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Essentials of Statistics for the Behavioral Sciences

8th edition

Frederick J Gravetter

State University of New York, Brockport

Larry B. Wallnau

State University of New York, Brockport

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Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has

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Chapter 1 Introduction to Statistics 3

Chapter 2 Frequency Distributions 37

Chapter 3 Measures of Central Tendency 59

Chapter 4 Measures of Variability 89

Chapter 5 z-Scores: Location of Scores and Standardized Distributions 123

Chapter 6 Probability 149

Chapter 7 Probability and Samples: The Distribution of Sample Means 175

Chapter 8 Introduction to Hypothesis Testing 203

Using t Statistics for Inferences About Population Means and Mean DifferencesPART III

Chapter 9 Introduction to the t Statistic 249

Chapter 10 The t Test for Two Independent Samples 279

Chapter 11 The t Test for Two Related Samples 313

Analysis of Variance: Tests for Differences Among Two or More Population MeansPART IV

Chapter 12 Introduction to Analysis of Variance 345

Chapter 13 Repeated-Measures and Two-Factor Analysis of Variance 393

iii

Contents in Brief

Introduction and Descriptive StatisticsPART I

Foundations of Inferential StatisticsPART II

Correlations and Nonparametric TestsPART V

Chapter 14 Correlation 449

Chapter 15 The Chi-Square Statistic: Tests for Goodness of Fit and Independence 509

iv CONTENTS IN BRIEF

Contents

Introduction to Statistics 3

1.1 Statistics, Science, and Observations 4 1.2 Populations and Samples 5 1.3 Data Structures, Research Methods, and Statistics 12 1.4 Variables and Measurement 20 1.5 Statistical Notation 26 Summary 30 Focus on Problem Solving 32 Demonstration 1.1 32 Problems 33

Frequency Distributions 37

2.1 Introduction to Frequency Distributions 38 2.2 Frequency Distribution Tables 38 2.3 Frequency Distribution Graphs 44 2.4 The Shape of a Frequency Distribution 50 Summary 52 Focus on Problem Solving 54 Demonstration 2.1 55 Problems 56

Measures of Central Tendency 59

3.1 Defining Central Tendency 60 3.2 The Mean 61 3.3 The Median 69 3.4 The Mode 73 3.5 Selecting a Measure of Central Tendency 74 3.6 Central Tendency and the Shape of the Distribution 80 Summary 82 Focus on Problem Solving 84 Demonstration 3.1 84 Problems 85

Measures of Variability 89

4.1 Defining Variability 90 4.2 The Range 91

Chapter 1

Chapter 2

Chapter 3

Chapter 4

•

Introduction and Descriptive StatisticsPART I

•

vi CONTENTS

4.3 Standard Deviation and Variance for a Population 92 4.4 Standard Deviation and Variance for a Sample 99 4.5 More About Variance and Standard Deviation 104 Summary 112 Focus on Problem Solving 114 Demonstration 4.1 115 Problems 115

• Part I Review 119 • Review Exercises 119

z-Scores: Location of Scores and Standardized Distributions 123

5.1 Introduction to z-Scores 124 5.2 z-Scores and Location in a Distribution 125 5.3 Using z-Scores to Standardize a Distribution 131 5.4 Other Standardized Distributions Based on z-Scores 136 5.5 Computing z-Scores for a Sample 138 5.6 Looking Ahead to Inferential Statistics 140 Summary 143 Focus on Problem Solving 144 Demonstration 5.1 145 Demonstration 5.2 145 Problems 146

Probability 149

6.1 Introduction to Probability 150 6.2 Probability and the Normal Distribution 155 6.3 Probabilities and Proportions for Scores from

a Normal Distribution 162 6.4 Looking Ahead to Inferential Statistics 169 Summary 171 Focus on Problem Solving 172 Demonstration 6.1 172 Problems 173

Probability and Samples: The Distribution of Sample Means 175

7.1 Samples and Populations 176 7.2 The Distribution of Sample Means 176 7.3 Probability and the Distribution of Sample Means 186 7.4 More About Standard Error 190

Chapter 5

Chapter 6

Chapter 7 •

•

Foundations of Inferential StatisticsPART II

CONTENTS vii

7.5 Looking Ahead to Inferential Statistics 194 Summary 198 Focus on Problem Solving 199 Demonstration 7.1 200 Problems 201

Introduction to Hypothesis Testing 203

8.1 The Logic of Hypothesis Testing 204 8.2 Uncertainty and Errors in Hypothesis Testing 213 8.3 More About Hypothesis Tests 217 8.4 Directional (One-Tailed) Hypothesis Tests 224 8.5 Concerns About Hypothesis Testing: Measuring Effect Size 227 8.6 Statistical Power 232 Summary 237 Focus on Problem Solving 239 Demonstration 8.1 240 Demonstration 8.2 240 Problems 241

• Part II Review 245 • Review Exercises 245

Chapter 8 •

•

Chapter 9

Chapter 10

Introduction to the t Statistic 249

9.1 The t Statistic: An Alternative to z 250 9.2 Hypothesis Tests with the t Statistic 255 9.3 Measuring Effect Size for the t Statistic 260 9.4 Directional Hypotheses and One-Tailed Tests 268 Summary 271 Focus on Problem Solving 273 Demonstration 9.1 273 Demonstration 9.2 274 Problems 275

The t Test for Two Independent Samples 279

10.1 Introduction to the Independent-Measures Design 280 10.2 The t Statistic for an Independent-Measures Research Design 281 10.3 Hypothesis Tests and Effect Size with the Independent-

Measures t Statistic 288 10.4 Assumptions Underlying the Independent-Measures

t Formula 300 Summary 303 Focus on Problem Solving 305 Demonstration 10.1 307 Demonstration 10.2 308 Problems 308

Using t Statistics for Interferences About Population Means and Mean DifferencesPART III

viii CONTENTS

The t Test for Two Related Samples 313

11.1 Introduction to Repeated-Measures Designs 314 11.2 The t Statistic for a Repeated-Measures Research Design 315 11.3 Hypothesis Tests and Effect Size for the Repeated-Measures Design 320 11.4 Uses and Assumptions for Repeated-Measures t Tests 328

Summary 331 Focus on Problem Solving 332 Demonstration 11.1 333 Demonstration 11.2 335 Problems 335

• Part III Review 341 • Review Exercises 341

Chapter 11 •

•

Analysis of Variance: Tests for Differences Among Two or More Population MeansPART IV

Introduction to Analysis of Variance 345

12.1 Introduction 346 12.2 The Logic of ANOVA 350 12.3 ANOVA Notation and Formulas 354 12.4 The Distribution of F-Ratios 362 12.5 Examples of Hypothesis Testing and Effect Size with ANOVA 364 12.6 Post Hoc Tests 375 12.7 The Relationships Between ANOVA and t Tests 379

Summary 381 Focus on Problem Solving 383 Demonstration 12.1 384 Demonstration 12.2 386 Problems 387

Repeated-Measures and Two-Factor Analysis of Variance 393

13.1 Overview 394 13.2 Repeated-Measures ANOVA 394 13.3 Two-Factor ANOVA (Independent Measures) 409 Summary 428 Focus on Problem Solving 432 Demonstration 13.1 434 Demonstration 13.2 435 Problems 438

• Part IV Review 445 • Review Exercises 445

Chapter 12

Chapter 13

CONTENTS ix

Correlations and Nonparametric TestsPART V

•

Correlation 449

14.1 Introduction 450 14.2 The Pearson Correlation 453 14.3 Using and Interpreting the Pearson Correlation 458 14.4 Hypothesis Tests with the Pearson Correlation 464 14.5 Alternatives to the Pearson Correlation 472 14.6 Introduction to Linear Equations and Regression 481 Summary 496 Focus on Problem Solving 500 Demonstration 14.1 502 Problems 503

The Chi-Square Statistic: Tests for Goodness of Fit and Independence 509

15.1 Parametric and Nonparametric Statistical Tests 510 15.2 The Chi-Square Test for Goodness of Fit 511 15.3 The Chi-Square Test for Independence 521 15.4 Measuring Effect Size for the Chi-Square Test for Independence 532

