the information in Table 4.3, we see that these events consist of the following sample points.

L 5 {(2, 6), (2, 7), (3, 6)} M 5 {(3, 8), (4, 7), (4, 8)}

A variety of additional events can be defined for the KP&L project, but in each case the event must be identified as a collection of sample points for the experiment. Given the probabilities of the sample points shown in Table 4.3, we can use the following definition to compute the probability of any event that KP&L management might want to consider.

Probability of an Event

The probability of any event is equal to the sum of the probabilities of the sample points in the event.

Using this definition, we calculate the probability of a particular event by adding the probabilities of the sample points (experimental outcomes) that make up the event. We can now compute the probability that the project will take 10 months or less to complete. Because this event is given by C 5 {(2, 6), (2, 7), (2, 8), (3, 6), (3, 7), (4, 6)}, the probability of event C, denoted P(C ), is given by

P(C ) 5 P(2, 6) 1 P(2, 7) 1 P(2, 8) 1 P(3, 6) 1 P(3, 7) 1 P(4, 6)

Refer to the sample point probabilities in Table 4.3; we have

P(C ) 5 .15 1 .15 1 .05 1 .10 1 .20 1 .05 5 .70

Similarly, because the event that the project is completed in less than 10 months is given by L 5 {(2, 6), (2, 7), (3, 6)}, the probability of this event is given by

P(L) 5 P(2, 6) 1 P(2, 7) 1 P(3, 6) 5 .15 1 .15 1 .10 5 .40

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Finally, for the event that the project is completed in more than 10 months, we have M 5 {(3, 8), (4, 7), (4, 8)} and thus P(M ) 5 P(3, 8) 1 P(4, 7) 1 P(4, 8) 5 .05 1 .10 1 .15 5 .30

Using these probability results, we can now tell KP&L management that there is a .70 probability that the project will be completed in 10 months or less, a .40 probability that the project will be completed in less than 10 months, and a .30 probability that the project will be completed in more than 10 months. This procedure of computing event probabilities can be repeated for any event of interest to the KP&L management. Any time that we can identify all the sample points of an experiment and assign proba bilities to each, we can compute the probability of an event using the definition. However, in many experiments the large number of sample points makes the identification of the sample points, as well as the determination of their associated probabilities, extremely cumbersome, if not impossible. In the remaining sections of this chapter, we present some basic probability relationships that can be used to compute the probability of an event without knowledge of all the sample point probabilities. NOTES AND COMMENTS 1. The sample space, S, is an event. Because it contains all the experimental outcomes, it has a probability of 1; that is, P(S) 5 1. 2. When the classical method is used to assign probabilities, the assumption is that the experimental

outcomes are equally likely. In such cases, the probability of an event can be computed by counting the number of experimental outcomes in the event and dividing the result by the total number of experimental outcomes.

Exercises

Methods 14. A n experiment has four equally likely outcomes: E1, E2, E3, and E4. a. What is the probability that E2 occurs? b. What is the probability that any two of the outcomes occur (e.g., E1 or E3)? c. What is the probability that any three of the outcomes occur (e.g., E1 or E2 or E4)? 15. C onsider the experiment of selecting a playing card from a deck of 52 playing cards. Each card corresponds to a sample point with a 1/52 probability. a. List the sample points in the event an ace is selected. b. List the sample points in the event a club is selected. c. List the sample points in the event a face card ( jack, queen, or king) is selected. d. Find the probabilities associated with each of the events in parts (a), ( b), and (c). 16. C onsider the experiment of rolling a pair of dice. Suppose that we are interested in the sum of the face values showing on the dice. a. How many sample points are possible? (Hint: Use the counting rule for multiple-step experiments.) b. List the sample points. c. What is the probability of obtaining a value of 7? d. What is the probability of obtaining a value of 9 or greater? e. Because each roll has six possible even values (2, 4, 6, 8, 10, and 12) and only five possible odd values (3, 5, 7, 9, and 11), the dice should show even values more often than odd values. Do you agree with this statement? Explain. f. What method did you use to assign the probabilities requested? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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4.2 Events and Their Probabilities

Applications 17. R efer to the KP&L sample points and sample point probabilities in Tables 4.2 and 4.3. a. The design stage (stage 1) will run over budget if it takes four months to complete. List the sample points in the event the design stage is over budget. b. What is the probability that the design stage is over budget? c. The construction stage (stage 2) will run over budget if it takes eight months to complete. List the sample points in the event the construction stage is over budget. d. What is the probability that the construction stage is over budget? e. What is the probability that both stages are over budget? 18. F ortune magazine publishes an annual list of the 500 largest companies in the United States. The corporate headquarters for the 500 companies are located in 38 different states The following table shows the eight states with the largest number of Fortune 500 companies (Money/CNN website, May 12, 2012).

State California Illinois New Jersey New York

Number of Companies State 53 32 21 50

Ohio Pennsylvania Texas Virginia

Number of Companies 28 23 52 24

Suppose one of the 500 companies is selected at random for a follow-up questionnaire. a. What is the probability that the company selected has its corporate headquarters in California? b. What is the probability that the company selected has its corporate headquarters in California, New York, or Texas? c. What is the probability that the company selected has its corporate headquarters in one of the eight states listed above? 19. D o you think global warming will have an impact on you during your lifetime? A CBS News/New York Times poll of 1000 adults in the United States asked this question (CBS News website, December 2014). Consider the responses by age groups shown below.

Age Response 18–29 301 Yes 134 293 No 131 432 Unsure 2 8

a. W hat is the probability that a respondent 18–29 years of age thinks that global warming will not pose a serious threat during his/her lifetime? b. What is the probability that a respondent 301 years of age thinks that global warming will not pose a serious threat during his/her lifetime? c. For a randomly selected respondent, what is the probability that a respondent answers yes? d. Based on the survey results, does there appear to be a difference between ages 18–29 and 30+ regarding concern over global warming? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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20. J unior Achievement USA and the Allstate Foundation surveyed teenagers aged 14 to 18 and asked at what age they think they will become financially independent. The responses of 944 teenagers who answered this survey question are as follows.

Age Financially Independent

Number of Responses

16 to 20 191 21 to 24 467 25 to 27 244 28 or older 42

Consider the experiment of randomly selecting a teenager from the population of teenagers aged 14 to 18. a. Compute the probability of being financially independent for each of the four age categories. b. What is the probability of being financially independent before the age of 25? c. What is the probability of being financially independent after the age of 24? d. Do the probabilities suggest that the teenagers may be somewhat unrealistic in their expectations about when they will become financially independent? 21. Data on U.S. work-related fatalities by cause follow (The World Almanac, 2012).

Cause of Fatality Transportation incidents Assaults and violent acts Contact with objects and equipment Falls Exposure to harmful substances or environments Fires and explosions

Number of Fatalities 1795 837 741 645 404 113

Assume that a fatality will be randomly chosen from this population. a. What is the probability the fatality resulted from a fall? b. What is the probability the fatality resulted from a transportation incident? c. What cause of fatality is least likely to occur? What is the probability the fatality resulted from this cause?

Some Basic Relationships of Probability 4.3 Complement of an Event Given an event A, the complement of A is defined to be the event consisting of all sample points that are not in A. The complement of A is denoted by Ac. Figure 4.4 is a diagram, known as a Venn diagram, that illustrates the concept of a complement. The rectangular area represents the sample space for the experiment and as such contains all possible sample points. The circle represents event A and contains only the sample points that belong to A. The shaded region of the rectangle contains all sample points not in event A and is by definition the complement of A. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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4.3 Some Basic Relationships of Probability

FIGURE 4.4 COMPLEMENT OF EVENT A IS SHADED

Sample Space S

Event A

Complement of Event A

In any probability application, either event A or its complement Ac must occur. Therefore, we have Psad 1 Psac d 5 1

Solving for P(A), we obtain the following result. Computing Probability Using the Complement

Psad 5 1 2 Psacd

(4.5)

Equation (4.5) shows that the probability of an event A can be computed easily if the proba bility of its complement, P(Ac), is known. As an example, consider the case of a sales manager who, after reviewing sales reports, states that 80% of new customer contacts result in no sale. By allowing A to denote the event of a sale and Ac to denote the event of no sale, the manager is stating that P(Ac) 5 .80. Using equation (4.5), we see that Psad 5 1 2 Psacd 5 1 2 .80 5 .20

We can conclude that a new customer contact has a .20 probability of resulting in a sale. In another example, a purchasing agent states a .90 probability that a supplier will send a shipment that is free of defective parts. Using the complement, we can conclude that there is a 1 2 .90 5 .10 probability that the shipment will contain defective parts.

Addition Law The addition law is helpful when we are interested in knowing the probability that at least one of two events occurs. That is, with events A and B we are interested in knowing the probability that event A or event B or both will occur. Before we present the addition law, we need to discuss two concepts related to the combination of events: the union of events and the intersection of events. Given two eventsA and B, the union of A and B is defined as follows.

Union of Two Events

The union of A and B is the event containing all sample points belonging to A or B or both. The union is denoted by A ø B.

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FIGURE 4.5 UNION OF EVENTS A AND B IS SHADED

Sample Space S

Event B

Event A

The Venn diagram in Figure 4.5 depicts the union of events A and B. Note that the two circles contain all the sample points in event A as well as all the sample points in event B. The fact that the circles overlap indicates that some sample points are contained in both A and B. The definition of the intersection of A and B follows. Intersection of Two Events

Given two events A and B, the intersection of A and B is the event containing the sample points belonging to both A and B. The intersection is denoted by A > B. The Venn diagram depicting the intersection of events A and B is shown in Figure 4.6. The area where the two circles overlap is the intersection; it contains the sample points that are in both A and B. Let us now continue with a discussion of the addition law. The addition law provides a way to compute the probability that event A or event B or both occur. In other words, the addition law is used to compute the probability of the union of two events. The addition law is written as follows.

Addition Law

Psa ø Bd 5 Psad 1 PsBd 2 Psa > Bd

(4.6)

FIGURE 4.6 INTERSECTION OF EVENTS A AND B IS SHADED

Sample Space S

Event A

Event B

4.3 Some Basic Relationships of Probability

199

To understand the addition law intuitively, note that the first two terms in the addition law, P(A) 1 P(B), account for all the sample points in A ø B. However, because the sample points in the intersection A > B are in both A and B, when we compute P(A) 1 P(B), we are in effect counting each of the sample points in A > B twice. We correct for this overcounting by subtracting P(A > B). As an example of an application of the addition law, let us consider the case of a small assembly plant with 50 employees. Each worker is expected to complete work assignments on time and in such a way that the assembled product will pass a final inspection. On occasion, some of the workers fail to meet the performance standards by completing work late or assembling a defective product. At the end of a performance evaluation period, the production manager found that 5 of the 50 workers completed work late, 6 of the 50 workers assembled a defective product, and 2 of the 50 workers both completed work late and assembled a defective product. Let

L 5 the event that the work is completed late D 5 the event that the assembled product is defective

The relative frequency information leads to the following probabilities. 5 5 .10 50 6 PsDd 5 5 .12 50 2 PsL > Dd 5 5 .04 50 PsLd 5

After reviewing the performance data, the production manager decided to assign a poor performance rating to any employee whose work was either late or defective; thus the event of interest is L ø D. What is the probability that the production manager assigned an employee a poor performance rating? Note that the probability question is about the union of two events. Specifically, we want to know P(L ø D). Using equation (4.6), we have

PsL ø Dd 5 PsLd 1 PsDd 2 PsL > Dd

Knowing values for the three probabilities on the right side of this expression, we can write

PsL ø Dd 5 .10 1 .12 2 .04 5 .18

This calculation tells us that there is a .18 probability that a randomly selected employee received a poor performance rating. As another example of the addition law, consider a recent study conducted by the personnel manager of a major computer software company. The study showed that 30% of the employees who left the firm within two years did so primarily because they were dissatisfied with their salary, 20% left because they were dissatisfied with their work assignments, and 12% of the former employees indicated dissatisfaction with both their salary and their work assignments. What is the probability that an employee who leaves

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within two years does so because of dissatisfaction with salary, dissatisfaction with the work assignment, or both? Let

S 5 the event that the employee leaves because of salary W 5 the event that the employee leaves because of work assignment

We have P(S ) 5 .30, P(W ) 5 .20, and P(S ù W ) 5 .12. Using equation (4.6), the addition law, we have

P(S ø W) 5 P(S) 1 P(W) 2 P(S > W) 5 .30 1 .20 2 .12 5 .38

We find a .38 probability that an employee leaves for salary or work assignment reasons. Before we conclude our discussion of the addition law, let us consider a special case that arises for mutually exclusive events.

Mutually Exclusive Events

Two events are said to be mutually exclusive if the events have no sample points in common.

Events A and B are mutually exclusive if, when one event occurs, the other cannot occur. Thus, a requirement for A and B to be mutually exclusive is that their intersection must contain no sample points. The Venn diagram depicting two mutually exclusive events A and B is shown in Figure 4.7. In this case P(A > B) 5 0 and the addition law can be written as follows.

Addition Law for Mutually Exclusive Events

Psa ø Bd 5 Psad 1 PsBd

FIGURE 4.7 MUTUALLY EXCLUSIVE EVENTS

Sample Space S

Event A

Event B

4.3 Some Basic Relationships of Probability

201

Exercises

Methods 22. S uppose that we have a sample space with five equally likely experimental outcomes: E1, E2, E3, E4, E5. Let

A 5 {E1, E2} B 5 {E3, E4} C 5 {E2, E3, E5} a. b. c. d. e.

Find P(A), P(B), and P(C ). Find P(A ø B). Are A and B mutually exclusive? Find Ac, C c, P(Ac), and P(C c). Find A ø B c and P(A ø B c). Find P(B ø C ).

23. Suppose that we have a sample space S 5 {E1, E2, E3, E4, E5, E6, E7}, where E1, E2, . . . , E7 denote the sample points. The following probability assignments apply: P(E1) 5 .05, P(E2) 5 .20, P(E3) 5 .20, P(E4) 5 .25, P(E5) 5 .15, P(E6) 5 .10, and P(E7) 5 .05. Let A 5 {E1, E4, E6} B 5 {E2, E4, E7} a. b. c. d. e.

C 5 {E2, E3, E5, E7}

Find P(A), P(B), and P(C ). Find A ø B and P(A ø B). Find A > B and P(A > B). Are events A and C mutually exclusive? Find B c and P(B c).

Applications 24. C larkson University surveyed alumni to learn more about what they think of Clarkson. One part of the survey asked respondents to indicate whether their overall experience at Clarkson fell short of expectations, met expectations, or surpassed expectations. The results showed that 4% of the respondents did not provide a response, 26% said that their experience fell short of expectations, and 65% of the respondents said that their experience met expectations. a. If we chose an alumnus at random, what is the probability that the alumnus would say their experience surpassed expectations? b. If we chose an alumnus at random, what is the probability that the alumnus would say their experience met or surpassed expectations? 25. T he Eco Pulse survey from the marketing communications firm Shelton Group asked individuals to indicate things they do that make them feel guilty (Los Angeles Times, August15, 2012). Based on the survey results, there is a .39 probability that a randomly selected person will feel guilty about wasting food and a .27 probability that a randomly selected person will feel guilty about leaving lights on when not in a room. Moreover, there is a .12 probability that a randomly selected person will feel guilty for both of these reasons. a. What is the probability that a randomly selected person will feel guilty for either wasting food or leaving lights on when not in a room? b. What is the probability that a randomly selected person will not feel guilty for either of these reasons? 26. Information about mutual funds provided by Morningstar Investment Research includes the type of mutual fund ( Domestic Equity, International Equity, or Fixed Income) and Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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the Morningstar rating for the fund. The rating is expressed from 1-star ( lowest rating) to 5-star ( highest rating). A sample of 25 mutual funds was selected from Morningstar Funds 500. The following counts were obtained: Sixteen mutual funds were Domestic Equity funds. Thirteen mutual funds were rated 3-star or less. Seven of the Domestic Equity funds were rated 4-star. Two of the Domestic Equity funds were rated 5-star. Assume that one of these 25 mutual funds will be randomly selected in order to learn more about the mutual fund and its investment strategy. a. What is the probability of selecting a Domestic Equity fund? b. What is the probability of selecting a fund with a 4-star or 5-star rating? c. What is the probability of selecting a fund that is both a Domestic Equity fund and a fund with a 4-star or 5-star rating? d. What is the probability of selecting a fund that is a Domestic Equity fund or a fund with a 4-star or 5-star rating? 27. A marketing firm would like to test-market the name of a new energy drink targeted at 18- to 29-year-olds via social media. A study by the Pew Research Center found that 35% of U.S. adults (18 and older) do not use social media (Pew Research Center website, October 2015). The percentage of U.S. young adults age 30 and older is 78%. Suppose that the percentage of the U.S. adult population that is either age 18–29 or uses social media is 67.2%. a. What is the probability that a randomly selected U.S. adult uses social media? b. What is the probability that a randomly selected U.S. adult is aged 18–29? c. What is the probability that a randomly selected U.S. adult is 18–29 and a user of social media? 28. A survey of magazine subscribers showed that 45.8% rented a car during the past 12 months for business reasons, 54% rented a car during the past 12 months for personal reasons, and 30% rented a car during the past 12 months for both business and personal reasons. a. What is the probability that a subscriber rented a car during the past 12 months for business or personal reasons? b. What is the probability that a subscriber did not rent a car during the past 12 months for either business or personal reasons? 29. H igh school seniors with strong academic records apply to the nation’s most selective colleges in greater numbers each year. Because the number of slots remains relatively stable, some colleges reject more early applicants. Suppose that for a recent admissions class, an Ivy League college received 2851 applications for early admission. Of this group, it a dmitted 1033 students early, rejected 854 outright, and deferred 964 to the regular admission pool for further consideration. In the past, this school has admitted 18% of the deferred early admission applicants during the regular admission process. Counting the students admitted early and the students a dmitted during the regular admission process, the total class size was 2375. Let E, R, and D represent the events that a student who applies for early admission is admitted early, rejected outright, or deferred to the regular admissions pool. a. Use the data to estimate P(E ), P(R), and P(D). b. Are events E and D mutually exclusive? Find P(E > D). c. For the 2375 students who were admitted, what is the probability that a randomly selected student was accepted during early admission? d. Suppose a student applies for early admission. What is the probability that the student will be admitted for early admission or be deferred and later admitted during the regular admission process?

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4.4 Conditional Probability

Conditional Probability 4.4 Often, the probability of an event is influenced by whether a related event already occurred. Suppose we have an event A with probability P(A). If we obtain new information and learn that a related event, denoted by B, already occurred, we will want to take advantage of this information by calculating a new probability for event A. This new probability of event A is called a conditional probability and is written P(A | B). We use the notation | to indicate that we are considering the probability of event A given the condition that event B has occurred. Hence, the notation P(A | B) reads “the probability of A given B.” As an illustration of the application of conditional probability, consider the situation of the promotion status of male and female officers of a major metropolitan police force in the eastern United States. The police force consists of 1200 officers, 960 men and 240 women. Over the past two years, 324 officers on the police force received promotions. The specific breakdown of promotions for male and female officers is shown in Table 4.4. After reviewing the promotion record, a committee of female officers raised a discrimi nation case on the basis that 288 male officers had received promotions, but only 36 female officers had received promotions. The police administration argued that the relatively low number of promotions for female officers was due not to discrimination, but to the fact that relatively few females are members of the police force. Let us show how conditional proba bility could be used to analyze the discrimination charge. Let M 5 event an officer is a man W 5 event an officer is a woman A 5 event an officer is promoted Ac 5 event an officer is not promoted Dividing the data values in Table 4.4 by the total of 1200 officers enables us to summarize the available information with the following probability values.

PsM > ad 5 288/1200 5 .24 probability that a randomly selected officer is a man and is promoted PsM > acd 5 672/1200 5 .56 probability that a randomly selected officer is a man and is not promoted PsW > ad 5 36/1200 5 .03 probability that a randomly selected officer is a woman and is promoted PsW > ac d 5 204/1200 5 .17 probability that a randomly selected officer is a woman and is not promoted

Because each of these values gives the probability of the intersection of two events, the probabilities are called joint probabilities. Table 4.5, which provides a summary of the probability information for the police officer promotion situation, is referred to as a joint probability table. TABLE 4.4 PROMOTION STATUS OF POLICE OFFICERS OVER THE PAST TWO YEARS

Men

Women

Total

Promoted 288 36 324 Not Promoted 672 204 876 Total

960

240

1200

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Chapter 4 Introduction to Probability

TABLE 4.5 JOINT PROBABILITY TABLE FOR PROMOTIONS Joint probabilities appear in the body of the table.

Men (M)

Women (W ) Total

Promoted (A) .24 .03 .27 Not Promoted (Ac ) .56 .17 .73 Total .80

.20

1.00

Marginal probabilities appear in the margins of the table.

The values in the margins of the joint probability table provide the probabilities of each event separately. That is, P(M ) 5 .80, P(W ) 5 .20, P(A) 5 .27, and P(Ac ) 5 .73. These probabilities are referred to as marginal probabilities because of their location in the margins of the joint probability table. We note that the marginal probabilities are found by summing the joint probabilities in the corresponding row or column of the joint probability table. For instance, the marginal probability of being promoted is P(A) 5 P(M > A) 1 P(W > A) 5 .24 1 .03 5 .27. From the marginal probabilities, we see that 80% of the force is male, 20% of the force is female, 27% of all officers received promotions, and 73% were not promoted. Let us begin the conditional probability analysis by computing the probability that an officer is promoted given that the officer is a man. In conditional probability notation, we are attempting to determine P(A | M ). To calculate P(A | M ), we first realize that this notation simply means that we are considering the probability of the event A (promotion) given that the condition designated as event M (the officer is a man) is known to exist. Thus P(A | M ) tells us that we are now concerned only with the promotion status of the 960 male officers. Because 288 of the 960 male officers received promotions, the proba bility of being promoted given that the officer is a man is 288/960 5 .30. In other words, given that an officer is a man, that officer had a 30% chance of receiving a promotion over the past two years. This procedure was easy to apply because the values in Table 4.4 show the number of officers in each category. We now want to demonstrate how conditional probabilities such as P(A | M ) can be computed directly from related event probabilities rather than the frequency data of Table 4.4. We have shown that P(A | M ) 5 288/960 5 .30. Let us now divide both the numerator and denominator of this fraction by 1200, the total number of officers in the study.

Psa | M d 5

288 288/1200 .24 5 5 5 .30 960 960/1200 .80

We now see that the conditional probability P(A | M ) can be computed as .24/.80. Refer to the joint probability table ( Table 4.5). Note in particular that .24 is the joint probability of A and M; that is, P(A > M ) 5 .24. Also note that .80 is the marginal probability that a randomly selected officer is a man; that is, P(M ) 5 .80. Thus, the conditional probability P(A | M ) can be computed as the ratio of the joint probability P(A > M ) to the marginal probability P(M ).

Psa | M d 5

Psa > M d .24 5 5 .30 PsM d .80

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4.4 Conditional Probability

The fact that conditional probabilities can be computed as the ratio of a joint probability to a marginal probability provides the following general formula for conditional probability calculations for two events A and B.

Conditional Probability

Psa | Bd 5

Psa > Bd PsBd

(4.7)

PsB | ad 5

Psa > Bd Psad

(4.8)

The Venn diagram in Figure 4.8 is helpful in obtaining an intuitive understanding of conditional probability. The circle on the right shows that event B has occurred; the portion of the circle that overlaps with event A denotes the event (A > B). We know that once event B has occurred, the only way that we can also observe event A is for the event (A > B) to occur. Thus, the ratio P(A > B)/P(B) provides the conditional probability that we will observe event A given that event B has already occurred. Let us return to the issue of discrimination against the female officers. The marginal probability in row 1 of Table 4.5 shows that the probability of promotion of an officer is P(A) 5 .27 (regardless of whether that officer is male or female). However, the critical issue in the discrimination case involves the two conditional probabilities P(A | M ) and P(A | W ). That is, what is the probability of a promotion given that the officer is a man, and what is the probability of a promotion given that the officer is a woman? If these two probabilities are equal, a discrimination argument has no basis because the chances of a promotion are the same for male and female officers. However, a difference in the two conditional probabilities will support the position that male and female officers are treated differently in promotion decisions. We already determined that P(A | M ) 5 .30. Let us now use the probability values in Table 4.5 and the basic relationship of conditional probability in equation (4.7) to compute FIGURE 4.8 CONDITIONAL PROBABILITY P(A | B) 5 P(A > B)/P(B) Event A > B

Event A

Event B

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the probability that an officer is promoted given that the officer is a woman; that is, P(A | W ). Using equation (4.7), with W replacing B, we obtain Psa | W d 5

Psa > W d .03 5 5 .15 PsW d .20

What conclusion do you draw? The probability of a promotion given that the officer is a man is .30, twice the .15 probability of a promotion given that the officer is a woman. Although the use of conditional probability does not in itself prove that discrimination exists in this case, the conditional probability values support the argument presented by the female officers.

Independent Events In the preceding illustration, P(A) 5 .27, P(A | M ) 5 .30, and P(A | W ) 5 .15. We see that the probability of a promotion (event A) is affected or influenced by whether the officer is a man or a woman. Particularly, because P(A | M ) Þ P(A), we would say that events A and M are dependent events. That is, the probability of event A (promotion) is altered or affected by knowing that event M (the officer is a man) exists. Similarly, with P(A | W ) Þ P(A), we would say that events A and W are dependent events. However, if the probability of event A is not changed by the existence of event M—that is, P(A | M ) 5 P(A)—we would say that events A and M are independent events. This situation leads to the following definition of the independence of two events.

Independent Events

Two events A and B are independent if

Psa | Bd 5 Psad

(4.9)

PsB | ad 5 PsBd

(4.10)

Otherwise, the events are dependent.

Multiplication Law Whereas the addition law of probability is used to compute the probability of a union of two events, the multiplication law is used to compute the probability of the intersection of two events. The multiplication law is based on the definition of conditional probability. Using equations (4.7) and (4.8) and solving for P(A > B), we obtain the multiplication law.

Multiplication Law

Psa > Bd 5 PsBdPsa | Bd

(4.11)

Psa > Bd 5 PsadPsB | ad

(4.12)

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4.4 Conditional Probability

To illustrate the use of the multiplication law, consider a newspaper circulation department where it is known that 84% of the households in a particular neighborhood subscribe to the daily edition of the paper. If we let D denote the event that a household subscribes to the daily edition, P(D) 5 .84. In addition, it is known that the probability that a household that already holds a daily subscription also subscribes to the Sunday edition (event S ) is .75; that is, P(S | D) 5 .75. What is the probability that a household subscribes to both the Sunday and daily editions of the newspaper? Using the multiplication law, we compute the desired P(S > D) as

PsS > Dd 5 PsDdPsS | Dd 5 .84s.75d 5 .63

We now know that 63% of the households subscribe to both the Sunday and daily editions. Before concluding this section, let us consider the special case of the multiplication law when the events involved are independent. Recall that events A and B are independent whenever P(A | B) 5 P(A) or P(B | A) 5 P(B). Hence, using equations (4.11) and (4.12) for the special case of independent events, we obtain the following multiplication law.

Multiplication Law for Independent Events

Psa > Bd 5 PsadPsBd

(4.13)

To compute the probability of the intersection of two independent events, we simply multiply the corresponding probabilities. Note that the multiplication law for independent events provides another way to determine whether A and B are independent. That is, if P(A > B) 5 P(A)P(B), then A and B are independent; if P(A > B) Þ P(A)P(B), then A and B are dependent. As an application of the multiplication law for independent events, consider the situation of a service station manager who knows from past experience that 80% of the customers use a credit card when they purchase gasoline. What is the probability that the next two customers purchasing gasoline will each use a credit card? If we let

a 5 the event that the first customer uses a credit card B 5 the event that the second customer uses a credit card

then the event of interest is A > B. Given no other information, we can reasonably assume that A and B are independent events. Thus,

Psa > Bd 5 PsadPsBd 5 s.80ds.80d 5 .64

To summarize this section, we note that our interest in conditional probability is motivated by the fact that events are often related. In such cases, we say the events are dependent and the conditional probability formulas in equations (4.7) and (4.8) must be used to compute the event probabilities. If two events are not related, they are independent; in this case neither event’s probability is affected by whether the other event occurred. NOTE AND COMMENT Do not confuse the notion of mutually exclusive events with that of independent events. Two events with nonzero probabilities cannot be both mutually exclusive and independent. If one mutually

exclusive event is known to occur, the other cannot occur; thus, the probability of the other event occurring is reduced to zero. They are therefore dependent.

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Exercises

Methods 30. Suppose that we have two events, A and B, with P(A) 5 .50, P(B) 5 .60, and P(A >B) 5 .40. a. Find P(A | B). b. Find P(B | A). c. Are A and B independent? Why or why not? 31. A ssume that we have two events, A and B, that are mutually exclusive. Assume further that we know P(A) 5 .30 and P(B) 5 .40. a. What is P(A > B)? b. W hat is P(A | B)? c. A student in statistics argues that the concepts of mutually exclusive events and independent events are really the same, and that if events are mutually exclusive they must be independent. Do you agree with this statement? Use the probability information in this problem to justify your answer. d. What general conclusion would you make about mutually exclusive and independent events given the results of this problem?

Applications 32. Consider the following example survey results of 18- to 34-year-olds in the United States, in response to the question “Are you currently living with your family?”:

Yes

Totals

Men Women Totals

106 92 198

141 161 302

247 253 500

a. Develop the joint probability table for these data and use it to answer the following questions. b. What are the marginal probabilities? c. What is the probability of living with family given you are an 18- to 34-year-old man in the U.S.? d. What is the probability of living with family given you are an 18- to 34-year-old woman in the U.S.? e. What is the probability of an 18- to 34-year-old in the U.S. living with family? f. If, in the U.S., 49.4% of 18- to 34-year-olds are male, do you consider this a good representative sample? Why? 33. Students taking the Graduate Management Admissions Test (GMAT ) were asked about their undergraduate major and intent to pursue their MBA as a full-time or part-time student. A summary of their responses follows.

Undergraduate Major

Intended Full-Time Enrollment Part-Time Status Totals

Business Engineering Other Totals 352 150

197 161

251 194

502

358

445

1305

800 505

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4.4 Conditional Probability

a. Develop a joint probability table for these data. b. Use the marginal probabilities of undergraduate major (business, engineering, or other) to comment on which undergraduate major produces the most potential MBA students. c. If a student intends to attend classes full-time in pursuit of an MBA degree, what is the probability that the student was an undergraduate engineering major? d. If a student was an undergraduate business major, what is the probability that the student intends to attend classes full-time in pursuit of an MBA degree? e. Let F denote the event that the student intends to attend classes full-time in pursuit of an MBA degree, and let B denote the event that the student was an undergraduate business major. Are events F and B independent? Justify your answer. 34. T he Bureau of Transportation Statistics reports on-time performance for airlines at major U.S. airports. JetBlue, United, and US Airways share Terminal C at Boston’s Logan Airport. The percentage of on-time flights reported for a sample month were 76.8% for JetBlue, 71.5% for United, and 82.2% for US Airways. Assume that 30% of the flights arriving at Terminal C are JetBlue flights, 32% are United flights, and 38% US Airways flights. a. Develop a joint probability table with three rows (the airlines) and two columns (ontime and late). b. An announcement is made that Flight 1382 will be arriving at gate 20 of Terminal C. What is the probability that Flight 1382 will arrive on time? c. What is the most likely airline for Flight 1382? What is the probability that Flight 1382 is by this airline? d. Suppose that an announcement is made saying that Flight 1382 will now be arriving late. What is the most likely airline for this flight? What is the probability that Flight 1382 is by this airline? 35. T o better understand how husbands and wives feel about their finances, Money Magazine conducted a national poll of 1010 married adults age 25 and older with household incomes of $50,000 or more (Money Magazine website, December 14, 2014). Consider the following example set of responses to the question, “Who is better at getting deals?”

Who Is Better? Respondent

I Am

My Spouse

We Are Equal

Husband Wife

278 290

127 111

102 102

a. D evelop a joint probability table and use it to answer the following questions. b. Construct the marginal probabilities for Who Is Better (I Am, My Spouse, We Are Equal). Comment. c. Given that the respondent is a husband, what is the probability that he feels he is better at getting deals than his wife? d. Given that the respondent is a wife, what is the probability that she feels she is better at getting deals than her husband? e. Given a response “My spouse” is better at getting deals, what is the probability that the response came from a husband? f. Given a response “We are equal,” what is the probability that the response came from a husband? What is the probability that the response came from a wife? 36. National Basketball Association player Jamal Crawford of the L.A. Clippers is an excellent free-throw shooter, making 93% of his shots (ESPN website). Assume that, late in a basketball game, Jamal Crawford is fouled and is awarded two shots. a. What is the probability that he will make both shots? b. What is the probability that he will make at least one shot? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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c. What is the probability that he will miss both shots? d. Late in a basketball game, a team often intentionally fouls an opposing player in order to stop the game clock. The usual strategy is to intentionally foul the other team’s worst free-throw shooter. Assume that the Los Angeles Clippers’ center makes 58% of his free-throw shots. Calculate the probabilities for the center as shown in parts (a), ( b), and (c), and show that intentionally fouling the Los Angeles Clippers’ center is a better strategy than intentionally fouling Jamal Crawford. Assume as in parts (a), ( b), and (c) that two shots will be awarded. 37. A joint survey by Parade magazine and Yahoo! found that 59% of American workers say that if they could do it all over again, they would choose a different career (USA Today, September 24, 2012). The survey also found that 33% of American workers say they plan to retire early and 67% say they plan to wait and retire at age 65 or older. Assume that the following joint probability table applies.

Retire Early

Yes No

Same .20 .21 .41 Career Different .13 .46 .59

.33 .67

a. What is the probability a worker would select the same career? b. What is the probability a worker who would select the same career plans to retire early? c. What is the probability a worker who would select a different career plans to retire early? d. What do the conditional probabilities in parts ( b) and (c) suggest about the reasons workers say they would select the same career? 38. T he Institute for Higher Education Policy, a Washington, DC–based research firm, studied the payback of student loans for 1.8 million college students who had student loans that began to become due six years prior. The study found that 50% of the student loans were being paid back in a satisfactory fashion, and 50% of the student loans were delinquent. The following joint probability table shows the probabilities of a student’s loan status and whether or not the student had received a college degree.

College Degree

Yes No

Loan Satisfactory .26 .24 .50 Status Delinquent .16 .34 .50

.42 .58

a. W hat is the probability that a student with a student loan had received a college degree? b. What is the probability that a student with a student loan had not received a college degree? c. Given the student had received a college degree, what is the probability that the student has a delinquent loan? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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d. G iven the student had not received a college degree, what is the probability that the student has a delinquent loan? e. What is the impact of dropping out of college without a degree for students who have a student loan?

Bayes’ Theorem 4.5 In the discussion of conditional probability, we indicated that revising probabilities when new information is obtained is an important phase of probability analysis. Often, we begin the analysis with initial or prior probability estimates for specific events of interest. Then, from sources such as a sample, a special report, or a product test, we obtain additional information about the events. Given this new information, we update the prior probability values by calculating revised probabilities, referred to as posterior probabilities. Bayes’ theorem provides a means for making these probability calculations. The steps in this probability revision process are shown in Figure 4.9. As an application of Bayes’ theorem, consider a manufacturing firm that receives shipments of parts from two different suppliers. Let A1 denote the event that a part is from supplier 1 and A2 denote the event that a part is from supplier 2. Currently, 65% of the parts purchased by the company are from supplier 1 and the remaining 35% are from supplier 2. Hence, if a part is selected at random, we would assign the prior probabilities P(A1) 5 .65 and P(A2) 5 .35. The quality of the purchased parts varies with the source of supply. Historical data suggest that the quality ratings of the two suppliers are as shown in Table 4.6. If we let G denote the event that a part is good and B denote the event that a part is bad, the information in Table 4.6 provides the following conditional probability values. Psg | a1d 5 .98 PsB | a1d 5 .02 Psg | a2d 5 .95 PsB | a2d 5 .05

The tree diagram in Figure 4.10 depicts the process of the firm receiving a part from one of the two suppliers and then discovering that the part is good or bad as a two-step experiment. We see that four experimental outcomes are possible; two correspond to the part being good and two correspond to the part being bad. FIGURE 4.9 PROBABILITY REVISION USING BAYES’ THEOREM

Prior Probabilities

New Information

Application of Bayes’ Theorem

Posterior Probabilities

TABLE 4.6 HISTORICAL QUALITY LEVELS OF TWO SUPPLIERS

Percentage Percentage Good Parts Bad Parts

Supplier 1 98 Supplier 2 95

2 5

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FIGURE 4.10 TREE DIAGRAM FOR TWO-SUPPLIER EXAMPLE Step 2 Condition

Step 1 Supplier

Experimental Outcome (A1, G)

G B

(A1, B)

(A2, G)

G B

(A2, B) Note: Step 1 shows that the part comes from one of two suppliers, and step 2 shows whether the part is good or bad.

Each of the experimental outcomes is the intersection of two events, so we can use the multiplication rule to compute the probabilities. For instance, Psa1, gd 5 Psa1 > gd 5 Psa1dPsg | a1d

The process of computing these joint probabilities can be depicted in what is called a probability tree (see Figure 4.11). From left to right through the tree, the probabilities for each branch at step 1 are prior probabilities and the probabilities for each branch at step 2 are conditional probabilities. To find the probabilities of each experimental outcome, we simply multiply the probabilities on the branches leading to the outcome. Each of these joint probabilities is shown in Figure 4.11 along with the known probabilities for each branch. Suppose now that the parts from the two suppliers are used in the firm’s manufacturing process and that a machine breaks down because it attempts to process a bad part. Given FIGURE 4.11 PROBABILITY TREE FOR TWO-SUPPLIER EXAMPLE Step 1 Supplier

Step 2 Condition P(G | A1)

Probability of Outcome P( A1 > G ) P( A1)P(G | A1) .6370

.98 P(A1)

P(B | A1) .02

P( A1 > B) P( A1)P( B | A1) .0130

P(A2)

P(G | A2)

P( A2 > G) P( A2)P(G | A2) .3325

.35

.95 P(B | A2)

.65

.05

P( A2 > B) P( A2)P( B | A2) .0175

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4.5 Bayes’ Theorem

the information that the part is bad, what is the probability that it came from supplier 1 and what is the probability that it came from supplier 2? With the information in the probability tree ( Figure 4.11), Bayes’ theorem can be used to answer these questions. Letting B denote the event that the part is bad, we are looking for the posterior probabilities P(A1 | B) and P(A2 | B). From the law of conditional probability, we know that

PsA1 | Bd 5

PsA1 > Bd PsBd

(4.14)

Referring to the probability tree, we see that PsA1 > Bd 5 PsA1dPsB | A1d

(4.15)

To find P(B), we note that event B can occur in only two ways: (A1 > B) and (A2 > B). Therefore, we have PsBd 5 PsA1 > Bd 1 PsA2 > Bd 5 PsA1dPsB | A1d 1 PsA2 dPsB | A2 d

(4.16)

Substituting from equations (4.15) and (4.16) into equation (4.14) and writing a similar result for P(A2 | B), we obtain Bayes’ theorem for the case of two events. The Reverend Thomas Bayes (1702–1761), a Presbyterian minister, is credited with the original work leading to the version of Bayes’ theorem in use today.

Bayes’ Theorem (Two-Event Case)

PsA1 | Bd 5

PsA1dPsB | A1d PsA1dPsB | A1d 1 PsA2 dPsB | A2 d

(4.17)

PsA2 | Bd 5

PsA2dPsB | A2d PsA1dPsB | A1d 1 PsA2 dPsB | A2 d

(4.18)

Using equation (4.17) and the probability values provided in the example, we have PsA1dPsB | A1d PsA1dPsB | A1d 1 PsA2 dPsB | A2 d

PsA1 | Bd 5

s.65ds.02d .0130 5 s.65ds.02d 1 s.35ds.05d .0130 1 .0175

.0130 5 .4262 .0305

In addition, using equation (4.18), we find P(A2 | B).

PsA2 | Bd 5 5

s.35ds.05d s.65ds.02d 1 s.35ds.05d

.0175 .0175 5 5 .5738 .0130 1 .0175 .0305

Note that in this application we started with a probability of .65 that a part selected at random was from supplier 1. However, given information that the part is bad, the probability Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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that the part is from supplier 1 drops to .4262. In fact, if the part is bad, there is more than a 50–50 chance that it came from supplier 2; that is, P(A2 | B) 5 .5738. Bayes’ theorem is applicable when the events for which we want to compute posterior probabilities are mutually exclusive and their union is the entire sample space.1 For the case of n mutually exclusive events A1, A2, . . . , An, whose union is the entire sample space, Bayes’ theorem can be used to compute any posterior probability P(Ai | B), as shown here.

Bayes’ Theorem

Psai | Bd 5

Psai dPsB | ai d

Psa1dPsB | a1d 1 Psa2 dPsB | a2 d 1 Á 1 Psan dPsB | an d

(4.19)

With prior probabilities P(A1), P(A2), . . . , P(An) and the appropriate conditional probabili ties P(B | A1), P(B | A2), . . . , P(B | An), equation (4.19) can be used to compute the posterior probability of the events A1, A2, . . . , An.

Tabular Approach A tabular approach is helpful in conducting the Bayes’ theorem calculations. Such an approach is shown in Table 4.7 for the parts supplier problem. The computations shown there are done in the following steps. Step 1. Prepare the following three columns: Column 1—The mutually exclusive events Ai for which posterior probabilities are desired Column 2—The prior probabilities P(Ai) for the events Column 3—The conditional probabilities P(B | Ai) of the new information B given each event Step 2. In column 4, compute the joint probabilities P(Ai > B) for each event and the new information B by using the multiplication law. These joint probabilities are found by multiplying the prior probabilities in column 2 by the corresponding conditional probabilities in column 3; that is, P(Ai > B) 5 P(Ai)P(B | Ai). Step 3. Sum the joint probabilities in column 4. The sum is the probability of the new information, P(B). Thus we see in Table 4.7 that there is a .0130 probability that TABULAR APPROACH TO BAYES’ THEOREM CALCULATIONS TABLE 4.7 FOR THE TWO-SUPPLIER PROBLEM (1) Events Ai

(2) Prior Probabilities P(Ai )

(3) Conditional Probabilities P(B | Ai )

A 1 A 2

.65 .35

.02 .05

1.00

(4) Joint Probabilities P(Ai > B)

(5) Posterior Probabilities P(Ai | B)

.0130 .0130/.0305 5 .4262 .0175 .0175/.0305 5 .5738

P(B) 5 .0305

1.0000

If the union of events is the entire sample space, the events are said to be collectively exhaustive.

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4.5 Bayes’ Theorem

the part came from supplier 1 and is bad and a .0175 probability that the part came from supplier 2 and is bad. Because these are the only two ways in which a bad part can be obtained, the sum .0130 1 .0175 shows an overall probability of .0305 of finding a bad part from the combined shipments of the two suppliers. Step 4. In column 5, compute the posterior probabilities using the basic relationship of conditional probability. Psai | Bd 5

Psai > Bd PsBd

Note that the joint probabilities P(Ai > B) are in column 4 and the probability P(B) is the sum of column 4. NOTES AND COMMENTS 1. Bayes’ theorem is used extensively in decision analysis. The prior probabilities are often subjective estimates provided by a decision maker. Sample information is obtained and posterior probabilities are computed for use in choosing the best decision.