15.5 Assumptions and Restrictions for Chi-Square Tests 534 Summary 535 Focus on Problem Solving 539 Demonstration 15.1 539 Demonstration 15.2 541 Problems 541

• Part V Review 547 • Review Exercises 547

Basic Mathematics Review 549

Statistical Tables 571

Solutions for Odd-Numbered Problems in the Text 583

General Instructions for Using SPSS 601

Statistics Organizer: Finding the Right Statistics for Your Data 605

References 619

Index 625

Chapter 14

Chapter 15

Appendix A

Appendix B

Appendix C

Appendix D

Many students in the behavioral sciences view the required statistics course as an intimi- dating obstacle that has been placed in the middle of an otherwise interesting curriculum. They want to learn about human behavior—not about math and science. As a result, the statistics course is seen as irrelevant to their education and career goals. However, as long as the behavioral sciences are founded in science, a knowledge of statistics will be neces- sary. Statistical procedures provide researchers with objective and systematic methods for describing and interpreting their research results. Scientific research is the system that we use to gather information, and statistics are the tools that we use to distill the informa- tion into sensible and justified conclusions. The goal of this book is not only to teach the methods of statistics but also to convey the basic principles of objectivity and logic that are essential for science and valuable for decision making in everyday life.

Those of you who are familiar with previous editions of Essentials of Statistics for the Behavioral Sciences will notice that some changes have been made. These changes are summarized in the section titled “To the Instructor.” In revising this text, our stu- dents have been foremost in our minds. Over the years, they have provided honest and useful feedback. Their hard work and perseverance has made our writing and teaching most rewarding. We sincerely thank them. Students who are using this edition should please read the section of the preface titled “To the Student.”

The book chapters are organized in the sequence that we use for our own statistics courses. We begin with descriptive statistics, and then examine a variety of statistical pro- cedures focused on sample means and variance before moving on to correlational methods and nonparametric statistics. Information about modifying this sequence is presented in the “To the Instructor” section for individuals who prefer a different organization. Each chapter contains numerous examples—many based on actual research studies—along with learning checks, a summary and list of key terms, and a set of 20 to 30 problems.

Those of you familiar with the previous edition of Essentials of Statistics for the Behavioral Sciences will notice a number of changes in the eighth edition. Throughout the book, research examples have been updated, real-world examples have been added, and the end-of-chapter problems have been extensively revised. The book has been separated into five sections to emphasize the similarities among groups of statistical methods. Each section contains two to four chapters and begins with an introduction and concludes with a review, including review exercises. Major revisions for this edition include:

• The former Chapter 12 on estimation has been eliminated. In its place, sec- tions on confidence intervals have been added to the three chapters presenting t statistics.

• A new appendix titled Statistics Organizer: Finding the Right Statistics for Your Data, discusses the process of selecting the correct statistics to be used with different categories of data and replaces the Statistics Organizer that appeared as an appendix in earlier editions.

Preface

T O T H E I N S T R U C T O R

Other specific and noteworthy revisions include:

Chapter 1 A separate section explains how statistical methods can be classified using the same categories that are used to group data structures and research methods.

Chapter 2 The discussion of histograms has been modified to differentiate discrete and continuous variables.

Chapter 3 A modified definition of the median acknowledges that this value is not algebraically defined and that determining the median, especially for discrete variables, can be somewhat subjective.

Chapter 4 Relatively minor editing for clarity. The section on variance and inferen- tial statistics has been simplified.

Chapter 5 Relatively minor editing for clarity.

Chapter 6 The concepts of random sample and independent random sample have been clarified with separate definitions. A new figure helps demonstrate the process of using the unit normal table to find proportions for negative z-scores.

Chapter 7 Relatively minor editing for clarity.

Chapter 8 The chapter has been shortened by substantial editing that eliminated several pages of unnecessary text, particularly in the sections on errors (Type I and II) and power.

Chapter 9 The section describing how sample size and sample variance influence the outcome of a hypothesis test has been moved so that it appears immediately after the hypothesis test example. A new section introduces confidence intervals in the context of describing effect size, describes how confidence intervals are reported in the litera- ture, and discusses factors affecting the width of a confidence interval.

Chapter 10 An expanded section discusses how sample variance and sample size influ- ence the outcome of an independent-measures hypothesis test and measures of effect size. A new section introduces confidence intervals as an alternative for describing effect size. The relationship between a confidence interval and a hypothesis test is also discussed.

Chapter 11 The description of repeated-measures and matched-subjects designs has been clarified and we increased emphasis on the concept that all calculations for the related-samples test are done with the difference scores. A new section introduces con- fidence intervals as an alternative for describing effect size and discusses the relation- ship between a confidence interval and a hypothesis test.

The former Chapter 12 has been deleted. The content from this chapter discussing con- fidence intervals has been added to Chapters 9, 10, and 11.

Chapter 12 (former Chapter 13, introducing ANOVA) The discussion of testwise alpha levels versus experimentwise alpha levels has been moved from a box into the text, and definitions of the two terms have been added. To emphasize the concepts of ANOVA rather than the formulas, SS

between treatments is routinely found by subtraction

instead of being computed directly. Two alternative equations for SS between treatments

have been moved from the text into a box.

xii PREFACE

Chapter 13 (former Chapter 14, introducing repeated-measures and two-factor ANOVA) A new section demonstrates the relationship between ANOVA and the t test when a repeated-measures study is comparing only two treatments. Extensive editing has shortened the chapter and simplified the presentation.

Chapter 14 (formerly Chapter 15, introducing correlation and regression) New sec- tions present the t statistic for testing hypotheses about the Pearson correlation and demonstrate how the t test for significance of a correlation is equivalent to the F-ratio used for analysis of regression.

Chapter 15 (formerly Chapter 16, introducing chi-square tests) Relatively minor editing has shortened and clarified the chapter.

Matching the Text to Your Syllabus We have tried to make separate chapters, and even sections of chapters, completely self-contained so that they can be deleted or reorganized to fit the syllabus for nearly any instructor. Some common examples are as follows:

• It is common for instructors to choose between emphasizing analysis of variance (Chapters 12 and 13) or emphasizing correlation/regression (Chapter 14). It is rare for a one-semester course to provide complete coverage of both topics.

• Although we choose to complete all the hypothesis tests for means and mean differences before introducing correlation (Chapter 14), many instructors prefer to place correlation much earlier in the sequence of course topics. To accommodate this, sections 14.1, 14.2, and 14.3 present the calculation and interpretation of the Pearson correlation and can be introduced immediately following Chapter 4 (variability). Other sections of Chapter 14 refer to hypothesis testing and should be delayed until the process of hypothesis testing (Chapter 8) has been introduced.

• It is also possible for instructors to present the chi-square tests (Chapter 15) much earlier in the sequence of course topics. Chapter 15, which presents hypothesis tests for proportions, can be presented immediately after Chapter 8, which introduces the process of hypothesis testing. If this is done, we also recommend that the Pearson correlation (Sections 14.1, 14.2, and 14.3) be pre- sented early to provide a foundation for the chi-square test for independence.

A primary goal of this book is to make the task of learning statistics as easy and pain- less as possible. Among other things, you will notice that the book provides you with a number of opportunities to practice the techniques you will be learning in the form of learning checks, examples, demonstrations, and end-of-chapter problems. We en- courage you to take advantage of these opportunities. Read the text rather than just memorize the formulas. We have taken care to present each statistical procedure in a conceptual context that explains why the procedure was developed and when it should be used. If you read this material and gain an understanding of the basic concepts un- derlying a statistical formula, you will find that learning the formula and how to use it will be much easier. In the following section, “Study Hints,” we provide advice that we give our own students. Ask your instructor for advice as well; we are sure that other instructors will have ideas of their own.

Over the years, the students in our classes and other students using our book have given us valuable feedback. If you have any suggestions or comments about this book, you can write to either Professor Emeritus Frederick Gravetter or Professor Emeritus

PREFACE xiii

T O T H E S T U D E N T

Larry Wallnau at the Department of Psychology, SUNY College at Brockport, 350 New Campus Drive, Brockport, New York 14420. You can also contact Professor Emeritus Gravetter directly at [email protected].

Study Hints You may find some of these tips helpful, as our own students have reported.

• The key to success in a statistics course is to keep up with the material. Each new topic builds on previous topics. If you have learned the previous mate- rial, then the new topic is just one small step forward. Without the proper background, however, the new topic can be a complete mystery. If you find that you are falling behind, get help immediately.

• You will learn (and remember) much more if you study for short periods sev- eral times per week rather than try to condense all of your studying into one long session. For example, it is far more effective to study half an hour every night than to have a single 3}12}-hour study session once a week. We cannot even work on writing this book without frequent rest breaks.

• Do some work before class. Keep a little ahead of the instructor by reading the appropriate sections before they are presented in class. Although you may not fully understand what you read, you will have a general idea of the topic, which will make the lecture easier to follow. Also, you can identify material that is particularly confusing and then be sure the topic is clarified in class.

• Pay attention and think during class. Although this advice seems obvious, often it is not practiced. Many students spend so much time trying to write down every example presented or every word spoken by the instructor that they do not actually understand and process what is being said. Check with your instructor—there may not be a need to copy every example presented in class, especially if there are many examples like it in the text. Sometimes, we tell our students to put their pens and pencils down for a moment and just listen.

• Test yourself regularly. Do not wait until the end of the chapter or the end of the week to check your knowledge. After each lecture, work some of the end- of-chapter problems and do the Learning Checks. Review the Demonstration Problems, and be sure you can define the Key Terms. If you are having trou- ble, get your questions answered immediately (reread the section, go to your instructor, or ask questions in class). By doing so, you will be able to move ahead to new material.

• Do not kid yourself! Avoid denial. Many students watch their instructor solve problems in class and think to themselves, “This looks easy—I understand it.” Do you really understand it? Can you really do the problem on your own without having to leaf through the pages of a chapter? Although there is noth- ing wrong with using examples in the text as models for solving problems, you should try working a problem with your book closed to test your level of mastery.

• We realize that many students are embarrassed to ask for help. It is our big- gest challenge as instructors. You must find a way to overcome this aversion. Perhaps contacting the instructor directly would be a good starting point, if asking questions in class is too anxiety-provoking. You could be pleasantly surprised to find that your instructor does not yell, scold, or bite! Also, your instructor might know of another student who can offer assistance. Peer tutor- ing can be very helpful.

xiv PREFACE

mailto:[email protected]

Ancillaries for this edition include the following:

• Aplia Statistics for Psychology and the Behavioral Sciences: An online inter- active learning solution that ensures students stay involved with their course- work and master the basic tools and concepts of statistical analysis. Created by a research psychologist to help students excel, Aplia’s content engages students with questions based on real-world scenarios that help students un- derstand how statistics applies to everyday life. At the same time, all chapter assignments are automatically graded and provide students with detailed explanations, making sure they learn from and improve with every question.

• Instructor’s Manual with Test Bank: Contains chapter outlines, annotated learning objectives, lecture suggestions, test items, and solutions to all end-of- chapter problems in the text. Test items are also available as a Word down- load or for ExamView computerized test bank software.

• PowerLecture with ExamView®: This CD includes the instructor’s manual, test bank, lecture slides with book figures, and more. Featuring automatic grading, ExamView, also available within PowerLecture, allows you to create, de- liver, and customize tests and study guides (both print and online) in minutes. Assessments appear onscreen exactly as they will print or display online; you can build tests of up to 250 questions using up to 12 question types, and you can enter an unlimited number of new questions or edit existing questions.

• WebTutor on Blackboard and WebCT: Jump-start your course with customiz- able, text-specific content within your Course Management System.

• Psychology CourseMate: Cengage Learning’s Psychology CourseMate brings course concepts to life with interactive learning, study, and exam prepara- tion tools that support the printed textbook. Go to www.cengagebrain.com. Psychology CourseMate includes:

• An interactive eBook;

• Interactive teaching and learning tools, including

• quizzes,

• flashcards,

• videos,

• and more; plus

• The Engagement Tracker, a first-of-its-kind tool that monitors student engage- ment in the course.

It takes a lot of good, hardworking people to produce a book. Our friends at Wadsworth/ Cengage Learning have made enormous contributions to this textbook. We thank: Jon-David Hague, Publisher; Tim Matray, Acquisitions Editor; Paige Leeds, Assistant Editor; Nicole Richards, Editorial Assistant; Charlene M. Carpentier, Content Project Manager; Jasmin Tokatlian, Media Editor; and Pam Galbreath, Art Director. Special thanks go to Liana Sarkisian and Arwen Petty, our Developmental Editors, and to Mike Ederer, who led us through production at Graphic World.

Reviewers play a very important role in the development of a manuscript. Accordingly, we offer our appreciation to the following colleagues for their assistance with the eighth edition: Patricia Tomich, Kent State University; Robert E. Wickham, University of Houston; Jessica Urschel, Western Michigan University; Wilson Chu, California State University, Long Beach; Melissa Platt, University of Oregon; Brian Detweiler-Bedell, Lewis and Clark College.

PREFACE xv

A N C I L L A R I E S

A C K N O W L E D G M E N T S

http://www.cengagebrain.com

Larry B. Wallnau is Professor Emeritus of Psychology at the State University of New York College at Brockport. At Brockport he taught courses relating to the biological basis of behavior and published numerous re- search articles, primarily in the field of biopsychology. With Dr. Gravetter, he co-authored Statistics for the Behavioral Sciences. He also has provided editorial consulting for a number of publishers and journals. He is an FCC-licensed amateur radio operator, and in his spare time he is seeking worldwide contacts with other radio enthusiasts.

About the Authors

Frederick J Gravetter is Professor Emeritus of Psychology at the State University of New York College at Brockport. While teaching at Brockport, Dr. Gravetter specialized in statistics, experimental design, and cognitive psychology. He received his bachelor’s degree in mathematics from M.I.T. and his Ph.D. in psychology from Duke University. In addition to publishing this textbook and several research articles, Dr. Gravetter co-authored Research Methods for the Behavioral Sciences and Statistics for the Behavioral Sciences.

xvi

P A R T

I Chapter 1 Introduction to Statistics 3

Chapter 2 Frequency Distributions 37

Chapter 3 Measures of Central Tendency 59

Chapter 4 Measures of Variability 89

W e have divided this book into five sections, each cover-

ing a general topic area of statistics. The first section,

consisting of Chapters 1 to 4, provides a broad over-

view of statistical methods and a more focused presentation of

those methods that are classified as descriptive statistics.

By the time you finish the four chapters in this part, you should

have a good understanding of the general goals of statistics and

you should be familiar with the basic terminology and notation

used in statistics. In addition, you should be familiar with the tech-

niques of descriptive statistics that help researchers organize and

summarize the results they obtain from their research. Specifically,

you should be able to take a set of scores and present them in a

table or in a graph that provides an overall picture of the complete

set. Also, you should be able to summarize a set of scores by cal-

culating one or two values (such as the average) that describe the

entire set.

At the end of this section, there is a brief summary and a set of

review problems that should help to integrate the elements from

the separate chapters.

Introduction and Descriptive Statistics

C H A P T E R

1 Introduction to Statistics

1.1 Statistics, Science, and Observations

1.2 Populations and Samples

1.3 Data Structures, Research Methods, and Statistics

1.4 Variables and Measurement

1.5 Statistical Notation

Summary

Focus on Problem Solving

Demonstration 1.1

Problems

Aplia for Essentials of Statistics for the Behavioral Sciences

After reading, go to “Resources” at the end of this chapter for

an introduction on how to use Aplia’s homework and learning

resources.