2. An event and its complement are mutually exclusive, and their union is the entire sample space. Thus, Bayes’ theorem is always applicable for computing posterior probabilities of an event and its complement.

Exercises

Methods 39. The prior probabilities for events A1 and A2 are P(A1) 5 .40 and P(A2) 5 .60. It is also known that P(A1 > A2) 5 0. Suppose P(B | A1) 5 .20 and P(B | A2) 5 .05. a. Are A1 and A2 mutually exclusive? Explain. b. Compute P(A1 > B) and P(A2 > B). c. Compute P(B). d. Apply Bayes’ theorem to compute P(A1 | B) and P(A2 | B). 40. T he prior probabilities for events A1, A2, and A3 are P(A1) 5 .20, P(A2) 5 .50, and P(A3) 5 .30, respectively. The conditional probabilities of event B given A1, A2, and A3 are P(B | A1) 5 .50, P(B | A2) 5 .40, and P(B | A3) 5 .30. a. Compute P(B > A1), P(B > A2), and P(B > A3). b. Apply Bayes’ theorem, equation (4.19), to compute the posterior probability P(A2 | B). c. Use the tabular approach to applying Bayes’ theorem to compute P(A1 | B), P(A2 | B), and P(A3 | B).

Applications 41. A consulting firm submitted a bid for a large research project. The firm’s management initially felt they had a 50–50 chance of getting the project. However, the agency to which the bid was submitted subsequently requested additional information on the bid. Past experience indicates that for 75% of the successful bids and 40% of the unsuccessful bids the agency requested additional information. a. What is the prior probability of the bid being successful (that is, prior to the request for additional information)? b. What is the conditional probability of a request for additional information given that the bid will ultimately be successful? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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c. C ompute the posterior probability that the bid will be successful given a request for additional information. 42. A local bank reviewed its credit card policy with the intention of recalling some of its credit cards. In the past approximately 5% of cardholders defaulted, leaving the bank unable to collect the outstanding balance. Hence, management established a prior probability of .05 that any particular cardholder will default. The bank also found that the probability of missing a monthly payment is .20 for customers who do not default. Of course, the probability of missing a monthly payment for those who default is 1. a. Given that a customer missed a monthly payment, compute the posterior probability that the customer will default. b. The bank would like to recall its credit card if the probability that a customer will default is greater than .20. Should the bank recall its credit card if the customer misses a monthly payment? Why or why not? 43. I n August 2012, tropical storm Isaac formed in the Caribbean and was headed for the Gulf of Mexico. There was an initial probability of .69 that Isaac would become a hurricane by the time it reached the Gulf of Mexico ( National Hurricane Center website, August 21, 2012). a. What was the probability that Isaac would not become a hurricane but remain a tropical storm when it reached the Gulf of Mexico? b. Two days later, the National Hurricane Center projected the path of Isaac would pass directly over Cuba before reaching the Gulf of Mexico. How did passing over Cuba alter the probability that Isaac would become a hurricane by the time it reached the Gulf of Mexico? Using the following probabilities to answer this question. Hurricanes that reach the Gulf of Mexico have a .08 probability of having passed over Cuba. Tropical storms that reach the Gulf of Mexico have a .20 probability of having passed over Cuba. c. What happens to the probability of becoming a hurricane when a tropical storm passes over a landmass such as Cuba? 44. P arFore created a website to market golf equipment and golf apparel. Management would like a special pop-up offer to appear for female website visitors and a different special pop-up offer to appear for male website visitors. From a sample of past website visitors, ParFore’s management learned that 60% of the visitors are male and 40% are female. a. What is the probability that a current visitor to the website is female? b. Suppose 30% of ParFore’s female visitors previously visited the Dillard’s Department Store website and 10% of ParFore’s male visitors previously visited the Dillard’s Department Store website. If the current visitor to ParFore’s website previously visited the Dillard’s website, what is the revised probability that the current visitor is female? Should the ParFore’s website display the special offer that appeals to female visitors or the special offer that appeals to male visitors? 45. The percentage of adult users of the Internet who use Facebook has increased over time (Pew Research Internet Project, 2013). Of adult Internet users age 18–49, 81% use Facebook. Of adult Internet users age 50 and older, 54% use Facebook. Assume that 52% of adult Internet users are age 18–49. a. What is the probability that a randomly selected adult user of the Internet is age 50 or older? b. Given that an adult Internet user uses Facebook, what is the probability that he/she is age 18–49?

Summary In this chapter we introduced basic probability concepts and illustrated how probability analysis can be used to provide helpful information for decision making. We described how probability can be interpreted as a numerical measure of the likelihood that an event will occur. In addition, we saw that the probability of an event can be computed either by Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Glossary

summing the probabilities of the experimental outcomes (sample points) comprising the event or by using the relationships established by the addition, conditional probability, and multiplication laws of probability. For cases in which additional information is available, we showed how Bayes’ theorem can be used to obtain revised or posterior probabilities.

Glossary Addition lawA probability law used to compute the probability of the union of two events. It is P(A > B) 5 P(A) 1 P(B) 2 P(A ø B). For mutually exclusive events, P(A > B) 5 0; in this case the addition law reduces to P(A ø B) 5 P(A) 1 P(B). Basic requirements for assigning probabilitiesTwo requirements that restrict the manner in which probability assignments can be made: (1) For each experimental outcome Ei we must have 0 # P(Ei) # 1; (2) considering all experimental outcomes, we must have P(E1) 1 P(E2) 1 Á 1 P(En) 5 1.0. Bayes’ theoremA method used to compute posterior probabilities. Classical method A method of assigning probabilities that is appropriate when all the experimental outcomes are equally likely. CombinationIn an experiment we may be interested in determining the number of ways n objects may be selected from among N objects without regard to the order in which the n objects are selected. Each selection of n objects is called a combination and the total N N! number of combinations of N objects taken n at a time is CnN 5 5 for n 5 n n!sN 2 nd! 0, 1, 2, . . . , N. Complement ofA The event consisting of all sample points that are not in A. Conditional probability The probability of an event given that another event already occurred. The conditional probability of A given B is P(A | B) 5 P(A > B)/P(B). EventA collection of sample points. ExperimentA process that generates well-defined outcomes. Independent eventsTwo events A and B where P(A | B) 5 P(A) or P(B | A) 5 P(B); that is, the events have no influence on each other. Intersection of A and BThe event containing the sample points belonging to both A and B. The intersection is denoted A > B. Joint probabilityThe probability of two events both occurring; that is, the probability of the intersection of two events. Marginal probabilityThe values in the margins of a joint probability table that provide the probabilities of each event separately. Multiple-step experimentAn experiment that can be described as a sequence of steps. If a multiple-step experiment has k steps with n1 possible outcomes on the first step, n2 possible outcomes on the second step, and so on, the total number of experimental outcomes is given by (n1)(n2) . . . (nk). Multiplication lawA probability law used to compute the probability of the intersection of two events. It is P(A > B) 5 P(B)P(A | B) or P(A > B) 5 P(A)P(B | A). For independent events it reduces to P(A > B) 5 P(A)P(B). Mutually exclusive eventsEvents that have no sample points in common; that is, A > B is empty and P(A > B) 5 0. PermutationIn an experiment we may be interested in determining the number of ways n objects may be selected from among N objects when the order in which the n objects are selected is important. Each ordering of n objects is called a permutation and the total n umber N N! of permutations of N objects taken n at a time is PnN 5 n! 5 for n 5 0, 1, n sN 2 nd! 2, . . . , N.

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Posterior probabilitiesRevised probabilities of events based on additional information. Prior probabilitiesInitial estimates of the probabilities of events. Probability A numerical measure of the likelihood that an event will occur. Relative frequency methodA method of assigning probabilities that is appropriate when data are available to estimate the proportion of the time the experimental outcome will occur if the experiment is repeated a large number of times. Sample pointAn element of the sample space. A sample point represents an experimental outcome. Sample spaceThe set of all experimental outcomes. Subjective methodA method of assigning probabilities on the basis of judgment. Tree diagram A graphical representation that helps in visualizing a multiple-step experiment. Union of A and BThe event containing all sample points belonging to A or B or both. The union is denoted A ø B. Venn diagramA graphical representation for showing symbolically the sample space and operations involving events in which the sample space is represented by a rectangle and events are represented as circles within the sample space.

Key Formulas Counting Rule for Combinations

CnN 5

1Nn 2 5 n!sNN!2 nd!

(4.1)

1Nn 2 5 sN N!2 nd!

(4.2)

Counting Rule for Permutations PnN 5 n!

Computing Probability Using the Complement PsAd 5 1 2 PsAc d

(4.5)

PsA ø Bd 5 PsAd 1 PsBd 2 PsA > Bd

(4.6)

Addition Law

Conditional Probability

PsA > Bd PsBd

(4.7)

PsA > Bd PsAd

(4.8)

PsA | Bd 5

PsB | Ad 5

219

Supplementary Exercises

Multiplication Law

(4.11) (4.12)

Psa > Bd 5 PsBdPsa | Bd Psa > Bd 5 PsadPsB | ad

Multiplication Law for Independent Events

(4.13)

Psa > Bd 5 PsadPsBd

Bayes’ Theorem

Psai | Bd 5

Psai dPsB | ai d

Psa1dPsB | a1d 1 Psa2 dPsB | a2 d 1 . . . 1 Psan dPsB | an d

(4.19)

Supplementary Exercises 46. A USA Today survey of adults aged 18 and older conducted by Princess Cruises asked how many days into a vacation it takes until respondents feel truly relaxed. The responses were as follows: 422—a day or less; 181—2 days; 80—3 days; 121—4 or more days; and 201—never feel relaxed. a. How many adults participated in the Princess Cruises survey? b. What response has the highest probability? What is the probability of this response? c. What is the probability a respondent never feels truly relaxed on a vacation? d. What is the probability it takes a respondent 2 or more days to feel truly relaxed? 47. A financial manager made two new investments—one in the oil industry and one in municipal bonds. After a one-year period, each of the investments will be classified as either successful or unsuccessful. Consider the making of the two investments as an experiment. a. How many sample points exist for this experiment? b. Show a tree diagram and list the sample points. c. Let O 5 the event that the oil industry investment is successful and M 5 the event that the municipal bond investment is successful. List the sample points in O and in M. d. List the sample points in the union of the events (O ø M ). e. List the sample points in the intersection of the events (O > M ). f. Are events O and M mutually exclusive? Explain. 48. Below are the results of a survey of 1364 individuals who were asked if they use social media and other websites to voice their opinions about television programs.

Female Male

Uses Social Media and Other Websites to Voice Opinions About Television Programs

Doesn’t Use Social Media and Other Websites to Voice Opinions About Television Programs

395 323

291 355

a. Show a joint probability table. b. What is the probability a respondent is female? c. What is the conditional probability a respondent uses social media and other websites to voice opinions about television programs given the respondent is female?

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d. Let F denote the event that the respondent is female and A denote the event that the respondent uses social media and other websites to voice opinions about television programs. Are events F and A independent? 49. A study of 31,000 hospital admissions in New York State found that 4% of the admissions led to treatment-caused injuries. One-seventh of these treatment-caused injuries resulted in death, and one-fourth were caused by negligence. Malpractice claims were filed in one out of 7.5 cases involving negligence, and payments were made in one out of every two claims. a. What is the probability a person admitted to the hospital will suffer a treatment-caused injury due to negligence? b. What is the probability a person admitted to the hospital will die from a treatmentcaused injury? c. In the case of a negligent treatment-caused injury, what is the probability a malpractice claim will be paid? 50. A telephone survey to determine viewer response to a new television show obtained the following data.

Rating Frequency Poor Below average Average Above average Excellent

4 8 11 14 13

a. W hat is the probability that a randomly selected viewer will rate the new show as average or better? b. What is the probability that a randomly selected viewer will rate the new show below average or worse? 51. T he U.S. Census Bureau serves as the leading source of quantitative data about the nation’s people and economy. The following crosstabulation shows the number of households (1000s) and the household income by the highest level of education for the head of household (U.S. Census Bureau website, 2013). Only households in which the head has a high school diploma or more are included. Household Income Highest Level of Education High school graduate Bachelor’s degree Master’s degree Doctoral degree Total

Under $25,000

$25,000 to $49,999

$50,000 to $99,999

$100,000 and Over

Total

9,880 2,484 685 79

9,970 4,164 1,205 160

9,441 7,666 3,019 422

3,482 7,817 4,094 1,076

32,773 22,131 9,003 1,737

13,128

15,499

20,548

16,469

65,644

a. Develop a joint probability table. b. What is the probability of the head of one of these households having a master’s degree or more education? c. What is the probability of a household headed by someone with a high school diploma earning $100,000 or more? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Supplementary Exercises

d. What is the probability of one of these households having an income below $25,000? e. What is the probability of a household headed by someone with a bachelor’s degree earning less than $25,000? f. Is household income independent of educational level? 52. An MBA new matriculants survey provided the following data for 2018 students.

Applied to More Than One School

Yes No

23 and under 207 24–26 299 27–30 185 31–35 66 36 and over 51

Age Group

201 379 268 193 169

a. F or a randomly selected MBA student, prepare a joint probability table for the experi ment consisting of observing the student’s age and whether the student applied to one or more schools. b. What is the probability that a randomly selected applicant is 23 or under? c. What is the probability that a randomly selected applicant is older than 26? d. What is the probability that a randomly selected applicant applied to more than one school? 53. R efer again to the data from the MBA new matriculants survey in exercise 52. a. Given that a person applied to more than one school, what is the probability that the person is 24–26 years old? b. Given that a person is in the 36-and-over age group, what is the probability that the person applied to more than one school? c. What is the probability that a person is 24–26 years old or applied to more than one school? d. Suppose a person is known to have applied to only one school. What is the probability that the person is 31 or more years old? e. Is the number of schools applied to independent of age? Explain. 54. T he Pew Internet & American Life project conducted a survey that included several questions about how Internet users feel about search engines and other websites collecting information about them and using this information either to shape search results or target advertising to them. In one question, participants were asked, “If a search engine kept track of what you search for, and then used that information to personalize your future search results, how would you feel about that?” Respondents could indicate either “Would not be okay with it because you feel it is an invasion of your privacy” or “Would be okay with it, even if it means they are gathering information about you.” Joint probabilities of responses and age groups are summarized in the following table.

Age

Not Okay

Okay

18–29 .1485 .0604 30–49 .2273 .0907 501 .4008 .0723

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Chapter 4 Introduction to Probability

a. W hat is the probability a respondent will not be okay with this practice? b. Given a respondent is 30–49 years old, what is the probability the respondent will be okay with this practice? c. Given a respondent is not okay with this practice, what is the probability the respondent is 501 years old? d. Is the attitude about this practice independent of the age of the respondent? Why or why not? e. Do attitudes toward this practice differ for respondents who are 18–29 years old and respondents who are 501 years old? 55. A large consumer goods company ran a television advertisement for one of its soap products. On the basis of a survey that was conducted, probabilities were assigned to the following events. B 5 individual purchased the product S 5 individual recalls seeing the advertisement B > S 5 individual purchased the product and recalls seeing the advertisement The probabilities assigned were P(B) 5 .20, P(S ) 5 .40, and P(B > S ) 5 .12. a. What is the probability of an individual’s purchasing the product given that the individual recalls seeing the advertisement? Does seeing the advertisement increase the probability that the individual will purchase the product? As a decision maker, would you recommend continuing the advertisement (assuming that the cost is reasonable)? b. Assume that individuals who do not purchase the company’s soap product buy from its competitors. What would be your estimate of the company’s market share? Would you expect that continuing the advertisement will increase the company’s market share? Why or why not? c. The company also tested another advertisement and assigned it values of P(S ) 5 .30 and P(B > S ) 5 .10. What is P(B | S ) for this other advertisement? Which advertisement seems to have had the bigger effect on customer purchases? 56. Cooper Realty is a small real estate company located in Albany, New York, specializing primarily in residential listings. They recently became interested in determining the likelihood of one of their listings being sold within a certain number of days. An analysis of company sales of 800 homes in previous years produced the following data.

Days Listed Until Sold

Under 30

31–90

Over 90

Total

Under $150,000 50 $150,000–$199,999 20 Initial Asking Price $200,000–$250,000 20 Over $250,000 10

40 150 280 30

10 100 80 250 100 400 10 50

Total

500

200

100

800

a. If A is defined as the event that a home is listed for more than 90 days before being sold, estimate the probability of A. b. If B is defined as the event that the initial asking price is under $150,000, estimate the probability of B. c. What is the probability of A > B? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Supplementary Exercises

d. A ssuming that a contract was just signed to list a home with an initial asking price of less than $150,000, what is the probability that the home will take Cooper Realty more than 90 days to sell? e. Are events A and B independent? 57. A company studied the number of lost-time accidents occurring at its Brownsville, Texas plant. Historical records show that 6% of the employees suffered lost-time accidents last year. Management believes that a special safety program will reduce such accidents to 5% during the current year. In addition, it estimates that 15% of employees who had lost-time accidents last year will experience a lost-time accident during the current year. a. What percentage of the employees will experience lost-time accidents in both years? b. What percentage of the employees will suffer at least one lost-time accident over the two-year period? 58. A ccording to the Open Doors Report, 9.5% of all full-time U.S. undergraduate students study abroad. Assume that 60% of the undergraduate students who study abroad are female and that 49% of the undergraduate students who do not study abroad are female. a. Given a female undergraduate student, what is the probability that she studies abroad? b. Given a male undergraduate student, what is the probability that he studies abroad? c. What is the overall percentage of full-time female undergraduate students? What is the overall percentage of full-time male undergraduate students? 59. A n oil company purchased an option on land in Alaska. Preliminary geologic studies assigned the following prior probabilities.

Pshigh { quality oild 5 .50 Psmedium { quality oild 5 .20 Psno oild 5 .30

a. W hat is the probability of finding oil? b. After 200 feet of drilling on the first well, a soil test is taken. The probabilities of finding the particular type of soil identified by the test follow.

Pssoil | high { quality oild 5 .20 Pssoil | medium { quality oild 5 .80 Pssoil | no oild 5 .20

How should the firm interpret the soil test? What are the revised probabilities, and what is the new probability of finding oil? 60. A study reported by Forbes indicated that the five most common words appearing in spam e-mails are shipping!, today!, here!, available, and fingertips! Many spam filters separate spam from ham (e-mail not considered to be spam) through application of Bayes’ theorem. Suppose that for one e-mail a ccount, 1 in every 10 messages is spam and the proportions of spam messages that have the five most common words in spam e-mail are given below. shipping! .051 today! .045 here! .034 available .014 fingertips! .014 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Chapter 4 Introduction to Probability

Also suppose that the proportions of ham messages that have these words are

shipping! .0015 today! .0022 here! .0022 available .0041 fingertips! .0011

a. I f a message includes the word shipping!, what is the probability the message is spam? If a message includes the word shipping!, what is the probability the message is ham? Should messages that include the word shipping! be flagged as spam? b. If a message includes the word today!, what is the probability the message is spam? If a message includes the word here!, what is the probability the message is spam? Which of these two words is a stronger indicator that a message is spam? Why? c. If a message includes the word available, what is the probability the message is spam? If a message includes the word fingertips!, what is the probability the message is spam? Which of these two words is a stronger indicator that a message is spam? Why? d. What insights do the results of parts ( b) and (c) yield about what enables a spam filter that uses Bayes’ theorem to work effectively?

Case Problem 1

Hamilton County Judges Hamilton County judges try thousands of cases per year. In an overwhelming majority of the cases disposed, the verdict stands as rendered. However, some cases are appealed, and of those appealed, some of the cases are reversed. Kristen DelGuzzi of The Cincinnati Enquirer conducted a study of cases handled by Hamilton County judges over a three-year period. Shown in Table 4.8 are the results for 182,908 cases handled (disposed) by 38 judges in Common Pleas Court, Domestic Relations Court, and Municipal Court. Two of the judges (Dinkelacker and Hogan) did not serve in the same court for the entire three-year period. The purpose of the newspaper’s study was to evaluate the performance of the judges. Appeals are often the result of mistakes made by judges, and the newspaper wanted to know which judges were doing a good job and which were making too many mistakes. You are called in to assist in the data analysis. Use your knowledge of probability and conditional probability to help with the ranking of the judges. You also may be able to analyze the likelihood of appeal and reversal for cases handled by different courts.

Managerial Report Prepare a report with your rankings of the judges. Also, include an analysis of the likelihood of appeal and case reversal in the three courts. At a minimum, your report should include the following: 1. 2. 3. 4. 5.

he probability of cases being appealed and reversed in the three different courts. T The probability of a case being appealed for each judge. The probability of a case being reversed for each judge. The probability of reversal given an appeal for each judge. Rank the judges within each court. State the criteria you used and provide a rationale for your choice.

Case Problem 1 Hamilton County Judges

TABLE 4.8 TOTAL CASES DISPOSED, APPEALED, AND REVERSED IN HAMILTON

COUNTY COURTS Common Pleas Court

Judge

Judge Fred Cartolano Thomas Crush Patrick Dinkelacker Timothy Hogan Robert Kraft William Mathews William Morrissey Norbert Nadel Arthur Ney, Jr. Richard Niehaus Thomas Nurre John O’Connor Robert Ruehlman J. Howard Sundermann Ann Marie Tracey Ralph Winkler Total

Total Cases Appealed Reversed Disposed Cases Cases 3,037 137 12 3,372 119 10 1,258 44 8 1,954 60 7 3,138 127 7 2,264 91 18 3,032 121 22 2,959 131 20 3,219 125 14 3,353 137 16 3,000 121 6 2,969 129 12 3,205 145 18 955 60 10 3,141 127 13 3,089 88 6 43,945 1762 199

Domestic Relations Court Judge Penelope Cunningham Patrick Dinkelacker Deborah Gaines Ronald Panioto Total

Total Cases Appealed Reversed Disposed Cases Cases 2,729 7 1 6,001 19 4 8,799 48 9 12,970 32 3 30,499 106 17

Municipal Court Judge Mike Allen Nadine Allen Timothy Black David Davis Leslie Isaiah Gaines Karla Grady Deidra Hair Dennis Helmick Timothy Hogan James Patrick Kenney Joseph Luebbers William Mallory Melba Marsh Beth Mattingly Albert Mestemaker Mark Painter Jack Rosen Mark Schweikert David Stockdale John A. West Total

Total Cases Appealed Reversed Disposed Cases Cases 6,149 43 4 7,812 34 6 7,954 41 6 7,736 43 5 5,282 35 13 5,253 6 0 2,532 5 0 7,900 29 5 2,308 13 2 2,798 6 1 4,698 25 8 8,277 38 9 8,219 34 7 2,971 13 1 4,975 28 9 2,239 7 3 7,790 41 13 5,403 33 6 5,371 22 4 2,797 4 2 108,464 500 104

225

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Chapter 4 Introduction to Probability

Case Problem 2

Rob’s Market Rob’s Market (RM) is a regional food store chain in the southwest United States. David White, director of Business Intelligence for RM, would like to initiate a study of the purchase behavior of customers who use the RM loyalty card (a card that customers scan at checkout to qualify for discounted prices). The use of the loyalty card allows RM to capture what is known as “point-of-sale” data, that is, a list of products purchased by a given customer as he/she checks out of the market. David feels that better understanding of which products tend to be purchased together could lead to insights for better pricing and display strategies as well as a better understanding of sales and the potential impact of different levels of coupon discounts. This type of analysis is known as market basket analysis, as it is a study of what different customers have in their shopping baskets as they check out of the store. As a prototype study, David wants to investigate customer buying behavior with regard to bread, jelly, and peanut butter. RM’s Information Technology (IT) group, at David’s request, has provided a data set of purchases by 1000 customers over a one-week period. The data set contains the following variables for each customer: ●● ●● ●●

Bread – wheat, white, or none Jelly – grape, strawberry, or none Peanut butter – creamy, natural, or none

The variables appear in the above order from left to right in the data set, where each row is a customer. For example, the first record of the data set is white grape none which means that customer #1 purchased white bread, grape jelly, and no peanut butter. The second record is white strawberry none which means that customer #2 purchased white bread, strawberry jelly, and no peanut butter. The sixth record in the data set is none none none which means that the sixth customer did not purchase bread, jelly, or peanut butter. Other records are interpreted in a similar fashion. David would like you to do an initial study of the data to get a better understanding of RM customer behavior with regard to these three products.

Managerial Report MarketBasket

Prepare a report that gives insight into the purchase behavior of customers who use the RM loyalty card. At a minimum your report should include estimates of the following: 1. The probability that a random customer does not purchase any of the three products (bread, jelly, or peanut butter). 2. The probability that a random customer purchases white bread. 3. The probability that a random customer purchases wheat bread. 4. The probability that a random customer purchases grape jelly given that he/she purchases white bread. 5. The probability that a random customer purchases strawberry jelly given that he/ she purchases white bread. 6. The probability that a random customer purchases creamy peanut butter given that he/she purchases white bread.

Case Problem 2 Rob’s Market

One way to answer these questions is to use pivot tables (discussed in Chapter 2) to obtain absolute frequencies and use the pivot table results to calculate the relevant probabilities.

227

7. The probability that a random customer purchases natural peanut butter given that he/she purchases white bread. 8. The probability that a random customer purchases creamy peanut butter given that he/she purchases wheat bread. 9. The probability that a random customer purchases natural peanut butter given that he/she purchases wheat bread. 10. The probability that a random customer purchases white bread, grape jelly, and creamy peanut butter.

CHAPTER

Discrete Probability Distributions CONTENTS STATISTICS IN PRACTICE: CITIBANK 5.1

RANDOM VARIABLES Discrete Random Variables Continuous Random Variables

5.2 Developing DISCRETE PROBABILITY DISTRIBUTIONS 5.3 EXPECTED VALUE AND VARIANCE Expected Value Variance Using Excel to Compute the Expected Value, Variance, and Standard Deviation 5.4

Bivariate distributions, covariance, and financial portfolios A Bivariate Empirical Discrete Probability Distribution Financial Applications Summary

5.5

BINOMIAL PROBABILITY DISTRIBUTION A Binomial Experiment Martin Clothing Store Problem Using Excel to Compute Binomial Probabilities Expected Value and Variance for the Binomial Distribution

5.6 POISSON PROBABILITY DISTRIBUTION An Example Involving Time Intervals An Example Involving Length or Distance Intervals Using Excel to Compute Poisson Probabilities 5.7 HYPERGEOMETRIC PROBABILITY DISTRIBUTION Using Excel to Compute Hypergeometric Probabilities

229

Statistics in Practice

STATISTICS in PRACTICE CITIBANK*

Citibank, the retail banking division of Citigroup, offers a wide range of financial services including checking and saving accounts, loans and mortgages, insurance, and investment services. It delivers these services through a unique system referred to as Citibanking. Citibank was one of the first banks in the United States to introduce automatic teller machines (ATMs). Citibank’s ATMs, located in Citicard Banking Centers (CBCs), let customers do all of their banking in one place with the touch of a finger, 24 hours a day, 7 days a week. More than 150 different banking functions—from deposits to managing investments—can be performed with ease. Citibank customers use ATMs for 80% of their transactions. Each Citibank CBC operates as a waiting line system with randomly arriving customers seeking service at one of the ATMs. If all ATMs are busy, the arriving customers wait in line. Periodic CBC capacity studies are used to analyze customer waiting times and to determine whether additional ATMs are needed. Data collected by Citibank showed that the random customer arrivals followed a probability distribution known as the Poisson distribution. Using the Poisson distribution, Citibank can compute probabilities for the number of customers arriving at a CBC during any time period and make decisions concerning the number of ATMs needed. For example, let x 5 the number of customers arriving during a one-minute period. Assuming that a particular CBC has a mean arrival rate of two customers per minute, the following table shows the *The authors are indebted to Ms. Stacey Karter, Citibank, for providing this Statistics in Practice.

Jeff Greenberg/Alamy Stock Photo

Long Island City, New York

Each Citicard Banking Center operates as a waiting line system with randomly arriving customers seeking service at an ATM. probabilities for the number of customers arriving during a one-minute period. x

Probability

.1353

1 2 3 4 5 or more

.2707 .2707 .1804 .0902 .0527

Discrete probability distributions, such as the one used by Citibank, are the topic of this chapter. In addition to the Poisson distribution, you will learn about the binomial and hypergeometric distributions and how they can be used to provide helpful probability information.

In this chapter we extend the study of probability by introducing the concepts of random variables and probability distributions. Random variables and probability distributions are models for populations of data. The focus of this chapter is on probability distributions for discrete data, that is, discrete probability distributions. We will introduce two types of discrete probability distributions. The first type is a table with one column for the values of the random variable and a second column for the associated probabilities. We will see that the rules for assigning probabilities to experimental outcomes introduced in Chapter 4 are used to assign probabilities for such a distribution. The second type of discrete probability distribution uses a special mathematical function to compute the probabilities for each value of the random variable. We present Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Chapter 5 Discrete Probability Distributions

three probability distributions of this type that are widely used in practice: the binomial, Poisson, and hypergeometric distributions.

Random Variables 5.1 In Chapter 4 we defined the concept of an experiment and its associated experimental outcomes. A random variable provides a means for describing experimental outcomes using numerical values. Random variables must assume numerical values. Random Variable Random variables must assume numerical values.

A random variable is a numerical description of the outcome of an experiment. In effect, a random variable associates a numerical value with each possible experimental outcome. The particular numerical value of the random variable depends on the outcome of the experiment. A random variable can be classified as being either discrete or continuous depending on the numerical values it assumes.

Discrete Random Variables A random variable that may assume either a finite number of values or an infinite sequence of values such as 0, 1, 2, . . . is referred to as a discrete random variable. For example, consider the experiment of an accountant taking the certified public accountant (CPA) examination. The examination has four parts. We can define a random variable as x 5 the number of parts of the CPA examination passed. It is a discrete random variable because it may assume the finite number of values 0, 1, 2, 3, or 4. As another example of a discrete random variable, consider the experiment of cars arriving at a tollbooth. The random variable of interest is x 5 the number of cars arriving during a one-day period. The possible values for x come from the sequence of integers 0, 1, 2, and so on. Hence, x is a discrete random variable assuming one of the values in this infinite sequence. Although the outcomes of many experiments can naturally be described by numerical values, others cannot. For example, a survey question might ask an individual to recall the message in a recent television commercial. This experiment would have two possible outcomes: The individual cannot recall the message and the individual can recall the message. We can still describe these experimental outcomes numerically by defining the discrete random variable x as follows: Let x 5 0 if the individual cannot recall the message and x 5 1 if the individual can recall the message. The numerical values for this random variable are arbitrary (we could use 5 and 10), but they are acceptable in terms of the definition of a random variable—namely, x is a random variable because it provides a numerical description of the outcome of the experiment. Table 5.1 provides some additional examples of discrete random variables. Note that in each example the discrete random variable assumes a finite number of values or an infinite sequence of values such as 0, 1, 2, . . . . These types of discrete random variables are discussed in detail in this chapter.

Continuous Random Variables A random variable that may assume any numerical value in an interval or collection of intervals is called a continuous random variable. Experimental outcomes based on Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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5.1 Random Variables

TABLE 5.1 EXAMPLES OF DISCRETE RANDOM VARIABLES

Experiment Random Variable (x) Contact five customers Number of customers who place an order Inspect a shipment of 50 radios Number of defective radios Operate a restaurant for one day Number of customers Sell an automobile Gender of the customer

Possible Values for the Random Variable 0, 1, 2, 3, 4, 5 0, 1, 2, . . . , 49, 50 0, 1, 2, 3, . . . 0 if male; 1 if female

measurement scales such as time, weight, distance, and temperature can be described by continuous random variables. For example, consider an experiment of monitoring incoming telephone calls to the claims office of a major insurance company. Suppose the random variable of interest is x 5 the time between consecutive incoming calls in minutes. This random variable may assume any value in the interval x $ 0. Actually, an infinite number of values are possible for x, including values such as 1.26 minutes, 2.751 minutes, 4.3333 minutes, and so on. As another example, consider a 90-mile section of interstate highway I-75 north of Atlanta, Georgia. For an emergency ambulance service located in Atlanta, we might define the random variable as x 5 number of miles to the location of the next traffic accident along this section of I-75. In this case, x would be a continuous random variable assuming any value in the interval 0 # x # 90. Additional examples of continuous random variables are listed in Table 5.2. Note that each example describes a random variable that may assume any value in an interval of values. Continuous random variables and their probability distributions will be the topic of Chapter 6.

TABLE 5.2 EXAMPLES OF CONTINUOUS RANDOM VARIABLES

Experiment Random Variable (x) Operate a bank Time between customer arrivals in minutes Fill a soft drink can Number of ounces (max 5 12.1 ounces) Construct a new library Percentage of project complete after six months Test a new chemical process Temperature when the desired reaction takes place (min 150° F; max 212° F)

Possible Values for the Random Variable x$0 0 # x # 12.1 0 # x # 100 150 # x # 212

NOTE AND COMMENT One way to determine whether a random variable is discrete or continuous is to think of the values of the random variable as points on a line segment. Choose two points representing values of the

random variable. If the entire line segment between the two points also represents possible values for the random variable, then the random variable is continuous.

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Chapter 5 Discrete Probability Distributions

Exercises

Methods 1. C onsider the experiment of tossing a coin twice. a. List the experimental outcomes. b. Define a random variable that represents the number of heads occurring on the two tosses. c. Show what value the random variable would assume for each of the experimental outcomes. d. Is this random variable discrete or continuous? 2. C onsider the experiment of a worker assembling a product. a. Define a random variable that represents the time in minutes required to assemble the product. b. What values may the random variable assume? c. Is the random variable discrete or continuous?

Applications 3. T hree students scheduled interviews for summer employment at the Brookwood Institute. In each case the interview results in either an offer for a position or no offer. Experimental outcomes are defined in terms of the results of the three interviews. a. List the experimental outcomes. b. Define a random variable that represents the number of offers made. Is the random variable continuous? c. Show the value of the random variable for each of the experimental outcomes. 4. I n January the U.S. unemployment rate dropped to 8.3% (U.S. Department of Labor website, February 10, 2012). The Census Bureau includes nine states in the Northeast region. Assume that the random variable of interest is the number of Northeastern states with an unemployment rate in January that was less than 8.3%. What values may this random variable assume? 5. T o perform a certain type of blood analysis, lab technicians must perform two procedures. The first procedure requires either one or two separate steps, and the second procedure requires one, two, or three steps. a. List the experimental outcomes associated with performing the blood analysis. b. If the random variable of interest is the total number of steps required to do the complete analysis (both procedures), show what value the random variable will assume for each of the experimental outcomes. 6. L isted is a series of experiments and associated random variables. In each case, identify the values that the random variable can assume and state whether the random variable is discrete or continuous.

Experiment

Random Variable (x)

a. Take a 20-question examination Number of questions answered correctly b. Observe cars arriving at a tollbooth Number of cars arriving at tollbooth for 1 hour c. Audit 50 tax returns Number of returns containing errors d. Observe an employee’s work Number of nonproductive hours in an eight-hour workday e. Weigh a shipment of goods Number of pounds

5.2 Developing Discrete Probability Distributions

233

Developing Discrete Probability Distributions 5.2 The probability distribution for a random variable describes how probabilities are distributed over the values of the random variable. For a discrete random variable x, a probability function, denoted by f(x), provides the probability for each value of the random variable. As such, you might suppose that the classical, subjective, and relative frequency methods of assigning probabilities introduced in Chapter 4 would be useful in developing discrete probability distributions. They are, and in this section we show how. Application of this methodology leads to what we call tabular discrete probability distributions, that is, probability distributions that are presented in a table. The classical method of assigning probabilities to values of a random variable is applicable when the experimental outcomes generate values of the random variable that are equally likely. For instance, consider the experiment of rolling a die and observing the number on the upward face. It must be one of the numbers 1, 2, 3, 4, 5, or 6, and each of these outcomes is equally likely. Thus, if we let x 5 number obtained on one roll of a die and f(x) 5 the probability of x, the probability distribution of x is given in Table 5.3. The subjective method of assigning probabilities can also lead to a table of values of the random variable together with the associated probabilities. With the subjective method the individual developing the probability distribution uses their best judgment to assign each probability. So, unlike probability distributions developed using the classical method, different people can be expected to obtain different probability distributions. The relative frequency method of assigning probabilities to values of a random variable is applicable when reasonably large amounts of data are available. We then treat the data as if they were the population and use the relative frequency method to assign probabilities to the experimental outcomes. The use of the relative frequency method to develop discrete probability distributions leads to what is called an empirical discrete distribution. With the large amounts of data available today (e.g., scanner data, credit card data), this type of probability distribution is becoming more widely used in practice. Let us illustrate by considering the sale of automobiles at a dealership. We will use the relative frequency method to develop a probability distribution for the number of cars sold per day at DiCarlo Motors in Saratoga, New York. Over the past 300 days, DiCarlo has experienced 54 days with no automobiles sold, 117 days with one automobile sold, 72 days with two automobiles sold, 42 days with three automobiles sold, 12 days with four automobiles sold, and 3 days with five automobiles sold. Suppose we consider the experiment of observing a day of operations at DiCarlo Motors and define the random variable of interest as x 5 the number of automobiles sold during a day. Using the relative frequencies to assign probabilities to the values of the random variable x, we can develop the probability distribution for x. TABLE 5.3 Probability Distribution for Number Obtained on One Roll

of a Die

Number Obtained Probability of x x f (x) 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6

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Chapter 5 Discrete Probability Distributions

TABLE 5.4 PROBABILITY DISTRIBUTION FOR THE NUMBER OF AUTOMOBILES SOLD

DURING A DAY AT DICARLO MOTORS x f(x) 0 .18 1 .39 2 .24 3 .14 4 .04 5 .01

Total 1.00

In probability function notation, f (0) provides the probability of 0 automobiles sold, f (1) provides the probability of 1 automobile sold, and so on. Because historical data show 54 of 300 days with 0 automobiles sold, we assign the relative frequency 54/300 5 .18 to f (0), indicating that the probability of 0 automobiles being sold during a day is .18. Similarly, because 117 of 300 days had 1 automobile sold, we assign the relative frequency 117/300 5 .39 to f (1), indicating that the probability of exactly 1 automobile being sold during a day is .39. Continuing in this way for the other values of the random variable, we compute the values for f (2), f (3), f (4), and f (5) as shown in Table 5.4. A primary advantage of defining a random variable and its probability distribution is that once the probability distribution is known, it is relatively easy to determine the probability of a variety of events that may be of interest to a decision maker. For example, using the probability distribution for DiCarlo Motors as shown in Table 5.4, we see that the most probable number of automobiles sold during a day is 1 with a probability of f (1) 5 .39. In addition, the probability of selling 3 or more automobiles during a day is f (3) 1 f (4) 1 f (5) 5 .14 1 .04 1 .01 5 .19. These probabilities, plus others the decision maker may ask about, provide information that can help the decision maker understand the process of selling automobiles at DiCarlo Motors. In the development of a probability function for any discrete random variable, the following two conditions must be satisfied. These conditions are the analogs to the two basic requirements for assigning probabilities to experimental outcomes presented in Chapter 4.

Required Conditions for a Discrete Probability Function

f sx $ 0 o f sxd 5 1

(5.1) (5.2)

Table 5.4 shows that the probabilities for the random variable x satisfy equation (5.1); f(x) is greater than or equal to 0 for all values of x. In addition, because the probabilities sum to 1, equation (5.2) is satisfied. Thus, the DiCarlo Motors probability function is a valid discrete probability function. We can also show the DiCarlo Motors probability distribution graphically. In Figure 5.1 the values of the random variable x for DiCarlo Motors are shown on the horizontal axis and the probability associated with these values is shown on the vertical axis. In addition to the probability distributions shown in tables, a formula that gives the probability function, f(x), for every value of x is often used to describe probability

235

5.2 Developing Discrete Probability Distributions

FIGURE 5.1 GRAPHICAL REPRESENTATION OF THE PROBABILITY DISTRIBUTION

FOR THE NUMBER OF AUTOMOBILES SOLD DURING A DAY AT DICARLO MOTORS f(x)

.40 Probability

.30 .20 .10 .00

0 1 2 3 4 5 Number of Automobiles Sold During a Day

distributions. The simplest example of a discrete probability distribution given by a formula is the discrete uniform probability distribution. Its probability function is defined by equation (5.3).

Discrete Uniform Probability Function

f sxd 5 1/n

(5.3)

where

n 5 the number of values the random variable may assume

For example, consider again the experiment of rolling a die. We define the random variable x to be the number of dots on the upward face. For this experiment, n 5 6 values are possible for the random variable; x 5 1, 2, 3, 4, 5, 6. We showed earlier how the probability distribution for this experiment can be expressed as a table. Since the probabilities are equally likely, the discrete uniform probability function can also be used. The probability function for this discrete uniform random variable is

ƒ(x) 5 1/6 x 5 1, 2, 3, 4, 5, 6

Several widely-used discrete probability distributions are specified by formulas. Three important cases are the binomial, Poisson, and hypergeometric distributions; these distributions are discussed later in the chapter.

236

Chapter 5 Discrete Probability Distributions

Exercises

Methods 7. The probability distribution for the random variable x follows.

x f(x)

20 .20 25 .15 30 .25 35 .40

a. b. c. d.

I s this probability distribution valid? Explain. What is the probability that x 5 30? What is the probability that x is less than or equal to 25? What is the probability that x is greater than 30?

Applications 8. T he following data were collected by counting the number of operating rooms in use at Tampa General Hospital over a 20-day period: On three of the days only one operating room was used, on five of the days two were used, on eight of the days three were used, and on four days all four of the hospital’s operating rooms were used. a. Use the relative frequency approach to construct an empirical discrete probability distribution for the number of operating rooms in use on any given day. b. Draw a graph of the probability distribution. c. Show that your probability distribution satisfies the required conditions for a valid discrete probability distribution. 9. E mployee retention is a major concern for many companies. A survey of Americans asked how long they have worked for their current employer (Bureau of Labor Statistics website, December 2015). Consider the following example of sample data of 2000 college graduates who graduated five years ago.

Time with Current Employer (years)

1 2 3 4 5

Number 506 390 310 218 576

Let x be the random variable indicating the number of years the respondent has worked for her/his current employer. a. Use the data to develop an empirical discrete probability distribution for x. b. Show that your probability distribution satisfies the conditions for a valid discrete probability distribution. c. What is the probability that a respondent has been at her/his current place of employment for more than 3 years? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

237

5.2 Developing Discrete Probability Distributions

10. T he percent frequency distributions of job satisfaction scores for a sample of information systems (IS) senior executives and middle managers are as follows. The scores range from a low of 1 (very dissatisfied) to a high of 5 (very satisfied).