4 CHAPTER 1 INTRODUCTION TO STATISTICS

STATISTICS, SCIENCE, AND OBSERVATIONS

Before we begin our discussion of statistics, we ask you to read the following paragraph

taken from the philosophy of Wrong Shui (Candappa, 2000).

The Journey to Enlightenment

In Wrong Shui, life is seen as a cosmic journey, a struggle to overcome unseen and

unexpected obstacles at the end of which the traveler will find illumination and

enlightenment. Replicate this quest in your home by moving light switches away from

doors and over to the far side of each room.*

Why did we begin a statistics book with a bit of twisted philosophy? Actually, the

paragraph is an excellent (and humorous) counterexample for the purpose of this book.

Specifically, our goal is to help you avoid stumbling around in the dark by providing

lots of easily available light switches and plenty of illumination as you journey through

the world of statistics. To accomplish this, we try to present sufficient background and

a clear statement of purpose as we introduce each new statistical procedure. Remember

that all statistical procedures were developed to serve a purpose. If you understand why

a new procedure is needed, you will find it much easier to learn.

As you read through the following chapters, keep in mind that the general topic of

statistics follows a well-organized, logically developed progression that leads from

basic concepts and definitions to increasingly sophisticated techniques. Thus, the mate-

rial presented in the early chapters of this book serves as a foundation for the material

that follows. The content of the first nine chapters, for example, provides an essential

background and context for the statistical methods presented in Chapter 10. If you turn

directly to Chapter 10 without reading the first nine chapters, you will find the material

confusing and incomprehensible. However, we should reassure you that the progression

from basic concepts to complex statistical techniques is a slow, step-by-step process.

As you learn the basic background material, you will develop a good frame of refer-

ence for understanding and incorporating new, more sophisticated concepts as they are

presented.

The objectives for this first chapter are to provide an introduction to the topic of

statistics and to give you some background for the rest of the book. We discuss the role

of statistics within the general field of scientific inquiry, and we introduce some of the

vocabulary and notation that are necessary for the statistical methods that follow.

Statistics are often defined as facts and figures, such as average income, crime rate,

birth rate, baseball batting averages, and so on. These statistics are usually informa-

tive and time saving because they condense large quantities of information into a few

simple figures. Later in this chapter we return to the notion of calculating statistics

(facts and figures) but, for now, we concentrate on a much broader definition of sta-

tistics. Specifically, we use the term statistics to refer to a set of mathematical proce-

dures. In this case, we are using the term statistics as a shortened version of statistical

procedures. For example, you are probably using this book for a statistics course in

which you will learn about the statistical techniques that are used for research in the

behavioral sciences.

1.1

OV E R V I E W

D E F I N I T I O N S O F STAT I ST I C S

*Candappa, R. (2000). The little book of wrong shui. Kansas City: Andrews McMeel Publishing. Reprinted

by permission.

SECTION 1.2 / POPULATIONS AND SAMPLES 5

Research in the behavioral sciences (and other fields) involves gathering informa-

tion. To determine, for example, whether college students learn better by reading mate-

rial on printed pages or on a computer screen, you would need to gather information

about students’ study habits and their academic performance. When researchers finish

the task of gathering information, they typically find themselves with pages and pages

of measurements such as IQ scores, personality scores, exam scores, and so on. In this

book, we present the statistics that researchers use to analyze and interpret the informa-

tion that they gather. Specifically, statistics serve two general purposes:

1. Statistics are used to organize and summarize the information so that the re-

searcher can see what happened in the research study and can communicate the

results to others.

2. Statistics help the researcher to answer the questions that initiated the research

by determining exactly what general conclusions are justified based on the

specific results that were obtained.

The term statistics refers to a set of mathematical procedures for organizing,

summarizing, and interpreting information.

Statistical procedures help to ensure that the information or observations are

presented and interpreted in an accurate and informative way. In somewhat gran-

diose terms, statistics help researchers bring order out of chaos. Statistics also

provide researchers with a set of standardized techniques that are recognized and

understood throughout the scientific community. Thus, the statistical methods used

by one researcher are familiar to other researchers, who can accurately interpret the

statistical analyses with a full understanding of how the analysis was done and what

the results signify.

POPULATIONS AND SAMPLES

Research in the behavioral sciences typically begins with a general question about a

specific group (or groups) of individuals. For example, a researcher may want to know

what factors are associated with academic dishonesty among college students. Or a

researcher may want to examine the amount of time spent in the bathroom for men

compared to women. In the first example, the researcher is interested in the group of

college students. In the second example, the researcher wants to compare the group

of men with the group of women. In statistical terminology, the entire group that a

researcher wishes to study is called a population.

A population is the entire set of the individuals of interest for a particular

research question.

As you can well imagine, a population can be quite large—for example, the entire

set of men on the planet Earth. A researcher might be more specific, limiting the popu-

lation for study to retired men who live in the United States. Perhaps the investigator

would like to study the population consisting of men who are professional basketball

players. Populations can obviously vary in size from extremely large to very small,

depending on how the researcher defines the population. The population being studied

should always be identified by the researcher. In addition, the population need not

D E F I N I T I O N

1.2

D E F I N I T I O N

6 CHAPTER 1 INTRODUCTION TO STATISTICS

consist of people—it could be a population of rats, corporations, parts produced in a

factory, or anything else a researcher wants to study. In practice, populations are typi-

cally very large, such as the population of college sophomores in the United States or

the population of small businesses.

Because populations tend to be very large, it usually is impossible for a researcher to

examine every individual in the population of interest. Therefore, researchers typically

select a smaller, more manageable group from the population and limit their studies to

the individuals in the selected group. In statistical terms, a set of individuals selected

from a population is called a sample. A sample is intended to be representative of its

population, and a sample should always be identified in terms of the population from

which it was selected.

A sample is a set of individuals selected from a population, usually intended to

represent the population in a research study.

Just as we saw with populations, samples can vary in size. For example, one study

might examine a sample of only 10 autistic children, and another study might use a

sample of more than 10,000 people who take a specific cholesterol medication.

So far we have talked about a sample being selected from a population. However,

this is actually only half of the full relationship between a sample and its population.

Specifically, when a researcher finishes examining the sample, the goal is to general-

ize the results back to the entire population. Remember that the research started with

a general question about the population. To answer the question, a researcher studies

a sample and then generalizes the results from the sample to the population. The full

relationship between a sample and a population is shown in Figure 1.1.

Typically, researchers are interested in specific characteristics of the individuals in

the population (and in the sample), or they are interested in outside factors that may

influence the individuals. For example, a researcher may be interested in the influence

D E F I N I T I O N

VA R I A B L E S A N D DATA

THE POPULATION All of the individuals of interest

THE SAMPLE The individuals selected to

participate in the research study

The results from the sample are generalized

to the population

The sample is selected from the population

FIGURE 1.1

The relationship between

a population and a sample.

SECTION 1.2 / POPULATIONS AND SAMPLES 7

of the weather on people’s moods. As the weather changes, do people’s moods also

change? Something that can change or have different values is called a variable.

A variable is a characteristic or condition that changes or has different values for

different individuals.

Once again, variables can be characteristics that differ from one individual to an-

other, such as height, weight, gender, or personality. Also, variables can be environmen-

tal conditions that change such as temperature, time of day, or the size of the room in

which the research is being conducted.

To demonstrate changes in variables, it is necessary to make measurements of the

variables being examined. The measurement obtained for each individual is called a

datum or, more commonly, a score or raw score. The complete set of scores is called

the data set, or simply the data.

Data (plural) are measurements or observations. A data set is a collection of

measurements or observations. A datum (singular) is a single measurement or

observation and is commonly called a score or raw score.

Before we move on, we should make one more point about samples, populations, and

data. Earlier, we defined populations and samples in terms of individuals. For example,

we discussed a population of college students and a sample of autistic children. Be fore-

warned, however, that we will also refer to populations or samples of scores. Because

research typically involves measuring each individual to obtain a score, every sample (or

population) of individuals produces a corresponding sample (or population) of scores.

When describing data, it is necessary to distinguish whether the data come from a popula-

tion or a sample. A characteristic that describes a population—for example, the average

score for the population—is called a parameter. A characteristic that describes a sample

is called a statistic. Thus, the average score for a sample is an example of a statistic.

Typically, the research process begins with a question about a population parameter.

However, the actual data come from a sample and are used to compute sample statistics.

A parameter is a value, usually a numerical value, that describes a population. A

parameter is usually derived from measurements of the individuals in the population.

A statistic is a value, usually a numerical value, that describes a sample. A statistic

is usually derived from measurements of the individuals in the sample.

Every population parameter has a corresponding sample statistic, and most research

studies involve using statistics from samples as the basis for answering questions about

population parameters. As a result, much of this book is concerned with the relationship

between sample statistics and the corresponding population parameters. In Chapter 7,

for example, we examine the relationship between the mean obtained for a sample and

the mean for the population from which the sample was obtained.

Although researchers have developed a variety of different statistical procedures to or-

ganize and interpret data, these different procedures can be classified into two general

categories. The first category, descriptive statistics, consists of statistical procedures

that are used to simplify and summarize data.

D E F I N I T I O N

D E F I N I T I O N S

PA R A M E T E R S A N D STAT I ST I C S

D E F I N I T I O N S

D E S C R I P T I V E A N D I N F E R E N T I A L

STAT I ST I CA L M E T H O D S

8 CHAPTER 1 INTRODUCTION TO STATISTICS

BOX

1.1 THE MARGIN OF ERROR BETWEEN STATISTICS AND PARAMETERS

The margin of error is the sampling error. In this

case, the percentages that are reported were obtained

from a sample and are being generalized to the whole

population. As always, you do not expect the statistics

from a sample to be perfect. There is always some

margin of error when sample statistics are used to

represent population parameters.

One common example of sampling error is the error

associated with a sample proportion. For example,

in newspaper articles reporting results from political

polls, you frequently find statements such as this:

Candidate Brown leads the poll with 51% of the

vote. Candidate Jones has 42% approval, and the

remaining 7% are undecided. This poll was taken

from a sample of registered voters and has a margin

of error of plus-or-minus 4 percentage points.

Descriptive statistics are statistical procedures used to summarize, organize, and

simplify data.

Descriptive statistics are techniques that take raw scores and organize or summarize

them in a form that is more manageable. Often the scores are organized in a table or a

graph so that it is possible to see the entire set of scores. Another common technique

is to summarize a set of scores by computing an average. Note that even if the data set

has hundreds of scores, the average provides a single descriptive value for the entire set.

The second general category of statistical techniques is called inferential statistics.

Inferential statistics are methods that use sample data to make general statements about

a population.

Inferential statistics consist of techniques that allow us to study samples and

then make generalizations about the populations from which they were selected.

Because populations are typically very large, it usually is not possible to measure

everyone in the population. Therefore, a sample is selected to represent the population.

By analyzing the results from the sample, we hope to answer general questions about

the population. Typically, researchers use sample statistics as the basis for drawing

conclusions about population parameters.

One problem with using samples, however, is that a sample provides only limited

information about the population. Although samples are generally representative of their

populations, a sample is not expected to give a perfectly accurate picture of the whole

population. Thus, there typically is some discrepancy between a sample statistic and the

corresponding population parameter. This discrepancy is called sampling error, and it

creates the fundamental problem that inferential statistics must always address (Box 1.1).

Sampling error is the naturally occurring discrepancy, or error, that exists

between a sample statistic and the corresponding population parameter.

The concept of sampling error is illustrated in Figure 1.2. The figure shows a

population of 1,000 college students and two samples, each with 5 students, who

have been selected from the population. Notice that each sample contains differ-

ent individuals who have different characteristics. Because the characteristics of

each sample depend on the specific people in the sample, statistics vary from one

D E F I N I T I O N

SECTION 1.2 / POPULATIONS AND SAMPLES 9

sample to another. For example, the five students in sample 1 have an average age of

19.8 years and the students in sample 2 have an average age of 20.4 years.

Also note that the statistics obtained for a sample are not identical to the parameters

for the entire population. In Figure 1.2, for example, neither sample has statistics that

are exactly the same as the population parameters. You should also realize that Figure

1.2 shows only two of the hundreds of possible samples. Each sample would contain

different individuals and would produce different statistics. This is the basic concept

of sampling error: sample statistics vary from one sample to another and typically are

different from the corresponding population parameters.

As a further demonstration of sampling error, imagine that your statistics class is

separated into two groups by drawing a line from front to back through the middle of

the room. Now imagine that you compute the average age (or height, or IQ) for each

group. Will the two groups have exactly the same average? Almost certainly they will

not. No matter what you chose to measure, you will probably find some difference

between the two groups. However, the difference you obtain does not necessarily mean

that there is a systematic difference between the two groups. For example, if the average

age for students on the right-hand side of the room is higher than the average for stu-

dents on the left, it is unlikely that some mysterious force has caused the older people

to gravitate to the right side of the room. Instead, the difference is probably the result of

random factors such as chance. The unpredictable, unsystematic differences that exist

from one sample to another are an example of sampling error.

FIGURE 1.2

A demonstration of sam-

pling error. Two samples

are selected from the

same population. Notice

that the sample statistics

are different from one

sample to another, and all

of the sample statistics

are different from the

corresponding population

parameters. The natural

differences that exist, by

chance, between a sample

statistic and a population

parameter are called

sampling error.

Population of 1000 college students

Population Parameters Average Age � 21.3 years

Average IQ � 112.5 65% Female, 35% Male

Sample #1

Eric Jessica Laura Karen Brian

Sample Statistics Average Age � 19.8 Average IQ � 104.6

60% Female, 40% Male

Sample #2

Tom Kristen Sara

Andrew John

Sample Statistics Average Age � 20.4 Average IQ � 114.2

40% Female, 60% Male

1 0 CHAPTER 1 INTRODUCTION TO STATISTICS

The following example shows the general stages of a research study and demonstrates

how descriptive statistics and inferential statistics are used to organize and interpret the

data. At the end of the example, note how sampling error can affect the interpretation

of experimental results, and consider why inferential statistical methods are needed to

deal with this problem.

Figure 1.3 shows an overview of a general research situation and demonstrates the roles

that descriptive and inferential statistics play. The purpose of the research study is to ad-

dress a question that we posed earlier: Do college students learn better by studying text

on printed pages or on a computer screen? Two samples are selected from the population

of college students. The students in sample A are given printed pages of text to study

for 30 minutes and the students in sample B study the same text on a computer screen.

STAT I ST I C S I N T H E CO N T E X T

O F R E S E A R C H

E X A M P L E 1 . 1

Step 1

Step 2

Step 3

Experiment:

Descriptive statistics:

Inferential statistics:

Compare two studying methods

Test scores for the students in each sample

Organize and simplify

Interpret results

Sample A Read from printed

pages

25 27 30 19 29

26 21 28 23 26

28 27 24 26 22

20 23 25 22 18

22 17 28 19 24

27 23 21 22 19

Sample B Read from computer

screen

Data

Average Score = 26

The sample data show a 4-point difference between the two methods of studying. However, there are two ways to interpret the results. 1. There actually is no difference between the two studying methods, and the sample difference is due to chance (sampling error). 2. There really is a difference between the two methods, and the sample data accurately reflect this difference. The goal of inferential statistics is to help researchers decide between the two interpretations.

Population of College Students

Average Score = 22

20 25 30 20 25 30

FIGURE 1.3

The role of statistics in research.

SECTION 1.2 / POPULATIONS AND SAMPLES 1 1

Next, all of the students are given a multiple-choice test to evaluate their knowledge of

the material. At this point, the researcher has two sets of data: the scores for sample A

and the scores for sample B (see Figure 1.3). Now is the time to begin using statistics.

First, descriptive statistics are used to simplify the pages of data. For example, the

researcher could draw a graph showing the scores for each sample or compute the aver-

age score for each sample. Note that descriptive methods provide a simplified, organized

description of the scores. In this example, the students who studied printed pages had an aver-

age score of 26 on the test, and the students who studied text on the computer averaged 22.