Job Satisfaction Score

IS Senior Executives (%)

IS Middle Managers (%)

1 5 4 2 9 10 3 3 12 4 42 46 5 41 28

a. b. c. d. e.

evelop a probability distribution for the job satisfaction score of a senior executive. D Develop a probability distribution for the job satisfaction score of a middle manager. What is the probability a senior executive will report a job satisfaction score of 4 or 5? What is the probability a middle manager is very satisfied? Compare the overall job satisfaction of senior executives and middle managers.

technician services mailing machines at companies in the Phoenix area. Depending on 11. A the type of malfunction, the service call can take 1, 2, 3, or 4 hours. The different types of malfunctions occur at about the same frequency. a. Develop a probability distribution for the duration of a service call. b. Draw a graph of the probability distribution. c. Show that your probability distribution satisfies the conditions required for a discrete probability function. d. What is the probability a service call will take 3 hours? e. A service call has just come in, but the type of malfunction is unknown. It is 3:00 p.m. and service technicians usually get off at 5:00 p.m. What is the probability the service technician will have to work overtime to fix the machine today? 12. Time Warner Cable provides television and Internet service to millions of customers. Suppose that the management of Time Warner Cable subjectively assesses a probability distribution for the number of new subscribers next year in the state of New York as follows.

x f(x) 100,000 .10 200,000 .20 300,000 .25 400,000 .30 500,000 .10 600,000 .05

a. I s this probability distribution valid? Explain. b. What is the probability Time Warner will obtain more than 400,000 new subscribers? c. What is the probability Time Warner will obtain fewer than 200,000 new subscribers? 13. A psychologist determined that the number of sessions required to obtain the trust of a new patient is either 1, 2, or 3. Let x be a random variable indicating the number of sessions required to gain the patient’s trust. The following probability function has been proposed.

x f (x) 5 for x 5 1, 2, or 3 6

238

Chapter 5 Discrete Probability Distributions

a. I s this probability function valid? Explain. b. What is the probability that it takes exactly 2 sessions to gain the patient’s trust? c. What is the probability that it takes at least 2 sessions to gain the patient’s trust? he following table is a partial probability distribution for the MRA Company’s pro14. T jected profits (x 5 profit in $1000s) for the first year of operation (the negative value denotes a loss).

x f(x)

2100 .10 0 .20 50 .30 100 .25 150 .10 200

a. W hat is the proper value for f(200)? What is your interpretation of this value? b. What is the probability that MRA will be profitable? c. What is the probability that MRA will make at least $100,000?

Expected Value and Variance 5.3 Expected Value The expected value, or mean, of a random variable is a measure of the central location for the random variable. The formula for the expected value of a discrete random variable x follows. The expected value is a weighted average of the values of the random variable where the weights are the probabilities.

The expected value does not have to be a value the random variable can assume.

Expected Value of a Discrete Random Variable

E(x) 5 5 o xf (x)

(5.4)

Both the notations E(x) and are used to denote the expected value of a random variable. Equation (5.4) shows that to compute the expected value of a discrete random variable, we must multiply each value of the random variable by the corresponding probability f (x) and then add the resulting products. Using the DiCarlo Motors automobile sales example from Section 5.2, we show the calculation of the expected value for the number of automobiles sold during a day in Table 5.5. The sum of the entries in the xf (x) column shows that the expected value is 1.50 automobiles per day. We therefore know that although sales of 0, 1, 2, 3, 4, or 5 automobiles are possible on any one day, over time DiCarlo can anticipate selling an average of 1.50 automobiles per day. Assuming 30 days of operation during a month, we can use the expected value of 1.50 to forecast average monthly sales of 30(1.50) 5 45 automobiles.

Variance The expected value provides a measure of central tendency for a random variable, but we often also want a measure of variability, or dispersion. Just as we used the variance in Chapter 3 to summarize the variability in data, we now use variance to summarize the variability in the values of a random variable. The formula for the variance of a discrete random variable follows. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

239

5.3 Expected Value and Variance

TABLE 5.5 CALCULATION OF THE EXPECTED VALUE FOR THE NUMBER

OF AUTOMOBILES SOLD DURING A DAY AT DICARLO MOTORS x f (x)

xf(x)

0 .18 0(.18) 5 .00 1 .39 1(.39) 5 .39 2 .24 2(.24) 5 .48 3 .14 3(.14) 5 .42 4 .04 4(.04) 5 .16 5 .01 5(.01) 5 .05

1.50

E(x) 5 5 oxƒ(x)

The variance is a weighted average of the squared deviations of a random variable from its mean. The weights are the probabilities.

Variance of a Discrete Random Variable

Var sxd 5 2 5 osx 2 d2f sxd

(5.5)

As equation (5.5) shows, an essential part of the variance formula is the deviation, x 2 , which measures how far a particular value of the random variable is from the expected value, or mean, . In computing the variance of a random variable, the deviations are squared and then weighted by the corresponding value of the probability function. The sum of these weighted squared deviations for all values of the random variable is referred to as the variance. The notations Var(x) and 2 are both used to denote the variance of a random variable. The calculation of the variance for the probability distribution of the number of automobiles sold during a day at DiCarlo Motors is summarized in Table 5.6. We see that the variance is 1.25. The standard deviation, , is defined as the positive square root of the variance. Thus, the standard deviation for the number of automobiles sold during a day is 5 Ï1.25 5 1.118

The standard deviation is measured in the same units as the random variable ( 5 1.118 automobiles) and therefore is often preferred in describing the variability of a random variable. The variance 2 is measured in squared units and is thus more difficult to interpret. TABLE 5.6 CALCULATION OF THE VARIANCE FOR THE NUMBER OF AUTOMOBILES

SOLD DURING A DAY AT DICARLO MOTORS x

x 2 m

(x 2 m)2

0 0 2 1.50 5 21.50 2.25 1 1 2 1.50 5 2.50 .25 2 2 2 1.50 5 .50 .25 3 3 2 1.50 5 1.50 2.25 4 4 2 1.50 5 2.50 6.25 5 5 2 1.50 5 3.50 12.25

f(x)

(x 2 m)2f(x)

.18 2.25(.18) 5 .4050 .39 .25(.39) 5 .0975 .24 .25(.24) 5 .0600 .14 2.25(.14) 5 .3150 .04 6.25(.04) 5 .2500 .01 12.25(.01) 5 .1225

2 5 o(x 2 )2ƒ(x)

1.2500

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Chapter 5 Discrete Probability Distributions

Using Excel to Compute the Expected Value, Variance, and Standard Deviation The calculations involved in computing the expected value and variance for a discrete random variable can easily be made in an Excel worksheet. One approach is to enter the formulas necessary to make the calculations in Tables 5.4 and 5.5. An easier way, however, is to make use of Excel’s SUMPRODUCT function. In this subsection we show how to use the SUMPRODUCT function to compute the expected value and variance for daily automobile sales at DiCarlo Motors. Refer to Figure 5.2 as we describe the tasks involved. The formula worksheet is in the background; the value worksheet is in the foreground. Enter/Access Data: The data needed are the values for the random variable and the corresponding probabilities. Labels, values for the random variable, and the corresponding probabilities are entered in cells A1:B7. Enter Functions and Formulas: The SUMPRODUCT function multiplies each value inone range by the corresponding value in another range and sums the products. To use the SUMPRODUCT function to compute the expected value of daily automobile sales at DiCarlo Motors, we entered the following formula into cell B9:

5SUMPRODUCT(A2:A7,B2:B7)

Note that the first range, A2:A7, contains the values for the random variable, daily automobile sales. The second range, B2:B7, contains the corresponding probabilities. Thus, the SUMPRODUCT function in cell B9 is computing A2*B2 1 A3*B3 1 A4*B4 1 A5*B5 1 A6*B6 1 A7*B7; hence, it is applying the formula in equation (5.4) to compute the expected value. The result, shown in cell B9 of the value worksheet, is 1.5. FIGURE 5.2 EXCEL WORKSHEET FOR EXPECTED VALUE, VARIANCE, AND STANDARD DEVIATION

241

5.3 Expected Value and Variance

The formulas in cells C2:C7 are used to compute the squared deviations from the expected value or mean of 1.5 (the mean is in cell B9). The results, shown in the value worksheet, are the same as the results shown in Table 5.5. The formula necessary to compute the variance for daily automobile sales was entered into cell B11. It uses the SUMPRODUCT function to multiply each value in the range C2:C7 by each corresponding value in the range B2:B7 and sums the products. The result, shown in the value worksheet, is 1.25. Because the standard deviation is the square root of the variance, we entered the formula 5SQRT(B11) into cell B13 to compute the standard deviation for daily automobile sales. The result, shown in the value worksheet, is 1.118.

Exercises

Methods 15. The following table provides a probability distribution for the random variable x. x f(x) 3 .25 6 .50 9 .25

a. C ompute E(x), the expected value of x. b. C ompute 2, the variance of x. c. Compute , the standard deviation of x. 16. The following table provides a probability distribution for the random variable y. y f ( y) 2 .20 4 .30 7 .40 8 .10

a. C ompute E(y). b. C ompute Var(y) and .

Applications

Coldstream12

17. D uring the summer of 2014, Coldstream Country Club in Cincinnati, Ohio collected data on 443 rounds of golf played from its white tees. The data for each golfer’s score on the twelfth hole are contained in the DATAfile Coldstream12. a. Construct an empirical discrete probability distribution for the player scores on the twelfth hole. b. A par is the score that a good golfer is expected to get for the hole. For hole number 12, par is four. What is the probability of a player scoring less than or equal to par on hole number 12? c. What is the expected score for hole number 12? d. What is the variance for hole number 12? e. What is the standard deviation for hole number 12? 18. T he American Housing Survey reported the following data on the number of times that owner-occupied and renter-occupied units had a water supply stoppage lasting 6 or more hours over a 3-month period.

242

Chapter 5 Discrete Probability Distributions

Number of Units (1000s) Number of Times Owner Occupied Renter Occupied 0 439 394 1 1100 760 2 249 221 3 98 92 4 times or more 120 111

a. D efine a random variable x 5 number of times that owner-occupied units had a water supply stoppage lasting 6 or more hours in the past 3 months and develop a probability distribution for the random variable. (Let x 5 4 represent 4 or more times.) b. Compute the expected value and variance for x. c. Define a random variable y 5 number of times that renter-occupied units had a water supply stoppage lasting 6 or more hours in the past 3 months and develop a probability distribution for the random variable. (Let y 5 4 represent 4 or more times.) d. Compute the expected value and variance for y. e. What observations can you make from a comparison of the number of water supply stoppages reported by owner-occupied units versus renter-occupied units? 19. W est Virginia has one of the highest divorce rates in the nation, with an annual rate of approximately 5 divorces per 1000 people (Centers for Disease Control and Prevention website, January 12, 2012). The Marital Counseling Center, Inc. (MCC) thinks that the high divorce rate in the state may require them to hire additional staff. Working with a consultant, the management of MCC has developed the following probability distribution for x 5 the number of new clients for marriage counseling for the next year.

a. b. c. d.

x ƒ(x) 10 .05 20 .10 30 .10 40 .20 50 .35 60 .20

this probability distribution valid? Explain. Is What is the probability MCC will obtain more than 30 new clients? What is the probability MCC will obtain fewer than 20 new clients? Compute the expected value and variance of x.

20. T he probability distribution for damage claims paid by the Newton Automobile Insurance Company on collision insurance follows.

Payment ($) Probability 0 .85 500 .04 1000 .04 3000 .03 5000 .02 8000 .01 10000 .01 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

243

5.3 Expected Value and Variance

a. U se the expected collision payment to determine the collision insurance premium that would enable the company to break even. b. The insurance company charges an annual rate of $520 for the collision coverage. What is the expected value of the collision policy for a policyholder? (Hint: It is the expected payments from the company minus the cost of coverage.) Why does the policyholder purchase a collision policy with this expected value? 21. T he following probability distributions of job satisfaction scores for a sample of information systems (IS) senior executives and middle managers range from a low of 1 (very dissatisfied) to a high of 5 (very satisfied). Probability Job Satisfaction IS Senior IS Middle Score Executives Managers 1 .05 .04 2 .09 .10 3 .03 .12 4 .42 .46 5 .41 .28

a. b. c. d. e.

hat is the expected value of the job satisfaction score for senior executives? W What is the expected value of the job satisfaction score for middle managers? Compute the variance of job satisfaction scores for executives and middle managers. Compute the standard deviation of job satisfaction scores for both probability distributions. Compare the overall job satisfaction of senior executives and middle managers.

22. T he demand for a product of Carolina Industries varies greatly from month to month. The probability distribution in the following table, based on the past two years of data, shows the company’s monthly demand.

Unit Demand Probability 300 .20 400 .30 500 .35 600 .15

a. I f the company bases monthly orders on the expected value of the monthly demand, what should Carolina’s monthly order quantity be for this product? b. Assume that each unit demanded generates $70 in revenue and that each unit ordered costs $50. How much will the company gain or lose in a month if it places an order based on your answer to part (a) and the actual demand for the item is 300 units? 23. I n Gallup’s Annual Consumption Habits Poll, telephone interviews were conducted for a random sample of 1014 adults aged 18 and over. One of the questions was, “How many cups of coffee, if any, do you drink on an average day?” The following table shows the results obtained (Gallup website, August 6, 2012).

Number of Cups per Day

Number of Responses

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Chapter 5 Discrete Probability Distributions

Define a random variable x 5 number of cups of coffee consumed on an average day. Let x 5 4 represent four or more cups. a. Develop a probability distribution for x. b. Compute the expected value of x. c. Compute the variance of x. d. Suppose we are only interested in adults who drink at least one cup of coffee on an average day. For this group, let y 5 the number of cups of coffee consumed on an average day. Compute the expected value of y and compare it to the expected value of x. 24. T he J. R. Ryland Computer Company is considering a plant expansion to enable the company to begin production of a new computer product. The company’s president must determine whether to make the expansion a medium- or large-scale project. Demand for the new product is uncertain, which for planning purposes may be low demand, medium demand, or high demand. The probability estimates for demand are .20, .50, and .30, respectively. Letting x and y indicate the annual profit in thousands of dollars, the firm’s planners developed the following profit forecasts for the medium- and large-scale expansion projects. Medium-Scale Large-Scale Expansion Profit Expansion Profit

x f(x) Demand

y f ( y)

Low 50 .20 0 .20 Medium 150 .50 100 .50 High 200 .30 300 .30

a. C ompute the expected value for the profit associated with the two expansion alternatives. Which decision is preferred for the objective of maximizing the expected profit? b. Compute the variance for the profit associated with the two expansion alternatives. Which decision is preferred for the objective of minimizing the risk or uncertainty?

Bivariate Distributions, Covariance, 5.4 and Financial Portfolios A probability distribution involving two random variables is called a bivariate probability distribution. In discussing bivariate probability distributions, it is useful to think of a bivariate experiment. Each outcome for a bivariate experiment consists of two values, one for each random variable. For example, consider the bivariate experiment of rolling a pair of dice. The outcome consists of two values, the number obtained with the first die and the number obtained with the second die. As another example, consider the experiment of observing the financial markets for a year and recording the percentage gain for a stock fund and a bond fund. Again, the experimental outcome provides a value for two random variables, the percent gain in the stock fund and the percent gain in the bond fund. When dealing with bivariate probability distributions, we are often interested in the relationship between the random variables. In this section, we introduce bivariate distributions and show how the covariance and correlation coefficient can be used as a measure of linear association between the random variables. We shall also see how bivariate probability distributions can be used to construct and analyze financial portfolios.

A Bivariate Empirical Discrete Probability Distribution Recall that in Section 5.2 we developed an empirical discrete distribution for daily sales at the DiCarlo Motors automobile dealership in Saratoga, New York. DiCarlo has another Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

5.4 Bivariate Distributions, Covariance, and Financial Portfolios

245

TABLE 5.7 Number of Automobiles Sold at DiCarlo’s Saratoga and

Geneva Dealerships Over 300 Days Saratoga Dealership Geneva Dealership 0 1 2 3 4 5 Total 0 1 2 3 Total

21 30 24 9 2 0 86 21 36 33 18 2 1 111 9 42 9 12 3 2 77 3 9 6 3 5 0 26 54 117 72 42 12 3 300

dealership in Geneva, New York. Table 5.7 shows the number of cars sold at each of the dealerships over a 300-day period. The numbers in the bottom row (Labeled Total) are the frequencies we used to develop an empirical probability distribution for daily sales at DiCarlo’s Saratoga dealership in Section 5.2. The numbers in the rightmost column (Labeled Total) are the frequencies of daily sales for the Geneva dealership. Entries in the body of the table give the number of days the Geneva dealership had a level of sales indicated by the row, when the Saratoga dealership had the level of sales indicated by the column. For example, the entry of 33 in the Geneva dealership row labeled 1 and the Saratoga column labeled 2 indicates that for 33 days out of the 300, the Geneva dealership sold 1 car and the Saratoga dealership sold 2 cars. Suppose we consider the bivariate experiment of observing a day of operations at DiCarlo Motors and recording the number of cars sold. Let us define x 5 number of cars sold at the Geneva dealership and y 5 the number of cars sold at the Saratoga dealership. We can now divide all of the frequencies in Table 5.7 by the number of observations (300) to develop a bivariate empirical discrete probability distribution for automobile sales at the two DiCarlo dealerships. Table 5.8 shows this bivariate discrete probability distribution. The probabilities in the lower margin provide the marginal distribution for the DiCarlo Motors Saratoga dealership. The probabilities in the right margin provide the marginal distribution for the DiCarlo Motors Geneva dealership. The probabilities in the body of the table provide the bivariate probability distribution for sales at both dealerships. Bivariate probabilities are often called joint probabilities. We see that the joint probability of selling 0 automobiles at Geneva and 1 automobile at Saratoga on a typical day is f (0, 1) 5 .1000, the joint probability of selling 1 automobile at Geneva and 4 automobiles at Saratoga on a typical day is .0067, and so on. Note that there is one bivariate probability for each experimental outcome. With TABLE 5.8 Bivariate Empirical Discrete Probability Distribution for

Daily Sales at DiCarlo Dealerships in Saratoga and Geneva, New York Saratoga Dealership Geneva Dealership 0 1 2 3 4 5 Total 0 1 2 3 Total

.0700 .1000 .0800 .0300 .0067 .0000 .2867 .0700 .1200 .1100 .0600 .0067 .0033 .3700 .0300 .1400 .0300 .0400 .0100 .0067 .2567 .0100 .0300 .0200 .0100 .0167 .0000 .0867 .18 .39 .24 .14 .04 .01 1.0000

246

Chapter 5 Discrete Probability Distributions

4 possible values for x and 6 possible values for y, there are 24 experimental outcomes and bivariate probabilities. Suppose we would like to know the probability distribution for total sales at both DiCarlo dealerships and the expected value and variance of total sales. We can define s 5 x 1 y as total sales for DiCarlo Motors. Working with the bivariate probabilities in Table 5.8, we see that f(s 5 0) 5 .0700, f(s 5 1) 5 .0700 1 .1000 5 .1700, f(s 5 2) 5 .0300 1 .1200 1 .0800 5 .2300, and so on. We show the complete probability distribution for s 5 x 1 y along with the computation of the expected value and variance in Table 5.9. The expected value is E(s) 5 2.6433 and the variance is Var(s) 5 2.3895. With bivariate probability distributions, we often want to know the relationship between the two random variables. The covariance and/or correlation coefficient are good measures of association between two random variables. We saw in Chapter 3 how to compute the covariance and correlation coefficient for sample data. The formula we will use for computing the covariance between two random variables x and y is given below. Covariance of Random Variables x and y1

xy 5 [Var (x 1 y) 2 Var (x) 2 Var (y)]y2

(5.6)

We have already computed Var (s) 5 Var(x 1 y) and, in Section 5.2, we computed Var(y). Now we need to compute Var(x) before we can use equation (5.6) to compute the covariance of x and y. Using the probability distribution for x (the right margin of Table 5.8), we compute E(x) and Var(x) in Table 5.10. We can now use equation (5.6) to compute the covariance of the random variables x and y. xy 5 [Varsx 1 yd 2 Varsxd 2 Varsyd]y2 5 s2.3895 2 .8696 2 1.25dy2 5 .1350 A covariance of .1350 indicates that daily sales at DiCarlo’s two dealerships have a positive relationship. To get a better sense of the strength of the relationship we can compute the TABLE 5.9 Calculation of the Expected Value and Variance for Total

Daily Sales at DiCarlo Motors

f(s)

0 1 2 3 4 5 6 7 8

sf(s)

s 2 E(s) (s 2 E(s))2

.0700 .0000 22.6433 6.9872 .1700 .1700 21.6433 2.7005 .2300 .4600 20.6433 0.4139 .2900 .8700 0.3567 0.1272 .1267 .5067 1.3567 1.8405 .0667 .3333 2.3567 5.5539 .0233 .1400 3.3567 11.2672 .0233 .1633 4.3567 18.9805 .0000 .0000 5.3567 28.6939 E(s) 5 2.6433

(s 2 E(s))2 f(s) .4891 .4591 .0952 .0369 .2331 .3703 .2629 .4429 .0000 Var(s) 5 2.3895

Another formula is often used to compute the covariance of x and y when Var(x 1 y) is not known. It is fxi 2 E(xi)][yi 2 Esyid] f(xi, yj). xy 5

o i,j

247

5.4 Bivariate Distributions, Covariance, and Financial Portfolios

TABLE 5.10 Calculation of the Expected Value and Variance of Daily

Automobile Sales at DiCarlo Motors’ Geneva Dealership x f(x) xf(x) x 2 E(x) [(x 2 E(x)]2 0 .2867 .0000 21.1435 1.3076 1 .3700 .3700 2.1435 0.0206 .8565 0.8565 2 .2567 .5134 1.8565 3 .0867 .2601 1.8565

E(x) 5 1.1435

[x 2 E(x)]2 f(x) .3749 .0076 .1883 .2988 Var(x) 5 .8696

correlation coefficient. The correlation coefficient for the two random variables x and y is given by equation (5.7). Correlation between Random Variables x and y

xy 5

xy xy

(5.7)

From equation (5.7), we see that the correlation coefficient for two random variables is the covariance divided by the product of the standard deviations for the two random variables. Let us compute the correlation coefficient between daily sales at the two DiCarlo dealerships. First we compute the standard deviations for sales at the Saratoga and Geneva dealerships by taking the square root of the variance.

x 5 Ï.8696 5 .9325

y 5 Ï1.25 5 1.1180

Now we can compute the correlation coefficient as a measure of the linear association between the two random variables.

xy 5

xy xy

.1350 5 .1295 s.9325ds1.1180d

In Chapter 3 we defined the correlation coefficient as a measure of the linear association between two variables. Values near 11 indicate a strong positive linear relationship; values near 21 indicate a strong negative linear relationship; and values near zero indicate a lack of a linear relationship. This interpretation is also valid for random variables. The correlation coefficient of .1295 indicates there is a weak positive relationship between the random variables representing daily sales at the two DiCarlo dealerships. If the correlation coefficient had equaled zero, we would have concluded that daily sales at the two dealerships were independent.

Financial Applications Let us now see how what we have learned can be useful in constructing financial portfolios that provide a good balance of risk and return. A financial advisor is considering four possible economic scenarios for the coming year and has developed a probability distribution showing the percent return, x, for investing in a large-cap stock fund and the percent return, y, for investing in a long-term government bond fund given each of the scenarios. The bivariate Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Chapter 5 Discrete Probability Distributions

TABLE 5.11 Probability Distribution of Percent Returns for Investing

in a Large-Cap Stock fund, x, and Investing in a Long-Term Government Bond Fund, y Economic Scenario

Probability f(x, y)

Large-Cap Stock Fund (x)

Long-Term Government Bond Fund (y)

Recession .10 240 30 Weak growth .25 5 5 Stable growth .50 15 4 Strong growth .15 30 2

probability distribution for x and y is shown in Table 5.11. Table 5.11 is simply a list with a separate row for each experimental outcome (economic scenario). Each row contains the joint probability for the experimental outcome and a value for each random variable. Since there are only 4 joint probabilities, the tabular form used in Table 5.11 is simpler than the one we used for DiCarlo Motors where there were (4)(6) 5 24 joint probabilities. Using the formula in Section 5.3 for computing the expected value of a single random variable, we can compute the expected percent return for investing in the stock fund, E(x), and the expected percent return for investing in the bond fund, E(y).

esxd 5 .10s240d 1 .25s5d 1 .5s15d 1 .15s30d 5 9.25 esyd 5 .10s30d 1 .25s5d 1 .5s4d 1 .15s2d 5 6.55

Using this information, we might conclude that investing in the stock fund is a better investment. It has a higher expected return, 9.25%. But financial analysts recommend that investors also consider the risk associated with an investment. The standard deviation of percent return is often used as a measure of risk. To compute the standard deviation, we must first compute the variance. Using the formula in Section 5.3 for computing the variance of a single random variable, we can compute the variance of the percent returns for the stock and bond fund investments. Var(x) 5 .1(2 40 2 9.25)2 1 .25(5 2 9.25)2 1 .50(15 2 9.25)2 1 .15(30 2 9.25)2 5 328.1875 Varsyd 5 .1s30 2 6.55d2 1 .25s5 2 6.55d2 1 .50s4 2 6.55d2 1 .15s2 2 6.55d2 5 61.9475 The standard deviation of the return from an investment in the stock fund is x 5 Ï328.1875 5 18.1159% and the standard deviation of the return from an investment in the bond fund is y 5 Ï61.9475 5 7.8707%. So, we can conclude that investing in the bond fund is less risky. It has the smaller standard deviation. We have already seen that the stock fund offers a greater expected return, so if we want to choose between investing in either the stock fund or the bond fund it depends on our attitude toward risk and return. An aggressive investor might choose the stock fund because of the higher expected return; a conservative investor might choose the bond fund because of the lower risk. But there are other options. What about the possibility of investing in a portfolio consisting of both an investment in the stock fund and an investment in the bond fund? Suppose we would like to consider three alternatives: investing solely in the largecap stock fund, investing solely in the long-term government bond fund, and splitting our funds equally between the stock fund and the bond fund (one-half in each). We have already computed the expected value and standard deviation for investing solely in the stock fund and the bond fund. Let us now evaluate the third alternative: constructing a Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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5.4 Bivariate Distributions, Covariance, and Financial Portfolios

portfolio by investing equal amounts in the large-cap stock fund and in the long-term government bond fund. To evaluate this portfolio, we start by computing its expected return. We have previously defined x as the percent return from an investment in the stock fund and y as the percent return from an investment in the bond fund so the percent return for our portfolio is r 5 .5x 1 .5y. To find the expected return for a portfolio with one-half invested in the stock fund and one-half invested in the bond fund, we want to compute E(r) 5 E(.5x 1 .5y). The expression .5x 1 .5y is called a linear combination of the random variables x and y. Equation (5.8) provides an easy method for computing the expected value of a linear combination of the random variables x and y when we already know E(x) and E(y). In equation (5.8), a represents the coefficient of x and b represents the coefficient of y in the linear combination. Expected Value of a Linear Combination of Random Variables x and y

e(ax 1 by) 5 ae(x) 1 be(y)

(5.8)

Since we have already computed E(x) 5 9.25 and E(y) 5 6.55, we can use equation (5.8) to compute the expected value of our portfolio.

es.5x 1 .5yd 5 .5esxd 1 .5esyd 5 .5s9.25d 1 .5s6.55d 5 7.9

We see that the expected return for investing in the portfolio is 7.9%. With $100 invested, we would expect a return of $100(.079) 5 $7.90; with $1000 invested we would expect a return of $1000(.079) 5 $79.00; and so on. But what about the risk? As mentioned previously, financial analysts often use the standard deviation as a measure of risk. Our portfolio is a linear combination of two random variables, so we need to be able to compute the variance and standard deviation of a linear combination of two random variables in order to assess the portfolio risk. When the covariance between two random variables is known, the formula given by equation (5.9) can be used to compute the variance of a linear combination of two random variables. Variance of a Linear Combination of Two Random Variables

Var(ax 1 by) 5 a2Var(x) 1 b2Var(y) 1 2abxy

(5.9)

where xy is the covariance of x and y

We computed Var(x 1 y) 5 119.46 the same way we did for DiCarlo Motors in the previous subsection.

From equation (5.9), we see that both the variance of each random variable individually and the covariance between the random variables are needed to compute the variance of a linear combination of two random variables and hence the variance of our portfolio. We have already computed the variance of each random variable individually: Var(x) 5 328.1875 and Var(y) 5 61.9475. Also, it can be shown that Var(x 1 y) 5 119.46. So, using equation (5.6), the covariance of the random variables x and y is xy 5 [Varsx 1 yd2 Varsxd 2Varsyd]y2 5 [119.462328.1875261.9475]y2 52135.3375 A negative covariance between x and y, such as this, means that when x tends to be above its mean, y tends to be below its mean and vice versa.

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Chapter 5 Discrete Probability Distributions

We can now use equation (5.9) to compute the variance of return for our portfolio.

Vars.5x 1 .5yd 5 .52s328.1875d 1 .52s61.9475d 1 2s.5ds.5ds2135.3375d 5 29.865

The standard deviation of our portfolio is then given by .5x 1 .5y 5 Ï29.865 5 5.4650%. This is our measure of risk for the portfolio consisting of investing 50% in the stock fund and 50% in the bond fund. Perhaps we would now like to compare the three investment alternatives: investing solely in the stock fund, investing solely in the bond fund, or creating a portfolio by dividing our investment amount equally between the stock and bond funds. Table 5.12 shows the expected returns, variances, and standard deviations for each of the three alternatives. Which of these alternatives would you prefer? The expected return is highest for investing 100% in the stock fund, but the risk is also highest. The standard deviation is 18.1159%. Investing 100% in the bond fund has a lower expected return, but a significantly smaller risk. Investing 50% in the stock fund and 50% in the bond fund (the portfolio) has an expected return that is halfway between that of the stock fund alone and the bond fund alone. But note that it has less risk than investing 100% in either of the individual funds. Indeed, it has both a higher return and less risk (smaller standard deviation) than investing solely in the bond fund. So we would say that investing in the portfolio dominates the choice of investing solely in the bond fund. Whether you would choose to invest in the stock fund or the portfolio depends on your attitude toward risk. The stock fund has a higher expected return. But the portfolio has significantly less risk and also provides a fairly good return. Many would choose it. It is the negative covariance between the stock and bond funds that has caused the portfolio risk to be so much smaller than the risk of investing solely in either of the individual funds. The portfolio analysis we just performed was for investing 50% in the stock fund and the other 50% in the bond fund. How would you calculate the expected return and the variance for other portfolios? Equations (5.8) and (5.9) can be used to make these calculations easily. Suppose we wish to create a portfolio by investing 25% in the stock fund and 75% in the bond fund. What are the expected value and variance of this portfolio? The percent return for this portfolio is r 5.25x 1 .75y, so we can use equation (5.8) to get the expected value of this portfolio:

es.25x 1 .75yd 5 .25esxd 1 .75esyd 5 .25s9.25d 1 .75s6.55d 5 7.225

TABLE 5.12 Expected Values, Variances, and Standard Deviations

for Three Investment Alternatives Investment Alternative 100% in Stock Fund 100% in Bond Fund Portfolio (50% in Stock fund, 50% in Bond fund)

Expected Return (%) 9.25 6.55 7.90

Variance of Return

Standard Deviation of Return (%)

328.1875 61.9475 29.865

18.1159 7.8707 5.4650

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5.4 Bivariate Distributions, Covariance, and Financial Portfolios

Likewise, we may calculate the variance of the portfolio using equation (5.9): Vars.25x 1 .75yd 5 s.25d2Varsxd 1 s.75d2 Varsyd 1 2s.25ds.75dxy 5 .0625s328.1875d 1 s.5625ds61.9475d 1 s.375ds2135.3375d 5 4.6056 The standard deviation of the new portfolio is .25x 1 .75y 5 Ï4.6056 5 2.1461.

Summary We have introduced bivariate discrete probability distributions in this section. Since such distributions involve two random variables, we are often interested in a measure of association between the variables. The covariance and the correlation coefficient are the two measures we introduced and showed how to compute. A correlation coefficient near 1 or 21 indicates a strong correlation between the two random variables, and a correlation coefficient near zero indicates a weak correlation between the variables. If two random variables are independent, the covariance and the correlation coefficient will equal zero. We also showed how to compute the expected value and variance of linear combinations of random variables. From a statistical point of view, financial portfolios are linear combinations of random variables. They are actually a special kind of linear combination called a weighted average. The coefficients are nonnegative and add to 1. The portfolio example we presented showed how to compute the expected value and variance for a portfolio consisting of an investment in a stock fund and a bond fund. The same methodology can be used to compute the expected value and variance of a portfolio consisting of any two financial assets. It is the effect of covariance between the individual random variables on the variance of the portfolio that is the basis for much of the theory of reducing portfolio risk by diversifying across investment alternatives. NOTES AND COMMENTS 1. Equations (5.8) and (5.9), along with their extensions to three or more random variables, are key building blocks in financial portfolio construction and analysis. 2. Equations (5.8) and (5.9) for computing the expected value and variance of a linear combination of two random variables can be extended to three or more random variables. The extension of equation (5.8) is straightforward; one more term is

added for each additional random variable. The extension of equation (5.9) is more complicated because a separate term is needed for the covariance between all pairs of random variables. We leave these extensions to more advanced books. 3. The covariance term of equation (5.9) shows why negatively correlated random variables (investment alternatives) reduce the variance and, hence, the risk of a portfolio.

Exercises

Methods 25. Given below is a bivariate distribution for the random variables x and y.

f(x, y)

.2 .5 .3

50 30 40

80 50 60

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Chapter 5 Discrete Probability Distributions

a. b. c. d.

Compute the expected value and the variance for x and y. Develop a probability distribution for x 1 y. Using the result of part (b), compute E(x 1 y) and Var(x 1 y). Compute the covariance and correlation for x and y. Are x and y positively related, negatively related, or unrelated? e. Is the variance of the sum of x and y bigger than, smaller than, or the same as the sum of the individual variances? Why?

26. A person is interested in constructing a portfolio. Two stocks are being considered. Let x 5 percent return for an investment in stock 1, and y 5 percent return for an investment in stock 2. The expected return and variance for stock 1 are E(x) 5 8.45% and Var(x) 5 25. The expected return and variance for stock 2 are E(y) 5 3.20% and Var(y) 5 1. The covariance between the returns is xy 5 23. a. What is the standard deviation for an investment in stock 1 and for an investment in stock 2? Using the standard deviation as a measure of risk, which of these stocks is the riskier investment? b. What is the expected return and standard deviation, in dollars, for a person who invests $500 in stock 1? c. What is the expected percent return and standard deviation for a person who constructs a portfolio by investing 50% in each stock? d. What is the expected percent return and standard deviation for a person who constructs a portfolio by investing 70% in stock 1 and 30% in stock 2? e. Compute the correlation coefficient for x and y and comment on the relationship between the returns for the two stocks. 27. T he Chamber of Commerce in a Canadian city has conducted an evaluation of 300 restaurants in its metropolitan area. Each restaurant received a rating on a 3-point scale on typical meal price (1 least expensive to 3 most expensive) and quality (1 lowest quality to 3 greatest quality). A crosstabulation of the rating data is shown. Forty-two of the restaurants received a rating of 1 on quality and 1 on meal price, 39 of the restaurants received a rating of 1 on quality and 2 on meal price, and so on. Forty-eight of the restaurants received the highest rating of 3 on both quality and meal price.

Meal Price (y) Quality (x) 1 2 3 Total

1 42 39 3 84 2 33 63 54 150 3 3 15 48 66 Total

78 117 105 300

a. D evelop a bivariate probability distribution for quality and meal price of a randomly selected restaurant in this Canadian city. Let x 5 quality rating and y 5 meal price. b. Compute the expected value and variance for quality rating, x. c. Compute the expected value and variance for meal price, y. d. The Var(x 1 y) 5 1.6691. Compute the covariance of x and y. What can you say about the relationship between quality and meal price? Is this what you would expect? e. Compute the correlation coefficient between quality and meal price? What is the strength of the relationship? Do you suppose it is likely to find a low-cost restaurant in this city that is also high quality? Why or why not? 28. P ortaCom has developed a design for a high-quality portable printer. The two key components of manufacturing cost are direct labor and parts. During a testing period, the Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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253

company has developed prototypes and conducted extensive product tests with the new printer. PortaCom’s engineers have developed the following bivariate probability distribution for the manufacturing costs. Parts cost (in dollars) per printer is represented by the random variable x and direct labor cost (in dollars) per printer is represented by the random variable y. Management would like to use this probability distribution to estimate manufacturing costs.

Direct Labor (y) Parts (x) 43 45 48 Total

85 95

0.05 0.2 0.2 0.45 0.25 0.2 0.1 0.55

Total 0.30 0.4 0.3 1.00

a. S how the marginal distribution of direct labor cost and compute its expected value, variance, and standard deviation. b. Show the marginal distribution of parts cost and compute its expected value, variance, and standard deviation. c. Total manufacturing cost per unit is the sum of direct labor cost and parts cost. Show the probability distribution for total manufacturing cost per unit. d. Compute the expected value, variance, and standard deviation of total manufacturing cost per unit. e. Are direct labor and parts costs independent? Why or why not? If you conclude that they are not, what is the relationship between direct labor and parts cost? f. PortaCom produced 1500 printers for its product introduction. The total manufacturing cost was $198,350. Is that about what you would expect? If it is higher or lower, what do you think may be the reason? 29. J .P. Morgan Asset Management publishes information about financial investments. Over a 10-year period, the expected return for the S&P 500 was 5.04% with a standard deviation of 19.45% and the expected return over that same period for a core bonds fund was 5.78% with a standard deviation of 2.13% (J.P. Morgan Asset Management, Guide to the Markets, 1st Quarter, 2012). The publication also reported that the correlation between the S&P 500 and core bonds is 2.32. You are considering portfolio investments that are composed of an S&P 500 index fund and a core bonds fund. a. Using the information provided, determine the covariance between the S&P 500 and core bonds. b. Construct a portfolio that is 50% invested in an S&P 500 index fund and 50% in a core bonds fund. In percentage terms, what are the expected return and standard deviation for such a portfolio? c. Construct a portfolio that is 20% invested in an S&P 500 index fund and 80% invested in a core bonds fund. In percentage terms, what are the expected return and standard deviation for such a portfolio? d. Construct a portfolio that is 80% invested in an S&P 500 index fund and 20% invested in a core bonds fund. In percentage terms, what are the expected return and standard deviation for such a portfolio? e. Which of the portfolios in parts (b), (c), and (d) has the largest expected return? Which has the smallest standard deviation? Which of these portfolios is the best investment alternative? f. Discuss the advantages and disadvantages of investing in the three portfolios in parts (b), (c), and (d). Would you prefer investing all your money in the S&P 500 index, the core bonds fund, or one of the three portfolios? Why? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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30. I n addition to the information in exercise 29 on the S&P 500 and core bonds, J.P. Morgan Asset Management reported that the expected return for real estate investment trusts (REITs) was 13.07% with a standard deviation of 23.17% (J.P. Morgan Asset Management, Guide to the Markets, 1st Quarter, 2012). The correlation between the S&P 500 and REITs is .74 and the correlation between core bonds and REITs is 2.04. You are considering portfolio investments that are composed of an S&P 500 index fund and REITs as well as portfolio investments composed of a core bonds fund and REITs. a. Using the information provided here and in exercise 29, determine the covariance between the S&P 500 and REITs and between core bonds and REITs. b. Construct a portfolio that is 50% invested in an S&P 500 fund and 50% invested in REITs. In percentage terms, what are the expected return and standard deviation for such a portfolio? c. Construct a portfolio that is 50% invested in a core bonds fund and 50% invested in REITs. In percentage terms, what are the expected return and standard deviation for such a portfolio? d. Construct a portfolio that is 80% invested in a core bonds fund and 20% invested in REITs. In percentage terms, what are the expected return and standard deviation for such a portfolio? e. Which of the portfolios in parts (b), (c), and (d) would you recommend to an aggressive investor? Which would you recommend to a conservative investor? Why?

Binomial Probability Distribution 5.5 The binomial probability distribution is a discrete probability distribution that has many applications. It is associated with a multiple-step experiment that we call the binomial experiment.

A Binomial Experiment A binomial experiment exhibits the following four properties.

Properties of a Binomial Experiment 1. The experiment consists of a sequence of n identical trials. 2. Two outcomes are possible on each trial. We refer to one outcome as a success

and the other outcome as a failure. 3. The probability of a success, denoted by p, does not change from trial to trial.

Consequently, the probability of a failure, denoted by 1 2 p, does not change from trial to trial. 4. The trials are independent.

Jakob Bernoulli (1654–1705), the first of the Bernoulli family of Swiss mathematicians, published a treatise on probability that contained the theory ofpermutations and combinations, as well as the binomial theorem.

If properties 2, 3, and 4 are present, we say the trials are generated by a Bernoulli process. If, in addition, property 1 is present, we say we have a binomial experiment. Figure 5.3 depicts one possible sequence of successes and failures for a binomial experiment involving eight trials. In a binomial experiment, our interest is in the number of successes occurring in the n trials. If we let x denote the number of successes occurring in the n trials, we see that x can assume the values of 0, 1, 2, 3, . . . , n. Because the number of values is finite, x is a discrete random variable. The probability distribution associated with this random variable is called the binomial probability distribution. For example, consider the experiment of tossing a

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5.5 Binomial Probability Distribution

FIGURE 5.3 ONE POSSIBLE SEQUENCE OF SUCCESSES AND FAILURES

FOR AN EIGHT-TRIAL BINOMIAL EXPERIMENT Property 1:

The experiment consists of n 5 8 identical trials.

Property 2:

Each trial results in either success (S) or failure (F).

Trials

Outcomes

coin five times and on each toss observing whether the coin lands with a head or a tail on its upward face. Suppose we want to count the number of heads appearing over the five tosses. Does this experiment show the properties of a binomial experiment? What is the random variable of interest? Note that 1. The experiment consists of five identical trials; each trial involves the tossing of one coin. 2. Two outcomes are possible for each trial: a head or a tail. We can designate head a success and tail a failure. 3. The probability of a head and the probability of a tail are the same for each trial, with p 5 .5 and 1 2 p 5 .5. 4. The trials or tosses are independent because the outcome on any one trial is not affected by what happens on other trials or tosses. Thus, the properties of a binomial experiment are satisfied. The random variable of interest is x 5 the number of heads appearing in the five trials. In this case, x can assume the values of 0, 1, 2, 3, 4, or 5. As another example, consider an insurance salesperson who visits 10 randomly selected families. The outcome associated with each visit is classified as a success if the family purchases an insurance policy and a failure if the family does not. From past experience, the salesperson knows the probability that a randomly selected family will purchase an insurance policy is .10. Checking the properties of a binomial experiment, we observe that 1. The experiment consists of 10 identical trials; each trial involves contacting one family. 2. Two outcomes are possible on each trial: the family purchases a policy (success) or the family does not purchase a policy (failure). 3. The probabilities of a purchase and a nonpurchase are assumed to be the same for each sales call, with p 5 .10 and 1 2 p 5 .90. 4. The trials are independent because the families are randomly selected. Because the four assumptions are satisfied, this example is a binomial experiment. The random variable of interest is the number of sales obtained in contacting the 10 families. In this case, x can assume the values of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. Property 3 of the binomial experiment is called the stationarity assumption and is sometimes confused with property 4, independence of trials. To see how they differ, consider again the case of the salesperson calling on families to sell insurance policies. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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If, as the day wore on, the salesperson got tired and lost enthusiasm, the probability of success (selling a policy) might drop to .05, for example, by the tenth call. In such a case, property 3 (stationarity) would not be satisfied, and we would not have a binomial experiment. Even if property 4 held—that is, the purchase decisions of each family were made independently—it would not be a binomial experiment if property 3 was not satisfied. In applications involving binomial experiments, a special mathematical formula, called the binomial probability function, can be used to compute the probability of x successes inthe n trials. Using probability concepts introduced in Chapter 4, we will show in the context of an illustrative problem how the formula can be developed.