Once the researcher has described the results, the next step is to interpret the

outcome. This is the role of inferential statistics. In this example, the researcher

has found a difference of 4 points between the two samples (sample A averaged 26

and sample B averaged 22). The problem for inferential statistics is to differentiate

between the following two interpretations:

1. There is no real difference between the two study methods, and the 4-point

difference between the samples is just an example of sampling error (like the

samples in Figure 1.2).

2. There really is a difference between the two study methods, and the 4-point differ-

ence between the samples was caused by the different methods of studying.

In simple English, does the 4-point difference between samples provide convincing

evidence of a difference between the two studying methods, or is the 4-point difference

just chance? The purpose of inferential statistics is to answer this question.

1. A researcher is interested in the texting habits of high school students in the

United States. If the researcher measures the number of text messages that each

individual sends each day and calculates the average number for the entire group of

high school students, the average number would be an example of a ___________.

2. A researcher is interested in how watching a reality television show featuring

fashion models influences the eating behavior of 13-year-old girls.

a. A group of 30 13-year-old girls is selected to participate in a research study.

The group of 30 13-year-old girls is an example of a ___________.

b. In the same study, the amount of food eaten in one day is measured for each

girl and the researcher computes the average score for the 30 13-year-old girls.

The average score is an example of a __________.

3. Statistical techniques are classified into two general categories. What are the two cat-

egories called, and what is the general purpose for the techniques in each category?

4. Briefly define the concept of sampling error.

1. parameter

2. a. sample

b. statistic

3. The two categories are descriptive statistics and inferential statistics. Descriptive techniques

are intended to organize, simplify, and summarize data. Inferential techniques use sample

data to reach general conclusions about populations.

4. Sampling error is the error, or discrepancy, between the value obtained for a sample statistic

and the value for the corresponding population parameter.

L E A R N I N G C H E C K

ANSWERS

1 2 CHAPTER 1 INTRODUCTION TO STATISTICS

DATA STRUCTURES, RESEARCH METHODS, AND STATISTICS

Some research studies are conducted simply to describe individual variables as they

exist naturally. For example, a college official may conduct a survey to describe the

eating, sleeping, and studying habits of a group of college students. When the results

consist of numerical scores, such as the number of hours spent studying each day, they

are typically described by the statistical techniques that are presented in Chapters 3

and 4. Non-numerical scores are typically described by computing the proportion or

percentage in each category. For example, a recent newspaper article reported that 61%

of the adults in the United States drink alcohol.

Most research, however, is intended to examine relationships between two or more

variables. For example, is there a relationship between the amount of violence that

children see on television and the amount of aggressive behavior they display? Is there a

relationship between the quality of breakfast and level of academic performance for el-

ementary school children? Is there a relationship between the number of hours of sleep

and grade point average for college students? To establish the existence of a relation-

ship, researchers must make observations—that is, measurements of the two variables.

The resulting measurements can be classified into two distinct data structures that also

help to classify different research methods and different statistical techniques. In the

following section we identify and discuss these two data structures.

Data structure I. Measuring two variables for each individual: The correlational

method One method for examining the relationship between variables is to observe

the two variables as they exist naturally for a set of individuals. That is, simply mea-

sure the two variables for each individual. For example, research has demonstrated a

relationship between sleep habits, especially wake-up time, and academic performance

for college students (Trockel, Barnes, and Egget, 2000). The researchers used a survey

to measure wake-up time and school records to measure academic performance for

each student. Figure 1.4 shows an example of the kind of data obtained in the study.

The researchers then look for consistent patterns in the data to provide evidence for a

relationship between variables. For example, as wake-up time changes from one student

to another, is there also a tendency for academic performance to change?

Patterns in the data are often easier to see if the scores are presented in a graph.

Figure 1.4 also shows the scores for the eight students in a graph called a scatter plot. In

the scatter plot, each individual is represented by a point so that the horizontal position

corresponds to the student’s wake-up time and the vertical position corresponds to the

student’s academic performance score. The scatter plot shows a clear relationship be-

tween wake-up time and academic performance: as wake-up time increases, academic

performance decreases.

A research study that simply measures two different variables for each individual

and produces the kind of data shown in Figure 1.4 is an example of the correlational

method, or the correlational research strategy.

In the correlational method, two different variables are observed to determine

whether there is a relationship between them.

1.3

I N D I V I D UA L VA R I A B L E S

R E L AT I O N S H I P S B E T W E E N VA R I A B L E S

D E F I N I T I O N

SECTION 1.3 / DATA STRUCTURES, RESEARCH METHODS, AND STATISTICS 1 3

Limitations of the correlational method The results from a correlational study

can demonstrate the existence of a relationship between two variables, but they do not

provide an explanation for the relationship. In particular, a correlational study cannot

demonstrate a cause-and-effect relationship. For example, the data in Figure 1.4 show

a systematic relationship between wake-up time and academic performance for a group

of college students; those who sleep late tend to have lower performance scores than

those who wake early. However, there are many possible explanations for the relation-

ship and we do not know exactly what factor (or factors) is responsible for late sleepers

having lower grades. In particular, we cannot conclude that waking students up earlier

would cause their academic performance to improve, or that studying more would cause

students to wake up earlier. To demonstrate a cause-and-effect relationship between

two variables, researchers must use the experimental method, which is discussed next.

Data structure II. Comparing two (or more) groups of scores: Experimental and

nonexperimental methods The second method for examining the relationship between

two variables involves the comparison of two or more groups of scores. In this situation,

the relationship between variables is examined by using one of the variables to define

the groups, and then measuring the second variable to obtain scores for each group. For

example, one group of elementary school children is shown a 30-minute action/adven-

ture television program involving numerous instances of violence, and a second group is

shown a 30-minute comedy that includes no violence. Both groups are then observed on

the playground and a researcher records the number of aggressive acts committed by each

child. An example of the resulting data is shown in Figure 1.5. The researcher compares

the scores for the violence group with the scores for the no-violence group. A systematic

difference between the two groups provides evidence for a relationship between viewing

television violence and aggressive behavior for elementary school children.

One specific research method that involves comparing groups of scores is known as

the experimental method or the experimental research strategy. The goal of an experi-

mental study is to demonstrate a cause-and-effect relationship between two variables.

T H E E X P E R I M E N TA L M E T H O D

3.8

3.6

3.4

3.2

3.0

2.8

2.6

2.4

2.2

2.0

7 8 9

Wake-up time

A c

a d

e m

ic p

e rf

o rm

a n

c e

10 11 12FIGURE 1.4

One of two data structures for studies evaluating the

relationship between variables. Note that there are

two separate measurements for each individual

(wake-up time and academic performance). The

same scores are shown in a table (a) and a graph (b).

A B C D E F G H

11 9 9

12 7

10 10

Student Wake-up

Time

2.4 3.6 3.2 2.2 3.8 2.2 3.0 3.0

Academic Performance

(a) (b)

1 4 CHAPTER 1 INTRODUCTION TO STATISTICS

Specifically, an experiment attempts to show that changing the value of one variable

causes changes to occur in the second variable. To accomplish this goal, the experi-

mental method has two characteristics that differentiate experiments from other types

of research studies:

1. Manipulation The researcher manipulates one variable by changing its value

from one level to another. A second variable is observed (measured) to deter-

mine whether the manipulation causes changes to occur.

2. Control The researcher must exercise control over the research situation to

ensure that other, extraneous variables do not influence the relationship being

examined.

To demonstrate these two characteristics, consider an experiment in which researchers

demonstrate the pain-killing effects of handling money (Zhou & Vohs, 2009). In the

experiment, a group of college students was told that they were participating in a

manual dexterity study. The researchers then created two treatment conditions by

manipulating the kind of material that each participant would be handling. Half of

the students were given a stack of money to count and the other half got a stack of

blank pieces of paper. After the counting task, the participants were asked to dip their

hands into bowls of painfully hot water (122° F) and rate how uncomfortable it was.

Participants who had counted money rated the pain significantly lower than those who

had counted paper. The structure of the experiment is shown in Figure 1.6.

To be able to say that the difference in pain perception is caused by the money, the

researcher must rule out any other possible explanation for the difference. That is, the

researchers must control any other variables that might affect pain tolerance. There are

two general categories of variables that researchers must consider:

1. Participant Variables These are characteristics such as age, gender, and

intelligence that vary from one individual to another.

In the money-counting experiment, for example, suppose that the participants

in the money condition were primarily females and those in the paper condition

were primarily males. In this case, there is an alternative explanation for any

difference in the pain ratings that exists between the two groups. Specifically, it

is possible that the difference was caused by the money, but it also is possible

that the difference was caused by the participants’ gender (females can tolerate

One variable (violence/no violence) is used to define groups

A second variable (aggressive behavior) is measured to obtain scores within each group

4 2 0 1 3 2 4 1 3

0 2 1 3 0 0 1 1 1

Violence No

Violence

Compare groups of scores

FIGURE 1.5

The second data structure

for studies evaluating

the relationship between

variables. Note that one

variable is used to define

the groups and the second

variable is measured to

obtain scores within each

group.

In more complex experi-

ments, a researcher may

systematically manipulate

more than one variable

and may observe more than

one variable. Here we are

considering the simplest

case, in which only one

variable is manipulated and

only one variable is observed.

SECTION 1.3 / DATA STRUCTURES, RESEARCH METHODS, AND STATISTICS 1 5

more pain than males can). Whenever a research study allows more than one

explanation for the results, the study is said to be confounded because it is im-

possible to reach an unambiguous conclusion.

2. Environmental Variables These are characteristics of the environment such

as lighting, time of day, and weather conditions. Using the money-counting

experiment (see Figure 1.6) as an example, suppose that the individuals in the

money condition were all tested in the morning and the individuals in the paper

condition were all tested in the evening. Again, this would produce a confounded

experiment because the researcher could not determine whether the differences

in the pain ratings were caused by the money or caused by the time of day.

Researchers typically use three basic techniques to control other variables. First,

the researcher could use random assignment, which means that each participant has an

equal chance of being assigned to each of the treatment conditions. The goal of random

assignment is to distribute the participant characteristics evenly between the two groups

so that neither group is noticeably smarter (or older, or faster) than the other. Random

assignment can also be used to control environmental variables. For example, partici-

pants could be assigned randomly for testing either in the morning or in the afternoon.

Second, the researcher can use matching to ensure equivalent groups or equivalent envi-

ronments. For example, the researcher could match groups by ensuring that every group

has exactly 60% females and 40% males. Finally, the researcher can control variables

by holding them constant. For example, if an experiment uses only 10-year-old children

as participants (holding age constant), then the researcher can be certain that one group

is not noticeably older than another.

In the experimental method, one variable is manipulated while another variable

is observed and measured. To establish a cause-and-effect relationship between

the two variables, an experiment attempts to control all other variables to prevent

them from influencing the results.

Terminology in the experimental method Specific names are used for the two

variables that are studied by the experimental method. The variable that is manipulated

D E F I N I T I O N

Variable #1: Counting money or blank paper (the independent variable) Manipulated to create two treatment conditions.

Variable #2: Pain rating (the dependent variable) Measured in each of the treatment conditions.

7 4 5 6 6 8 6 5 5 6

8 10 8 9 8

10 7 8 8 7

Money Paper

Compare groups of scores

FIGURE 1.6

The structure of an

experiment. Participants

are randomly assigned

to one of two treatment

conditions: counting

money or counting blank

pieces of paper. Later,

each participant is tested

by placing one hand in

a bowl of hot (122º F)

water and rating the level

of pain. A difference be-

tween the ratings for the

two groups is attributed

to the treatment (paper

versus money).

1 6 CHAPTER 1 INTRODUCTION TO STATISTICS

by the experimenter is called the independent variable. It can be identified as the

treatment conditions to which participants are assigned. For the example in Figure 1.6,

money versus blank paper is the independent variable. The variable that is observed and

measured to obtain scores within each condition is the dependent variable. For the

example in Figure 1.6, the level of pain is the dependent variable.

The independent variable is the variable that is manipulated by the researcher.

In behavioral research, the independent variable usually consists of the two (or

more) treatment conditions to which subjects are exposed. The independent vari-

able consists of the antecedent conditions that are manipulated prior to observing

the dependent variable.

The dependent variable is the variable that is observed to assess the effect of the

treatment.

An experimental study evaluates the relationship between two variables by manipu-

lating one variable (the independent variable) and measuring one variable (the depen-

dent variable). Note that in an experiment only one variable is actually measured. You

should realize that this is different from a correlational study, in which both variables

are measured and the data consist of two separate scores for each individual.

Control Conditions in an Experiment Often an experiment will include a condition

in which the participants do not receive any treatment. The scores from these individu-

als are then compared with scores from participants who do receive the treatment. The

goal of this type of study is to demonstrate that the treatment has an effect by showing

that the scores in the treatment condition are substantially different from the scores in

the no-treatment condition. In this kind of research, the no-treatment condition is called

the control condition, and the treatment condition is called the experimental condition.

Individuals in a control condition do not receive the experimental treatment.

Instead, they either receive no treatment or they receive a neutral, placebo treat-

ment. The purpose of a control condition is to provide a baseline for comparison

with the experimental condition. The individuals in the control condition are

often called the control group.

Individuals in the experimental condition do receive the experimental treatment

and are often called the experimental group.

Note that the independent variable always consists of at least two values. (Something

must have at least two different values before you can say that it is “variable.”) For the

money-counting experiment (see Figure 1.6), the independent variable is money versus

plain paper. For an experiment with an experimental group and a control group, the

independent variable is treatment versus no treatment.

In informal conversation, there is a tendency for people to use the term experiment to

refer to any kind of research study. You should realize, however, that the term only ap-

plies to studies that satisfy the specific requirements outlined earlier. In particular, a real

experiment must include manipulation of an independent variable and rigorous control

of other, extraneous variables. As a result, there are a number of other research designs

that compare groups of scores but are not true experiments. Two examples are shown

in Figure 1.7 and are discussed in the following paragraphs. This type of research study

is classified as nonexperimental.

D E F I N I T I O N S

N O N E X P E R I M E N TA L M E T H O D S :

N O N E Q U I VA L E N T G R O U P S A N D

P R E – P O ST ST U D I E S

SECTION 1.3 / DATA STRUCTURES, RESEARCH METHODS, AND STATISTICS 1 7

The top part of Figure 1.7 shows an example of a nonequivalent groups study com-

paring boys and girls. Notice that this study involves comparing two groups of scores

(like an experiment). However, the researcher has no ability to control the assignment

of participants to groups—the males automatically go in the boy group and the females

go in the girl group. Because this type of research compares preexisting groups, the

researcher cannot control the assignment of participants to groups and cannot ensure

equivalent groups. Other examples of nonequivalent group studies include comparing

8-year-old children and 10-year-old children or comparing people with an eating disor-

der and those with no disorder. Because it is impossible to use techniques like random

assignment to control participant variables and ensure equivalent groups, this type of

research is not a true experiment.

The bottom part of Figure 1.7 shows an example of a pre–post study comparing

depression scores before therapy and after therapy. The two groups of scores are

obtained by measuring the same variable (depression) twice for each participant; once

before therapy and again after therapy. In a pre–post study, however, the researcher has

no control over the passage of time. The “before” scores are always measured earlier

than the “after” scores. Although a difference between the two groups of scores may be

caused by the treatment, it is always possible that the scores simply change as time goes

by. For example, the depression scores may decrease over time in the same way that the

symptoms of a cold disappear over time. In a pre–post study, the researcher also has

Variable #1: Subject gender (the quasi-independent variable) Not manipulated, but used to create two groups of subjects

Variable #2: Verbal test scores (the dependent variable) Measured in each of the two groups

17 19 16 12 17 18 15 16

12 10 14 15 13 12 11 13

Boys Girls

Any difference?