Martin Clothing Store Problem Let us consider the purchase decisions of the next three customers who enter the Martin Clothing Store. On the basis of past experience, the store manager estimates the probability that any one customer will make a purchase is .30. What is the probability that two of the next three customers will make a purchase? Using a tree diagram (Figure 5.4), we can see that the experiment of observing the three customers each making a purchase decision has eight possible outcomes. Using S to denote success (a purchase) and F to denote failure (no purchase), we are interested in experimental outcomes involving two successes in the three trials (purchase decisions). Next, let us verify that the experiment involving the sequence of three purchase decisions

FIGURE 5.4 TREE DIAGRAM FOR THE MARTIN CLOTHING STORE PROBLEM First Customer

Second Customer

Third Customer

Experimental Outcome

Value of x

(S, S, S)

(S, S, F)

(S, F, S)

(S, F, F)

(F, S, S)

(F, S, F)

(F, F, S)

(F, F, F)

S 5 Purchase F 5 No purchase x 5 Number of customers making a purchase

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5.5 Binomial Probability Distribution

can be viewed as a binomial experiment. Checking the four requirements for a binomial experiment, we note that 1. The experiment can be described as a sequence of three identical trials, one trial for each of the three customers who will enter the store. 2. Two outcomes—the customer makes a purchase (success) or the customer does not make a purchase (failure)—are possible for each trial. 3. The probability that the customer will make a purchase (.30) or will not make a purchase (.70) is assumed to be the same for all customers. 4. The purchase decision of each customer is independent of the decisions of the other customers. Hence, the properties of a binomial experiment are present. The number of experimental outcomes resulting in exactly x successes in n trials can be computed using the following formula.2

Number of Experimental Outcomes Providing Exactly x Successes in n Trials

n n! 5 x x!(n 2 x)!

(5.10)

where n! 5 n(n 2 1)(n 2 2) Á (2)(1)

and, by definition,

0! 5 1

Now let us return to the Martin Clothing Store experiment involving three customer purchase decisions. Equation (5.10) can be used to determine the number of experimental outcomes involving two purchases; that is, the number of ways of obtaining x 5 2 successes in the n 5 3 trials. From equation (5.10) we have

SD SD

s3ds2ds1d 6 n 3 3! 5 5 5 5 53 x 2 2!s3 2 2d! s2ds1ds1d 2

Equation (5.10) shows that three of the experimental outcomes yield two successes. From Figure 5.3 we see that these three outcomes are denoted by (S, S, F ), (S, F, S), and (F, S, S). Using equation (5.10) to determine how many experimental outcomes have three successes (purchases) in the three trials, we obtain

SD SD

s3ds2ds1d n 3 3! 3! 6 5 5 5 5 5 51 x 3 3!s3 2 3d! 3!0! 3s2ds1ds1d 6

From Figure 5.4 we see that the one experimental outcome with three successes is identified by (S, S, S).

This formula, introduced in Chapter 4, determines the number of combinations of n objects selected x at a time. For the binomial experiment, this combinatorial formula provides the number of experimental outcomes (sequences of n trials) resulting in x successes.

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We know that equation (5.10) can be used to determine the number of experimental outcomes that result in x successes in n trials. If we are to determine the probability of x successes in n trials, however, we must also know the probability associated with each of these experimental outcomes. Because the trials of a binomial experiment are independent, we can simply multiply the probabilities associated with each trial outcome to find the probability of a particular sequence of successes and failures. The probability of purchases by the first two customers and no purchase by the third customer, denoted (S, S, F), is given by

pps1 2 pd

With a .30 probability of a purchase on any one trial, the probability of a purchase on the first two trials and no purchase on the third is given by s.30ds.30ds.70d 5 s.30d2s.70d 5 .063

Two other experimental outcomes also result in two successes and one failure. The probabilities for all three experimental outcomes involving two successes follow.

1st Customer

Trial Outcomes 2nd Customer

3rd Customer

Experimental Outcome

Probability of Experimental Outcome

Purchase Purchase No purchase (S, S, F ) pp(1 2 p) 5 p2(1 2 p) 5 (.30)2(.70) 5 .063 Purchase No purchase Purchase (S, F, S) p(1 2 p)p 5 p2(1 2 p) 5 (.30)2(.70) 5 .063 No purchase Purchase Purchase (F, S, S) (1 2 p)pp 5 p2(1 2 p) 5 (.30)2(.70) 5 .063

Observe that all three experimental outcomes with two successes have exactly the same probability. This observation holds in general. In any binomial experiment, all sequences of trial outcomes yielding x successes in n trials have the same probability of occurrence. The probability of each sequence of trials yielding x successes in n trials follows. Probability of a particular sequence of trial outcomes 5 pxs1 2 pdsn2xd with x successes in n trials

(5.11)

For the Martin Clothing Store, this formula shows that any experimental outcome with two successes has a probability of p 2(1 2 p)(322) 5 p 2(1 2 p)1 5 (.30)2(.70)1 5 .063. Because equation (5.10) shows the number of outcomes in a binomial experiment with x successes and equation (5.11) gives the probability for each sequence involving x successes, we combine equations (5.10) and (5.11) to obtain the following binomial probability function. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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5.5 Binomial Probability Distribution

Binomial Probability Function

f (x) 5

where

n x p (1 2 p)(n2x) x

(5.12)

x 5 the number of successes p 5 the probability of a success on one trial n 5 the number of trials f (x) 5 the probability of x successes in n trials

n n! 5 x x!(n 2 x)!

For the binomial probability distribution, x is a discrete random variable with the probability function f(x) applicable for values of x 5 0, 1, 2, . . . , n. In the Martin Clothing Store example, let us use equation (5.12) to compute the probability that no customer makes a purchase, exactly one customer makes a purchase, exactly two customers make a purchase, and all three customers make a purchase. The calculations are summarized in Table 5.13, which gives the probability distribution of the number of customers making a purchase. Figure 5.5 is a graph of this probability distribution. The binomial probability function can be applied to any binomial experiment. If we are satisfied that a situation demonstrates the properties of a binomial experiment and if we know the values of n and p, we can use equation (5.12) to compute the probability of x successes in the n trials. If we consider variations of the Martin experiment, such as 10 customers rather than 3 entering the store, the binomial probability function given by equation (5.12) is still applicable. Suppose we have a binomial experiment with n 5 10, x 5 4, and p 5 .30. The probability of making exactly four sales to 10 customers entering the store is f s4d 5

10! s.30d4s.70d6 5 .2001 4!6!

TABLE 5.13 PROBABILITY DISTRIBUTION FOR THE NUMBER OF CUSTOMERS

MAKING A PURCHASE x f (x) 0

3! (.30)0(.70)3 5 .343 0!3!

3! (.30)1(.70)2 5 .441 1!2!

3! (.30)2(.70)1 5 .189 2!1!

3! .027 (.30)3(.70)0 5 3!0! 1.000

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Chapter 5 Discrete Probability Distributions

FIGURE 5.5 GRAPHICAL REPRESENTATION OF THE PROBABILITY DISTRIBUTION

FOR THE NUMBER OF CUSTOMERS MAKING A PURCHASE f (x)

.50

Probability

.40 .30 .20 .10 .00

1 2 3 Number of Customers Making a Purchase

Using Excel to Compute Binomial Probabilities For many probability functions that can be specified as formulas, Excel provides functions for computing probabilities and cumulative probabilities. In this section, we show how Excel’s BINOM.DIST function can be used to compute binomial probabilities and cumulative binomial probabilities. We begin by showing how to compute the binomial probabilities for the Martin Clothing Store example shown in Table 5.13. Refer to Figure 5.6 as we describe the tasks involved. The formula worksheet is in the background; the value worksheet is in the foreground. FIGURE 5.6 EXCEL WORKSHEET FOR COMPUTING BINOMIAL PROBABILITIES OF NUMBER OF

CUSTOMERS MAKING A PURCHASE

5.5 Binomial Probability Distribution

261

Enter/Access Data: In order to compute a binomial probability we must know the number of trials (n), the probability of success ( p), and the value of the random variable (x). For the Martin Clothing Store example, the number of trials is 3; this value has been entered into cell D1. The probability of success is .3; this value has been entered into cell D2. Because we want to compute the probability for x 5 0, 1, 2, and 3, these values were entered into cells B5:B8. Enter Functions and Formulas: The BINOM.DIST function has four inputs: The first is the value of x, the second is the value of n, the third is the value of p, and the fourth is FALSE or TRUE. We choose FALSE for the fourth input if a probability is desired and TRUE if a cumulative probability is desired. The formula 5BINOM.DIST(B5,$D$1,$D$2,FALSE) has been entered into cell C5 to compute the probability of 0 successes in 3 trials. Note in the value worksheet that the probability computed for f (0), .343, is the same as that shown in Table 5.13. The formula in cell C5 is copied to cells C6:C8 to compute the probabilities for x 5 1, 2, and 3 successes, respectively. We can also compute cumulative probabilities using Excel’s BINOM.DIST function. To illustrate, let us consider the case of 10 customers entering the Martin Clothing Store and compute the probabilities and cumulative probabilities for the number of customers making a purchase. Recall that the cumulative probability for x 5 1 is the probability of 1 or fewer purchases, the cumulative probability for x 5 2 is the probability of 2 or fewer purchases, and so on. So, the cumulative probability for x 5 10 is 1. Refer to Figure 5.7 as we describe the tasks involved in computing these cumulative p robabilities. The formula worksheet is in the background; the value worksheet is in the foreground. Enter/Access Data: We entered the number of trials (10) into cell D1, the probability of success (.3) into cell D2, and the values for the random variable into cells B5:B15. Enter Functions and Formulas: The binomial probabilities for each value of the random variable are computed in column C and the cumulative probabilities are computed in column D. We entered the formula 5BINOM.DIST(B5,$D$1,$D$2,FALSE) into cell C5 to compute the probability of 0 successes in 10 trials. Note that we used FALSE as the FIGURE 5.7 EXCEL WORKSHEET FOR COMPUTING PROBABILITIES AND CUMULATIVE PROBABILI-

TIES FOR NUMBER OF PURCHASES WITH 10 CUSTOMERS

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Chapter 5 Discrete Probability Distributions

fourth input in the BINOM.DIST function. The probability (.0282) is shown in cell C5 of the value worksheet. The formula in cell C5 is simply copied to cells C6:C15 to compute the remaining probabilities. To compute the cumulative probabilities we start by entering the formula 5BINOM. DIST(B5,$D$1,$D$2,TRUE) into cell D5. Note that we used TRUE as the fourth input in the B INOM.DIST function. The formula in cell D5 is then copied to cells D6:D15 to compute the remaining cumulative probabilities. In cell D5 of the value worksheet we see that the cumulative probability for x 5 0 is the same as the probability for x 5 0. Each of the remaining cumulative probabilities is the sum of the previous cumulative probability and the individual probability in column C. For instance, the cumulative probability for x 5 4 is given by .6496 1 .2001 5 .8497. Note also that the cumulative probability for x 5 10 is 1. The cumulative probability of x 5 9 is also 1 because the probability of x 5 10 is zero (to four decimal places of accuracy).

Expected Value and Variance for the Binomial Distribution In Section 5.3 we provided formulas for computing the expected value and variance of a discrete random variable. In the special case where the random variable has a binomial distribution with a known number of trials n and a known probability of success p, the general formulas for the expected value and variance can be simplified. The results follow. Expected Value and Variance for the Binomial Distribution

e(x) 5 5 np 2

Var(x) 5 5 np(1 2 p)

(5.13) (5.14)

For the Martin Clothing Store problem with three customers, we can use equation (5.13) to compute the expected number of customers who will make a purchase.

esxd 5 np 5 3s.30d 5 .9

Suppose that for the next month the Martin Clothing Store forecasts 1000 customers will enter the store. What is the expected number of customers who will make a purchase? The answer is 5 np 5 s1000ds.3d 5 300. Thus, to increase the expected number of purchases, Martin’s must induce more customers to enter the store and/or somehow increase the probability that any individual customer will make a purchase after entering. For the Martin Clothing Store problem with three customers, we see that the variance and standard deviation for the number of customers who will make a purchase are

2 5 nps1 2 pd 5 3s.3ds.7d 5 .63 5 Ï.63 5 .79

For the next 1000 customers entering the store, the variance and standard deviation for the number of customers who will make a purchase are

2 5 nps1 2 pd 5 1000s.3ds.7d 5 210 5 Ï210 5 14.49

5.5 Binomial Probability Distribution

263

Exercises

Methods 31. C onsider a binomial experiment with two trials and p 5 .4. a. Draw a tree diagram for this experiment (see Figure 5.3). b. Compute the probability of one success, f (1). c. Compute f (0). d. Compute f (2). e. Compute the probability of at least one success. f. Compute the expected value, variance, and standard deviation. 32. C onsider a binomial experiment with n 5 10 and p 5 .10. a. Compute f(0). b. Compute f (2). c. Compute P(x # 2). d. Compute P(x $ 1). e. Compute E(x). f. Compute Var(x) and . 33. C onsider a binomial experiment with n 5 20 and p 5 .70. a. Compute f(12). b. Compute f (16). c. Compute P(x $ 16). d. Compute P(x # 15). e. Compute E(x). f. Compute Var (x) and .

Applications 34. For its Music 360 survey, Nielsen Co. asked teenagers how they listened to music in the past 12 months. Nearly two-thirds of U.S. teenagers under the age of 18 say they use Google Inc.’s video-sharing site to listen to music and 35% of the teenagers said they use Pandora Media Inc.’s custom online radio service (The Wall Street Journal, August 14, 2012). Suppose 10 teenagers are selected randomly to be interviewed about how they listen to music. a. Is randomly selecting 10 teenagers and asking whether or not they use Pandora Media Inc.’s online service a binomial experiment? b. What is the probability that none of the 10 teenagers uses Pandora Media Inc.’s online radio service? c. What is the probability that 4 of the 10 teenagers use Pandora Media Inc.’s online radio service? d. What is the probability that at least 2 of the 10 teenagers use Pandora Media Inc.’s online radio service? 35. T he Center for Medicare and Medical Services reported that there were 295,000 appeals for hospitalization and other Part A Medicare service. For this group, 40% of first-round appeals were successful (The Wall Street Journal, October 22, 2012). Suppose 10 firstround appeals have just been received by a Medicare appeals office. a. Compute the probability that none of the appeals will be successful. b. Compute the probability that exactly one of the appeals will be successful. c. What is the probability that at least two of the appeals will be successful? d. What is the probability that more than half of the appeals will be successful? 36. W hen a new machine is functioning properly, only 3% of the items produced are defective. Assume that we will randomly select two parts produced on the machine and that we are interested in the number of defective parts found. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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a. b. c. d.

escribe the conditions under which this situation would be a binomial experiment. D Draw a tree diagram similar to Figure 5.4 showing this problem as a two-trial experiment. How many experimental outcomes result in exactly one defect being found? Compute the probabilities associated with finding no defects, exactly one defect, and two defects.

37. A ccording to a study by the Pew Research Center, 15% of adults in the United States do not use the Internet (Pew Research Center website, December, 15, 2014). Suppose that 10 adults in the United States are selected randomly. a. Is the selection of the 10 adults a binomial experiment? Explain. b. What is the probability that none of the adults use the Internet? c. What is the probability that 3 of the adults use the Internet? d. What is the probability that at least 1 of the adults uses the Internet? 38. M ilitary radar and missile detection systems are designed to warn a country of an enemy attack. A reliability question is whether a detection system will be able to identify an attack and issue a warning. Assume that a particular detection system has a .90 probability of detecting a missile attack. Use the binomial probability distribution to answer the following questions. a. What is the probability that a single detection system will detect an attack? b. If two detection systems are installed in the same area and operate independently, what is the probability that at least one of the systems will detect the attack? c. If three systems are installed, what is the probability that at least one of the systems will detect the attack? d. Would you recommend that multiple detection systems be used? Explain. 39. M arket-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in the summer of 2014, Google’s Chrome browser exceeded a 20% market share for the first time, with a 20.37% share of the browser market (Forbes website, December 15, 2014). For a randomly selected group of 20 Internet browser users, answer the following questions. a. Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser. b. Compute the probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser. c. For the sample of 20 Internet browser users, compute the expected number of Chrome users. d. For the sample of 20 Internet browser users, compute the variance and standard deviation for the number of Chrome users. 40. A study conducted by the Pew Research Center showed that 75% of 18- to 34-year-olds living with their parents say they contribute to household expenses. Suppose that a random sample of fifteen 18- to 34-year-olds living with their parents is selected and asked if they contribute to household expenses. a. Is the selection of the fifteen 18- to 34-year-olds living with their parents a binomial experiment? Explain. b. If the sample shows that none of the fifteen 18- to 34-year-olds living with their parents contributes to household expenses, would you question the results of the Pew Research study? Explain. c. What is the probability that at least 10 of the fifteen 18- to 34-year-olds living with their parents contribute to household expenses? 41. A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 students registered for the course. a. Compute the probability that 2 or fewer will withdraw. b. Compute the probability that exactly 4 will withdraw. c. Compute the probability that more than 3 will withdraw. d. Compute the expected number of withdrawals. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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5.6 Poisson Probability Distribution

42. A Gallup Poll showed that 30% of Americans are satisfied with the way things are going in the United States (Gallup website, September 12, 2012). Suppose a sample of 20 Americans is selected as part of a study of the state of the nation. a. Compute the probability that exactly 4 of the 20 Americans surveyed are satisfied with the way things are going in the United States. b. Compute the probability that at least 2 of the Americans surveyed are satisfied with the way things are going in the United States. c. For the sample of 20 Americans, compute the expected number of Americans who are satisfied with the way things are going in the United States. d. For the sample of 20 Americans, compute the variance and standard deviation of the number of Americans who are satisfied with the way things are going in the United States. 43. A ccording to a study conducted by the Toronto-based social media analytics firm Sysomos, 71% of all tweets get no reaction. That is, these are tweets that are not replied to or retweeted (Sysomos website, January 5, 2015). Suppose we randomly select 100 tweets. a. What is the expected number of these tweets with no reaction? b. What are the variance and standard deviation for the number of these tweets with no reaction?

Poisson Probability Distribution 5.6 The Poisson probability distribution is often used to model random arrivals in waiting line situations.

In this section, we consider a discrete random variable that is often useful in estimating the number of occurrences over a specified interval of time or space. For example, the random variable of interest might be the number of arrivals at a car wash in one hour, the number of repairs needed in 10 miles of highway, or the number of leaks in 100 miles of pipeline. If the following two properties are satisfied, the number of occurrences is a random variable described by the Poisson probability distribution.

Properties of a Poisson Experiment

1. The probability of an occurrence is the same for any two intervals of equal length. 2. The occurrence or nonoccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval.

The Poisson probability function is defined by equation (5.15).

Siméon Poisson taught mathematics at the Ecole Polytechnique in Paris from 1802 to 1808. In 1837, he published a work entitled “Researches on the Probability of Criminal and Civil Verdicts,” which includes a discussion of what later became known as the Poisson distribution.

Poisson Probability Function

f(x) 5

xe2 x!

(5.15)

where f(x) 5 the probability of x occurrences in an interval 5 expected value or mean number of occurrences in an interval e 5 2.71828

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Chapter 5 Discrete Probability Distributions

For the Poisson probability distribution, x is a discrete random variable indicating the number of occurrences in the interval. Since there is no stated upper limit for the number of occurrences, the probability function f (x) is applicable for values x 5 0, 1, 2, . . . without limit. In practical applications, x will eventually become large enough so that f (x) is approximately zero and the probability of any larger values of x becomes negligible.

An Example Involving Time Intervals

Bell Labs used the Poisson distribution to model the arrival of telephone calls.

Suppose that we are interested in the number of arrivals at the drive-up teller window of a bank during a 15-minute period on weekday mornings. If we can assume that the probability of a car arriving is the same for any two time periods of equal length and that the arrival or nonarrival of a car in any time period is independent of the arrival or nonarrival in any other time period, the Poisson probability function is applicable. Suppose these assumptions are satisfied and an analysis of historical data shows that the average number of cars arriving in a 15-minute period of time is 10; in this case, the following probability function applies. f sxd 5

10xe210 x!

The random variable here is x 5 number of cars arriving in any 15-minute period. If management wanted to know the probability of exactly five arrivals in 15 minutes, we would set x 5 5 and thus obtain

A property of the Poisson distribution is that the mean and variance are equal.

105e210 Probability of exactly 5 f s5d 5 5 .0378 5 arrivals in 15 minutes 5!

The probability of five arrivals in 15 minutes was obtained by using a calculator to evaluate the probability function. Excel also provides a function called POISSON.DIST for computing Poisson probabilities and cumulative probabilities. This function is easier to use when numerous probabilities and cumulative probabilities are desired. At the end of this section, we show how to compute these probabilities with Excel. In the preceding example, the mean of the Poisson distribution is 5 10 arrivals per 15-minute period. A property of the Poisson distribution is that the mean of the distribution and the variance of the distribution are equal. Thus, the variance for the number of arrivals during 15-minute periods is 2 5 10. The standard deviation is 5 Ï10 5 3.16. Our illustration involves a 15-minute period, but other time periods can be used. Suppose we want to compute the probability of one arrival in a 3-minute period. Because 10 is the expected number of arrivals in a 15-minute period, we see that 10/15 5 2/3 istheexpected number of arrivals in a 1-minute period and that (2/3)(3 minutes) 5 2 is the expected number of arrivals in a 3-minute period. Thus, the probability of x arrivals in a 3-minute time period with 5 2 is given by the following Poisson probability f unction. f sxd 5

2xe22 x!

The probability of one arrival in a 3-minute period is calculated as follows: 21e22 Probability of exactly 5 f s1d 5 5 .2707 1 arrival in 3 minutes 1! Earlier we computed the probability of five arrivals in a 15-minute period; it was .0378. Note that the probability of one arrival in a three-minute period (.2707) is not the same. When computing a Poisson probability for a different time interval, we must first convert the mean arrival rate to the time period of interest and then compute the probability. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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267

An Example Involving Length or Distance Intervals Let us illustrate an application not involving time intervals in which the Poisson distribution is useful. Suppose we are concerned with the occurrence of major defects in a highway one month after resurfacing. We will assume that the probability of a defect is the same for any two highway intervals of equal length and that the occurrence or nonoccurrence of a defect in any one interval is independent of the occurrence or nonoccurrence of a defect in any other interval. Hence, the Poisson distribution can be applied. Suppose we learn that major defects one month after resurfacing occur at the average rate of two per mile. Let us find the probability of no major defects in a particular 3-mile section of the highway. Because we are interested in an interval with a length of 3 miles, 5 (2 defects/mile)(3 miles) 5 6 represents the expected number of major defects over the 3-mile section of highway. Using equation (5.11), the probability of no major defects is f (0) 5 60e26/0! 5 .0025. Thus, it is unlikely that no major defects will occur in the 3-mile section. In fact, this example indicates a 1 2 .0025 5 .9975 probability of at least one major defect in the 3-mile highway section.

Using Excel to Compute Poisson Probabilities The Excel function for computing Poisson probabilities and cumulative probabilities is called POISSON.DIST. It works in much the same way as the Excel function for computing binomial probabilities. Here we show how to use it to compute Poisson probabilities and cumulative probabilities. To illustrate, we use the example introduced earlier in this section: cars arrive at a bank drive-up teller window at the mean rate of 10 per 15-minute time interval. Refer to Figure 5.8 as we describe the tasks involved.

FIGURE 5.8 EXCEL WORKSHEET FOR COMPUTING POISSON PROBABILITIES

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Chapter 5 Discrete Probability Distributions

Enter/Access Data: In order to compute a Poisson probability, we must know the mean number of occurrences sd per time period and the number of occurrences for which we want to compute the probability (x). For the drive-up teller window example, the occurrences of interest are the arrivals of cars. The mean arrival rate is 10, which has been entered into cell D1. Earlier in this section, we computed the probability of 5 arrivals. But suppose we now want to compute the probability of 0 up through 20 arrivals. To do so, we enter the values 0, 1, 2, . . . , 20 into cells A4:A24. Enter Functions and Formulas: The POISSON.DIST function has three inputs: The first is the value of x, the second is the value of , and the third is FALSE or TRUE. We choose FALSE for the third input if a probability is desired and TRUE if a cumulative probability is desired. The formula 5POISSON.DIST(A4,$D$1,FALSE) has been entered into cell B4 to compute the probability of 0 arrivals in a 15-minute period. The value worksheet in the foreground shows that the probability of 0 arrivals is 0.0000. The formula in cell B4 is copied to cells B5:B24 to compute the probabilities for 1 through 20 arrivals. Note, in cell B9 of the value worksheet, that the probability of 5 arrivals is .0378. This r esult is the same as we calculated earlier in the text. Notice how easy it was to compute all the probabilities for 0 through 20 arrivals using the POISSON.DIST function. These calculations would take quite a bit of work using a calculator. We have also used Excel’s chart tools to develop a graph of the Poisson probability distribution of arrivals. See the value worksheet in Figure 5.8. This chart gives a nice graphical presentation of the probabilities for the various number of arrival possibilities in a 15-minute interval. We can quickly see that the most likely number of arrivals is 9 or 10 and that the probabilities fall off rather smoothly for smaller and larger values. Let us now see how cumulative probabilities are generated using Excel’s POISSON .DIST function. It is really a simple extension of what we have already done. We again use the example of arrivals at a drive-up teller window. Refer to Figure 5.9 as we describe the tasks involved. Enter/Access Data: To compute cumulative Poisson probabilities we must provide the mean number of occurrences ( ) per time period and the values of x that we are interested in. The mean arrival rate (10) has been entered into cell D1. Suppose we want to compute the cumulative probabilities for a number of arrivals ranging from zero up through 20. To do so, we enter the values 0, 1, 2, . . . , 20 into cells A4:A24. Enter Functions and Formulas: Refer to the formula worksheet in the background ofFigure 5.8. The formulas we enter into cells B4:B24 of Figure 5.9 are the same as in Figure 5.8 with one exception. Instead of FALSE for the third input, we enter the word TRUE to obtain cumulative probabilities. After entering these formulas into cells B4:B24 of the worksheet in Figure 5.9, the cumulative probabilities shown were obtained. Note, in Figure 5.9, that the probability of 5 or fewer arrivals is .0671 and that the probability of 4 or fewer arrivals is .0293. Thus, the probability of exactly 5 arrivals is the difference in these two numbers: f (5) 5 .0671 2 .0293 5 .0378. We computed this probability earlier in this section and in Figure 5.8. Using these cumulative probabilities, it is easy to compute the probability that a random variable lies within a certain interval. For instance, suppose we wanted to know the probability of more than 5 and fewer than 16 arrivals. We would just find the cumulative probability of 15 arrivals and subtract from that the cumulative probability for 5 arrivals. Referring to Figure 5.9 to obtain the appropriate probabilities, we obtain .9513 2 .0671 5 .8842. With such a high probability, we could conclude that 6 to 15 cars will arrive in most 15-minute intervals. Using the cumulative probability for 20 arrivals, we can also conclude that the probability of more than 20 arrivals in a 15-minute period is 1 2 .9984 5 .0016; thus, there is almost no chance of more than 20 cars arriving. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

5.6 Poisson Probability Distribution

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FIGURE 5.9 EXCEL WORKSHEET FOR COMPUTING CUMULATIVE POISSON PROBABILITIES

Exercises

Methods 44. C onsider a Poisson distribution with 5 3. a. Write the appropriate Poisson probability function. b. Compute f (2). c. Compute f (1). d. Compute P(x $ 2). 45. C onsider a Poisson distribution with a mean of two occurrences per time period. a. Write the appropriate Poisson probability function. b. What is the expected number of occurrences in three time periods? c. Write the appropriate Poisson probability function to determine the probability of x occurrences in three time periods. d. Compute the probability of two occurrences in one time period. e. Compute the probability of six occurrences in three time periods. f. Compute the probability of five occurrences in two time periods.

Applications 46. P hone calls arrive at the rate of 48 per hour at the reservation desk for Regional Airways. a. Compute the probability of receiving three calls in a 5-minute interval of time. b. Compute the probability of receiving exactly 10 calls in 15 minutes. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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c. S uppose no calls are currently on hold. If the agent takes 5 minutes to complete the current call, how many callers do you expect to be waiting by that time? What is the probability that none will be waiting? d. If no calls are currently being processed, what is the probability that the agent can take 3 minutes for personal time without being interrupted by a call? 47. D uring the period of time that a local university takes phone-in registrations, calls come in at the rate of one every two minutes. a. What is the expected number of calls in one hour? b. What is the probability of three calls in five minutes? c. What is the probability of no calls in a five-minute period? 48. I n a one-year period, New York City had a total of 11,232 motor vehicle accidents that occurred on Monday through Friday between the hours of 3 p.m. and 6 p.m. (New York State Department of Motor Vehicles website). This corresponds to a mean of 14.4 accidents per hour. a. Compute the probability of no accidents in a 15-minute period. b. Compute the probability of at least one accident in a 15-minute period. c. Compute the probability of four or more accidents in a 15-minute period. 49. A irline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute. a. Compute the probability of no arrivals in a one-minute period. b. Compute the probability that three or fewer passengers arrive in a one-minute period. c. Compute the probability of no arrivals in a 15-second period. d. Compute the probability of at least one arrival in a 15-second period. 50. A ccording to the National Oceanic and Atmospheric Administration (NOAA), the state of Colorado averages 18 tornadoes every June (NOAA website, November 8, 2012). (Note: There are 30 days in June.) a. Compute the mean number of tornadoes per day. b. Compute the probability of no tornadoes during a day. c. Compute the probability of exactly one tornado during a day. d. Compute the probability of more than one tornado during a day. 51. O ver 500 million tweets are sent per day (Digital Marketing Ramblings website, December 15, 2014). Assume that the number of tweets per hour follows a Poisson distribution and that Bob receives on average 7 tweets during his lunch hour. a. What is the probability that Bob receives no tweets during his lunch hour? b. What is the probability that Bob receives at least 4 tweets during his lunch hour? c. What is the expected number of tweets Bob receives during the first 30 minutes of his lunch hour? d. What is the probability that Bob receives no tweets during the first 30 minutes of his lunch hour?

Hypergeometric Probability Distribution 5.7 The hypergeometric probability distribution is closely related to the binomial distribution. The two probability distributions differ in two key ways. With the hypergeometric distribution, the trials are not independent, and the probability of success changes from trial to trial. In the usual notation for the hypergeometric distribution, r denotes the number of elements in the population of size N labeled success, and N 2 r denotes the number of elements in the population labeled failure. The hypergeometric probability function is used to compute the probability that in a random selection of n elements, selected Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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5.7 Hypergeometric Probability Distribution

without replacement, we obtain x elements labeled success and n 2 x elements labeled failure. For this outcome to occur, we must obtain x successes from the r successes in the population and n 2 x failures from the N 2 r failures. The following hypergeometric probability function provides f(x), the probability of obtaining x successes in n trials.

Hypergeometric Probability Function

f(x) 5

S DS D SD r x

n2r n2x

n n

(5.16)

where

Note that

x 5 the number of successes n 5 the number of trials f(x) 5 the probability of x successes in n trials n 5 the number of elements in the population r 5 the number of elements in the population labeled success

SD n n

represents the number of ways n elements can be selected from a

r represents the number of ways that x successes can be selected x n2r from a total of r successes in the population; and represents the number of ways n2x that n 2 x failures can be selected from a total of N 2 r failures in the population. For the hypergeometric probability distribution, x is a discrete random variable and the probability function f (x) given by equation (5.16) is usually applicable for values of x 5 0, 1, 2, . . . , n. However, only values of x where the number of observed successes is less than or equal to the number of successes in the population (x # r) and where the number of observed failures is less than or equal to the number of failures in the population (n 2 x # N 2 r) are valid. If these two conditions do not hold for one or more values of x, the corresponding f (x) 5 0 indicating that the probability of this value of x is zero. To illustrate the computations involved in using equation (5.16), let us consider the following quality control application. Electric fuses produced by Ontario Electric are packaged in boxes of 12 units each. Suppose an inspector randomly selects 3 of the 12 fuses in a box for testing. If the box contains exactly 5 defective fuses, what is the probability that the inspector will find exactly one of the 3 fuses defective? In this application, n 5 3 and N 5 12. With r 5 5 defective fuses in the box the probability of finding x 5 1 defective fuse is population of size N;

f s1d 5

S D

S DS D S DS D SD S D 5 1

7 2

12 3

5! 1!4!

7! 2!5!

12! 3!9!

s5ds21d 5 .4773 220

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Chapter 5 Discrete Probability Distributions

Now suppose that we wanted to know the probability of finding at least 1 defective fuse. The easiest way to answer this question is to first compute the probability that the inspector does not find any defective fuses. The probability of x 5 0 is

f s0d 5

S DS D S DS D SD S D 5 0

7 3

12 3

5! 0!5!

7! 3!4!

12! 3!9!

s1ds35d 5 .1591 220

With a probability of zero defective fuses f (0) 5 .1591, we conclude that the probability of finding at least 1 defective fuse must be 1 2 .1591 5 .8409. Thus, there is a reasonably high probability that the inspector will find at least 1 defective fuse. The mean and variance of a hypergeometric distribution are as follows.

SD S DS DS D

esxd 5 5 n

Varsxd 5 2 5 n

r n

n2n n21

(5.17)

(5.18)

In the preceding example n 5 3, r 5 5, and N 5 12. Thus, the mean and variance for the number of defective fuses are

SD S D S DS DS D S DS DS D 5n

2 5 n

r n

r 5 53 5 1.25 n 12

n2n 5 53 n21 12

5 12

12 2 3 5 .60 12 2 1

The standard deviation is 5 Ï.60 5 .77.

Using Excel to Compute Hypergeometric Probabilities The Excel function for computing hypergeometric probabilities is HYPGEOM.DIST. It has five inputs: the first is the value of x, the second is the value of n, the third is the value of r, the fourth is the value of N, and the fifth is FALSE or TRUE. We choose FALSE if a probability is desired and TRUE if a cumulative probability is desired. This function’s usage is similar to that of BINOM.DIST for the binomial distribution and POISSON.DIST for the Poisson distribution, so we dispense with showing a worksheet figure and just explain how to use the function. Let us reconsider the example of selecting 3 fuses for inspection from a fuse box containing 12 fuses, 5 of which are defective. We want to find the probability that 1 of the 3 fuses selected is defective. In this case, the five inputs are x 5 1, n 5 3, r 5 5, N 5 12, and FALSE. So, the appropriate formula to place in a cell of an Excel worksheet is 5HYPGEOM.DIST(1,3,5,12,FALSE). Placing this formula in a cell of an Excel worksheet provides a hypergeometric probability of .4773. If we want to know the probability that none of the 3 fuses selected is defective, the five function inputs are x 5 0, n 5 3, r 5 5, N 5 12, and FALSE. So, using the HYPGEOM. DIST function to compute the probability of randomly selecting 3 fuses w ithout any being defective, we would enter the following formula into an Excel worksheet: 5HYPGEOM. DIST(0,3,5,12,FALSE). The probability is .1591. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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5.7 Hypergeometric Probability Distribution

Cumulative probabilities can be obtained in a similar fashion by using TRUE for the fifth input. For instance, to compute the probability of finding at most 1 defective fuse, the appropriate formula is =HYPGEOM.DIST(1,3,5,12,TRUE). Placing this formula in a cell of an Excel worksheet provides a hypergeometric cumulative probability of .6364. NOTE AND COMMENT onsider a hypergeometric distribution with n C trials. Let p 5 (r/N ) denote the probability of a success on the first trial. If the population size is large, the term (N 2 n)/(N 2 1) in equation (5.18) approaches 1. As a result, the expected value and variance can be written E(x) 5 np and Var(x) 5 np(1 2 p). Note that these expressions

are the same as the expressions used to compute the expected value and variance of a binomial distribution, as in equations (5.13) and (5.14). When the population size is large, a hypergeometric distribution can be approximated by a binomial distribution with n trials and a probability of success p 5 (r/N ).

Exercises

Methods 52. Suppose N 5 10 and r 5 3. Compute the hypergeometric probabilities for the following values of n and x. a. n 5 4, x 5 1. b. n 5 2, x 5 2. c. n 5 2, x 5 0. d. n 5 4, x 5 2. e. n 5 4, x 5 4. 53. Suppose N 5 15 and r 5 4. What is the probability of x 5 3 for n 5 10?

Applications 54. A recent survey showed that a majority of Americans plan on doing their holiday shopping online because they don’t want to spend money on gas driving from store to store (SOASTA website, October 24, 2012). Suppose we have a group of 10 shoppers; 7 prefer to do their holiday shopping online and 3 prefer to do their holiday shopping in stores. A random sample of 3 of these 10 shoppers is selected for a more in-depth study of how the economy has impacted their shopping behavior. a. What is the probability that exactly 2 prefer shopping online? b. What is the probability that the majority (either 2 or 3) prefer shopping online? 55. B lackjack, or twenty-one as it is frequently called, is a popular gambling game played in Las Vegas casinos. A player is dealt two cards. Face cards (jacks, queens, and kings) and tens have a point value of 10. Aces have a point value of 1 or 11. A 52-card deck contains 16 cards with a point value of 10 (jacks, queens, kings, and tens) and four aces. a. What is the probability that both cards dealt are aces or 10-point cards? b. What is the probability that both of the cards are aces? c. What is the probability that both of the cards have a point value of 10? d. A blackjack is a 10-point card and an ace for a value of 21. Use your answers to parts (a), (b), and (c) to determine the probability that a player is dealt blackjack. (Hint: Part (d) is not a hypergeometric problem. Develop your own logical relationship as to how the hypergeometric probabilities from parts (a), (b), and (c) can be combined to answer this question.) Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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56. A xline Computers manufactures personal computers at two plants, one in Texas and the other in Hawaii. The Texas plant has 40 employees; the Hawaii plant has 20. A random sample of 10 employees is asked to fill out a benefits questionnaire. a. What is the probability that none of the employees in the sample work at the plant in Hawaii? b. What is the probability that 1 of the employees in the sample works at the plant in Hawaii? c. What is the probability that 2 or more of the employees in the sample work at the plant in Hawaii? d. What is the probability that 9 of the employees in the sample work at the plant in Texas? 57. T he Zagat Restaurant Survey provides food, decor, and service ratings for some of the top restaurants across the United States. For 15 restaurants located in Boston, the average price of a dinner, including one drink and tip, was $48.60. You are leaving on a business trip to Boston and will eat dinner at three of these restaurants. Your company will reimburse you for a maximum of $50 per dinner. Business associates familiar with these restaurants have told you that the meal cost at one-third of these restaurants will exceed $50. Suppose that you randomly select three of these restaurants for dinner. a. What is the probability that none of the meals will exceed the cost covered by your company? b. What is the probability that one of the meals will exceed the cost covered by your company? c. What is the probability that two of the meals will exceed the cost covered by your company? d. What is the probability that all three of the meals will exceed the cost covered by your company? 58. S uppose that a shipment of 100 boxes of apples has 8 boxes in which the apples show signs of spoilage. A quality control inspection selects 10 boxes at random, opens these selected boxes, and counts the number of boxes out of 10 in which the apples show signs of spoilage. What is the probability that exactly 2 boxes in the sample show signs of spoilage?

Summary A random variable provides a numerical description of the outcome of an experiment. The probability distribution for a random variable describes how the probabilities are distributed over the values the random variable can assume. For any discrete random variable x, the probability distribution is defined by a probability function, denoted by f (x), which provides the probability associated with each value of the random variable. We introduced two types of discrete probability distributions. One type involved providing a list of the values of the random variable and the associated probabilities in a table. We showed how the relative frequency method of assigning probabilities could be used to develop empirical discrete probability distributions of this type. Bivariate empirical distributions were also discussed. With bivariate distributions, interest focuses on the relationship between two random variables. We showed how to compute the covariance and correlation coefficient as measures of such a relationship. We also showed how bivariate distributions involving market returns on financial assets could be used to create financial portfolios. The second type of discrete probability distribution we discussed involved the use of a mathematical function to provide the probabilities for the random variable. The binomial, Poisson, and hypergeometric distributions discussed were all of this type. The binomial distribution can be used to determine the probability of x successes in n trials whenever the experiment has the following properties: Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Glossary

275

1. The experiment consists of a sequence of n identical trials. 2. Two outcomes are possible on each trial, one called success and the other failure. 3. The probability of a success p does not change from trial to trial. Consequently, the probability of failure, 1 2 p, does not change from trial to trial. 4. The trials are independent. When the four properties hold, the binomial probability function can be used to determine the probability of obtaining x successes in n trials. Formulas were also presented for the mean and variance of the binomial distribution. The Poisson distribution is used when it is desirable to determine the probability of obtaining x occurrences over an interval of time or space. The following assumptions are necessary for the Poisson distribution to be applicable: 1. The probability of an occurrence of the event is the same for any two intervals of equal length. 2. The occurrence or nonoccurrence of the event in any interval is independent of the occurrence or nonoccurrence of the event in any other interval. A third discrete probability distribution, the hypergeometric, was introduced in Section 5.7. Like the binomial, it is used to compute the probability of x successes in n trials. But, in contrast to the binomial, the probability of success changes from trial to trial.

Glossary Binomial experiment An experiment having the four properties stated at the beginning of Section 5.5. Binomial probability distribution A probability distribution showing the probability of x successes in n trials of a binomial experiment. Binomial probability function The function used to compute binomial probabilities. Bivariate probability distribution A probability distribution involving two random variables. A discrete bivariate probability distribution provides a probability for each pair of values that may occur for the two random variables. Continuous random variable A random variable that may assume any numerical value in an interval or collection of intervals. Discrete random variable A random variable that may assume either a finite number of values or an infinite sequence of values. Discrete uniform probability distribution A probability distribution for which each possible value of the random variable has the same probability. Empirical discrete distribution A discrete probability distribution for which the relative frequency method is used to assign the probabilities. Expected value A measure of the central location, or mean, of a random variable. Hypergeometric probability distribution A probability distribution showing the probability of x successes in n trials from a population with r successes and N 2 r failures. Hypergeometric probability function The function used to compute hypergeometric probabilities. Poisson probability distribution A probability distribution showing the probability of x occurrences of an event over a specified interval of time or space. Poisson probability function The function used to compute Poisson probabilities. Probability distribution A description of how the probabilities are distributed over the values of the random variable. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Probability function A function, denoted by f (x), that provides the probability that x assumes a particular value for a discrete random variable. Random variable A numerical description of the outcome of an experiment. Standard deviation The positive square root of the variance. Variance A measure of the variability, or dispersion, of a random variable.

Key Formulas Discrete Uniform Probability Function f sxd 5 1/n

(5.3)

Expected Value of a Discrete Random Variable esxd 5 5 oxf sxd

(5.4)

Variance of a Discrete Random Variable

Varsxd 5 2 5 osx 2 d2f sxd

(5.5)

Covariance of Random Variables x and y

(5.6)

xy 5 [Varsx 1 yd 2 Varsxd 2 Varsyd]/2

Correlation between Random Variables x and y

xy 5

xy xy

(5.7)

Expected Value of a Linear Combination of Random Variables x and y

(5.8)

esax 1 byd 5 aesxd 1 besyd

Variance of a Linear Combination of Two Random Variables

Varsax 1 byd 5 a2Varsxd 1 b2Varsyd 1 2abxy

(5.9)

where xy is the covariance of x and y Number of Experimental Outcomes Providing Exactly x Successes in n Trials

n n! 5 x x!sn 2 xd!

(5.10)

Binomial Probability Function

f sxd 5

n x p s1 2 pdsn 2 xd x

(5.12)

277

Supplementary Exercises

Expected Value for the Binomial Distribution

(5.13)

esxd 5 5 np

Variance for the Binomial Distribution Varsxd 5 2 5 nps1 2 pd

(5.14)

Poisson Probability Function f sxd 5

Hypergeometric Probability Function

f sxd 5

xe2 x!

S DS D SD r x

n2r n2x

n n

(5.15)

(5.16)

Expected Value for the Hypergeometric Distribution

esxd 5 5 n

r n

(5.17)

Variance for the Hypergeometric Distribution

S DS DS D

Varsxd 5 2 5 n

r n

n2n n21

(5.18)

Supplementary Exercises 59. T he U.S. Coast Guard (USCG) provides a wide variety of information on boating accidents including the wind condition at the time of the accident. The following table shows the results obtained for 4401 accidents (USCG website, November 8, 2012).

Wind Condition

Percentage of Accidents

None 9.6 Light 57.0 Moderate 23.8 Strong 7.7 Storm 1.9

Let x be a random variable reflecting the known wind condition at the time of each accident. Set x 5 0 for none, x 5 1 for light, x 5 2 for moderate, x 5 3 for strong, and x 5 4 for storm. a. Develop a probability distribution for x. b. Compute the expected value of x. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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c. C ompute the variance and standard deviation for x. d. Comment on what your results imply about the wind conditions during boating accidents. 60. T he Car Repair Ratings website provides consumer reviews and ratings for garages in the United States and Canada. The time customers wait for service to be completed is one of the categories rated. The following table provides a summary of the wait-time ratings (1 5 Slow/Delays; 10 5 Quick/On Time) for 40 randomly selected garages located in the province of Ontario, Canada.

Wait-Time Rating 1 2 3 4 5 6 7 8 9 10

Number of Garages 6 2 3 2 5 2 4 5 5 6

a. D evelop a probability distribution for x 5 wait-time rating. b. Any garage that receives a wait-time rating of at least 9 is considered to provide outstanding service. If a consumer randomly selects one of the 40 garages for their next car service, what is the probability the garage selected will provide outstanding wait-time service? c. What is the expected value and variance for x? d. Suppose that seven of the 40 garages reviewed were new car dealerships. Of the seven new car dealerships, two were rated as providing outstanding wait-time service. Compare the likelihood of a new car dealership achieving an outstanding wait-time service rating as compared to other types of service providers. 61. T he budgeting process for a midwestern college resulted in expense forecasts for the coming year (in $ millions) of $9, $10, $11, $12, and $13. Because the actual expenses are unknown, the following respective probabilities are assigned: .3, .2, .25, .05, and .2. a. Show the probability distribution for the expense forecast. b. What is the expected value of the expense forecast for the coming year? c. What is the variance of the expense forecast for the coming year? d. If income projections for the year are estimated at $12 million, comment on the financial position of the college. 62. A bookstore at the Hartsfield-Jackson Airport in Atlanta sells reading materials (paperback books, newspapers, magazines) as well as snacks (peanuts, pretzels, candy, etc.). A point-of-sale terminal collects a variety of information about customer purchases. The following table shows the number of snack items and the number of items of reading material purchased by the most recent 600 customers.

Reading Material Snacks 0 1 2 0 0 60 18 1 240 90 30 2 120 30 12

Supplementary Exercises

279

a. U sing the data in the table construct an empirical discrete bivariate probability distribution for x 5 number of snack items and y 5 number of reading materials for a randomly selected customer purchase. What is the probability of a customer purchase consisting of one item of reading materials and two snack items? What is the probability of a customer purchasing one snack item only? Why is the probability f(x 5 0, y 5 0) 5 0? b. Show the marginal probability distribution for the number of snack items purchased. Compute the expected value and variance. c. What is the expected value and variance for the number of reading materials purchased by a customer? d. Show the probability distribution for t 5 total number of items for a randomly selected customer purchase. Compute its expected value and variance. e. Compute the covariance and correlation coefficient between x and y. What is the relationship, if any, between the number of reading materials and number of snacks purchased? 63. T he Knowles/Armitage (KA) group at Merrill Lynch advises clients on how to create a diversified investment portfolio. One of the investment alternatives they make available to clients is the All World Fund composed of global stocks with good dividend yields. One of their clients is interested in a portfolio consisting of investment in the All World Fund and a treasury bond fund. The expected percent return of an investment in the All World Fund is 7.80% with a standard deviation of 18.90%. The expected percent return of an investment in a treasury bond fund is 5.50% and the standard deviation is 4.60%. The covariance of an investment in the All World Fund with an investment in a treasury bond fund is 212.4. a. Which of the funds would be considered the more risky? Why? b. If KA recommends that the client invest 75% in the All World Fund and 25% in the treasury bond fund, what is the expected percent return and standard deviation for such a portfolio? What would be the expected return and standard deviation, in dollars, for a client investing $10,000 in such a portfolio? c. If KA recommends that the client invest 25% in the All World Fund and 75% in the treasury bond fund, what is the expected return and standard deviation for such a portfolio? What would be the expected return and standard deviation, in dollars, for a client investing $10,000 in such a portfolio? d. W hich of the portfolios in parts (b) and (c) would you recommend for an aggressive investor? Which would you recommend for a conservative investor? Why? 64. The Pew Research Center surveyed adults who own/use the following technologies: Internet, smartphone, e-mail, and land-line phone (USA Today, March 26, 2014) and asked which of these technologies would be “very hard” to give up. The following responses were obtained: Internet 53%, smartphone 49%, e-mail 36%, and land-line phone 28%. a. If 20 adult Internet users are surveyed, what is the probability that 3 users will report that it would be very hard to give it up? b. If 20 adults who own a land-line phone are surveyed, what is the probability that 5 or fewer will report that it would be very hard to give it up? c. If 2000 owners of smartphones were surveyed, what is the expected number that will report that it would be very hard to give it up? d. If 2000 users of e-mail were surveyed, what is expected number that will report that it would be very hard to give it up? What is the variance and standard deviation? 65. T he following table shows the percentage of individuals in each age group who use an online tax program to prepare their federal income tax return (CompleteTax website, November 9, 2012). Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Chapter 5 Discrete Probability Distributions

Age

Online Tax Program (%)

18–34 16 35–44 12 45–54 10 55–64 8 651 2

Suppose a follow-up study consisting of personal interviews is to be conducted to determine the most important factors in selecting a method for filing taxes. a. How many 18- to 34-year-olds must be sampled to find an expected number of at least 25 who use an online tax program to prepare their federal income tax return? b. How many 35- to 44-year-olds must be sampled to find an expected number of at least 25 who use an online tax program to prepare their federal income tax return? c. How many 651-year-olds must be sampled to find an expected number of at least 25 who use an online tax program to prepare their federal income tax return? d. If the number of 18- to 34-year-olds sampled is equal to the value identified in part (a), what is the standard deviation of the percentage who use an online tax program? e. If the number of 35- to 44-year-olds sampled is equal to the value identified in part (b), what is the standard deviation of the percentage who use an online tax program? 66. M any companies use a quality control technique called acceptance sampling to monitor incoming shipments of parts, raw materials, and so on. In the electronics industry, component parts are commonly shipped from suppliers in large lots. Inspection of a sample of n components can be viewed as the n trials of a binomial experiment. The outcome for each component tested (trial) will be that the component is classified as good or defective. Reynolds Electronics accepts a lot from a particular supplier if the defective components in the lot do not exceed 1%. Suppose a random sample of five items from a recent shipment is tested. a. Assume that 1% of the shipment is defective. Compute the probability that no items in the sample are defective. b. Assume that 1% of the shipment is defective. Compute the probability that exactly one item in the sample is defective. c. What is the probability of observing one or more defective items in the sample if 1% of the shipment is defective? d. Would you feel comfortable accepting the shipment if one item was found to be defective? Why or why not? 67. PBS News Hour reported that 39.4% of Americans between the ages of 25 and 64 have at least a two-year college degree (PBS website, December 15, 2014). Assume that 50 Americans between the ages of 25 and 64 are selected randomly. a. What is the expected number of people with at least a two-year college-degree? b. What are the variance and standard deviation for the number of people with at least a two-year college degree? 68. M ahoney Custom Home Builders, Inc. of Canyon Lake, Texas asked visitors to their website what is most important when choosing a home builder. Possible responses were quality, price, customer referral, years in business, and special features. Results showed that 23.5% of the respondents chose price as the most important factor (Mahoney Custom Homes website, November 13, 2012). Suppose a sample of 200 potential home buyers in the Canyon Lake area was selected. a. How many people would you expect to choose price as the most important factor when choosing a home builder? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

Case Problem 1 Go Bananas!

281

b. W hat is the standard deviation of the number of respondents who would choose price as the most important factor in selecting a home builder? c. What is the standard deviation of the number of respondents who do not list price as the most important factor in selecting a home builder? 69. C ars arrive at a car wash randomly and independently; the probability of an arrival is the same for any two time intervals of equal length. The mean arrival rate is 15 cars per hour. What is the probability that 20 or more cars will arrive during any given hour of operation? 70. A new automated production process averages 1.5 breakdowns per day. Because of the cost associated with a breakdown, management is concerned about the possibility of having 3 or more breakdowns during a day. Assume that breakdowns occur randomly, that the probability of a breakdown is the same for any two time intervals of equal length, and that breakdowns in one period are independent of breakdowns in other periods. What is the probability of having 3 or more breakdowns during a day? 71. A regional director responsible for business development in the state of Pennsylvania is concerned about the number of small business failures. If the mean number of small business failures per month is 10, what is the probability that exactly 4 small businesses will fail during a given month? Assume that the probability of a failure is the same for any two months and that the occurrence or nonoccurrence of a failure in any month is independent of failures in any other month. 72. C ustomer arrivals at a bank are random and independent; the probability of an arrival in any one-minute period is the same as the probability of an arrival in any other one-minute period. Answer the following questions, assuming a mean arrival rate of 3 customers per minute. a. What is the probability of exactly 3 arrivals in a one-minute period? b. What is the probability of at least 3 arrivals in a one-minute period?

Case Problem 1

Go Bananas! Great Grasslands Grains, Inc. (GGG) manufactures and sells a wide variety of breakfast cereals. GGG’s product development lab recently created a new cereal that consists of rice flakes and banana-flavored marshmallows. The company’s marketing research department has tested the new cereal extensively and has found that consumers are enthusiastic about the cereal when 16-ounce boxes contain at least 1.6 ounces and no more than 2.4 ounces of the banana-flavored marshmallows. As GGG prepares to begin producing and selling 16-ounce boxes of the new cereal, which it has named Go Bananas!, management is concerned about the amount of bananaflavored marshmallows. It wants to be careful not to include less than 1.6 ounces or more than 2.4 ounces of banana-flavored marshmallows in each 16-ounce box of Go Bananas!. Tina Finkel, VP of Production for GGG, has suggested that the company measure the weight of banana-flavored marshmallows in a random sample of 25 boxes of Go Bananas! on a weekly basis. Each week, GGG can count the number of boxes out of the 25 boxes in the sample that contain less than 1.6 ounces or more than 2.4 ounces of banana-flavored marshmallows; if the number of boxes that fail to meet the standard weight of banana-flavored marshmallows is too high, production will be shut down and inspected. Ms. Finkel and her staff have designed the production process so that only 8% of all 16-ounce boxes of Go Bananas! fail to meet the standard weight of banana-flavored marshmallows. After much debate, GGG management has decided to shut down production of Go Bananas! if at least five boxes in a weekly sample fail to meet the standard weight of banana-flavored marshmallows.

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Managerial Report Prepare a managerial report that addresses the following issues. 1. C alculate the probability that a weekly sample will result in a shutdown of production if the production process is working properly. Comment on GGG management’s policy for deciding when to shut down production of Go Bananas!. 2. GGG management wants to shut down production of Go Bananas! no more than 1% of the time when the production process is working properly. Suggest the appropriate number of boxes in the weekly sample that must fail to meet the standard weight of banana-flavored marshmallows in order for production to be shut down if this goal is to be achieved. 3. Ms. Finkel has suggested that if given sufficient resources, she could redesign the production process to reduce the percentage of 16-ounce boxes of Go Bananas! that fail to meet the standard weight of banana-flavored marshmallows when the process is working properly. To what level must Ms. Finkel reduce the percentage of 16-ounce boxes of Go Bananas! that fail to meet the standard weight of banana-flavored marshmallows when the process is working properly in order for her to reduce the probability at least five of the sampled boxes fail to meet the standard to .01 or less?

Case Problem 2

McNeil’s Auto Mall Harriet McNeil, proprietor of McNeil’s Auto Mall, believes that it is good business for her automobile dealership to have more customers on the lot than can be served, as she believes this creates an impression that demand for the automobiles on her lot is high. However, she also understands that if there are far more customers on the lot than can be served by her salespeople, her dealership may lose sales to customers who become frustrated and leave without making a purchase. Ms. McNeil is primarily concerned about the staffing of salespeople on her lot on Saturday mornings (8:00 a.m. to noon), which are the busiest time of the week for McNeil’s Auto Mall. On Saturday mornings, an average of 6.8 customers arrive per hour. The customers arrive randomly at a constant rate throughout the morning, and a salesperson spends an average of one hour with a customer. Ms. McNeil’s experience has led her to conclude that if there are two more customers on her lot than can be served at any time on a Saturday morning, her automobile dealership achieves the optimal balance of creating an impression of high demand without losing too many customers who become frustrated and leave without making a purchase. Ms. McNeil now wants to determine how many salespeople she should have on her lot on Saturday mornings in order to achieve her goal of having two more customers on her lot than can be served at any time. She understands that occasionally the number of customers on her lot will exceed the number of salespersons by more than two, and she is willing to accept such an occurrence no more than 10% of the time.

Managerial Report Ms. McNeil has asked you to determine the number of salespersons she should have on her lot on Saturday mornings in order to satisfy her criteria. In answering Ms. McNeil’s question, consider the following three questions: 1. How is the number of customers who arrive on the lot on a Saturday morning distributed? 2. Suppose Ms. McNeil currently uses five salespeople on her lot on Saturday mornings. Using the probability distribution you identified in (1), what is the probability that the number of customers who arrive on her lot will exceed the number of salespersons by more than two? Does her current Saturday morning employment strategy satisfy her stated objective? Why or why not? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Case Problem 4 Sagittarius Casino

3. What is the minimum number of salespeople Ms. McNeil should have on her lot on Saturday mornings to achieve her objective?

Case Problem 3

Grievance Committee at Tuglar Corporation Several years ago, management at Tuglar Corporation established a grievance committee composed of employees who volunteered to work toward the amicable resolution of disputes between Tuglar management and its employees. Each year management issues a call for volunteers to serve on the grievance committee, and 10 of the respondents are randomly selected to serve on the committee for the upcoming year. Employees in the Accounting Department are distressed because no member of their department has served on the Tuglar grievance committee in the past five years. Management has assured its employees in the Accounting Department that the selections have been made randomly, but these assurances have not quelled suspicions that management has intentionally omitted accountants from the committee. The table below summarizes the total number of volunteers and the number of employees from the Accounting Department who have volunteered for the grievance committee in each of the past five years:

Total Number of Volunteers Number of Volunteers from the Accounting Department

2013

2014

2015

2016

2017

In its defense, management has provided these numbers to the Accounting Department. Given these numbers, is the lack of members of the Accounting Department on the grievance committee for the past five years suspicious (i.e., unlikely)?

Managerial Report In addressing the issue of whether or not the committee selection process is random, consider the following questions: 1. How is the number of members of the Accounting Department who are selected to serve on the grievance committee distributed? 2. Using the probability distribution you identified in (1), what is the probability for each of these five years that no members of the Accounting Department have been selected to serve? 3. Using the probabilities you identified in (2), what is the probability that no members of the Accounting Department have been selected to serve during the past five years? 4. What is the cause of the lack of Accounting Department representation on the grievance committee over the past five years? What can be done to increase the probability that a member of the Accounting Department will be selected to serve on the grievance committee using the current selection method?

Case Problem 4

Sagittarius Casino The Sagittarius Casino’s strategy for establishing a competitive advantage over its competitors is to periodically create unique and interesting new games for its customers to play. Sagittarius management feels it is time for the casino to once again introduce a new game to

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excite its customer base, and Sagittarius’s Director of Research and Development, Lou Zerbit, believes he and his staff have developed a new game that will accomplish this goal. The game, which they have named POSO! (an acronym for Payouts On Selected Outcomes), is to be played in the following manner. A player will select two different values from 1, 2, 3, 4, 5, and 6. Two dice are then rolled. If the first number the player selected comes up on at least one of the two dice, the player wins $5.00; if the second number the player selected comes up on both of the dice, the player wins $10.00. If neither of these events occurs, the player wins nothing. For example, suppose a player fills out the following card for one game of POSO!

POSO! First Number (select one number from 1, 2, 3, 4, 5, or 6)

4 If this number comes up on at least one die, you win $5.00!

Second Number (select a different number from 1, 2, 3, 4, 5, or 6)

2 If this number comes up on both dice, you win $10.00!

When the two dice are rolled, if at least one die comes up 4 the player will win $5.00, if both dice come up 2 the player will win $10.00, and if any other outcome occurs the player wins nothing.

Managerial Report Sagittarius management now has three questions about POSO! These questions should be addressed in your report. 1. Although they certainly do not want to pay out more than they take in, casinos like to offer games in which players win frequently; casino managers believe this keeps players excited about playing the game. What is the probability a player will win if she or he plays a single game of POSO!? 2. What is the expected amount a player will win when playing one game of POSO!? 3. Sagittarius managers want to take in more than they pay out on average for a game of POSO!. Furthermore, casinos such as Sagittarius are often looking for games that provide their gamers with an opportunity to play for a small bet, and Sagittarius management would like to charge players $2.00 to play one game of POSO!. What will be the expected profit earned by Sagittarius Casino on a single play if a player has to pay $2.00 for a single play of POSO!? Will Sagittarius Casino expect to earn or lose money on POSO! if a player pays $2.00 for a single play? What is the minimum amount Sagittarius Casino can charge a player for a single play of POSO! and still expect to earn money? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

CHAPTER

Continuous Probability Distributions CONTENTS STATISTICS IN PRACTICE: PROCTER & GAMBLE 6.1

UNIFORM PROBABILITY DISTRIBUTION Area as a Measure of Probability

6.2 NORMAL PROBABILITY DISTRIBUTION Normal Curve Standard Normal Probability Distribution Computing Probabilities for Any Normal Probability Distribution Grear Tire Company Problem Using Excel to Compute Normal Probabilities

6.3

XPONENTIAL PROBABILITY E DISTRIBUTION Computing Probabilities for the Exponential Distribution Relationship Between the Poisson and Exponential Distributions Using Excel to Compute Exponential Probabilities

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Chapter 6 Continuous Probability Distributions

STATISTICS in PRACTICE PROCTER & GAMBLE* Procter & Gamble (P&G) produces and markets such products as detergents, disposable diapers, over-the-counter pharmaceuticals, dentifrices, bar soaps, mouthwashes, and paper towels. Worldwide, it has the leading brand in more categories than any other c onsumer products company. Since its merger with Gillette, P&G also produces and markets razors, blades, and many other personal care products. As a leader in the application of statistical methods in decision making, P&G employs people with diverse academic backgrounds: engineering, statistics, operations research, analytics and business. The major quantitative technologies for which these people provide support are probabilistic decision and risk analysis, advanced simulation, quality improvement, and quantitative methods (e.g., linear programming, regression analysis, probability analysis). The Industrial Chemicals Division of P&G is a major supplier of fatty alcohols derived from natural substances such as coconut oil and from petroleum-based derivatives. The division wanted to know the economic risks and opportunities of expanding its fatty-alcohol production facilities, so it called in P&G’s experts in probabilistic decision and risk analysis to help. After structuring and modeling the problem, they determined that the key to profitability was the cost difference between the petroleumand coconut-based raw materials. Future costs were unknown, but the analysts were able to approximate them with the following continuous random variables.

x 5 the coconut oil price per pound of fatty alcohol

and

y 5 the petroleum raw material price per pound of fatty alcohol

Because the key to profitability was the difference between these two random variables, a third random variable, d 5 x 2 y, was used in the analysis. Experts were interviewed to determine the probability distributions for x and y. In turn, this information was used to develop a probability distribution for the difference in *The authors are indebted to Joel Kahn of Procter & Gamble for providing this Statistics in Practice.

CINCINNATI, OHIO

Procter & Gamble is a leader in the application of statistical methods in decision making. prices d. This continuous probability distribution showed a .90 probability that the price difference would be $.0655 or less and a .50 probability that the price difference would be $.035 or less. In addition, there was only a .10 probability that the price difference would be $.0045 or less.† The Industrial Chemicals Division thought that being able to quantify the impact of raw material price differences was key to reaching a consensus. The probabilities obtained were used in a sensitivity analysis of the raw material price difference. The analysis yielded s ufficient insight to form the basis for a recommendation to management. The use of continuous random variables and their probability distributions was helpful to P&G in analyzing the economic risks associated with its fatty-alcohol production. In this chapter, you will gain an understanding of continuous random variables and their probability distributions, including one of the most important probability distributions in statistics, the normal distribution. †

The price differences stated here have been modified to protect proprietary data.

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6.1 Uniform Probability Distribution

In the preceding chapter we discussed discrete random variables and their probability distributions. In this chapter we turn to the study of continuous random variables. Specifically, we discuss three continuous probability distributions: the uniform, the normal, and the exponential. A fundamental difference separates discrete and continuous random variables in terms of how probabilities are computed. For a discrete random variable, the probability function f(x) provides the probability that the random variable assumes a particular value. With continuous random variables, the counterpart of the probability function is the probability density function, also denoted by f(x). The difference is that the probability density function does not directly provide probabilities. However, the area under the graph of f(x) corresponding to a given interval does provide the probability that the continuous random variable x assumes a value in that interval. So when we compute probabilities for continuous random variables, we are computing the probability that the random variable assumes any value in an interval. Because the area under the graph of f(x) at any particular point is zero, one of the implications of the definition of probability for continuous random variables is that the probability of any particular value of the random variable is zero. In Section 6.1 we demonstrate these concepts for a continuous random variable that has a uniform distribution. Much of the chapter is devoted to describing and showing applications of the normal distribution. The normal distribution is of major importance because of its wide applica bility and its extensive use in statistical inference. The chapter closes with a discussion of the exponential distribution. The exponential distribution is useful in applications involving such factors as waiting times and service times.

Uniform Probability Distribution 6.1

Whenever the probability is proportional to the length of the interval, the random variable is uniformly distributed.

Consider the random variable x representing the flight time of an airplane traveling from Chicago to New York. Suppose the flight time can be any value in the interval from 120 minutes to 140 minutes. Because the random variable x can assume any value in that interval, x is a continuous rather than a discrete random variable. Let us assume that sufficient actual flight data are available to conclude that the probability of a flight time within any 1-minute interval is the same as the probability of a flight time within any other 1-minute interval contained in the larger interval from 120 to 140 minutes. With every 1-minute interval being equally likely, the random variable x is said to have a uniform probability distribution. The probability density function, which defines the uniform distribution for the flight-time random variable, is ƒ(x) 5

51/20 0

for 120 # x # 140 elsewhere

Figure 6.1 is a graph of this probability density function. In general, the uniform proba bility density function for a random variable x is defined by the following formula. Uniform Probability Density Function

ƒ(x) 5

1 b2a 0

for a # x # b elsewhere

(6.1)

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Chapter 6 Continuous Probability Distributions

FIGURE 6.1 UNIFORM PROBABILITY DistributioN FOR FLIGHT TIME f (x)

1 20

120

125

130 Flight Time in Minutes

135

140

As noted in the introduction, for a continuous random variable, we consider proba bility only in terms of the likelihood that a random variable assumes a value within a specified interval. In the flight time example, an acceptable probability question is: What is the probability that the flight time is between 120 and 130 minutes? That is, what is P(120 # x # 130)? Because the flight time must be between 120 and 140 minutes and because the probability is described as being uniform over this interval, we feel comfortable saying that P(120 # x # 130) 5 .50. In the following subsection we show that this probability can be computed as the area under the graph of f(x) from 120 to 130 (see Figure 6.2).

Area as a Measure of Probability Let us make an observation about the graph in Figure 6.2. Consider the area under the graph of f(x) in the interval from 120 to 130. The area is rectangular, and the area of a rectangle is simply the width multiplied by the height. With the width of the interval equal to 130 2 120 5 10 and the height equal to the value of the probability density function f(x) 5 1/20, we have area 5 width 3 height 5 10(1/20) 5 10/20 5 .50. What observation can you make about the area under the graph of f(x) and probability? They are identical! Indeed, this observation is valid for all continuous random variables.

FIGURE 6.2 AREA PROVIDES PROBABILITY OF a FLIGHT TIME BETWEEN 120

AND 130 MINUTES

f (x) P(120 # x #130) 5 Area 5 1/20(10) 5 10/20 5 .50 1 20 10 120

125

130

135

140

Flight Time in Minutes

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6.1 Uniform Probability Distribution

Once a probability density function f(x) is identified, the probability that x takes a value between some lower value x1 and some higher value x2 can be found by computing the area under the graph of f(x) over the interval from x1 to x2. Given the uniform distribution for flight time and using the interpretation of area as probability, we can answer any number of probability questions about flight times. For example, what is the probability of a flight time between 128 and 136 minutes? The width of the interval is 136 2 128 5 8. With the uniform height of f(x) 5 1/20, we see that P(128 # x # 136) 5 8(1/20) 5 .40. Note that P(120 # x # 140) 5 20(1/20) 5 1; that is, the total area under the graph of f(x) is equal to 1. This property holds for all continuous probability distributions and is the analog of the condition that the sum of the probabilities must equal 1 for a discrete probability function. For a continuous probability density function, we must also require that f(x) $ 0 for all values of x. This requirement is the analog of the requirement that f(x) $ 0 for discrete probability functions. Two major differences stand out between the treatment of continuous random variables and the treatment of their discrete counterparts. 1. We no longer talk about the probability of the random variable assuming a particular value. Instead, we talk about the probability of the random variable assuming a value within some given interval. 2. The probability of a continuous random variable assuming a value within some given interval from x1 to x 2 is defined to be the area under the graph of the probability density function between x1 and x 2. Because a single point is an interval of zero width, this implies that the probability of a continuous random variable assuming any particular value exactly is zero. It also means that the probability of a continuous random variable assuming a value in any interval is the same whether or not the endpoints are included.

To see that the probability of any single point is 0, refer to Figure 6.2 and compute the probability of a single point, say, x 5 125. P(x 5 125) 5 P(125 # x # 125) 5 0(1/20) 5 0.

The calculation of the expected value and variance for a continuous random variable is analogous to that for a discrete random variable. However, because the computational procedure involves integral calculus, we leave the derivation of the appropriate formulas to more advanced texts. For the uniform continuous probability distribution introduced in this section, the formulas for the expected value and variance are E(x) 5 Var(x) 5

a1b 2 (b 2 a)2 12

In these formulas, a is the smallest value and b is the largest value that the random variable may assume. Applying these formulas to the uniform distribution for flight times from Chicago to New York, we obtain

E(x) 5

Var (x) 5

(120 1 140) 5 130 2 (140 2 120)2 5 33.33 12

The standard deviation of flight times can be found by taking the square root of the variance. Thus, 5 5.77 minutes. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Chapter 6 Continuous Probability Distributions

NOTE AND COMMENT To see more clearly why the height of a probability density function is not a probability, think about a random variable with the following uniform proba bility distribution.

ƒ(x) 5

520

for 0 # x # .5 elsewhere

The height of the probability density function, f(x), is 2 for values of x between 0 and .5. However, we know probabilities can never be greater than 1. Thus, we see that f(x) cannot be interpreted as the probability of x.

Exercises

Methods 1. T he random variable x is known to be uniformly distributed between 1.0 and 1.5. a. Show the graph of the probability density function. b. Compute P(x 5 1.25). c. Compute P(1.0 # x # 1.25). d. Compute P(1.20 , x , 1.5). 2. The random variable x is known to be uniformly distributed between 10 and 20. a. Show the graph of the probability density function. b. Compute P(x , 15). c. Compute P(12 # x # 18). d. Compute E(x). e. Compute Var(x).

Applications 3. Delta Airlines quotes a flight time of 2 hours, 5 minutes for its flights from Cincinnati to Tampa. Suppose we believe that actual flight times are uniformly distributed between 2 hours and 2 hours, 20 minutes. a. Show the graph of the probability density function for flight time. b. What is the probability that the flight will be no more than 5 minutes late? c. What is the probability that the flight will be more than 10 minutes late? d. What is the expected flight time? 4. Most computer languages include a function that can be used to generate random numbers. In Excel, the RAND function can be used to generate random numbers between 0 and 1. If we let x denote a random number generated using RAND, then x is a continuous random variable with the following probability density function.

f(x) 5

510

for 0 # x # 1 elsewhere

a. Graph the probability density function. b. What is the probability of generating a random number between .25 and .75? c. What is the probability of generating a random number with a value less than or equal to .30? d. What is the probability of generating a random number with a value greater than .60? e. Generate 50 random numbers by entering 5 RAND() into 50 cells of an Excel worksheet. f. Compute the mean and standard deviation for the random numbers in part (e). Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

6.2 Normal Probability Distribution

291

5. In October 2012, Apple introduced a much smaller variant of the Apple iPad, known as the iPad Mini. Weighing less than 11 ounces, it was about 50% lighter than the standard iPad. Battery tests for the iPad Mini showed a mean life of 10.25 hours (The Wall Street Journal, October 31, 2012). Assume that battery life of the iPad Mini is uniformly distributed between 8.5 and 12 hours. a. Give a mathematical expression for the probability density function of battery life. b. What is the probability that the battery life for an iPad Mini will be 10 hours or less? c. What is the probability that the battery life for an iPad Mini will be at least 11 hours? d. What is the probability that the battery life for an iPad Mini will be between 9.5 and 11.5 hours? e. In a shipment of 100 iPad Minis, how many should have a battery life of at least 9 hours? 6. A Gallup Daily Tracking Survey found that the mean daily discretionary spending by Americans earning over $90,000 per year was $136 per day (USA Today, July 30, 2012). The discretionary spending excluded home purchases, vehicle purchases, and regular monthly bills. Let x 5 the discretionary spending per day and assume that a uniform probability density function applies with f (x) 5 .00625 for a # x # b. a. Find the values of a and b for the probability density function. b. What is the probability that consumers in this group have daily discretionary spending between $100 and $200? c. What is the probability that consumers in this group have daily discretionary spending of $150 or more? d. What is the probability that consumers in this group have daily discretionary spending of $80 or less? 7. Suppose we are interested in bidding on a piece of land and we know one other bidder is interested.1 The seller announced that the highest bid in excess of $10,000 will be accepted. Assume that the competitor’s bid x is a random variable that is uniformly distributed between $10,000 and $15,000. a. Suppose you bid $12,000. What is the probability that your bid will be accepted? b. Suppose you bid $14,000. What is the probability that your bid will be accepted? c. What amount should you bid to maximize the probability that you get the property? d. Suppose you know someone who is willing to pay you $16,000 for the property. Would you consider bidding less than the amount in part (c)? Why or why not?

Normal Probability Distribution 6.2 Abraham de Moivre, a French mathematician, published The Doctrine of Chances in 1733. He derived the normal distribution.

The most important probability distribution for describing a continuous random variable is the normal probability distribution. The normal distribution has been used in a wide variety of practical applications in which the random variables are heights and weights of people, test scores, scientific measurements, amounts of rainfall, and other similar values. It is also widely used in statistical inference, which is the major topic of the remainder of this book. In such applications, the normal distribution provides a description of the likely results obtained through sampling.

Normal Curve The form, or shape, of the normal distribution is illustrated by the bell-shaped normal curve in Figure 6.3. The probability density function that defines the bell-shaped curve of the normal distribution follows. 1

This exercise is based on a problem suggested to us by Professor Roger Myerson of Northwestern University.

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Chapter 6 Continuous Probability Distributions

FIGURE 6.3 BELL-SHAPED CURVE FOR THE NORMAL DISTRIBUTION

Standard Deviation

Mean

Normal Probability Density Function 1 x2 1 2 1 2 e 2 Ï2 2

f(x) 5

(6.2)

where 5 mean 5 standard deviation 5 3.14159 e 5 2.71828

We make several observations about the characteristics of the normal distribution. The normal curve has two parameters, and . They determine the location and shape of the normal distribution.

1. The entire family of normal distributions is differentiated by two parameters: the mean and the standard deviation . 2. The highest point on the normal curve is at the mean, which is also the median and mode of the distribution. 3. The mean of the distribution can be any numerical value: negative, zero, or positive. Three normal distributions with the same standard deviation but three different means (210, 0, and 20) are shown here.

210

4. T he normal distribution is symmetric, with the shape of the normal curve to the left of the mean a mirror image of the shape of the normal curve to the right of the mean. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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6.2 Normal Probability Distribution

The tails of the normal curve extend to infinity in both directions and theoretically never touch the horizontal axis. Because it is symmetric, the normal distribution is not skewed; its skewness measure is zero. 5. The standard deviation determines how flat and wide the normal curve is. Larger values of the standard deviation result in wider, flatter curves, showing more variability in the data. Two normal distributions with the same mean but with different standard deviations are shown here.

5 10

6. P robabilities for the normal random variable are given by areas under the normal curve. The total area under the curve for the normal distribution is 1. Because the distribution is symmetric, the area under the curve to the left of the mean is .50 and the area under the curve to the right of the mean is .50. 7. The percentage of values in some commonly used intervals are a. 68.3% of the values of a normal random variable are within plus or minus one standard deviation of its mean; b. 95.4% of the values of a normal random variable are within plus or minus two standard deviations of its mean; c. 99.7% of the values of a normal random variable are within plus or minus three standard deviations of its mean.

These percentages are the basis for the empirical rule introduced in Section 3.3.

Figure 6.4 shows properties (a), (b), and (c) graphically.

Standard Normal Probability Distribution A random variable that has a normal distribution with a mean of zero and a standard deviation of one is said to have a standard normal probability distribution. The letter z is commonly used to designate this particular normal random variable. Figure 6.5 is the graph of the standard normal distribution. It has the same general appearance as other normal distributions, but with the special properties of 5 0 and 5 1. Because 5 0 and 5 1, the formula for the standard normal probability density function is a simpler version of equation (6.2). Standard Normal Density Function

f(z) 5

1 Ï2

z2 2

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Chapter 6 Continuous Probability Distributions

FIGURE 6.4 AREAS UNDER THE CURVE FOR ANY NORMAL DISTRIBUTION 99.7% 95.4% 68.3%

2 1

2 3

2 2

1 1

1 3

1 2

FIGURE 6.5 THE STANDARD NORMAL DISTRIBUTION

For the normal probability density function, the height of the normal curve varies and more advanced mathematics is required to compute the areas that represent probability.

Because the standard normal random variable is continuous, P(z # 1.00) 5 P(z , 1.00).

As with other continuous random variables, probability calculations with any normal distribution are made by computing areas under the graph of the probability density function. Thus, to find the probability that a normal random variable is within any specific interval, we must compute the area under the normal curve over that interval. For the standard normal distribution, areas under the normal curve have been computed and are available in tables that can be used to compute probabilities. Such a table appears on the two pages inside the front cover of the text. The table on the left-hand page contains areas, or cumulative probabilities, for z values less than or equal to the mean of zero. The table on the right-hand page contains areas, or cumulative probabilities, for z values greater than or equal to the mean of zero. The three types of probabilities we need to compute include (1) the probability that the standard normal random variable z will be less than or equal to a given value; (2) the probability that z will be between two given values; and (3) the probability that z will be greater than or equal to a given value. To see how the cumulative probability table for the standard normal distribution can be used to compute these three types of probabilities, let us consider some examples. We start by showing how to compute the probability that z is less than or equal to 1.00; that is, P(z # 1.00). This cumulative probability is the area under the normal curve to the left of z 5 1.00 in the following graph.

295

6.2 Normal Probability Distribution

P(z # 1.00)

Refer to the right-hand page of the standard normal probability table inside the front cover of the text. The cumulative probability corresponding to z 5 1.00 is the table value located at the intersection of the row labeled 1.0 and the column labeled .00. First we find 1.0 in the left column of the table and then find .00 in the top row of the table. By looking in the body of the table, we find that the 1.0 row and the .00 column intersect at the value of .8413; thus, P(z # 1.00) 5 .8413. The following excerpt from the probability table shows these steps.

z .00 .01 .02 . . . .9 .8159 .8186 .8212 1.0 .8413 .8438 .8461 1.1 .8643 .8665 .8686 1.2 .8849 .8869 .8888 . . .

P(z # 1.00)

To illustrate the second type of probability calculation, we show how to compute the probability that z is in the interval between 2.50 and 1.25; that is, P(2.50 # z # 1.25). The following graph shows this area, or probability.

P(2.50 # z # 1.25) P(z , 2.50)

2.50 0

1.25

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Chapter 6 Continuous Probability Distributions

Three steps are required to compute this probability. First, we find the area under the normal curve to the left of z 5 1.25. Second, we find the area under the normal curve to the left of z 5 2.50. Finally, we subtract the area to the left of z 5 2.50 from the area to theleft of z 5 1.25 to find P(2.50 # z # 1.25). To find the area under the normal curve to the left of z 5 1.25, we first locate the 1.2 row in the standard normal probability table and then move across to the .05 column. Because the table value in the 1.2 row and the .05 column is .8944, P(z # 1.25) 5 .8944. Similarly, to find the area under the curve to the left of z 5 2.50, we use the left-hand page of the table to locate the table value in the 2.5 row and the .00 column; with a table value of .3085, P(z # 2.50) 5 .3085. Thus, P(2.50 # z # 1.25) 5 P(z # 1.25) 2 P(z # 2.50) 5 .8944 2 .3085 5 .5859. Let us consider another example of computing the probability that z is in the interval between two given values. Often it is of interest to compute the probability that a normal random variable assumes a value within a certain number of standard deviations of the mean. Suppose we want to compute the probability that the standard normal random variable is within one standard deviation of the mean; that is, P(21.00 # z # 1.00). To compute this probability we must find the area under the curve between 21.00 and 1.00. Earlier we found that P(z # 1.00) 5 .8413. Referring again to the table inside the front cover of the book, we find that the area under the curve to the left of z 5 21.00 is .1587, so P(z # 21.00) 5 .1587. Therefore, P(21.00 # z # 1.00) 5 P(z # 1.00) 2 P(z # 21.00) 5 .8413 2 .1587 5 .6826. This probability is shown graphically in the following figure.

P(21.00 # z # 1.00) 5 .8413 2 .1587 5 .6826

P(z # 2 –1.00) 5 .1587

21.00

1.00

To illustrate how to make the third type of probability computation, suppose we want to compute the probability of obtaining a z value of at least 1.58; that is, P(z $ 1.58). The value in the z 5 1.5 row and the .08 column of the cumulative normal table is .9429; thus, P(z , 1.58) 5 .9429. However, because the total area under the normal curve is 1, P(z $ 1.58) 5 1 2 .9429 5 .0571. This probability is shown in the following figure.

P(z , 1.58) 5 .9429 P(z $ 1.58) 5 1.0000 2 .9429 5 .0571

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6.2 Normal Probability Distribution

In the preceding illustrations, we showed how to compute probabilities given specified z values. In some situations, we are given a probability and are interested in working backward to find the corresponding z value. Suppose we want to find a z value such that the probability of obtaining a larger z value is .10. The following figure shows this situation graphically.

Probability 5 .10

What is this z value?

Given a probability, we can use the standard normal table in an inverse fashion to find the corresponding z value.

This problem is the inverse of those in the preceding examples. Previously, we specified the z value of interest and then found the corresponding probability, or area. In this exa mple, we are given the probability, or area, and asked to find the corresponding z value. To do so, we use the standard normal probability table somewhat differently. Recall that the standard normal probability table gives the area under the curve to the left of a particular z value. We have been given the information that the area in the upper tail of the curve is .10. Hence, the area under the curve to the left of the unknown z value must equal .9000. Scanning the body of the table, we find that .8997 is the cumulative probability value closest to .9000. The section of the table providing this result follows.

z .06 .07 .08 .09 . . . 1.0 .8554 .8577 .8599 .8621 1.1 .8770 .8790 .8810 .8830 1.2 .8962 .8980 .9015 .8997 1.3 .9131 .9147 .9162 .9177 1.4 .9279 .9292 .9306 .9319 . . Cumulative probability value . closest to .9000

Reading the z value from the leftmost column and the top row of the table, we find that the corresponding z value is 1.28. Thus, an area of approximately .9000 (actually Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Chapter 6 Continuous Probability Distributions

.8997) will be to the left of z 5 1.28.2 In terms of the question originally asked, there is an approximately .10 probability of a z value larger than 1.28. The examples illustrate that the table of cumulative probabilities for the standard normal probability distribution can be used to find probabilities associated with values of the standard normal random variable z. Two types of questions can be asked. The first type of question specifies a value, or values, for z and asks us to use the table to determine the corresponding areas or probabilities. The second type of question provides an area, or probability, and asks us to use the table to determine the corresponding z value. Thus, we need to be flexible in using the standard normal probability table to answer the desired probability question. In most cases, sketching a graph of the standard normal probability distribution and shading the appropriate area will help to visualize the situation and aid in determining the correct answer.

Computing Probabilities for Any Normal Probability Distribution The reason for discussing the standard normal distribution so extensively is that probabilities for all normal distributions are computed by using the standard normal distribution. That is, when we have a normal distribution with any mean and any standard deviation , we answer probability questions about the distribution by first converting to the standard normal distribution. Then we can use the standard normal probability table and the appropriate z values to find the desired probabilities. The formula used to convert any normal random variable x with mean and standard deviation to the standard normal random variable z follows. The formula for the standard normal random variable is similar to the formula we introduced in Chapter 3 for computing z-scores for a data set.

Converting to the Standard Normal Random Variable

(6.3)

A value of x equal to its mean results in z 5 ( 2 )/ 5 0. Thus, we see that a value of x equal to its mean corresponds to z 5 0. Now suppose that x is one standard deviation above its mean; that is, x 5 1 . Applying equation (6.3), we see that the corresponding z value is z 5 [( 1 ) 2 ]/ 5 / 5 1. Thus, an x value that is one standard deviation above its mean corresponds to z 5 1. In other words, we can interpret z as the number of standard deviations that the normal random variable x is from its mean . To see how this conversion enables us to compute probabilities for any normal distribu tion, suppose we have a normal distribution with 5 10 and 5 2. What is the proba bility that the random variable x is between 10 and 14? Using equation (6.3), we see that at x 5 10, z 5 (x 2 )/ 5 (10 2 10)/2 5 0 and that at x 5 14, z 5 (14 2 10)/2 5 4/2 5 2. Thus, the answer to our question about the probability of x being between 10 and 14 is given by the equivalent probability that z is between 0 and 2 for the standard normal distribution. In other words, the probability that we are seeking is the probability that the random variable x is b etween its mean and two standard deviations above the mean. Using z 5 2.00 and the standard normal probability table inside the front cover of the text, we see 2 We could use interpolation in the body of the table to get a better approximation of the z value that corresponds to an area of .9000. Doing so to provide one more decimal place of accuracy would yield a z value of 1.282. However, in most practical situations, sufficient accuracy is obtained simply by using the table value closest to the desired probability.

299

6.2 Normal Probability Distribution

that P(z # 2) 5 .9772. Because P(z # 0) 5 .5000, we can compute P(.00 # z # 2.00) 5 P(z # 2) 2 P(z # 0) 5 .9772 2 .5000 5 .4772. Hence the probability that x is between 10 and 14 is .4772.

Grear Tire Company Problem We turn now to an application of the normal probability distribution. Suppose the Grear Tire Company developed a new steel-belted radial tire to be sold through a national chain of discount stores. Because the tire is a new product, Grear’s managers believe that the mileage guarantee offered with the tire will be an important factor in the acceptance of the product. Before finalizing the tire mileage guarantee policy, Grear’s managers want probability information about the number of miles the tires will last. Let x denote the number of miles the tire lasts. From actual road tests with the tires, Grear’s engineering group estimated that the mean tire mileage is 5 36,500 miles and that the standard deviation is 5 5000. In addition, the data collected indicate that a normal distribution is a reasonable assumption. What percentage of the tires can be expected to last more than 40,000 miles? In other words, what is the probability that the tire mileage, x, will exceed 40,000? This question can be answered by finding the area of the darkly shaded region in Figure 6.6. At x 5 40,000, we have

x 2 40,000 2 36,500 3500 5 5 5 .70 5000 5000

Refer now to the bottom of Figure 6.6. We see that a value of x 5 40,000 on the Grear Tire normal distribution corresponds to a value of z 5 .70 on the standard normal distribution. Using the standard normal probability table, we see that the area under the standard normal curve to the left of z 5 .70 is .7580. Thus, 1.000 2 .7580 5 .2420 is the probability that z will exceed .70 and hence x will exceed 40,000. We can conclude that about 24.2% of the tires will exceed 40,000 in mileage. Let us now assume that Grear is considering a guarantee that will provide a discount on replacement tires if the original tires do not provide the guaranteed mileage. What should FIGURE 6.6 GREAR TIRE COMPANY MILEAGE DISTRIBUTION

P(x , 40,000)

5 5000

P(x $ 40,000) 5 ?

40,000

5 36,500 0 Note: z 5 0 corresponds to x 5 5 36,500

.70

z Note: z 5 .70 corresponds to x 5 40,000

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Chapter 6 Continuous Probability Distributions

FIGURE 6.7 GREAR’S DISCOUNT GUARANTEE

5 5000 10% of tires eligible for discount guarantee

x Guarantee mileage 5 ?

5 36,500

the guaranteed mileage be if Grear wants no more than 10% of the tires to be eligible for the discount guarantee? This question is interpreted graphically in Figure 6.7. According to Figure 6.7, the area under the curve to the left of the unknown guaranteed mileage must be .10. So, we must first find the z value that cuts off an area of .10 in the left tail of a standard normal distribution. Using the standard normal probability table, we see that z 5 21.28 cuts off an area of .10 in the lower tail. Hence, z 5 21.28 is the value of thestandard normal random variable corresponding to the desired mileage guarantee on the Grear Tire normal distribution. To find the value of x corresponding to z 5 21.28, wehave The guaranteed mileagewe need to find is 1.28 standard deviationsbelow the mean. Thus, x 5 2 1.28.

With 5 36,500 and 5 5000,

With the guarantee set at 30,000 miles, the actual percentage eligible for the guarantee will be 9.68%.

x2 5 21.28 x 2 5 21.28 x 5 2 1.28

x 5 36,500 2 1.28(5000) 5 30,100

Thus, a guarantee of 30,100 miles will meet the requirement that approximately 10% of the tires will be eligible for the guarantee. Perhaps, with this information, the firm will set its tire mileage guarantee at 30,000 miles. Again, we see the important role that probability distributions play in providing decision-making information. Namely, once a probability distribution is established for a particular application, it can be used to obtain probability information about the problem. Probability does not make a decision recommendation directly, but it provides information that helps the decision maker better understand the risks and uncertainties associated with the problem. Ultimately, this information may assist the decision maker in reaching a good decision.

Using Excel to Compute Normal Probabilities Excel provides two functions for computing probabilities and z values for a standard normal probability distribution: NORM.S.DIST and NORM.S.INV. The NORM.S.DIST function computes the cumulative probability given a z value, and the NORM.S.INV function Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

6.2 Normal Probability Distribution

The letter S that appears in the name of the NORM.S.DIST and NORM.S.INV functions reminds us that these functions relate to the standard normal probability distribution.

301

computes the z value given a cumulative probability. Two similar functions, NORM.DIST and NORM.INV, are available for computing the cumulative probability and the x value for any normal distribution. We begin by showing how to use the NORM.S.DIST and NORM.S.INV functions. The NORM.S.DIST function provides the area under the standard normal curve to the left of a given z value; thus, it provides the same cumulative probability we would obtain if we used the standard normal probability table inside the front cover of the text. Using the NORM.S.DIST function is just like having Excel look up cumulative normal probabilities for you. The NORM.S.INV function is the inverse of the NORM.S.DIST function; it takes a cumulative probability as input and provides the z value corresponding to that cumulative probability. Let’s see how both of these functions work by computing the probabilities and z values obtained earlier in this section using the standard normal probability table. Refer to Figure 6.8 as we describe the tasks involved. The formula worksheet is in the background; the value worksheet is in the foreground. Enter/Access Data: Open a blank worksheet. No data are entered in the worksheet. We will simply enter the appropriate z values and probabilities directly into the formulas as needed. Enter Functions and Formulas: The NORM.S.DIST function has two inputs: the z value and a value of TRUE or FALSE. For the second input we enter TRUE if a cumulative probability is desired, and we enter FALSE if the height of the standard normal curve is desired. Because we will always be using NORM.S.DIST to compute cumulative probabilities, we always choose TRUE for the second input. To illustrate the use of the NORM.S.DIST function, we compute the four probabilities shown in cells D3:D6 of Figure 6.8.

The probabilities in cells D4, 0.5858, and D5, 0.6827, differ from what we computed earlier due to rounding.

To compute the cumulative probability to the left of a given z value (area in lower tail), we simply evaluate NORM.S.DIST at the z value. For instance, to compute P(z # 1) we entered the formula 5NORM.S.DIST(1,TRUE) into cell D3. The result, .8413, is the same as o btained using the standard normal probability table. To compute the probability of z being in an interval we compute the value of NORM.S.DIST at the upper endpoint of the interval and subtract the value of NORM.S.DIST

FIGURE 6.8 EXCEL WORKSHEET FOR COMPUTING PROBABILITIES AND z VALUES

FOR THE STANDARD NORMAL DISTRIBUTION

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Chapter 6 Continuous Probability Distributions

at the lower endpoint of the interval. For instance, to find P(2.50 # z # 1.25), we entered the formula 5NORM.S.DIST(1.25,TRUE)-NORM.S.DIST(-.50,TRUE) into cell D4. The interval probability in cell D5 is computed in a similar fashion. To compute the probability to the right of a given z value (upper tail area), we must subtract the cumulative probability represented by the area under the curve below the z value (lower tail area) from 1. For example, to compute P(z $ 1.58) we entered the formula 51-NORM.S.DIST(1.58,TRUE) into cell D6. To compute the z value for a given cumulative probability (lower tail area), we use the NORM.S.INV function. To find the z value corresponding to an upper tail probability of .10, we note that the corresponding lower tail area is .90 and enter the formula 5NORM.S.INV(0.9) into cell D11. Actually, NORM.S.INV(0.9) gives us the z value providing a cumulative probability (lower tail area) of .9. But it is also the z value associated with an upper tail area of .10. Two other z values are computed in Figure 6.8. These z values will be used extensively in succeeding chapters. To compute the z value corresponding to an upper tail probability of .025, we entered the formula 5NORM.S.INV(0.975) into cell D12. To compute the z value corresponding to a lower tail probability of .025, we entered the formula 5NORM.S.INV(0.025) into cell D13. We see that z 5 1.96 corresponds to an upper tail probability of .025, and z 5 21.96 corresponds to a lower tail probability of .025. Let us now turn to the Excel functions for computing cumulative probabilities and x values for any normal distribution. The NORM.DIST function provides the area under the normal curve to the left of a given value of the random variable x; thus it provides cumulative probabilities. The NORM.INV function is the inverse of the NORM.DIST function; it takes a cumulative probability as input and provides the value of x corresponding to that cumulative probability. The NORM.DIST and NORM.INV functions do the same thing for any normal distribution that the NORM.S.DIST and NORM.S.INV functions do for the standard normal distribution. Let’s see how both of these functions work by computing probabilities and x values for the Grear Tire Company example introduced earlier in this section. Recall that the lifetime of a Grear tire has a mean of 36,500 miles and a standard deviation of 5000 miles. Refer to Figure 6.9 as we describe the tasks involved. The formula worksheet is in the background; the value worksheet is in the foreground. FIGURE 6.9 EXCEL WORKSHEET FOR COMPUTING PROBABILITIES AND x VALUES

FOR THE NORMAL DISTRIBUTION

6.2 Normal Probability Distribution

303

Enter/Access Data: Open a blank worksheet. No data are entered in the worksheet. We simply enter the appropriate x values and probabilities directly into the formulas as needed. Enter Functions and Formulas: The NORM.DIST function has four inputs: (1) the x value we want to compute the cumulative probability for, (2) the mean, (3) the standard deviation, and (4) a value of TRUE or FALSE. For the fourth input, we enter TRUE if a cumulative probability is desired, and we enter FALSE if the height of the curve is desired. Because we will always be using NORM.DIST to compute cumulative probabilities, we will always choose TRUE for the fourth input. To compute the cumulative probability to the left of a given x value (lower tail area), we simply evaluate NORM.DIST at the x value. For instance, to compute the probability that a Grear tire will last 20,000 miles or less, we entered the formula 5NORM.DIST(20000,36500,5000,TRUE) into cell D3. The value worksheet shows that this cumulative probability is .0005. So, we can conclude that almost all Grear tires will last at least 20,000 miles. To compute the probability of x being in an interval we compute the value of NORM. DIST at the upper endpoint of the interval and subtract the value of NORM.DIST at the lower endpoint of the interval. The formula in cell D4 provides the probability that a tire’s lifetime is between 20,000 and 40,000 miles, P(20,000 # x # 40,000). In the value worksheet, we see that this probability is .7576. To compute the probability to the right of a given x value (upper tail area), we must subtract the cumulative probability represented by the area under the curve below the x value (lower tail area) from 1. The formula in cell D5 computes the probability that a Grear tire will last for at least 40,000 miles. We see that this probability is .2420. To compute the x value for a given cumulative probability, we use the NORM.INV function. The NORM.INV function has only three inputs. The first input is the cumulative probability; the second and third inputs are the mean and standard deviation. For instance, to compute the tire mileage corresponding to a lower tail area of .1 for Grear Tire, we enter the formula 5NORM.INV(0.1,36500,5000) into cell D9. From the value worksheet, we see that 10% of the Grear tires will last for 30,092.24 miles or less. To compute the minimum tire mileage for the top 2.5% of Grear tires, we want to find the value of x corresponding to an area of .025 in the upper tail. This calculation is the same as finding the x value that provides a cumulative probability of .975. Thus we e ntered the formula 5NORM.INV(0.975,36500,5000) into cell D10 to compute this tire mileage. From the value worksheet, we see that 2.5% of the Grear tires will last at least 46,299.82 miles.

Exercises

Methods 8. Using Figure 6.4 as a guide, sketch a normal curve for a random variable x that has a mean of 5 100 and a standard deviation of 5 10. Label the horizontal axis with values of 70, 80, 90, 100, 110, 120, and 130. 9. A random variable is normally distributed with a mean of 5 50 and a standard deviation of 5 5. a. Sketch a normal curve for the probability density function. Label the horizontal axis with values of 35, 40, 45, 50, 55, 60, and 65. Figure 6.4 shows that the normal curve almost touches the horizontal axis at three standard deviations below and at three standard deviations above the mean (in this case at 35 and 65). b. What is the probability that the random variable will assume a value between 45 and 55? c. What is the probability that the random variable will assume a value between 40 and 60? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Chapter 6 Continuous Probability Distributions

10. Draw a graph for the standard normal distribution. Label the horizontal axis at values of 23, 22, 21, 0, 1, 2, and 3. Then compute the following probabilities. a. P(z # 1.5) b. P(z # 1) c. P(1 # z # 1.5) d. P(0 , z , 2.5) 11. Given that z is a standard normal random variable, compute the following probabilities. a. P(z # 21.0) b. P(z $ 21) c. P(z $ 21.5) d. P(22.5 # z) e. P(23 , z # 0) 12. Given that z is a standard normal random variable, compute the following probabilities. a. P(0 # z # .83) b. P(21.57 # z # 0) c. P(z . .44) d. P(z $ 2.23) e. P(z , 1.20) f. P(z # 2.71) 13. Given that z is a standard normal random variable, compute the following probabilities. a. P(21.98 # z # .49) b. P(.52 # z # 1.22) c. P(21.75 # z # 21.04) 14. Given that z is a standard normal random variable, find z for each situation. a. The area to the left of z is .9750. b. The area between 0 and z is .4750. c. The area to the left of z is .7291. d. The area to the right of z is .1314. e. The area to the left of z is .6700. f. The area to the right of z is .3300. 15. Given that z is a standard normal random variable, find z for each situation. a. The area to the left of z is .2119. b. The area between 2z and z is .9030. c. The area between 2z and z is .2052. d. The area to the left of z is .9948. e. The area to the right of z is .6915. 16. Given that z is a standard normal random variable, find z for each situation. a. The area to the right of z is .01. b. The area to the right of z is .025. c. The area to the right of z is .05. d. The area to the right of z is .10.

Applications 17. T he mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation Statistics website, November 2, 2012). Airfares were based on the total ticket value, which consisted of the price charged by the airlines plus any additional taxes and fees. Assume domestic airfares are normally distributed with a standard deviation of $110. a. What is the probability that a domestic airfare is $550 or more? b. What is the probability that a domestic airfare is $250 or less? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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c. W hat if the probability that a domestic airfare is between $300 and $500? d. What is the cost for the highest 3% of domestic airfares? 18. T he average return for large-cap domestic stock funds over the three years 2009–2011 was 14.4% (AAII Journal, February 2012). Assume the three-year returns were normally distributed across funds with a standard deviation of 4.4%. a. What is the probability an individual large-cap domestic stock fund had a three-year return of at least 20%? b. What is the probability an individual large-cap domestic stock fund had a three-year return of 10% or less? c. How big does the return have to be to put a domestic stock fund in the top 10% for the three-year period? 19. A utomobile repair costs continue to rise with the average cost now at $367 per repair (U.S. News & World Report website, January 5, 2015). Assume that the cost for an automobile repair is normally distributed with a standard deviation of $88. Answer the following questions about the cost of automobile repairs. a. What is the probability that the cost will be more than $450? b. What is the probability that the cost will be less than $250? c. What is the probability that the cost will be between $250 and $450? d. If the cost for your car repair is in the lower 5% of automobile repair charges, what is your cost? 20. T he average price for a gallon of gasoline in the United States is $3.73 and in Russia it is $3.40 (Bloomberg Businessweek, March 5–March 11, 2012). Assume these averages are the population means in the two countries and that the probability distributions are normally distributed with a standard deviation of $.25 in the United States and a standard deviation of $.20 in Russia. a. What is the probability that a randomly selected gas station in the United States charges less than $3.50 per gallon? b. What percentage of the gas stations in Russia charge less than $3.50 per gallon? c. What is the probability that a randomly selected gas station in Russia charged more than the mean price in the United States? 21. A person must score in the upper 2% of the population on an IQ test to qualify for membership in Mensa, the international high-IQ society. There are 110,000 Mensa members in 100 countries throughout the world (Mensa International website, January 8, 2013). If IQ scores are normally distributed with a mean of 100 and a standard deviation of 15, what score must a person have to qualify for Mensa? 22. T elevision viewing reached a new high when the Nielsen Company reported a mean daily viewing time of 8.35 hours per household (USA Today, November 11, 2009). Use a normal probability distribution with a standard deviation of 2.5 hours to answer the following questions about daily television viewing per household. a. What is the probability that a household views television between 5 and 10 hours a day? b. How many hours of television viewing must a household have in order to be in the top 3% of all television viewing households? c. What is the probability that a household views television more than 3 hours a day? 23. T he time needed to complete a final examination in a particular college course is normally distributed with a mean of 80 minutes and a standard deviation of 10 minutes. Answer the following questions. a. What is the probability of completing the exam in one hour or less? b. What is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes? c. Assume that the class has 60 students and that the examination period is 90 minutes in length. How many students do you expect will be unable to complete the exam in the allotted time? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Chapter 6 Continuous Probability Distributions

24. T he American Automobile Association (AAA) reported that families planning to travel over the Labor Day weekend would spend an average of $749 (The Associated Press, August 12, 2012). Assume that the amount spent is normally distributed with a standard deviation of $225. a. W hat the probability of family expenses for the weekend being less that $400? hat is the probability of family expenses for the weekend being $800 or more? b. W c. W hat is the probability that family expenses for the weekend will be between $500 and $1000? hat would the Labor Day weekend expenses have to be for the 5% of the families d. W with the most expensive travel plans? 25. N ew York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night (USA Today, April 30, 2012). Assume that room rates are normally distributed with a standard deviation of $55. a. What is the probability that a hotel room costs $225 or more per night? b. What is the probability that a hotel room costs less than $140 per night? c. What is the probability that a hotel room costs between $200 and $300 per night? d. What is the cost of the most expensive 20% of hotel rooms in New York City?

Exponential Probability Distribution 6.3 The exponential probability distribution may be used for random variables such as the time between arrivals at a car wash, the time required to load a truck, the distance between major defects in a highway, and so on. The exponential probability density function follows.

Exponential Probability Density Function

ƒ(x) 5

1 2x/ e

for x $ 0

(6.4)

where 5 expected value or mean e 5 2.71828

As an example of the exponential distribution, suppose that x represents the loading time for a truck at the Schips loading dock and follows such a distribution. If the mean, or average, loading time is 15 minutes ( 5 15), the appropriate probability density function for x is

ƒ(x) 5

1 2x/15 e 15

Figure 6.10 is the graph of this probability density function.

307

6.3 Exponential Probability Distribution

FIGURE 6.10 EXPONENTIAL DISTRIBUTION FOR THE SCHIPS LOADING DOCK EXAMPLE f (x) .07 P(x # 6) .05 P(6 # x # 18) .03 .01 0

12 18 24 Loading Time

Computing Probabilities for the Exponential Distribution In waiting line applications, the exponential distribution is often used for service time.

As with any continuous probability distribution, the area under the curve corresponding to an interval provides the probability that the random variable assumes a value in that interval. In the Schips loading dock example, the probability that loading a truck will take 6 minutes or less P(x # 6) is defined to be the area under the curve in Figure 6.10 from x 5 0 to x 5 6. Similarly, the probability that the loading time will be 18 minutes or less P(x # 18) is the area under the curve from x 5 0 to x 5 18. Note also that the probability that the loading time will be between 6 minutes and 18 minutes P(6 # x # 18) is given by the area under the curve from x 5 6 to x 5 18. To compute exponential probabilities such as those just described, we use the following formula. It provides the cumulative probability of obtaining a value for the exponential random variable of less than or equal to some specific value denoted by x0.

Exponential Distribution: Cumulative Probabilities

P(x # x0) 5 1 2 e2x0 /

(6.5)

For the Schips loading dock example, x 5 loading time in minutes and 5 15 minutes. Using equation (6.5),

P(x # x0) 5 1 2 e2x0 /15

Hence, the probability that loading a truck will take 6 minutes or less is

P(x # 6) 5 1 2 e26 /15 5 .3297

Using equation (6.5), we calculate the probability of loading a truck in 18 minutes or less.

P(x # 18) 5 1 2 e218 /15 5 .6988

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Chapter 6 Continuous Probability Distributions

A property of the exponential distribution is that the mean and standard deviation are equal.

Thus, the probability that loading a truck will take between 6 minutes and 18 minutes is equal to .6988 2 .3297 5 .3691. Probabilities for any other interval can be computed similarly. In the preceding example, the mean time it takes to load a truck is 5 15 minutes. A property of the exponential distribution is that the mean of the distribution and the standard deviation of the distribution are equal. Thus, the standard deviation for the time it takes to load a truck is 5 15 minutes. The variance is 2 5 (15)2 5 225.

Relationship Between the Poisson and Exponential Distributions In Section 5.5 we introduced the Poisson distribution as a discrete probability distribution that is often useful in examining the number of occurrences of an event over a specified interval of time or space. Recall that the Poisson probability function is

f (x) 5

xe2 x!

where If arrivals follow a Poisson distribution, the time between arrivals must follow an exponential distribution.

5 expected value or mean number of occurrences over a specified interval

The continuous exponential probability distribution is related to the discrete Poisson dis tribution. If the Poisson distribution provides an appropriate description of the number of occurrences per interval, the exponential distribution provides a description of the length of the interval between occurrences. To illustrate this relationship, suppose the number of cars that arrive at a car wash during one hour is described by a Poisson probability distribution with a mean of 10 cars per hour. The Poisson probability function that gives the probability of x arrivals per hour is

f (x) 5

10 xe210 x!

Because the average number of arrivals is 10 cars per hour, the average time between cars arriving is

1 hour 5 .1 hour/car 10 cars

Thus, the corresponding exponential distribution that describes the time between the arrivals has a mean of 5 .1 hour per car; as a result, the appropriate exponential proba bility density function is

f (x) 5

1 2x/.1 e 5 10e210x .1

Using Excel to Compute Exponential Probabilities Excel’s EXPON.DIST function can be used to compute exponential probabilities. We will illustrate by computing probabilities associated with the time it takes to load a truck at the Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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6.3 Exponential Probability Distribution

FIGURE 6.11 EXCEL WORKSHEET FOR COMPUTING PROBABILITIES FOR THE EXPONENTIAL

PROBABILITY DISTRIBUTION

Schips loading dock. This example was introduced at the beginning of the section. Refer to Figure 6.11 as we describe the tasks involved. The formula worksheet is in the background; the value worksheet is in the foreground. Enter/Access Data: Open a blank worksheet. No data are entered in the worksheet. We simply enter the appropriate values for the exponential random variable into the formulas as needed. The random variable is x 5 loading time. Enter Functions and Formulas: The EXPON.DIST function has three inputs: The first is the value of x, the second is 1/, and the third is TRUE or FALSE. We choose TRUE for the third input if a cumulative probability is desired and FALSE if the height of the probability density function is desired. We will always use TRUE because we will be computing cumulative probabilities. The first probability we compute is the probability that the loading time is 18 minutes or less. For the Schips problem, 1/ 5 1/15, so we enter the formula 5EXPON.DIST (18,1/15,TRUE) into cell D3 to compute the desired cumulative probability. From the value worksheet, we see that the probability of loading a truck in 18 minutes or less is .6988. The second probability we compute is the probability that the loading time is between 6 and 18 minutes. To find this probability we first compute the cumulative probability for the upper endpoint of the time interval and subtract the cumulative probability for the lower endpoint of the interval. The formula we have entered into cell D4 calculates this probability. The value worksheet shows that this probability is .3691. The last probability we calculate is the probability that the loading time is at least 8 minutes. Because the EXPON.DIST function computes only cumulative (lower tail) probabilities, we compute this probability by entering the formula 51-EXPON .DIST(8,1/15,TRUE) into cell D5. The value worksheet shows that the probability of a loading time of 8 minutes or more is .5866.

NOTE AND COMMENT As we can see in Figure 6.10, the exponential distribution is skewed to the right. Indeed, the skewness measure for the exponential

distributions is 2. The exponential distribution gives us a good idea what a skewed distribution looks like.

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Chapter 6 Continuous Probability Distributions

Exercises

Methods 26. Consider the following exponential probability density function.

f(x) 5 a. b. c. d.

1 2x/8 e 8

for x $ 0

Find P(x # 6). Find P(x # 4). Find P(x $ 6). Find P(4 # x # 6).

27. Consider the following exponential probability density function. a. b. c. d. e.

f(x) 5

1 2x/3 e 3

for x $ 0

rite the formula for P(x # x0 ). W Find P(x # 2). Find P(x $ 3). Find P(x # 5). Find P(2 # x # 5).

Applications 28. B attery life between charges for the Motorola Droid Razr Maxx is 20 hours when the primary use is talk time (The Wall Street Journal, March 7, 2012). The battery life drops to 7 hours when the phone is primarily used for Internet applications over cellular. Assume that the battery life in both cases follows an exponential distribution. a. Show the probability density function for battery life for the Droid Razr Maxx phone when its primary use is talk time. b. What is the probability that the battery charge for a randomly selected Droid Razr Maxx phone will last no more than 15 hours when its primary use is talk time? c. What is the probability that the battery charge for a randomly selected Droid Razr Maxx phone will last more than 20 hours when its primary use is talk time? d. What is the probability that the battery charge for a randomly selected Droid Razr Maxx phone will last no more than 5 hours when its primary use is Internet applications? 29. T he time between arrivals of vehicles at a particular intersection follows an exponential probability distribution with a mean of 12 seconds. a. Sketch this exponential probability distribution. b. What is the probability that the arrival time between vehicles is 12 seconds or less? c. What is the probability that the arrival time between vehicles is 6 seconds or less? d. What is the probability of 30 or more seconds between vehicle arrivals? 30. C omcast Corporation is the largest cable television company, the second largest Internet service provider, and the fourth largest telephone service provider in the United States. Generally known for quality and reliable service, the company periodically experiences unexpected service interruptions. On January 14, 2009, such an interruption occurred for the Comcast customers living in southwest Florida. When customers called the Comcast office, a recorded message told them that the company was aware of the service outage and that it was anticipated that service would be restored in two hours. Assume that two hours is the mean time to do the repair and that the repair time has an exponential probability d istribution. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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a. W hat is the probability that the cable service will be repaired in one hour or less? b. What is the probability that the repair will take between one hour and two hours? c. For a customer who calls the Comcast office at 1:00 p.m., what is the probability that the cable service will not be repaired by 5:00 p.m.? 31. W endy’s restaurant has been recognized for having the fastest average service time among fast food restaurants. In a benchmark study, Wendy’s average service time of 2.2 minutes was less than those of Burger King, Chick-fil-A, Krystal, McDonald’s, Taco Bell, and Taco John’s (QSR Magazine website, December 2014). Assume that the service time for Wendy’s has an exponential distribution. a. What is the probability that a service time is less than or equal to one minute? b. What is the probability that a service time is between 30 seconds and one minute? c. Suppose a manager of a Wendy’s is considering instituting a policy such that if the time it takes to serve you exceeds five minutes, your food is free. What is the probability that you will get your food for free? Comment. 32. T he Boston Fire Department receives 911 calls at a mean rate of 1.6 calls per hour (Mass.gov website, November 2012). Suppose the number of calls per hour follows a Poisson probability distribution. a. What is the mean time between 911 calls to the Boston Fire Department in minutes? b. Using the mean in part (a), show the probability density function for the time between 911 calls in minutes. c. What is the probability that there will be less than one hour between 911 calls? d. What is the probability that there will be 30 minutes or more between 911 calls? e. What is the probability that there will be more than 5 minutes but less than 20 minutes between 911 calls?

Summary This chapter extended the discussion of probability distributions to the case of continuous random variables. The major conceptual difference between discrete and continuous probability distributions involves the method of computing probabilities. With discrete distributions, the probability function f(x) provides the probability that the random variable x assumes various values. With continuous distributions, the probability density function f(x) does not provide probability values directly. Instead, probabilities are given by areas under the curve or graph of the probability density function f(x). Because the area under the curve above a single point is zero, we observe that the probability of any particular value is zero for a continuous random variable. Three continuous probability distributions—the uniform, normal, and exponential distributions—were treated in detail. The normal distribution is used widely in statistical inference and will be used extensively throughout the remainder of the text.

Glossary Exponential probability distribution A continuous probability distribution that is useful in computing probabilities for the time it takes to complete a task. Normal probability distribution A continuous probability distribution. Its probability density function is bell shaped and determined by its mean and standard deviation . Probability density function A function used to compute probabilities for a continuous random variable. The area under the graph of a probability density function over an interval represents probability. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Standard normal probability distribution A normal distribution with a mean of zero and a standard deviation of one. Uniform probability distribution A continuous probability distribution for which the probability that the random variable will assume a value in any interval is the same for each interval of equal length.

Key Formulas Uniform Probability Density Function

f(x) 5

1 b2a 0

for a # x # b

(6.1)

elsewhere

Normal Probability Density Function 1 x2 2 2 1 1 e 2 Ï2 2

f(x) 5

(6.2)

Converting to the Standard Normal Random Variable

(6.3)

Exponential Probability Density Function

f(x) 5

1 2 x/ e

for x $ 0

(6.4)

Exponential Distribution: Cumulative Probabilities

P(x # x0) 5 1 2 e2x0 /

(6.5)

Supplementary Exercises 33. A business executive, transferred from Chicago to Atlanta, needs to sell her house in Chicago quickly. The executive’s employer has offered to buy the house for $210,000, but the offer expires at the end of the week. The executive does not currently have a better offer but can afford to leave the house on the market for another month. From conversations with her realtor, the executive believes the price she will get by leaving the house on the market for another month is uniformly distributed between $200,000 and $225,000. a. If she leaves the house on the market for another month, what is the mathematical expression for the probability density function of the sales price? b. If she leaves it on the market for another month, what is the probability that she will get at least $215,000 for the house? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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c. I f she leaves it on the market for another month, what is the probability that she will get less than $210,000? d. Should the executive leave the house on the market for another month? Why or why not? 34. T he NCAA estimates that the yearly value of a full athletic scholarship at in-state public universities is $19,000 (The Wall Street Journal, March 12, 2012). Assume the scholarship value is normally distributed with a standard deviation of $2100. a. For the 10% of athletic scholarships of least value, how much are they worth? b. What percentage of athletic scholarships are valued at $22,000 or more? c. For the 3% of athletic scholarships that are most valuable, how much are they worth? 35. M otorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process p roduces items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected number of defects for a 1000-unit production run in the following situations. a. The process standard deviation is .15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects. b. Through process design improvements, the process standard deviation can be reduced to .05. Assume the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects. c. What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean? 36. D uring early 2012, economic hardship was stretching the limits of France’s welfare system. One indicator of the level of hardship was the increase in the number of people bringing items to a Paris pawnbroker. That number had risen to 658 per day (Bloomberg Businessweek, March 5–March 11, 2012). Assume the number of people bringing items to the pawnshop per day in 2012 is normally distributed with a mean of 658. a. Suppose you learn that on 3% of the days, 610 or fewer people brought items to the pawnshop. What is the standard deviation of the number of people bringing items to the pawnshop per day? b. On any given day, what is the probability that between 600 and 700 people bring items to the pawnshop? c. How many people bring items to the pawnshop on the busiest 3% of days? 37. T he port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest bulk cargo port in the world. The U.S. Army Corps of Engineers reports that the port handles a mean of 4.5 million tons of cargo per week (USA Today, September 25, 2012). Assume that the number of tons of cargo handled per week is normally distributed with a standard deviation of .82 million tons. a. What is the probability that the port handles less than 5 million tons of cargo per week? b. What is the probability that the port handles 3 million or more tons of cargo per week? c. What is the probability that the port handles between 3 million and 4 million tons of cargo per week? d. Assume that 85% of the time the port can handle the weekly cargo volume without extending operating hours. What is the number of tons of cargo per week that will require the port to extend its operating hours? 38. W ard Doering Auto Sales is considering offering a special service contract that will cover the total cost of any service work required on leased vehicles. From experience, the company manager estimates that yearly service costs are approximately normally distributed, with a mean of $150 and a standard deviation of $25. a. If the company offers the service contract to customers for a yearly charge of $200, what is the probability that any one customer’s service costs will exceed the contract price of $200? b. What is Ward’s expected profit per service contract? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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39. T he XO Group Inc. conducted a survey of 13,000 brides and grooms married in the United States and found that the average cost of a wedding is $29,858 (XO Group website, January 5, 2015). Assume that the cost of a wedding is normally distributed with a mean of $29,858 and a standard deviation of $5600. a. What is the probability that a wedding costs less than $20,000? b. What is the probability that a wedding costs between $20,000 and $30,000? c. For a wedding to be among the 5% most expensive, how much would it have to cost? 40. Assume that the test scores from a college admissions test are normally distributed, with a mean of 450 and a standard deviation of 100. a. What percentage of the people taking the test score between 400 and 500? b. Suppose someone receives a score of 630. What percentage of the people taking the test score better? What percentage score worse? c. If a particular university will not admit anyone scoring below 480, what percentage of the persons taking the test would be acceptable to the university? 41. According to the National Association of Colleges and Employers, the average starting salary for new college graduates in health sciences is $51,541. The average starting salary for new college graduates in business is $53,901 (National Association of Colleges and Employers website, January 5, 2015). Assume that starting salaries are normally distributed and that the standard deviation for starting salaries for new college graduates in health sciences is $11,000. Assume that the standard deviation for starting salaries for new college graduates in business is $15,000. a. What is the probability that a new college graduate in business will earn a starting salary of at least $65,000? b. What is the probability that a new college graduate in health sciences will earn a starting salary of at least $65,000? c. What is the probability that a new college graduate in health sciences will earn a starting salary less than $40,000? d. How much would a new college graduate in business have to earn in order to have a starting salary higher than 99% of all starting salaries of new college graduates in the health sciences? 42. A machine fills containers with a particular product. The standard deviation of filling weights is known from past data to be .6 ounce. If only 2% of the containers hold less than 18 ounces, what is the mean filling weight for the machine? That is, what must equal? Assume the filling weights have a normal distribution. 43. The American Time Use Study is a survey of people in the United States regarding the amount of time they spend on various activities in a typical day. On a given day, 22% of American men participate in sports, exercise, or recreation. On days they participated in these activities, men participated on average 1.8 hours in sports, exercise, or recreation (Bureau of Labor Statistics website, December 2015). Suppose that the time spent on sports, exercise, and recreation for men on days of participation is distributed exponentially. a. What is the probability that when a man participates in sports, exercise, or recreation, he does so for less than or equal to 1 hour? b. What is the probability that when a man participates in sports, exercise, or recreation, he does so for more than 2 hours? 44. A website for bed and breakfast inns gets approximately seven visitors per minute. Suppose the number of website visitors per minute follows a Poisson probability distribution. a. What is the mean time between visits to the website? b. Show the exponential probability density function for the time between website visits. c. What is the probability that no one will access the website in a 1-minute period? d. What is the probability that no one will access the website in a 12-second period? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Case Problem 1 Specialty Toys

45. Do you dislike waiting in line? Supermarket chain Kroger has used computer simulation and information technology to reduce the average waiting time for customers at 2300 stores. Using a new system called QueVision, which allows Kroger to better predict when shoppers will be checking out, the company was able to decrease average customer waiting time to just 26 seconds (InformationWeek website and The Wall Street Journal website, January 5, 2015). a. Assume that Kroger waiting times are exponentially distributed. Show the probability density function of waiting time at Kroger. b. What is the probability that a customer will have to wait between 15 and 30 seconds? c. What is the probability that a customer will have to wait more than 2 minutes? 46. The time (in minutes) between telephone calls at an insurance claims office has the following exponential probability distribution. f(x) 5 .50e2.50x

a. b. c. d.

for x $ 0

What is the mean time between telephone calls? What is the probability of having 30 seconds or less between telephone calls? What is the probability of having 1 minute or less between telephone calls? What is the probability of having 5 or more minutes without a telephone call?

Case Problem 1 Specialty Toys Specialty Toys, Inc. sells a variety of new and innovative children’s toys. Management learned that the preholiday season is the best time to introduce a new toy, because many families use this time to look for new ideas for December holiday gifts. When Specialty discovers a new toy with good market potential, it chooses an October market entry date. In order to get toys into its stores by October, Specialty places one-time orders with its manufacturers in June or July of each year. Demand for children’s toys can be highly volatile. If a new toy catches on, a sense of shortage in the marketplace often increases the demand to high levels and large profits can be realized. However, new toys can also flop, leaving Specialty stuck with high levels of inventory that must be sold at reduced prices. The most important question the company faces is deciding how many units of a new toy should be purchased to meet anticipated sales demand. If too few are purchased, sales will be lost; if too many are purchased, profits will be reduced because of low prices realized in clearance sales. For the coming season, Specialty plans to introduce a new product called Weather Teddy. This variation of a talking teddy bear is made by a company in Taiwan. When a child presses Teddy’s hand, the bear begins to talk. A built-in barometer selects one of five responses that predict the weather conditions. The responses range from “It looks to be a very nice day! Have fun” to “I think it may rain today. Don’t forget your umbrella.” Tests with the product show that, even though it is not a perfect weather predictor, its predictions are surprisingly good. Several of Specialty’s managers claimed Teddy gave predictions of the weather that were as good as those of many local television weather forecasters. As with other products, Specialty faces the decision of how many Weather Teddy units to order for the coming holiday season. Members of the management team suggested o rder quantities of 15,000, 18,000, 24,000, or 28,000 units. The wide range of order quantities suggested indicates considerable disagreement concerning the market potential. Theproduct management team asks you for an analysis of the stock-out probabilities for various order quantities, an estimate of the profit potential, and help with making an order quantity recommendation. Specialty expects to sell Weather Teddy for $24 based on a cost of $16 per unit. If inventory remains after the holiday season, Specialty will sell all surplus inventory for $5 per unit. After reviewing the sales history of similar products, Specialty’s senior sales forecaster predicted an expected demand of 20,000 units with a .95 probability that demand would be between 10,000 units and 30,000 units. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Managerial Report Prepare a managerial report that addresses the following issues and recommends an order quantity for the Weather Teddy product. 1. U se the sales forecaster’s prediction to describe a normal probability distribution that can be used to approximate the demand distribution. Sketch the distribution and show its mean and standard deviation. 2. Compute the probability of a stock-out for the order quantities suggested by members of the management team. 3. Compute the projected profit for the order quantities suggested by the management team under three scenarios: worst case in which sales 5 10,000 units, most likely case in which sales 5 20,000 units, and best case in which sales 5 30,000 units. 4. One of Specialty’s managers felt that the profit potential was so great that the order quantity should have a 70% chance of meeting demand and only a 30% chance of any stock-outs. What quantity would be ordered under this policy, and what is the projected profit under the three sales scenarios? 5. Provide your own recommendation for an order quantity and note the associated profit projections. Provide a rationale for your recommendation.

Case Problem 2

Gebhardt Electronics Gebhardt Electronics produces a wide variety of transformers that it sells directly to manufacturers of electronics equipment. For one component used in several models of its transformers, Gebhardt uses a 3-foot length of .20 mm diameter solid wire made of pure Oxygen-Free Electronic (OFE) copper. A flaw in the wire reduces its conductivity and increases the likelihood it will break, and this critical component is difficult to reach and repair after a transformer has been constructed. Therefore, Gebhardt wants to use primarily flawless lengths of wire in making this component. The company is willing to accept no more than a 1 in 20 chance that a 3-foot length taken from a spool will be flawless. Gebhardt also occasionally uses smaller pieces of the same wire in the manufacture of other components, so the 3-foot segments to be used for this component are essentially taken randomly from a long spool of .20 mm diameter solid OFE copper wire. Gebhardt is now considering a new supplier for copper wire. This supplier claims that its spools of .20 mm diameter solid OFE copper wire average 50 inches between flaws. Gebhardt now must determine whether the new supply will be satisfactory if the supplier’s claim is valid.

Managerial Report In making this assessment for Gebhardt Electronics, consider the following three questions: 1. If the new supplier does provide spools of .20 mm solid OFE copper wire that average 50 inches between flaws, how is the length of wire between two consecutive flaws distributed? 2. Using the probability distribution you identified in (1), what is the probability that Gebhardt’s criteria will be met (i.e., a 1 in 20 chance that a randomly selected 3-foot segment of wire provided by the new supplier will be flawless)? 3. In inches, what is the minimum mean length between consecutive flaws that would result in satisfaction of Gebhardt’s criteria? 4. In inches, what is the minimum mean length between consecutive flaws that would result in a 1 in 100 chance that a randomly selected 3-foot segment of wire provided by the new supplier will be flawless? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

CHAPTER

Sampling and Sampling Distributions CONTENTS STATISTICS IN PRACTICE: MEADWESTVACO CORPORATION 7.1 THE ELECTRONICS ASSOCIATES SAMPLING PROBLEM 7.2

SELECTING A SAMPLE Sampling from a Finite Population Sampling from an Infinite Population

7.3 POINT ESTIMATION Practical Advice 7.4 INTRODUCTION TO SAMPLING DISTRIBUTIONS 7.5 SAMPLING DISTRIBUTION OF x Expected Value of x Standard Deviation of x Form of the Sampling Distribution of x Sampling Distribution of x for the EAI Problem Practical Value of the Sampling Distribution of x Relationship Between the Sample Size and the Sampling Distribution of x

7.6

SAMPLING DISTRIBUTION OF p Expected Value of p Standard Deviation of p Form of the Sampling Distribution of p Practical Value of the Sampling Distribution of p

7.7 OTHER SAMPLING METHODS Stratified Random Sampling Cluster Sampling Systematic Sampling Convenience Sampling Judgment Sampling 7.8 PRACTICAL ADVICE: BIG DATA AND ERRORS IN SAMPLING Sampling Error Nonsampling Error Implications of Big Data

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STATISTICS in PRACTICE MEADWESTVACO CORPORATION* MeadWestvaco Corporation, a leading producer of pack aging, coated and specialty papers, and specialty chemicals, employs more than 17,000 people. It operates worldwide in 30 countries and serves customers located in approximately 100 countries. MeadWestvaco’s inter nal consulting group uses sampling to provide a variety of information that enables the company to obtain signif icant productivity benefits and remain competitive. For example, MeadWestvaco maintains large wood land holdings, which supply the trees, or raw material, for many of the company’s products. Managers need r eliable and accurate information about the timberlands and for ests to evaluate the company’s ability to meet its future raw material needs. What is the present volume inthe for ests? What is the past growth of the forests? What is the projected future growth of the forests? With answers to these important questions MeadWestvaco’s managers can develop plans for the future, including long-term planting and harvesting schedules for the trees. How does MeadWestvaco obtain the information it needs about its vast forest holdings? Data collected from sample plots throughout the forests are the basis for learning about the population of trees owned by the company. To identify the sample plots, the timberland holdings are first divided into three sections based on location and types of trees. Using maps and random numbers, MeadWestvaco analysts identify random sam ples of 1/5- to 1/ 7-acre plots in each section of the forest. MeadWestvaco foresters collect data from these sample plots to learn about the forest population. *The authors are indebted to Dr. Edward P. Winkofsky for providing this Statistics in Practice.

STAMFORD, CONNECTICUT

Random sampling of its forest holdings enables MeadWestvaco Corporation to meet future raw material needs. Foresters throughout the organization participate in the field data collection process. Periodically, twoperson teams gather information on each tree in every sample plot. The sample data are entered into the company’s continuous forest inventory (CFI) computer system. Reports from the CFI system include a number of frequency distribution summaries containing statistics on types of trees, present forest volume, past forest growth rates, and projected future forest growth and volume. Sampling and the associated statistical summaries of the sample data provide the reports essential for the effective management of MeadWestvaco’s forests and timberlands. In this chapter you will learn about simple random sampling and the sample selection process. In addition, you will learn how statistics such as the sample mean and sample proportion are used to estimate the population mean and population proportion. The important concept of sampling distribution is also introduced.

In Chapter 1 we presented the following definitions of an element, a population, and a sample. An element is the entity on which data are collected. A population is the collection of all the elements of interest. A sample is a subset of the population. The reason we select a sample is to collect data to make an inference and answer research questions about a population. Let us begin by citing two examples in which sampling was used to answer a research question about a population. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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1. Members of a political party in Texas were considering supporting a particular candidate for election to the U.S. Senate, and party leaders wanted to estimate the proportion of registered voters in the state favoring the candidate. A sample of 400 registered voters in Texas was selected and 160 of the 400 voters indicated a pref erence for the candidate. Thus, an estimate of the proportion of the population of registered voters favoring the candidate is 160/400 5 .40. 2. A tire manufacturer is considering producing a new tire designed to provide an increase in mileage over the firm’s current line of tires. To estimate the mean useful life of the new tires, the manufacturer produced a sample of 120 tires for testing. The test results provided a sample mean of 36,500 miles. Hence, an estimate of the mean useful life for the population of new tires was 36,500 miles. A sample mean provides an estimate of a population mean, and a sample proportion provides an estimate of a population proportion. With estimates such as these, some estimation error can be expected. This chapter provides the basis for determining how large that error might be.

It is important to realize that sample results provide only estimates of the values of the corresponding population characteristics. We do not expect exactly .40, or 40%, of the population of registered voters to favor the candidate, nor do we expect the sample mean of 36,500 miles to exactly equal the mean mileage for the population of all new tires produced. The reason is simply that the sample contains only a portion of the population. Some sampling error is to be expected. With proper sampling methods, the sample results will provide “good” estimates of the population parameters. But how good can we expect the sample results to be? Fortunately, statistical procedures are available for answering this question. Let us define some of the terms used in sampling. The sampled population is the population from which the sample is drawn, and a frame is a list of the elements from which the sample will be selected. In the first example, the sampled population is all registered voters in Texas, and the frame is a list of all the registered voters. Because the number of registered voters in Texas is a finite number, the first example is an illustration of sampling from a finite population. In Section 7.2, we discuss how a simple random sample can be selected when sampling from a finite population. The sampled population for the tire mileage example is more difficult to define because the sample of 120 tires was obtained from a production process at a particular point in time. We can think of the sampled population as the conceptual population of all the tires that could have been made by the production process at that particular point in time. In this sense the sampled population is considered infinite, making it impossible to construct a frame from which to draw the sample. In Section 7.2, we discuss how to select a random sample in such a situation. In this chapter, we show how simple random sampling can be used to select a sample from a finite population and describe how a random sample can be taken from an infinite population that is generated by an ongoing process. We then show how data obtained from a sample can be used to compute estimates of a p opulation mean, a population standard deviation, and a population proportion. In addition, we introduce the important concept of a sampling distribution. As we will show, knowledge of the appropriate sampling distribution enables us to make statements about how close the sample estimates are to the correspond ing population parameters. In section 7.7, we discuss some alternatives to simple random sampling that are often employed in practice. In the last section, we discuss sampling and nonsampling error, and how these relate to large samples.

The Electronics Associates Sampling Problem 7.1 The director of personnel for Electronics Associates, Inc. (EAI), has been assigned the task of developing a profile of the company’s 2500 employees. The characteristics to be identi fied include the mean annual salary for the employees and the proportion of employees having completed the company’s management training program. Using the 2500 employees as the population for this study, we can find the annual sal ary and the training program status for each individual by referring to the firm’s personnel Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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EAI

records. The data set containing this information for all 2500 employees in the population is in the DATAfile named EAI. Using the EAI data and the formulas presented in Chapter 3, we compute the population mean and the population standard deviation for the annual salary data. Population mean: 5 $51,800 Population standard deviation: 5 $4000

Often the cost of collecting information from a sample is substantially less than from a population, especially when personal interviews must be conducted to collect the information.

The data for the training program status show that 1500 of the 2500 employees completed the training program. Numerical characteristics of a population are called parameters. Letting p denote the proportion of the population that completed the training program, we see that p 5 1500/2500 5 .60. The population mean annual salary ( 5 $51,800), the population standard deviation of annual salary ( 5 $4000), and the population proportion that com pleted the training program ( p 5 .60) are parameters of the population of EAI employees. Now, suppose that the necessary information on all the EAI employees was not readily available in the company’s database. The question we now consider is how the firm’s director of personnel can obtain estimates of the population parameters by using a sample of employees rather than all 2500 employees in the population. Suppose that a sample of 30employees will be used. Clearly, the time and the cost of developing a profile would be substantially less for 30 employees than for the entire population. If the personnel director could be assured that a sample of 30 employees would provide adequate information about the population of 2500 employees, working with a sample would be preferable to working with the entire population. Let us explore the possibility of using a sample for the EAI study by first considering how we can identify a sample of 30 employees.

Selecting a Sample 7.2 In this section we describe how to select a sample. We first describe how to sample from a finite population and then describe how to select a sample from an infinite population.

Sampling from a Finite Population

Other methods of probability sampling are described in Section 7.7.

Statisticians recommend selecting a probability sample when sampling from a finite population because a probability sample allows them to make valid statistical infer ences about the population. The simplest type of probability sample is one in which each sample of size n has the same probability of being selected. It is called a simple random sample. A simple random sample of size n from a finite population of size N is defined as follows.

simple random sample (finite population)

A simple random sample of size n from a finite population of size N is a sample selected such that each possible sample of size n has the same probability of being selected. The random numbers generated using Excel’s RAND function follow a uniform probability distribution between 0 and 1.

The procedures used to select a simple random sample from a finite population are based upon the use of random numbers. We can use Excel’s RAND function to generate a random number between 0 and 1 by entering the formula 5RAND() into any cell in a worksheet. The number generated is called a random number because the mathematical procedure used by the RAND function guarantees that every number between 0 and 1 has

7.2 Selecting a Sample

321

the same probability of being selected. Let us see how these random numbers can be used to select a simple random sample. Our procedure for selecting a simple random sample of size n from a population of size N involves two steps. Step 1. Assign a random number to each element of the population. Step 2. Select the n elements corresponding to the n smallest random numbers. Because each set of n elements in the population has the same probability of being assigned the n smallest random numbers, each set of n elements has the same probability of being selected for the sample. If we select the sample using this two-step procedure, every sample of size n has the same probability of being selected; thus, the sample selected satisfies the definition of a simple random sample. Let us consider an example involving selecting a simple random sample of size n 5 5 from a population of size N 5 16. Table 7.1 contains a list of the 16 teams in the 2012 National Baseball League. Suppose we want to select a simple random sample of 5 teams to conduct in-depth interviews about how they manage their minor league franchises. Step 1 of our simple random sampling procedure requires that we assign a random number to each of the 16 teams in the population. Figure 7.1 shows a worksheet used to generate a random number corresponding to each of the 16 teams in the population. The names of the baseball teams are in column A, and the random numbers generated are in column B. From the formula worksheet in the background we see that the for mula 5RAND() has been entered into cells B2:B17 to generate the random numbers between 0 and 1. From the value worksheet in the foreground we see that Arizona is assigned the random number .850862, Atlanta has been assigned the random number .706245, and so on. The second step is to select the five teams corresponding to the five smallest random numbers as our sample. Looking through the random numbers in Figure 7.1, we see that the team corresponding to the smallest random number (.066942) is St. Louis, and that the four teams corresponding to the next four smallest random numbers are Washington, Houston, San Diego, and San Francisco. Thus, these five teams make up the simple random sample. Searching through the list of random numbers in Figure 7.1 to find the five smallest ran dom numbers is tedious, and it is easy to make mistakes. Excel’s Sort procedure simplifies this step. We illustrate by sorting the list of baseball teams in Figure 7.1 to find the five teams corresponding to the five smallest random numbers. Refer to the foreground worksheet in Figure 7.1 as we describe the steps involved. Step 1. Select any cell in the range B2:B17 Step 2. Click the Home tab on the Ribbon Step 3. In the Editing group, click Sort & Filter Step 4. Choose Sort Smallest to Largest TABLE 7.1 2012 NATIONAL BASEBALL LEAGUE TEAMS Arizona Milwaukee Atlanta New York Chicago Philadelphia Cincinnati Pittsburgh Colorado San Diego Florida San Francisco Houston St. Louis Los Angeles Washington

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FIGURE 7.1 WORKSHEET USED TO GENERATE A RANDOM NUMBER

CORRESPONDING TO EACH TEAM

National League

The Excel Sort procedure for identifying the employees associated with the 30 smallest random numbers is especially valuable with such a large population.

After completing these steps we obtain the worksheet shown in Figure 7.2.1 The teams listed in rows 2–6 are the ones corresponding to the smallest five random numbers; they are our simple random sample. Note that the random numbers shown in Figure 7.2 are in ascending order, and that the teams are not in their original order. For instance, St. Louis is the next to last team listed in Figure 7.1, but it is the first team selected in the simple random sample. Washington, the second team in our sample, is the sixteenth team in the original list, and so on. We now use this simple random sampling procedure to select a simple random sample of 30 EAI employees from the population of 2500 EAI employees. We begin by generat ing 2500 random numbers, one for each employee in the population. Then we select the 30 employees corresponding to the 30 smallest random numbers as our sample. Refer to Figure 7.3 as we describe the steps involved. Enter/Access Data: Open the DATAfile named EAI. The first three columns of the worksheet in the background show the annual salary data and training program status for the first 30 employees in the population of 2500 EAI employees. (The complete worksheet contains all 2500 employees.) 1 In order to show the random numbers from Figure 7.1 in ascending order in this worksheet, we turned off the automatic recalculation option prior to sorting for illustrative purposes. If the recalculation option were not turned off, a new set of random numbers would have been generated when the sort was completed. But the same five teams would be selected.

7.2 Selecting a Sample

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FIGURE 7.2 USING EXCEL’S SORT PROCEDURE TO SELECT THE SIMPLE RANDOM

SAMPLE OF FIVE TEAMS

Enter Functions and Formulas: In the background worksheet, the label Random Numbers has been entered into cell D1 and the formula 5RAND() has been entered into cells D2:D2501 to generate a random number between 0 and 1 for each of the 2500 EAI employees. The random number generated for the first employee is 0.613872, the random number generated for the second employee is 0.473204, and so on. Apply Tools: All that remains is to find the employees associated with the 30 smallest random numbers. To do so, we sort the data in columns A through D into ascending order by the random numbers in column D. Step 1. Select any cell in the range D2:D2501 Step 2. Click the Home tab on the Ribbon Step 3. In the Editing group, click Sort & Filter Step 4. Choose Sort Smallest to Largest After completing these steps we obtain the worksheet shown in the foreground of Figure 7.3. The employees listed in rows 2–31 are the ones corresponding to the smallest 30 random numbers that were generated. Hence, this group of 30 employees is a simple random sample. Note that the random numbers shown in the foreground of Figure 7.3 are in ascend ing order, and that the employees are not in their original order. For instance, employee 812 in the population is associated with the smallest random number and is the first element in the sample, and employee 13 in the population (see row 14 of the background worksheet) has been included as the 22nd observation in the sample (row 23 of the foreground worksheet).

Sampling from an Infinite Population Sometimes we want to select a sample from a population, but the population is infinitely large or the elements of the population are being generated by an ongoing process for which Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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FIGURE 7.3 USING EXCEL TO SELECT A SIMPLE RANDOM SAMPLE

Note: Rows 32–2501 are not shown.

there is no limit on the number of elements that can be generated. Thus, it is not possible to develop a list of all the elements in the population. This is considered the infinite popula tion case. With an infinite population, we cannot select a simple random sample because we cannot construct a frame consisting of all the elements. In the infinite population case, statisticians recommend selecting what is called a random sample. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

7.2 Selecting a Sample

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Random Sample (Infinite Population)

A random sample of size n from an infinite population is a sample selected such that the following conditions are satisfied. 1. E ach element selected comes from the same population. 2. Each element is selected independently. Care and judgment must be exercised in implementing the selection process for o btaining a random sample from an infinite population. Each case may require a different selection procedure. Let us consider two examples to see what we mean by the conditions: (1) Each element selected comes from the same population and (2) each element is selected independently. A common quality control application involves a production process where there is no limit on the number of elements that can be produced. The conceptual population we are sampling from is all the elements that could be produced (not just the ones that are produced) by the ongoing production process. Because we cannot develop a list of all the elements that could be produced, the population is considered infinite. To be more specific, let us consider a pro duction line designed to fill boxes of a breakfast cereal with a mean weight of 24 ounces of breakfast cereal per box. Samples of 12 boxes filled by this process are periodically selected by a quality control inspector to determine if the process is operating properly or if, perhaps, a machine malfunction has caused the process to begin underfilling or overfilling the boxes. With a production operation such as this, the biggest concern in selecting a random sample is to make sure that condition 1 (the sampled elements are selected from the same population) is satisfied. To ensure that this condition is satisfied, the boxes must be selected at approxi mately the same point in time. This way the inspector avoids the possibility of selecting some boxes when the process is operating properly and other boxes when the process is not operating properly and is underfilling or overfilling the boxes. With a production process such as this, the second condition (each element is selected independently) is satisfied by designing the produc tion process so that each box of cereal is filled independently. With this assumption, the quality control inspector only needs to worry about satisfying the same population condition. As another example of selecting a random sample from an infinite population, consider the population of customers arriving at a fast-food restaurant. Suppose an employee is asked to select and interview a sample of customers in order to develop a profile of customers who visit the restaurant. The customer arrival process is ongoing and there is no way to obtain a list of all customers in the population. So, for practical purposes, the population for this ongoing process is considered infinite. As long as a sampling procedure is designed so that all the elements in the sample are customers of the restaurant and they are selected independently, a random sample will be obtained. In this case, the employee collecting the sample needs to select the sample from people who come into the restaurant and make a purchase to ensure that the same population condition is satisfied. If, for instance, the employee selected some one for the sample who came into the restaurant just to use the restroom, that p erson would not be a customer and the same population condition would be violated. So, as long as the interviewer selects the sample from people making a purchase at the restaurant, condition 1 is satisfied. Ensuring that the customers are selected independently can be more difficult. The purpose of the second condition of the random sample selection procedure (each element is selected independently) is to prevent selection bias. In this case, selection bias would occur if the interviewer were free to select customers for the sample arbitrarily. The interviewer might feel more comfortable selecting customers in a particular age group and might avoid customers in other age groups. Selection bias would also occur if the interviewer selected a group of five customers who entered the restaurant together and Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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asked all of them to participate in the sample. Such a group of customers would be likely to exhibit similar characteristics, which might provide misleading information about the population of customers. Selection bias such as this can be avoided by ensuring that the selection of a particular customer does not influence the selection of any other customer. In other words, the elements (customers) are selected independently. McDonald’s, the fast-food restaurant leader, implemented a random sampling proce dure for this situation. The sampling procedure was based on the fact that some customers presented discount coupons. Whenever a customer presented a discount coupon, the next customer served was asked to complete a customer profile questionnaire. Because arriving customers presented discount coupons randomly and independently of other customers, this sampling procedure ensured that customers were selected independently. As a result, the sample satisfied the requirements of a random sample from an infinite population. Situations involving sampling from an infinite population are usually associated with a process that operates over time. Examples include parts being manufactured on a pro duction line, repeated experimental trials in a laboratory, transactions occurring at a bank, telephone calls arriving at a technical support center, and customers entering a retail store. In each case, the situation may be viewed as a process that generates elements from an infinite population. As long as the sampled elements are selected from the same popula tion and are selected independently, the sample is considered a random sample from an infinite population.

NOTES AND COMMENTS 1. In this section we have been careful to define two types of samples: a simple random sample from a finite population and a random sample from an infinite population. In the remainder of the text, we will generally refer to both of these as either a random sample or simply a sample. We will not make a distinction of the sample being a “simple” random sample unless it is necessary for the exercise or discussion. 2. Statisticians who specialize in sample surveys from finite populations use sampling methods that provide probability samples. With a proba bility sample, each possible sample has a known probability of selection and a random process is used to select the elements for the sample. Sim ple random sampling is one of these methods. In Section 7.7, we describe some other probability sampling methods: stratified random sampling,

cluster sampling, and systematic sampling. We use the term simple in simple random sampling to clarify that this is the probability sampling method that assures each sample of size n has the same probability of being selected. 3. The number of different simple random samples of size n that can be selected from a finite popu lation of size N is N! n!(N 2 n)!

In this formula, N! and n! are the factorial formulas discussed in Chapter 4. For the EAI problem with N 5 2500 and n 5 30, this expression can be used to show that approxi mately 2.75 3 1069 different simple random samples of 30 EAI employees can be obtained.

Exercises

Methods 1. Consider a finite population with five elements labeled A, B, C, D, and E. Ten possible simple random samples of size 2 can be selected. a. List the 10 samples beginning with AB, AC, and so on. b. Using simple random sampling, what is the probability that each sample of size 2 is selected? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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7.3 Point Estimation

c. Suppose we use Excel’s RAND function to assign random numbers to the five ele ments: A (.7266), B (.0476), C (.2459), D (.0957), E (.9408). List the simple random sample of size 2 that will be selected by using these random numbers. 2. Assume a finite population has 10 elements. Number the elements from 1 to 10 and use the following 10 random numbers to select a sample of size 4. .7545 .0936 .0341 .3242 .1449 .9060 .2420 .9773 .5428 .0729 3. The 2012 American League consists of 14 baseball teams. Suppose a sample of 5 teams is to be selected to conduct player interviews. The following table lists the 14 teams and the random numbers assigned by Excel’s RAND function. Use these random numbers to select a sample of size 5.

Random Random Team Number Team Number

American League

New York 0.178624 Baltimore 0.578370 Toronto 0.965807 Chicago 0.562178 Detroit 0.253574 Oakland 0.288287 Texas 0.500879

Boston 0.290197 Tampa Bay 0.867778 Minnesota 0.811810 Cleveland 0.960271 Kansas City 0.326836 Anaheim 0.895267 Seattle 0.839071

4. The U.S. Golf Association has instituted a ban on long and belly putters. This has caused a great deal of controversy among both amateur golfers and members of the Professional Golf Association (PGA). Shown below are the names of the top 10 finishers in the recent PGA Tour McGladrey Classic golf tournament. 1. Tommy Gainey 2. David Toms 3. Jim Furyk 4. Brendon de Jonge 5. D. J. Trahan

Select a simple random sample of 3 of these players to assess their opinions on the use of long and belly putters. 5. In this section we used a two-step procedure to select a simple random sample of 30 EAI employees. Use this procedure to select a simple random sample of 50 EAI employees. 6. Indicate which of the following situations involve sampling from a finite population and which involve sampling from an infinite population. In cases where the sampled population is finite, describe how you would construct a frame. a. Select a sample of licensed drivers in the state of New York. b. Select a sample of boxes of cereal off the production line for the Breakfast Choice Company. c. Select a sample of cars crossing the Golden Gate Bridge on a typical weekday. d. Select a sample of students in a statistics course at Indiana University. e. Select a sample of the orders being processed by a mail-order firm.

EAI

6. Davis Love III 7. Chad Campbell 8. Greg Owens 9. Charles Howell III 10. Arjun Atwal

7.3

Point Estimation Now that we have described how to select a simple random sample, let us return to the EAI problem. A simple random sample of 30 employees and the corresponding data on annual salary and management training program participation are as shown in Table 7.2. The notation

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Chapter 7 Sampling and Sampling Distributions

TABLE 7.2 ANNUAL SALARY AND TRAINING PROGRAM STATUS FOR A SIMPLE

RANDOM SAMPLE OF 30 EAI Employees

Annual Management Training Salary ($) Program

x1 5 49,094.30 x2 5 53,263.90 x3 5 49,643.50 x4 5 49,894.90 x5 5 47,621.60 x6 5 55,924.00 x7 5 49,092.30 x8 5 51,404.40 x9 5 50,957.70 x10 5 55,109.70 x11 5 45,922.60 x12 5 57,268.40 x13 5 55,688.80 x14 5 51,564.70 x15 5 56,188.20

Yes Yes Yes Yes No Yes Yes Yes Yes Yes Yes No Yes No No

Annual Management Training Salary ($) Program x16 5 51,766.00 Yes x17 5 52,541.30 No x18 5 44,980.00 Yes x19 5 51,932.60 Yes x20 5 52,973.00 Yes x21 5 45,120.90 Yes x22 5 51,753.00 Yes x23 5 54,391.80 No x24 5 50,164.20 No x25 5 52,973.60 No x26 5 50,241.30 No x27 5 52,793.90 No x28 5 50,979.40 Yes x29 5 55,860.90 Yes x30 5 57,309.10 No

x1, x 2, and so on is used to denote the annual salary of the first employee in the sample, the annual salary of the second employee in the sample, and so on. Participation in the manage ment training program is indicated by Yes in the management training program column. To estimate the value of a population parameter, we compute a corresponding character istic of the sample, referred to as a sample statistic. For example, to estimate the population mean and the population standard deviation for the annual salary of EAI employees, we use the data in Table 7.2 to calculate the corresponding sample statistics: the sample mean and the sample standard deviation s. Using the formulas for a sample mean and a sample standard deviation presented in Chapter 3, the sample mean is

oxi 1,554,420 5 $51,814 x 5 n 5 30 and the sample standard deviation is s 5

o(xi 2 x)2 n21

325,009,260 5 $3348 29

To estimate p, the proportion of employees in the population who completed the manage ment training program, we use the corresponding sample proportion p. Let x denote thenum ber of employees in the sample who completed the management training program. The data in Table 7.2 show that x 5 19. Thus, with a sample size of n 5 30, the sample proportion is

x 19 5 5 .63 n 30

By making the preceding computations, we perform the statistical procedure called point estimation. We refer to the sample mean x as the point estimator of the population mean , the sample standard deviation s as the point estimator of the population standard devia tion , and the sample proportion p as the point estimator of the population proportion p. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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7.3 Point Estimation

TABLE 7.3 SUMMARY OF POINT ESTIMATES OBTAINED FROM A SIMPLE RANDOM

SAMPLE OF 30 EAI Employees

Population Parameter 5 Population mean annual salary 5 Population standard deviation

Parameter Value $51,800

Point Estimate $51,814

$4000

s 5 Sample standard deviation for annual salary

$3348

.60

p 5 Sample proportion having

.63

for annual salary p 5 Population proportion having completed the management training program

Point Estimator x 5 Sample mean annual salary

completed the management training program

The numerical value obtained for x, s, or p is called the point estimate. Thus, for the simple r andom sample of 30 EAI employees shown in Table 7.2, $51,814 is the point estimate of , $3348 is the point estimate of , and .63 is the point estimate of p. Table 7.3 summarizes thesample results and compares the point estimates to the actual values of the population parameters. As is evident from Table 7.3, the point estimates differ somewhat from the correspond ing population parameters. This difference is to be expected because a sample, and not a census of the entire population, is being used to develop the point estimates. In the next chapter, we will show how to construct an interval estimate in order to provide information about how close the point estimate is to the population parameter.

Practical Advice The subject matter of most of the rest of the book is concerned with statistical inference. Point estimation is a form of statistical inference. We use a sample statistic to make an infer ence about a population parameter. When making inferences about a population based on a sample, it is important to have a close correspondence between the sampled population and the target population. The target population is the population we want to make inferences about, while the sampled population is the population from which the sample is actually taken. In this section, we have described the process of drawing a simple random sample from the population of EAI employees and making point estimates of characteristics of that same population. So the sampled population and the target population are identical, which is the desired situation. But in other cases, it is not as easy to obtain a close correspondence between the sampled and target populations. Consider the case of an amusement park selecting a sample of its customers to learn about characteristics such as age and time spent at the park. Suppose all the sample elements were selected on a day when park attendance was restricted to employees of a large company. Then the sampled population would be composed of employees of that company and members of their families. If the target population we wanted to make inferences about were typical park customers over a typical summer, then we might encounter a significant difference between the sampled population and the target population. In such a case, we would question the validity of the point estimates being made. Park management would be in the best position to know whether a sample taken on a particular day was likely to be representative of the target population. In summary, whenever a sample is used to make inferences about a population, we should make sure that the study is designed so that the sampled population and the target population are in close agreement. Good judgment is a necessary ingredient of sound statistical practice. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Exercises

Methods 7. The following data are from a simple random sample. 5

a. What is the point estimate of the population mean? b. What is the point estimate of the population standard deviation? 8. A survey question for a sample of 150 individuals yielded 75 Yes responses, 55 No r esponses, and 20 No Opinions. a. What is the point estimate of the proportion in the population who respond Yes? b. What is the point estimate of the proportion in the population who respond No?

Applications 9. A simple random sample of 5 months of sales data provided the following information: Month: 1 2 3 4 5 Units Sold: 94 100 85 94 92 a. Develop a point estimate of the population mean number of units sold per month. b. Develop a point estimate of the population standard deviation.

Morningstar

10. Morningstar publishes ratings data on 1208 company stocks. A sample of 40 of these stocks is contained in the DATAfile named Morningstar. Use the data set to answer the following questions. a. Develop a point estimate of the proportion of the stocks that receive Morningstar’s highest rating of 5 Stars. b. Develop a point estimate of the proportion of the Morningstar stocks that are rated Above Average with respect to business risk. c. Develop a point estimate of the proportion of the Morningstar stocks that are rated 2 Stars or less. 11. The National Football League (NFL) polls fans to develop a rating for each football game (NFL website, October 24, 2012). Each game is rated on a scale from 0 (forgettable) to 100 (memorable). The fan ratings for a random sample of 12 games follow. 57 61 86 74 72 73 20 57 80 79 83 74 a. Develop a point estimate of mean fan rating for the population of NFL games. b. Develop a point estimate of the standard deviation for the population of NFL games. 12. A sample of 426 U.S. adults age 50 and older were asked how important a variety of issues were in choosing whom to vote for in the most recent presidential election. a. What is the sampled population for this study? b. Social Security and Medicare were cited as “very important” by 350 respondents. Estimate the proportion of the population of U.S. adults age 50 and over who believe this issue is very important. c. Education was cited as “very important” by 74% of the respondents. Estimate the num ber of respondents who believe this issue is very important. d. Job Growth was cited as “very important” by 354 respondents. Estimate the propor tion of U.S. adults age 50 and over who believe job growth is very important. e. What is the target population for the inferences being made in parts (b) and (d)? Is it the same as the sampled population you identified in part (a)? Suppose you later learn

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7.4 Introduction to Sampling Distributions

that the sample was restricted to members of the AARP. Would you still feel the inferences being made in parts (b) and (d) are valid? Why or why not? 13. One of the questions in the Pew Internet & American Life Project asked adults if they used the Internet, at least occasionally (Pew website, October 23, 2012). The results showed that 454 out of 478 adults aged 18–29 answered Yes; 741 out of 833 adults aged 30–49 answered Yes; 1058 out of 1644 adults aged 50 and over answered Yes. a. Develop a point estimate of the proportion of adults aged 18–29 who use the Internet. b. Develop a point estimate of the proportion of adults aged 30–49 who use the Internet. c. Develop a point estimate of the proportion of adults aged 50 and over who use the Internet. d. Comment on any relationship between age and Internet use that seems apparent. e. Suppose your target population of interest is that of all adults (18 years of age and over). Develop an estimate of the proportion of that population who use the Internet.

EAI

14. In this section we showed how a simple random sample of 30 EAI employees can be used to develop point estimates of the population mean annual salary, the population standard deviation for annual salary, and the population proportion having completed the management training program. a. Use Excel to select a simple random sample of 50 EAI employees. b. Develop a point estimate of the mean annual salary. c. Develop a point estimate of the population standard deviation for annual salary. d. Develop a point estimate of the population proportion having completed the manage ment training program.

Introduction to Sampling Distributions 7.4 In the preceding section we said that the sample mean x is the point estimator of the popula tion mean , and the sample proportion p is the point estimator of the population proportion p. For the simple random sample of 30 EAI employees shown in Table 7.2, the point esti mate of is x 5 $51,814 and the point estimate of p is p 5 .63. Suppose we select another simple random sample of 30 EAI employees and obtain the following point estimates: Sample mean: x 5 $52,670 Sample proportion: p 5 .70

Note that different values of x and p were obtained. Indeed, a second simple random s ample of 30 EAI employees cannot be expected to provide the same point estimates as the first sample. Now, suppose we repeat the process of selecting a simple random sample of 30 EAI employees over and over again, each time computing the values of x and p. Table 7.4 TABLE 7.4 VALUES OF x and p FROM 500 SIMPLE RANDOM SAMPLES

OF 30 EAI Employees

Sample Sample Mean Sample Proportion Number (x) ( p) 1 2 3 4 . . . 500

51,814 52,670 51,780 51,588 . . . 51,752

.63 .70 .67 .53 . . . .50

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Chapter 7 Sampling and Sampling Distributions

TABLE 7.5 FREQUENCY and Relative frequency DISTRIBUTIONs OF x FROM

500 SIMPLE RANDOM SAMPLES OF 30 EAI Employees

Mean Annual Salary ($)

Frequency

Relative Frequency

49,500.00 – 49,999.99 2 50,000.00 –50,499.99 16 50,500.00–50,999.99 52 51,000.00 –51,499.99 101 51,500.00 –51,999.99 133 52,000.00 –52,499.99 110 52,500.00 –52,999.99 54 53,000.00 –53,499.99 26 53,500.00 –53,999.99 6

.004 .032 .104 .202 .266 .220 .108 .052 .012

1.000

Totals500

contains a portion of the results obtained for 500 simple random samples, and Table 7.5 shows the frequency and relative frequency distributions for the 500 x values. Figure 7.4 shows the relative frequency histogram for the x values. In Chapter 5 we defined a random variable as a numerical description of the outcome of an experiment. If we consider the process of selecting a simple random sample as an experiment, the sample mean x is the numerical description of the outcome of the experi ment. Thus, the sample mean x is a random variable. As a result, just like other random vari ables, x has a mean or expected value, a standard deviation, and a probability distribution.

FIGURE 7.4 RELATIVE FREQUENCY HISTOGRAM OF x VALUES FROM 500 SIMPLE

RANDOM SAMPLES OF SIZE 30 EACH

.30

Relative Frequency

.25 .20 .15 .10 .05

50,000

51,000

52,000 Values of x

53,000

54,000

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7.4 Introduction to Sampling Distributions

The ability to understand the material in subsequent chapters depends heavily on the ability to understand and use the sampling distributions presented in this chapter.

Because the various possible values of x are the result of different simple random samples, the probability distribution of x is called the sampling distribution of x. Knowledge of this sampling distribution and its properties will enable us to make probability statements about how close the sample mean x is to the population mean . Let us return to Figure 7.4. We would need to enumerate every possible sample of 30 employees and compute each sample mean to completely determine the sampling dis tribution of x. However, the histogram of 500 x values gives an approximation of this sampling distribution. From the approximation we observe the bell-shaped appearance of the distribution. We note that the largest concentration of the x values and the mean of the 500 x values is near the population mean 5 $51,800. We will describe the properties of the sampling distribution of x more fully in the next section. The 500 values of the sample proportion p are summarized by the relative frequency histogram in Figure 7.5. As in the case of x, p is a random variable. If every possible s ample of size 30 were selected from the population and if a value of p were computed for each sample, the resulting probability distribution would be the sampling distribution of p. The relative frequency histogram of the 500 sample values in Figure 7.5 provides a general idea of the appearance of the sampling distribution of p. In practice, we select only one simple random sample from the population. We repeated the sampling process 500 times in this section simply to illustrate that many different samples are possible and that the different samples generate a variety of values for the sample statistics x and p. The probability distribution of any particular sample statistic is called the sampling distribution of the statistic. In Section 7.5 we discuss the

FIGURE 7.5 RELATIVE FREQUENCY HISTOGRAM OF p VALUES FROM 500 SIMPLE

RANDOM SAMPLES OF SIZE 30 EACH

.40 .35

Relative Frequency

.30 .25 .20 .15 .10 .05

.32

.40

.48

.56 .64 Values of p

.72

.80

.88

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Chapter 7 Sampling and Sampling Distributions

characteristics of the sampling distribution of x. In Section 7.6 we discuss the character istics of the sampling distribution of p.

Sampling Distribution of x– 7.5 In the previous section we said that the sample mean x is a random variable and its prob ability distribution is called the sampling distribution of x.

Sampling Distribution of x

The sampling distribution of x is the probability distribution of all possible values of the sample mean x.

This section describes the properties of the sampling distribution of x. Just as with other probability distributions we studied, the sampling distribution of x has an expected value or mean, a standard deviation, and a characteristic shape or form. Let us begin by considering the mean of all possible x values, which is referred to as the expected value of x.

Expected Value of − x In the EAI sampling problem we saw that different simple random samples result in a vari ety of values for the sample mean x. Because many different values of the random variable x are possible, we are often interested in the mean of all possible values of x that can be generated by the various simple random samples. The mean of the x random variable is the expected value of x. Let E(x) represent the expected value of x and represent the mean of the population from which we are selecting a simple random sample. It can be shown that with simple random sampling, E(x) and are equal.

Expected Value of x The expected value of x equals the mean of the population from which the sample is selected.

E(x) 5

(7.1)

where

E(x) 5 the expected value of x 5 the population mean

This result shows that with simple random sampling, the expected value or mean of the sampling distribution of x is equal to the mean of the population. In Section 7.1 we saw that the mean annual salary for the population of EAI employees is 5 $51,800. Thus, according to equation (7.1), the mean of all possible sample means for the EAI study is also $51,800. When the expected value of a point estimator equals the population parameter, we say the point estimator is unbiased. Thus, equation (7.1) shows that x is an unbiased estimator of the population mean . Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

7.5 Sampling Distribution of x–

335

Standard Deviation of − x Let us define the standard deviation of the sampling distribution of x. We will use the fol lowing notation.

x 5 the standard deviation of x 5 the standard deviation of the population n 5 the sample size N 5 the population size It can be shown that the formula for the standard deviation of x depends on whether the population is finite or infinite. The two formulas for the standard deviation of x follow.

Standard Deviation of x

Finite Population x 5

Infinite Population

Î S D N2n N21

Ïn

x 5

Ïn

(7.2)

In comparing the two formulas in equation (7.2), we see that the factor Ï(N 2 n)y(N 2 1) is required for the finite population case but not for the infinite population case. This fac tor is commonly referred to as the finite population correction factor. In many practical sampling situations, we find that the population involved, although finite, is “large,” whereas the sample size is relatively “small.” In such cases the finite population correction factor Ï(N 2 n)y(N 2 1) is close to 1. As a result, the difference between the values of the stan dard deviation of x for the finite and infinite population cases becomes negligible. Then, x 5 yÏn becomes a good approximation to the standard deviation of x even though the population is finite. This observation leads to the following general guideline, or rule of thumb, for computing the standard deviation of x.

Use the Following Expression to Compute the Standard Deviation of x

x 5

Ïn

(7.3)

whenever 1. The population is infinite; or 2. The population is finite and the sample size is less than or equal to 5% of the population size; that is, n/N # .05.

Exercise 17 shows that when n /N # .05, the finite population correction factor has little effect on the value of x.

In cases where n /N . .05, the finite population version of formula (7.2) should be used in the computation of x. Unless otherwise noted, throughout the text we will assume that the population size is “large,” n /N # .05, and expression (7.3) can be used to compute x.

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Chapter 7 Sampling and Sampling Distributions

The term standard error is used throughout statistical inference to refer to the standard deviation of a point estimator.

To compute x, we need to know , the standard deviation of the population. To further emphasize the difference between x and , we refer to the standard deviation of x, x, as the standard error of the mean. In general, the term standard error refers to the standard deviation of a point estimator. Later we will see that the value of the standard error of the mean is helpful in determining how far the sample mean may be from the population mean. Let us now return to the EAI example and compute the standard error of the mean associated with simple random samples of 30 EAI employees. In Section 7.1 we saw that the standard deviation of annual salary for the population of 2500 EAI employees is 5 4000. In this case, the population is finite, with N 5 2500. However, with a sample size of 30, we have n /N 5 30/2500 5 .012. Because the sample size is less than 5% of the population size, we can ignore the finite population correction factor and use equation (7.3) to compute the standard error. x 5

Ïn

4000 5 730.3 Ï30

Form of the Sampling Distribution of − x The preceding results concerning the expected value and standard deviation for the sam pling distribution of x are applicable for any population. The final step in identifying the characteristics of the sampling distribution of x is to determine the form or shape of the sampling distribution. We will consider two cases: (1) The population has a normal distrib ution; and (2) the population does not have a normal distribution. Population has a normal distribution In many situations it is reasonable to assume that the population from which we are selecting a random sample has a normal, or nearly normal, distribution. When the population has a normal distribution, the sampling distri bution of x is normally distributed for any sample size. Population does not have a normal distribution When the population from which we are selecting a random sample does not have a normal distribution, the central limit theorem is helpful in identifying the shape of the sampling distribution of x. A statement of the central limit theorem as it applies to the sampling distribution of x follows.

Central Limit Theorem

In selecting random samples of size n from a population, the sampling distribution of the sample mean x can be approximated by a normal distribution as the sample size becomes large.

Figure 7.6 shows how the central limit theorem works for three different populations; each column refers to one of the populations. The top panel of the figure shows that none of the populations are normally distributed. Population I follows a uniform distribution. Population II is often called the rabbit-eared distribution. It is symmetric, but the more likely values fall in the tails of the distribution. Population III is shaped like the exponential distribution; it is skewed to the right. The bottom three panels of Figure 7.6 show the shape of the sampling distribution for samples of size n 5 2, n 5 5, and n 5 30. When the sample size is 2, we see that the shape of each sampling distribution is different from the shape of the corresponding population distribution. For samples of size 5, we see that the shapes of the sampling distributions Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

7.5 Sampling Distribution of x–

337

FIGURE 7.6 ILLUSTRATION OF THE CENTRAL LIMIT THEOREM

FOR THREE POPULATIONS Population I

Population II

Population III

Values of x

Population Distribution

Sampling Distribution of x (n 5 2)

Sampling Distribution of x (n 5 5)

Sampling Distribution of x (n 5 30)

for populations I and II begin to look similar to the shape of a normal distribution. Even though the shape of the sampling distribution for population III begins to look similar to the shape of a normal distribution, some skewness to the right is still present. Finally, for samples of size 30, the shapes of each of the three sampling distributions are approximately normal. From a practitioner standpoint, we often want to know how large the sample size needs to be before the central limit theorem applies and we can assume that the shape of the sampling distribution is approximately normal. Statistical researchers have inves tigated this question by studying the sampling distribution of x for a variety of popula tions and a variety of sample sizes. General statistical practice is to assume that, for most Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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applications, the sampling distribution of x can be approximated by a normal distribution whenever the sample is size 30 or more. In cases where the population is highly skewed or outliers are present, samples of size 50 may be needed. Finally, if the population is discrete, the sample size needed for a normal approximation often depends on the popula tion proportion. We say more about this issue when we discuss the sampling distribution of p in Section 7.6.

Sampling Distribution of − x for the EAI Problem Let us return to the EAI problem where we previously showed that E(x) 5 $51,800 and x 5 730.3. At this point, we do not have any information about the population distribution; it may or may not be normally distributed. If the population has a normal d istribution, the sampling distribution of x is normally distributed. If the population does not have a normal distribution, the simple random sample of 30 employees and the central limit theorem enable us to conclude that the sampling distribution of x can be approximated by a normal distribution. In either case, we are comfortable proceeding with the conclusion that the sampling distribution of x can be described by the normal distribution shown in Figure 7.7.

Practical Value of the Sampling Distribution of − x Whenever a simple random sample is selected and the value of the sample mean is used to estimate the value of the population mean , we cannot expect the sample mean to exactly equal the population mean. The practical reason we are interested in the sampling distribu tion of x is that it can be used to provide probability information about the difference between the sample mean and the population mean. To demonstrate this use, let us return to the EAI problem. Suppose the personnel director believes the sample mean will be an acceptable estimate of the population mean if the sample mean is within $500 of the population mean. However, it is not possible to guarantee that the sample mean will be within $500 of the population mean. Indeed, Table 7.5 and Figure 7.4 show that some of the 500 sample means differed by more than $2000 from the population mean. So we must think of the personnel director’s

FIGURE 7.7 SAMPLING DISTRIBUTION OF x FOR THE MEAN ANNUAL SALARY

OF A SIMPLE RANDOM SAMPLE OF 30 EAI Employees

Sampling distribution of x

x 5

4000 5 730.3 5 n 30

51,800 E(x)

7.5 Sampling Distribution of x–

339

request in probability terms. That is, the personnel director is concerned with the following question: What is the probability that the sample mean computed using a simple random sample of 30 EAI employees will be within $500 of the population mean? Because we have identified the properties of the sampling distribution of x (see Figure 7.7), we will use this distribution to answer the probability question. Refer to the sam pling distribution of x shown again in Figure 7.8. With a population mean of $51,800, the personnel director wants to know the probability that x is between $51,300 and $52,300. This probability is given by the darkly shaded area of the sampling distribution shown in Figure 7.8. Because the sampling distribution is normally distributed, with mean 51,800 and standard error of the mean 730.3, we can use the standard normal probability table to find the area or probability. We first calculate the z value at the upper endpoint of the interval (52,300) and use the table to find the cumulative probability at that point (left tail area). Then we compute the z value at the lower endpoint of the interval (51,300) and use the table to find the area under the curve to the left of that point (another left tail area). Subtracting the second tail area from the first gives us the desired probability. At x 5 52,300, we have

52,300 2 51,800 5 .68 730.30

Referring to the standard normal probability table, we find a cumulative probability (area to the left of z 5 .68) of .7517. At x 5 51,300, we have

51,300 2 51,800 5 2.68 730.30

The area under the curve to the left of z 5 2.68 is .2483. Therefore, P(51,300 # x # 52,300) 5 P(z # .68) 2 P(z , 2.68) 5 .7517 2 .2483 5 .5034.

FIGURE 7.8 PROBABILITY OF A SAMPLE MEAN BEING WITHIN $500

OF THE POPULATION MEAN for A SIMPLE RANDOM SAMPLE OF 30 EAI Employees

Sampling distribution of x

x 5 730.30 P(51,300 # x # 52,300)

P(x , 51,300)

51,300

51,800

52,300

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Using Excel’s NORM.DIST function is easier and provides more accurate results than using the tables with rounded values for z.

The desired probability can also be computed using Excel’s NORM.DIST function. The advantage of using the NORM.DIST function is that we do not have to make a sepa rate computation of the z value. Evaluating the NORM.DIST function at the upper end point ofthe interval provides the cumulative probability at 52,300. Entering the formula 5NORM.DIST(52300,51800,730.30,TRUE) into a cell of an Excel worksheet provides .7532 for this cumulative probability. Evaluating the NORM.DIST function at the lower endpoint of the interval provides the area under the curve to the left of 51,300. Entering the formula 5NORM.DIST(51300,51800,730.30,TRUE) into a cell of an Excel worksheet provides .2468 for this cumulative probability. The probability of x being in the interval from 51,300 to 52,300 is then given by .7532 2 .2468 5 .5064. We note that this result is slightly different from the probability obtained using the table, because in using the normal table we rounded to two decimal places of accuracy when computing the z value. The result obtained using NORM.DIST is thus more accurate. The preceding computations show that a simple random sample of 30 EAI employees has a .5064 probability of providing a sample mean x that is within $500 of the population mean. Thus, there is a 1 2 .5064 5 .4936 probability that the sampling error will be more than $500. In other words, a simple random sample of 30 EAI employees has roughly a 50–50 chance of providing a sample mean within the allowable $500. Perhaps a larger sample size should be considered. Let us explore this possibility by considering the relation ship between the sample size and the sampling distribution of x.

The sampling distribution of x can be used to provide probability information about how close the sample mean x is to the population mean .

Relationship Between the Sample Size and the Sampling Distribution of − x Suppose that in the EAI sampling problem we select a simple random sample of 100 EAI employees instead of the 30 originally considered. Intuitively, it would seem that with more data provided by the larger sample size, the sample mean based on n 5 100 should provide a better estimate of the population mean than the sample mean based on n 5 30. To see how much better, let us consider the relationship between the sample size and the sampling distribution of x. First note that E(x) 5 regardless of the sample size. Thus, the mean of all possible values of x is equal to the population mean regardless of the sample size n. However, note that the standard error of the mean, x 5 yÏn, is related to the square root of the sample size. Whenever the sample size is increased, the standard error of the mean x decreases. With n 5 30, the standard error of the mean for the EAI problem is 730.3. However, with the increase in the sample size to n 5 100, the standard error of the mean is decreased to

x 5

4000 5 400 5 Ïn Ï100

The sampling distributions of x with n 5 30 and n 5 100 are shown in Figure 7.9. Because the sampling distribution with n 5 100 has a smaller standard error, the values of x have less variation and tend to be closer to the population mean than the values of x with n 5 30. We can use the sampling distribution of x for the case with n 5 100 to compute the probability that a simple random sample of 100 EAI employees will provide a sample mean that is within $500 of the population mean. In this case the sampling distribution is normal with a mean of 51,800 and a standard deviation of 400 (see Figure 7.10). Again, we could compute the appropriate z values and use the standard normal probability distribution table to make this probability calculation. However, Excel’s NORM.DIST function is easier to use and provides more accurate results. Entering the formula 5NORM.DIST(52300,51800,400,TRUE) into a cell of an Excel worksheet provides the cumulative probability corresponding to x 5 52,300. The value provided Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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FIGURE 7.9 A COMPARISON OF THE SAMPLING DISTRIBUTIONS OF x FOR SIMPLE

RANDOM SAMPLES OF n 5 30 AND n 5 100 EAI Employees

With n 5 100, x 5 400

With n 5 30, x 5 730.3

51,800

FiGURE 7.10 PROBABILITY OF A SAMPLE MEAN BEING WITHIN $500

OF THE POPULATION MEAN FOR A SIMPLE RANDOM SAMPLE OF 100 EAI Employees

Sampling distribution of x

x 5 400

P(51,300 # x # 52,300) 5 .7888

51,800 51,300

52,300

by Excel is .8944. Entering the formula 5 NORM.DIST (51300,51800,400,TRUE) into a cell of an Excel worksheet provides the cumulative p robability corresponding to x 5 51,300. The value provided by Excel is .1056. Thus, the probability of x being in the interval from 51,300 to 52,300 is given by .8944 2 .1056 5 .7888. By increasing the sample size from 30 to 100 EAI employees, we increase the probability that the sampling error will be $500 or less; that is, the probability of obtaining a sample mean within $500 of the population mean increases from .5064 to .7888. The important point in this discussion is that as the sample size increases, the standard error of the mean decreases. As a result, a larger sample size will provide a higher prob ability that the sample mean falls within a specified distance of the population mean. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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NOTE AND COMMENT In presenting the sampling distribution of x for the EAI problem, we took advantage of the fact that the population mean 5 51,800 and the population standard deviation 5 4000 were known. However, usually the values of the population mean and the

population standard deviation that are needed to determine the sampling distribution of x will be unknown. In Chapter 8 we show how the sample mean x and the sample standard deviation s are used when and are unknown.

Exercises

Methods 15. A population has a mean of 200 and a standard deviation of 50. Suppose a simple random sample of size 100 is selected and x is used to estimate . a. What is the probability that the sample mean will be within 65 of the population mean? b. What is the probability that the sample mean will be within 610 of the population mean? 16. Assume the population standard deviation is 5 25. Compute the standard error of the mean, x, for sample sizes of 50, 100, 150, and 200. What can you say about the size of the standard error of the mean as the sample size is increased? 17. Suppose a random sample of size 50 is selected from a population with 5 10. Find the value of the standard error of the mean in each of the following cases (use the f inite population correction factor if appropriate). a. The population size is infinite. b. The population size is N 5 50,000. c. The population size is N 5 5000. d. The population size is N 5 500.

Applications 18. Refer to the EAI sampling problem. Suppose a simple random sample of 60 employees is used. a. Sketch the sampling distribution of x when simple random samples of size 60 are used. b. What happens to the sampling distribution of x if simple random samples of size 120 are used? c. What general statement can you make about what happens to the sampling distri bution of x as the sample size is increased? Does this generalization seem logical? Explain. 19. In the EAI sampling problem (see Figure 7.8), we showed that for n 5 30, there was .5064 probability of obtaining a sample mean within 6$500 of the population mean. a. What is the probability that x is within $500 of the population mean if a sample of size 60 is used? b. Answer part (a) for a sample of size 120. 20. B arron’s reported that the average number of weeks an individual is unemployed is 17.5 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 17.5 weeks and that the population standard deviation is 4 weeks. Suppose you would like to select a random sample of 50 unemployed individuals for a follow-up study. a. Show the sampling distribution of x, the sample mean average for a sample of 50 unemployed individuals. b. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1 week of the population mean? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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c. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1/2 week of the population mean? 21. The average full-sized front-loading Energy Star clothes washer uses 15 gallons of water per load (greenbuildingadvisor.com, December 4, 2015). Assume that the popula tion standard deviation for the number of gallons of water used to wash a load by these machines is 3.0. a. What is the probability that a random sample of 90 loads washed in a full-sized frontloading Energy Star clothes washer will provide a sample mean gallons of water used that is within one-half gallon of the population mean of 15 gallons? b. What is the probability that a random sample of 50 loads washed in a full-sized frontloading Energy Star clothes washer will provide a sample mean gallons of water used that is at least three-quarters of a gallon greater than the population mean of 15 gallons? c. What is the probability that a random sample of 75 loads washed in a full-sized frontloading Energy Star clothes washer will provide a sample mean gallons of water used that is no more than one-quarter of a gallon less than the population mean of 15 gallons? 22. T he Wall Street Journal reported that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return. The mean amount of deductions for this population of taxpayers was $16,642. Assume the standard deviation is 5 $2400. a. What is the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for each of the following sample sizes: 30, 50, 100, and 400? b. What is the advantage of a larger sample size when attempting to estimate the popula tion mean? 23. The Economic Policy Institute periodically issues reports on wages of entry-level workers. The institute reported that entry-level wages for male college graduates were $21.68 per hour and for female college graduates were $18.80 per hour in 2011 (Economic Policy Institute website, March 30, 2012). Assume the standard deviation for male graduates is$2.30, and for female graduates it is $2.05. a. What is the probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68? b. What is the probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80? c. In which of the preceding two cases, part (a) or part (b), do we have a higher proba bility of obtaining a sample estimate within $.50 of the population mean? Why? d. What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean? 24. According to the Current Results website, the state of California has a mean annual rainfall of 22 inches, whereas the state of New York has a mean annual rainfall of 42 inches. Assume that the standard deviation for both states is 4 inches. A sample of 30 years of rainfall for California and a sample of 45 years of rainfall for New York have been taken. a. Show the probability distribution of the sample mean annual rainfall for California. b. What is the probability that the sample mean is within 1 inch of the population mean for California? c. What is the probability that the sample mean is within 1 inch of the population mean for New York? d. In which case, part (b) or part (c), is the probability of obtaining a sample mean within 1 inch of the population mean greater? Why? Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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25. The mean preparation fee H&R Block charged retail customers in 2012 was $183 (The Wall Street Journal, March 7, 2012). Use this price as the population mean and assume the population standard deviation of preparation fees is $50. a. What is the probability that the mean price for a sample of 30 H&R Block retail customers is within $8 of the population mean? b. What is the probability that the mean price for a sample of 50 H&R Block retail customers is within $8 of the population mean? c. What is the probability that the mean price for a sample of 100 H&R Block retail customers is within $8 of the population mean? d. Which, if any, of the sample sizes in parts (a), (b), and (c) would you recommend to have at least a .95 probability that the sample mean is within $8 of the population mean? 26. To estimate the mean age for a population of 4000 employees, a simple random sample of 40 employees is selected. a. Would you use the finite population correction factor in calculating the standard error of the mean? Explain. b. If the population standard deviation is 5 8.2 years, compute the standard error both with and without the finite population correction factor. What is the rationale for ignoring the finite population correction factor whenever n/N # .05? c. What is the probability that the sample mean age of the employees will be within 62 years of the population mean age?

Sampling Distribution of p– 7.6 The sample proportion p is the point estimator of the population proportion p. The formula for computing the sample proportion is x p5 n where x 5 the number of elements in the sample that possess the characteristic of interest n 5 sample size

As noted in Section 7.4, the sample proportion p is a random variable and its probability distribution is called the sampling distribution of p. Sampling Distribution of p

The sampling distribution of p is the probability distribution of all possible values of the sample proportion p. To determine how close the sample proportion p is to the population proportion p, we need to understand the properties of the sampling distribution of p: the expected value of p, the standard deviation of p, and the shape or form of the sampling distribution of p.

Expected Value of p The expected value of p, the mean of all possible values of p, is equal to the population proportion p. Expected Value of p

E(p) 5 p

(7.4)

7.6 Sampling Distribution of p–

345

where E(p) 5 the expected value of p p 5 the population proportion

Because E( p) 5 p, p is an unbiased estimator of p. Recall from Section 7.1 we noted that p 5 .60 for the EAI population, where p is the proportion of the population of employees who participated in the company’s management training program. Thus, the expected value of p for the EAI sampling problem is .60.

Standard Deviation of p Just as we found for the standard deviation of x, the standard deviation of p depends on whether the population is finite or infinite. The two formulas for computing the standard deviation of p follow.

Standard Deviation of p

Finite Population p 5

Î Î N2n N21

Infinite Population

p(1 2 p) n

p 5

p(1 2 p) n

(7.5)

Comparing the two formulas in equation (7.5), we see that the only difference is the use of the finite population correction factor Ï(N 2 n)y(N 2 1). As was the case with the sample mean x, the difference between the expressions for the finite population and the infinite population becomes negligible if the size of the finite popu lation is large in comparison to the sample size. We follow the same rule of thumb that we recommended for the sample mean. That is, if the population is finite with n/N # .05, we will use p 5 Ïp(1 2 p)yn. However, if the population is finite with n/N . .05, the finite population correction factor should be used. Again, unless specifically noted, throughout the text we will assume that the population size is large in relation to the sample size and thus the finite population correction factor is unnecessary. In Section 7.5 we used the term standard error of the mean to refer to the standard deviation of x. We stated that in general the term standard error refers to the standard deviation of a point estimator. Thus, for proportions we use standard error of the proportion torefer to the standard deviation of p. Let us now return to the EAI example and compute the standard e rror of the proportion associated with simple random samples of 30 EAI employees. For the EAI study we know that the population proportion of employees who partici pated in the management training program is p 5 .60. With n/N 5 30/2500 5 .012, we can ignore the finite population correction factor when we compute the standard error of the proportion. For the simple random sample of 30 employees, p is

p 5

p(1 2 p) 5 n

.60(1 2 .60) 5 .0894 30

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Form of the Sampling Distribution of p Now that we know the mean and standard deviation of the sampling distribution of p, the final step is to determine the form or shape of the sampling distribution. The sample proportion is p 5 x/n. For a simple random sample from a large population, the value of x is a binomial random variable indicating the number of elements in the sample with the characteristic of interest. Because n is a constant, the probability of x/n is the same as the binomial probability of x, which means that the sampling distribution of p is also a discrete probability distribution and that the probability for each value of x/n is the same as the probability of x. Statisticians have shown that a binomial distribution can be approximated by a nor mal distribution whenever the sample size is large enough to satisfy the following two conditions:

np $ 5 and n(1 2 p) $ 5

Assuming these two conditions are satisfied, the probability distribution of x in the sample proportion, p 5 x/n, can be approximated by a normal distribution. And because n is a constant, the sampling distribution of p can also be approximated by a normal distribution. This approximation is stated as follows: The sampling distribution of p can be approximated by a normal distribution whenever np $ 5 and n(1 2 p) $ 5. In practical applications, when an estimate of a population proportion is desired, we find that sample sizes are almost always large enough to permit the use of a normal approxima tion for the sampling distribution of p. Recall that for the EAI sampling problem we know that the population proportion of employees who participated in the training program is p 5 .60. With a simple random sample of size 30, we have np 5 30(.60) 5 18 and n(1 2 p) 5 30(.40) 5 12. Thus, the sampling distribution of p can be approximated by the normal distribution shown in Figure 7.11. FIGURE 7.11 SAMPLING DISTRIBUTION OF p FOR THE PROPORTION OF EAI Employees

WHO PARTICIPATED IN THE MANAGEMENT TRAINING PROGRAM

Sampling distribution of p

p 5 .0894

.60 E( p)

7.6 Sampling Distribution of p–

347

Practical Value of the Sampling Distribution of p The practical value of the sampling distribution of p is that it can be used to provide proba bility information about the difference between the sample proportion and the population proportion. For instance, suppose that in the EAI problem the personnel director wants to know the probability of obtaining a value of p that is within .05 of the population proportion of EAI employees who participated in the training program. That is, what is the probability of obtaining a sample with a sample proportion p between .55 and .65? The darkly shaded area in Figure 7.12 shows this probability. Using the fact that the sampling distribution of p can be approximated by a normal probability distribution with a mean of .60 and a standard error of p 5 .0894, we can use Excel’s NORM.DIST function to make this calculation. Entering the formula 5NORM.DIST(.65,.60,.0894,TRUE) into a cell of an Excel worksheet provides the cumulative probability corresponding to p 5 .65. The value calculated by Excel is .7120. Entering the formula 5NORM.DIST(.55,.60,.0894,TRUE) into a cell of an Excel worksheet provides the cumulative probability corresponding to p 5 .55. The value calculated by Excel is .2880. Thus, the probability of p being in the interval from .55 to .65 is given by .7120 2 .2880 5 .4240. If we consider increasing the sample size to n 5 100, the standard error of the propor tion becomes

p 5

.60(1 2 .60) 5 .0490 100

With a sample size of 100 EAI employees, the probability of the sample proportion having a value within .05 of the population proportion can now be computed. Because the sampling distribution is approximately normal, with mean .60 and standard deviation .0490, we can use Excel’s NORM.DIST function to make this calculation. Entering the formula 5NORM.DIST(.65,.60,.0490,TRUE) into a cell of an Excel worksheet p rovides the cumulative probability corresponding to p 5 .65. The value calculated by Excel is .8462. Entering the formula 5NORM.DIST(.55,.60,.0490,TRUE) into a cell of an Excel worksheet provides the cumulative probability corresponding to p 5 .55. The value cal culated by Excel is .1538. Thus, the probability of p being in the interval from .55 to .65 is given by .8462 2 .1538 5 .6924. Increasing the sample size increases the probability that the sampling error will be less than or equal to .05 by .2684 (from .4240 to .6924). FiGURE 7.12 PROBABILITY OF OBTAINING p BETWEEN .55 AND .65

Sampling distribution of p

p 5 .0894

P(.55 # p # .65) 5 .4240 5 .7120 2 .2880

P( p # .55) 5 .2880

.55 .60

.65

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Exercises

Methods 27. A random sample of size 100 is selected from a population with p 5 .40. a. What is the expected value of p? b. What is the standard error of p? c. Show the sampling distribution of p. d. What does the sampling distribution of p show? 28. A population proportion is .40. A random sample of size 200 will be taken and the sample proportion p will be used to estimate the population proportion. a. What is the probability that the sample proportion will be within 6.03 of the popula tion proportion? b. What is the probability that the sample proportion will be within 6.05 of the popula tion proportion? 29. Assume that the population proportion is .55. Compute the standard error of the proportion, p, for sample sizes of 100, 200, 500, and 1000. What can you say about the size of the stan dard error of the proportion as the sample size is increased? 30. The population proportion is .30. What is the probability that a sample proportion will be within 6.04 of the population proportion for each of the following sample sizes? a. n 5 100 b. n 5 200 c. n 5 500 d. n 5 1000 e. What is the advantage of a larger sample size?

Applications 31. The president of Doerman Distributors, Inc. believes that 30% of the firm’s orders come from first-time customers. A random sample of 100 orders will be used to estimate the proportion of first-time customers. a. Assume that the president is correct and p 5 .30. What is the sampling distribution of p for this study? b. What is the probability that the sample proportion p will be between .20 and .40? c. What is the probability that the sample proportion will be between .25 and .35? 32. The Wall Street Journal reported that the age at first startup for 55% of entrepreneurs was 29 years or less and the age at first startup for 45% of entrepreneurs was 30 years or more. a. Suppose a sample of 200 entrepreneurs will be taken to learn about the most impor tant qualities of entrepreneurs. Show the sampling distribution of p where p is the sample proportion of entrepreneurs whose first startup was at 29 years of age or less. b. What is the probability that the sample proportion in part (a) will be within 6.05 of its population proportion? c. Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampling distribution of p where p is now the sample proportion of entrepreneurs whose first startup was at 30 years of age or more. d. What is the probability that the sample proportion in part (c) will be within 6.05 of its population proportion? e. Is the probability different in parts (b) and (d)? Why? f. Answer part (b) for a sample of size 400. Is the probability smaller? Why? 33. Seventy two percent of American adults have read a book within the past year (pewresearch.org, December 3, 2014). Assume this is the true population proportion Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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and that you plan to take a sample survey of 540 American adults to further investi gate their behavior. a. Show the sampling distribution of p, the proportion of your sample respondents who have read a book in the past year. b. What is the probability that your survey will provide a sample proportion within 6.03 of the population proportion? c. What is the probability that your survey will provide a sample proportion within 6.015 of the population proportion? 34. According to Reader’s Digest, 42% of primary care doctors think their patients receive unnec essary medical care. a. Suppose a sample of 300 primary care doctors was taken. Show the sampling distribution of the proportion of the doctors who think their patients receive unnecessary medical care. b. What is the probability that the sample proportion will be within 6.03 of the population proportion? c. What is the probability that the sample proportion will be within 6.05 of the popula tion proportion? d. What would be the effect of taking a larger sample on the probabilities in parts (b) and (c)? Why? 35. Thirty-six percent of all Americans drink bottled water more than once a week (Natural Resources Defense Council, December 4, 2015). Suppose you have been hired by the Natural Resources Defense Council to investigate bottled water consumption in St. Paul. You plan to select a sample of St. Paulites to estimate the proportion who drink bottled water more than once a week. Assume the population proportion of St. Paulites who drink bottled water more than once a week is .36, the same as the overall proportion of Americans who drink bottled water more than once a week. a. Suppose you select a sample of 540 St. Paulites. Show the sampling distribution of p. b. Based upon a sample of 540 St. Paulites, what is the probability that the sample pro portion will be within .04 of the population proportion? c. Suppose you select a sample of 200 St. Paulites. Show the sampling distribution of p. d. Based upon the smaller sample of only 200 St. Paulites, what is the probability that the sample proportion will be within .04 of the population proportion? e. As measured by the increase in probability, how much do you gain in precision by taking the larger sample in parts (a) and (b) rather than the smaller sample in parts (c) and (d)? 36. The Grocery Manufacturers of America reported that 76% of consumers read the ingredi ents listed on a product’s label. Assume the population proportion is p 5 .76 and a sample of 400 consumers is selected from the population. a. Show the sampling distribution of the sample proportion p, where p is the pro portion of the sampled consumers who read the ingredients listed on a product’s label. b. What is the probability that the sample proportion will be within 6.03 of the popula tion proportion? c. Answer part (b) for a sample of 750 consumers. 37. The Food Marketing Institute shows that 17% of households spend more than $100 per week on groceries. Assume the population proportion is p 5 .17 and a simple random s ample of 800 households will be selected from the population. a. Show the sampling distribution of p, the sample proportion of households spending more than $100 per week on groceries. b. What is the probability that the sample proportion will be within 6.02 of the popula tion proportion? c. Answer part (b) for a sample of 1600 households. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Other Sampling Methods 7.7 This section provides a brief introduction to survey sampling methods other than simple random sampling.

We described simple random sampling as a procedure for sampling from a finite population and discussed the properties of the sampling distributions of x and p when simple random sampling is used. Other methods such as stratified random sampling, cluster sampling, and systematic sampling provide advantages over simple random sampling in some of these situations. In this section we briefly introduce these alternative sampling methods.

Stratified Random Sampling Stratified random sampling works best when the variance among elements in each stratum is relatively small.

In stratified random sampling, the elements in the population are first divided into groups called strata, such that each element in the population belongs to one and only one stratum. The basis for forming the strata, such as department, location, age, industry type, and so on, is at the discretion of the designer of the sample. However, the best results are obtained when the elements within each stratum are as much alike as possible. Figure 7.13 is a dia gram of a population divided into H strata. After the strata are formed, a simple random sample is taken from each stratum. For mulas are available for combining the results for the individual stratum samples into one estimate of the population parameter of interest. The value of stratified random sampling depends on how homogeneous the elements are within the strata. If elements within strata are alike, the strata will have low variances. Thus relatively small sample sizes can be used to obtain good estimates of the strata characteristics. If strata are homogeneous, the stratified random sampling procedure provides results just as precise as those of simple random sampling by using a smaller total sample size.

Cluster Sampling Cluster sampling works best when each cluster provides a small-scale representation of the population.

In cluster sampling, the elements in the population are first divided into separate groups called clusters. Each element of the population belongs to one and only one cluster (see Figure 7.14). A simple random sample of the clusters is then taken. All elements within each sampled cluster form the sample. Cluster sampling tends to provide the best results when the elements within the clusters are not alike. In the ideal case, each cluster is a representative small-scale version of the entire population. The value of cluster sampling depends on how representative each cluster is of the entire population. If all clusters are alike in this regard, sampling a small number of clusters will provide good estimates of the population parameters.

FIGURE 7.13 DIAGRAM FOR STRATIFIED RANDOM SAMPLING

Population

Stratum 1

Stratum 2

. . .

Stratum H

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7.7 Other Sampling Methods

FIGURE 7.14 DIAGRAM FOR cluster SAMPLING

Population

Cluster 1

Cluster 2

. . .

Cluster K

One of the primary applications of cluster sampling is area sampling, where clusters are city blocks or other well-defined areas. Cluster sampling generally requires a larger total sample size than either simple random sampling or stratified random sampling. However, it can result in cost savings because when an interviewer is sent to a sampled cluster (e.g., a city-block location), many sample observations can be obtained in a rela tively short time. Hence, a larger sample size may be obtainable with a significantly lower total cost.

Systematic Sampling In some sampling situations, especially those with large populations, it is time-consuming to select a simple random sample by first finding a random number and then counting or searching through the list of the population until the corresponding element is found. An alternative to simple random sampling is systematic sampling. For example, if a sample size of 50 is desired from a population containing 5000 elements, we will sample one element for every 5000/50 5 100 elements in the population. A systematic sample for this case involves selecting randomly one of the first 100 elements from the population list. Other sample elements are identified by starting with the first sampled element and then selecting every 100th element that follows in the population list. In effect, the sample of 50 is identified by moving systematically through the population and identi fying every 100th element after the first randomly selected element. The sample of 50 usually will be easier to identify in this way than it would be if simple random sampling were used. Because the first element selected is a random choice, a systematic sample is usually assumed to have the properties of a simple random sample. This assumption is especially applicable when the list of elements in the population is a random ordering of the elements.

Convenience Sampling The sampling methods discussed thus far are referred to as probability sampling tech niques. Elements selected from the population have a known probability of being included in the sample. The advantage of probability sampling is that the sampling distribution of the appropriate sample statistic generally can be identified. Formulas such as the ones for simple random sampling presented in this chapter can be used to determine the properties of the sampling distribution. Then the sampling distribution can be used to make probability statements about the error associated with using the sample results to make inferences about the population. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Convenience sampling is a nonprobability sampling technique. As the name implies, the sample is identified primarily by convenience. Elements are included in the sample without prespecified or known probabilities of being selected. For example, a professor conducting research at a university may use student volunteers to constitute a sample simply because they are readily available and will participate as subjects for little or no cost. Similarly, an inspector may sample a shipment of oranges by selecting oranges haphazardly from among several crates. Labeling each orange and using a probability method of sampling would be impractical. Samples such as wildlife captures and volunteer panels for c onsumer research are also con venience samples. Convenience samples have the advantage of relatively easy sample selection and data collection; however, it is impossible to evaluate the “goodness” of the sample in terms of its representativeness of the population. A convenience sample may provide good results or it may not; no statistically justified procedure allows a probability analysis and inference about the quality of the sample results. Sometimes researchers apply statistical methods d esigned for probability samples to a convenience sample, arguing that the convenience sample can be treated as though it were a probability sample. However, this argument c annot be supported, and we should be cautious in interpreting the results of convenience samples that are used to make inferences about populations.

Judgment Sampling One additional nonprobability sampling technique is judgment sampling. In this approach, the person most knowledgeable on the subject of the study selects elements of the popula tion that he or she feels are most representative of the population. Often this method is a relatively easy way of selecting a sample. For example, a reporter may sample two or three senators, judging that those senators reflect the general opinion of all senators. However, the quality of the sample results depends on the judgment of the person selecting the sample. Again, great caution is warranted in drawing conclusions based on judgment samples used to make inferences about populations.

NOTE AND COMMENT We recommend using probability sampling methods when sampling from finite populations: simple random sampling, stratified random sampling, cluster sampling, or systematic sampling. For these methods, formulas are available for evaluat ing the “goodness” of the sample results in terms

7.8

of the closeness of the results to the population parameters being estimated. An evaluation of the goodness cannot be made with convenience or judgment sampling. Thus, great care should be used in interpreting the results based on nonprob ability sampling methods.

Practical Advice: Big Data and Errors in Sampling The purpose of collecting a sample is to make inferences and answer research questions about a population. Therefore, it is important that the sample look like, or be representative of, the population being investigated. In practice, a sample may fail to be representative of the population of interest because of sampling error and/or nonsampling error.

7.8 Practical Advice: Big Data and Errors in Sampling

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Sampling Error Whenever we use a sample to estimate the value of a population parameter, it is highly unlikely that the value of the sample statistic will be equal to the population parameter being estimated. For example, we would not expect the sample mean x to be exactly equal to the population mean m. The difference between the value of the sample statistic and the value of the corresponding population parameter is called the sampling error. If repeated independent random samples of the same size are collected from the population of interest using any of the four previously discussed probability sampling methods (simple random sampling, stratified random sampling, cluster sampling, or systematic sampling), on average the samples will be representative of the population. However, the use of random sampling does not ensure that any single sample will be representative of the population of interest. As discussed in the introduction to this chapter, sampling error is unavoidable when collecting a random sample; it is the risk we must accept when we chose to collect a random sample rather than incur the costs associated with taking a census of the population. Note that the standard errors of the sampling distributions of the sample mean x (shown in formula 7.2) and the sample proportion of p (shown in formula 7.5) reflect the potential for sampling error when using sample data to estimate the population mean m and the population proportion p, respectively. As the sample size increases, the standard errors of these sampling distributions decrease. Because these standard errors reflect the potential for sampling error when using sample data to estimate the population mean m and the popula tion proportion p, we see that for an extremely large sample there may be little potential for sampling error. For example, consider the online news service PenningtonDailyTimes.com (PDT). PDT’s primary source of revenue is the sale of advertising. Because prospective adver tisers are willing to pay a premium to advertise on websites that have long visit times, PDT’s management is keenly interested in the amount of time customers spend during their visits to PDT’s website. Based on historical data, PDT assumes s 5 20 seconds for the population standard deviation of the times spent by customers when they visit PDT’s website. Table 7.6 shows how the standard error of the sampling distribution of the sample mean time spent by customers when they visit PDT’s website decreases as the sample size increases when s 5 20. PDT could also collect information from this sample on whether a visitor to its web site clicked on any of the ads featured on the website. From its historical data, PDT knows that 50% of visitors to its website clicked on an ad featured on the website. Based upon this data, PDT now assumes a known value of p 5 .50. Table 7.7 shows how the standard error of the sampling distribution of the proportion of the sample that clicked on any of the ads featured on PenningtonDailyTimes.com decreases as the sample size increases when p 5 .50. The PDT example illustrates the relationship between standard errors and the sample size. We see in Table 7.6 that the standard error of the sample mean decreases as the sample size increases. For a sample of n 5 10, the standard error of the sample mean is 6.32456; when we increase the sample size to n 5 100,000, the standard error of the sample mean decreases to 0.06325; and at a sample size of n 5 1,000,000,000, the standard error of the sample mean decreases to only 0.00063. In Table 7.7 we see that the standard error of the sample proportion also decreases as the sample size increases. For a sample of n 5 10, the standard error of the sample proportion is .15811; when we increase the sample size to n 5 100,000, the standard error of the sample proportion decreases to .00158; and at a sample size of n 5 1,000,000,000, the standard error of the sample mean decreases to only .00002. In both instances, the standard error when n 5 1,000,000,000 is only one ten thousandth of the standard error when n 5 10! Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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A sample of one million or more visitors might seem unrealistic, but keep in mind that amazon.com had over 91 million visitors in March of 2016 (quantcast. com, May 13, 2016).

TABLE 7.6 STANDARD ERROR OF

THE SAMPLE MEAN x AT VARIOUS SAMPLE SIZES n Sample Size n

Standard Error sx 5 syÏn

10 6.32456 100 2.00000 500 0.89443 1,000 0.63246 10,000 0.20000 100,000 0.06325 1,000,000 0.02000 10,000,000 0.00632 100,000,000 0.00200 1,000,000,000 0.00063

TABLE 7.7 STANDARD ERROR OF

THE SAMPLE PROPORTION p AT VARIOUS SAMPLE SIZES n Sample Size n

Standard Error sp 5 Ïp(1 2 p)yn

10 .15811 100 .05000 500 .02236 1,000 .01581 10,000 .00500 100,000 .00158 1,000,000 .00050 10,000,000 .00016 100,000,000 .00005 1,000,000,000 .00002

Nonsampling Error

Nonsampling error can occur in a sample or a census.

The standard error of a sampling distribution decreases as the sample size n increases because a larger sample tends to better represent the population from which it has been drawn. However, this does not mean that we can conclude that an extremely large sample will always provide reliable information about the population of interest; this is because sampling error is not the sole reason a sample may fail to represent the target population. Deviations of the sample from the population that occur for reasons other than random sampling are referred to as nonsampling errors. Nonsampling errors are introduced into the sample data systematically and can occur for a variety of reasons. For example, the data in the sample may not have been drawn from the population of interest. From whom should PDT collect its data? Should it collect data on current visits to PenningtonDailyTimes.com? Should it attempt to attract new visitors and collect data on these visits? If so, should it measure the time spent at its website by visitors it has attracted from competitors’ websites or visitors who do not routinely visit online news sites? The answers depend on PDT’s research objectives. Is the company attempting to evaluate its current market, assess the potential of customers it can attract from competitors, or explore the potential of an entirely new market (individuals who do not routinely obtain their news from online news services)? If the research objective and the population from which the sample is to be drawn are not aligned, the data that PDT collects will not help the company accomplish its research objective. This type of error is referred to as a coverage error. Even when the sample is selected from the appropriate population, nonsampling error can occur when segments of the target population are systematically underrepresented or overrepresented in the sample. This may occur because the study design is flawed or because some segments of the population are either more likely or less likely to respond. Suppose PDT implements a pop-up questionnaire that opens when the visitor leaves PenningtonDailyTimes.com. Visitors to PenningtonDailyTimes.com who have installed pop-up blockers will likely be underrepresented, and visitors to PenningtonDailyTimes.com who have not installed pop-up blockers will likely be overrepresented. If the behavior of PenningtonDailyTimes.com visitors who have installed pop-up blockers differs from the behaviors of PenningtonDailyTimes.com visitors who have not installed pop-up blockers, attempting to draw conclusions from this sample about how all visitors to the PDT website behave may be misleading. This type of error is referred to as a non-response error.

7.8 Practical Advice: Big Data and Errors in Sampling

Errors that are introduced by interviewers or during the recording and preparation of the data are other types of nonresponse error. These types of error are referred to as interviewer errors and processing errors, respectively.

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Another potential source of nonsampling error is incorrect measurement of the char acteristic of interest. If PDT asks questions that are ambiguous or difficult for respondents to understand, the responses may not accurately reflect how the respondents intended to respond. For example, if PDT asks Are the news stories on PenningtonDailyTimes.com compelling and accurate?, respondents may be unsure how to respond. How should a visi tor respond if she or he feels the news stories on PenningtonDailyTimes.com are compelling but not accurate? What response is appropriate if the respondent feels the news stories on PenningtonDailyTimes.com are accurate but not compelling? A similar issue can arise if a questions is asked in a biased or leading way. If PDT says, Many readers find the news stories on PenningtonDailyTimes.com to be compelling and accurate! Do you find the news stories on PenningtonDailyTimes.com to be to be compelling and accurate?, the qualifying statement PDT makes prior to the actual question will likely result in a bias toward positive responses. Note that incorrect measurement of the characteristic of interest can also occur when respondents provide incorrect answers; this may be due to a respondent’s poor recall, unwillingness to respond honestly, or desire to provide an “acceptable” response. This type of error is referred to as a measurement error.

Implications of Big Data Suppose that last year the mean time spent by all visitors to the PDT website was 84 sec onds. Further suppose that the mean time has not changed since last year and that PDT now collects a sample of 1,000,000 visitors to its website (recall from chapter 1 that we often refer to very large or complex data sets as big data). With n 5 1,000,000 and the assumed known value of s 5 20 seconds for the population standard deviation, the standard error will be x 5 yÏn 5 20yÏ1,000,000 5 0.02. Note that because of the very large sample size, the sampling distribution of the sample mean x will be normally distributed. PDT can use this information to calculate the probability the sample mean x will fall within .15 of the population mean, or within the interval 84 6 .15. At x 5 83.85 z5

83.85 2 84 5 27.5 .02

and at x 5 84.15, z5

84.15 2 84 5 7.5 .02

Because P(z # 27.5) ø .0000 and P(z # 7.5) ø 1.0000, the probability the sample mean x will fall within 1.5 of the population mean is P(27.5 # z # 7.5) ø 1.0000 2 .0000 5 1.0000 Using Excel we find NORM.DIST(84.15,84,.02,TRUE) 2 NORM.DIST(83.85,84,.02,TRUE) 5 1.000 Now, suppose that for the new sample of 1,000,000 visitors the sample mean time spent by all visitors to the PDT website is 88 seconds. What could PDT conclude? Its calculations show that the probability the sample mean will be between 83.85 and 84.15 is approxi mately 1.0000, and yet the mean of PDT’s sample is 88. There are three possible reasons that PDT’s sample mean differs so substantially from the population mean for last year: ●● sampling error ●● nonsampling error ●● the population mean has changed since last year Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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Chapter 7 Sampling and Sampling Distributions

Because the sample size is extremely large, the sample should have very little sampling error and so sampling error cannot explain the substantial difference between PDT’s sample mean of x 5 88 seconds and the population mean of m 5 84 seconds for last year. Nonsam pling error is a possible explanation and should be investigated. If PDT determines that it introduced little or no nonsampling error into its sample data, the only remaining plausible explanation for the substantial difference between PDT’s sample mean and the population mean for last year is that the population mean has changed since last year. If the sample was collected properly, it provides evidence of a potentially important change in behavior of visitors to the PDT website that could have tremendous implications for PDT. No matter how small or large the sample, we must contend with the limitations of sam pling whenever we use sample data to learn about a population of interest. Although sampling error decreases as the size of the sample increases, an extremely large sample can indeed suffer from failure to be representative of the population of interest because of nonsampling error. When sampling, care must be taken to ensure that we minimize the introduction of non sampling error into the data collection process. This can be done by taking the following steps: ●●

●● ●●

●●

arefully define the target population before collecting sample data, and subse C quently design the data collection procedure so that a probability sample is drawn from this target population. Carefully design the data collection process and train the data collectors. Pretest the data collection procedure to identify and correct for potential sources of nonsampling error prior to final data collection. Use stratified random sampling when population-level information about an im portant qualitative variable is available to ensure the sample is representative of the population for that qualitative characteristic. Use systematic sampling when population-level information about an important quantitative variable is available to ensure the sample is representative of the popu lation for that quantitative characteristic.

Finally, recognize that every random sample (even an extremely large sample) will suffer from some degree of sampling error, and eliminating all potential sources of nonsampling error may be impractical. Understanding these limitations of sampling will enable us to be more realistic when interpreting sample data and using sample data to draw conclusions about the target population. In the next several chapters we will explore statistical methods for dealing with these issues in greater detail. NOTES AND COMMENTS 1. In the previous section of this chapter, we explained that one reason probability sampling methods are generally preferred over nonprob ability sampling methods is that formulas are available for evaluating the “goodness” of the sample results in terms of the closeness of the results to the population parameters being esti mated for sample data collected using prob ability sampling methods. An evaluation of the goodness cannot be made with convenience or judgment sampling. Another reason for the gen eral preference of probability sampling methods over nonprobability sampling methods is that probability sampling methods are less likely than nonprobability sampling methods to intro duce nonsampling error. Although nonsampling

can occur when either a probability sampling method or a nonprobability sampling method is used, nonprobability sampling methods such as convenience sampling and judgement sampling frequently introduce nonsampling error into the sample data. This is because of the manner in which sample data are collected when using a nonprobability sampling method. 2. Several approaches to statistical inference (interval estimation and hypothesis testing) are introduced in subsequent chapters of this book. These approaches assume nonsampling error has not been introduced into the sample data. The reliability of the results of statistical infer ence decreases as greater nonsampling error is introduced into the sample data.

Summary

357

Exercises

Applications 38. Martina Levitt, director of marketing for the messaging app Spontanversation, has been assigned the task of profiling users of this app. Assume that individuals who have downloaded Spontanversation use the app an average of 30 times per day with a standard deviation of 6. a. What is the sampling distribution of x if a random sample of 50 individuals who have downloaded Spontanversation is used? b. What is the sampling distribution of x if a random sample of 500,000 individuals who have downloaded Spontanversation is used? c. What general statement can you make about what happens to the sampling distribu tion of x as the sample size becomes extremely large? Does this generalization seem logical? Explain. 39. Consider the Spontanversation sampling problem for which individuals who have down loaded Spontanversation use the app an average of 30 times per day with a standard deviation of 6. a. What is the probability that x is within .5 of the population mean if a random sample of 50 individuals who have downloaded Spontanversation is used? b. What would you conclude if the mean of the sample collected in part (a) is 30.2? c. What is the probability that x is within .5 of the population mean if a sample of 500,000 individuals who have downloaded Spontanversation is used? d. What would you conclude if the mean of the sample collected in part (c) is 30.2? 40. The Wall Street Journal reported that 37% of all entrepreneurs who opened new U.S. busi nesses in the previous year were female (The Wall Street Journal, May 13, 2015). a. Suppose a random sample of 300 entrepreneurs who opened new U.S. businesses in the previous year will be taken to learn about which industries are most appealing to entrepreneurs. Show the sampling distribution of p, where p is the sample proportion of entrepreneurs who opened new U.S. businesses in the previous year that are female. b. What is the probability that the sample proportion in part (a) will be within 6.05 of its population proportion? c. Suppose a random sample of 30,000 entrepreneurs who opened new U.S. businesses in the previous year will be taken to learn about which industries are most appealing to entrepreneurs. Show the sampling distribution of p where p is the sample proportion of entrepreneurs who opened new U.S. businesses last year that are female. d. What is the probability that the sample proportion in part (c) will be within 6.05 of its population proportion? e. Is the probability different in parts (b) and (d)? Why? 41. The vice president of sales for Blasterman Cosmetics, Inc. believes that 40% of the compa ny’s orders come from customers who are less than 30 years old. A random sample of 10,000 orders will be used to estimate the proportion of customers who are less than 30 years old. a. Assume that the vice president of sales is correct and p 5 .40. What is the sampling distribution of p for this study? b. What is the probability that the sample proportion will be between .37 and .43? c. What is the probability that the sample proportion will be between .39 and .41? d. What would you conclude if the sample proportion is .36?

Summary In this chapter we presented the concepts of sampling and sampling distributions. We dem onstrated how a simple random sample can be selected from a finite population and how a random sample can be selected from an infinite population. The data collected from such Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

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samples can be used to develop point estimates of population parameters. Because d ifferent samples provide different values for the point estimators, point estimators such as x and p are random variables. The probability distribution of such a random variable is called a sampling distribution. In particular, we described in detail the sampling distributions of the sample mean x and the sample proportion p. In considering the characteristics of the sampling distributions of x and p, we stated that E(x) 5 and E( p) 5 p. After developing the standard deviation or standard error formulas for these estimators, we described the conditions necessary for the sampling distributions of x and p to follow a normal distribution. Other sampling methods including stratified random sampling, cluster sampling, systematic sampling, convenience sampli