Variable #1: Time (the quasi-independent variable) Not manipulated, but used to create two groups of scores

Variable #2: Depression scores (the dependent variable) Measured at each of the two different times

17 19 16 12 17 18 15 16

12 10 14 15 13 12 11 13

Before Therapy

After Therapy

Any difference?

(a)

(b)

FIGURE 1.7

Two examples of nonex-

perimental studies that

involve comparing

two groups of scores. In

(a), a participant variable

(gender) is used to

create groups, and then

the dependent variable

(verbal score) is measured

in each group. In

(b), time is the variable

used to define the two

groups, and the dependent

variable (depression) is

measured at each of the

two times.

Correlational studies are

also examples of nonex-

perimental research. In this

section, however, we are

discussing nonexperimental

studies that compare two or

more groups of scores.

1 8 CHAPTER 1 INTRODUCTION TO STATISTICS

no control over other variables that change with time. For example, the weather could

change from dark and gloomy before therapy to bright and sunny after therapy. In this

case, the depression scores could improve because of the weather and not because of

the therapy. Because the researcher cannot control the passage of time or other variables

related to time, this study is not a true experiment.

Terminology in nonexperimental research Although the two research studies

shown in Figure 1.7 are not true experiments, you should notice that they produce the

same kind of data that are found in an experiment (see Figure 1.6). In each case, one

variable is used to create groups, and a second variable is measured to obtain scores

within each group. In an experiment, the groups are created by manipulation of the

independent variable, and the participants’ scores are the dependent variable. The same

terminology is often used to identify the two variables in nonexperimental studies. That

is, the variable that is used to create groups is the independent variable and the scores

are the dependent variable. For example, the top part of Figure 1.7, gender (boy/girl), is

the independent variable and the verbal test scores are the dependent variable. However,

you should realize that gender (boy/girl) is not a true independent variable because it is

not manipulated. For this reason, the “independent variable” in a nonexperimental study

is often called a quasi-independent variable.

In a nonexperimental study, the “independent” variable that is used to create the

different groups of scores is often called the quasi-independent variable.

The two general data structures that we used to classify research methods can also be

used to classify statistical methods.

I. One group with two variables measured for each individual Recall that the data

from a correlational study consist of two scores, representing two different variables,

for each individual. The scores can be listed in a table or displayed in a scatter plot as

in Figure 1.5. The relationship between the two variables is usually measured and de-

scribed using a statistic called a correlation. Correlations and the correlational method

are discussed in detail in Chapter 14.

Occasionally, the measurement process used for a correlational study simply clas-

sifies individuals into categories that do not correspond to numerical values. For

example, Greitemeyer and Osswald (2010) examine the effect of prosocial video

games on prosocial behavior. One group of participants played a prosocial game and

a second group played a neutral game. After the game was finished, the experimenter

accidentally knocked a cup of pencils onto the floor and recorded whether the par-

ticipants helped to pick them up. Note that the researcher has two scores for each

individual (type of game and helping behavior) but neither of the scores is a numerical

value. This type of data is typically summarized in a table showing how many indi-

viduals are classified into each of the possible categories. Table 1.1 is an example

of this kind of summary table showing results similar to those obtained in the study.

The table shows, for example, that 12 of the 18 participants playing the prosocial

game helped to pick up pencils. This type of data can be coded with numbers (for

example, neutral � 0 and prosocial � 1) so that it is possible to compute a correlation.

However, the relationship between variables for non-numerical data, such as the data

in Table 1.1, is usually evaluated using a statistical technique known as a chi-square

test. Chi-square tests are presented in Chapter 15.

D E F I N I T I O N

DATA ST R UC T U R E S A N D STAT I ST I CA L

M E T H O D S

SECTION 1.3 / DATA STRUCTURES, RESEARCH METHODS, AND STATISTICS 1 9

II. Comparing two or more groups of scores Most of the statistical procedures

presented in this book are designed for research studies that compare groups of scores,

like the experimental study in Figure 1.6 and the nonexperimental studies in Figure 1.7.

Specifically, we examine descriptive statistics that summarize and describe the scores

in each group, and we examine inferential statistics that allow us to use the groups, or

samples, to generalize to the entire population.

When the measurement procedure produces numerical scores, the statistical evalu-

ation typically involves computing the average score for each group and then compar-

ing the averages. The process of computing averages is presented in Chapter 3, and a

variety of statistical techniques for comparing averages are presented in Chapters 8–13.

If the measurement process simply classifies individuals into non-numerical categories,

the statistical evaluation usually consists of computing proportions for each group and

then comparing proportions. Previously, in Table 1.1, we presented an example of non-

numerical data examining the relationship between type of video game and helping

behavior. The same data can be used to compare the proportions for prosocial game

players with the proportions for neutral game players. For example, 67% of those who

played the prosocial game helped the researcher compared to 33% of those who played

the neutral game. As mentioned before, these data are evaluated using a chi-square test,

which is presented in Chapter 15.

Type of Video Game

Prosocial Neutral

Helped 12 6

Did not Help 6 12

TABLE 1.1

Correlational data consisting of

non-numerical scores. Note that

there are two measurements for

each individual: type of game

played and helping behavior.

The numbers indicate how many

people are in each category. For

example, out of the 18 partici-

pants who played a prosocial

game, 12 helped the researcher.

1. A research study comparing alcohol use for college students in the United States

and Canada reports that more Canadian students drink but American students drink

more (Kuo, Adlaf, Lee, Gliksman, Demers, and Wechsler, 2002). Is this study an

example of an experiment? Explain why or why not.

2. What two elements are necessary for a research study to be an experiment?

3. Stephens, Atkins, and Kingston (2009) conducted an experiment in which

participants were able to tolerate more pain when they shouted their favorite

swear words over and over than when they shouted neutral words. Identify the

independent and dependent variables for this study.

1. This study is nonexperimental. The researcher is simply observing, not manipulating, two

nonequivalent groups of participants.

2. First, the researcher must manipulate one of the two variables being studied. Second, all

other variables that might influence the results must be controlled.

3. The independent variable is the type of word being shouted and the dependent variable is

the amount of pain tolerated by each participant.

L E A R N I N G C H E C K

ANSWERS

2 0 CHAPTER 1 INTRODUCTION TO STATISTICS

VARIABLES AND MEASUREMENT

The scores that are obtained in a research study are the result of observing and mea-

suring variables. For example, a researcher may finish a study with a set of IQ scores,

personality scores, or reaction-time scores. In this section, we take a closer look at the

variables that are being measured and the process of measurement.

Some variables, such as height, weight, and eye color are well-defined, concrete entities

that can be observed and measured directly. On the other hand, many variables studied

by behavioral scientists are internal characteristics that cannot be observed or measured

directly. However, we all assume that these variables exist and we use them to help

describe and explain behavior. For example, we say that a student does well in school

because he or she is intelligent. Or we say that someone is anxious in social situations,

or that someone seems to be hungry. Variables like intelligence, anxiety, and hunger are

called constructs, and because they are intangible and cannot be directly observed, they

are often called hypothetical constructs.

Although constructs such as intelligence are internal characteristics that cannot be

directly observed, it is possible to observe and measure behaviors that are representative

of the construct. For example, we cannot “see” intelligence but we can see examples of

intelligent behavior. The external behaviors can then be used to create an operational

definition for the construct. An operational definition measures and defines a construct

in terms of external behaviors. For example, we can measure performance on an IQ test

and then use the test scores as a definition of intelligence. Or hunger can be measured

and defined by the number of hours since last eating.

Constructs, also known as hypothetical constructs, are internal attributes or

characteristics that cannot be directly observed but are useful for describing and

explaining behavior.

An operational definition identifies a measurement procedure (a set of operations)

for measuring an external behavior and uses the resulting measurements as a

definition and a measurement of an internal construct. Note that an operational

definition has two components: First, it describes a set of operations for measuring

a construct. Second, it defines the construct in terms of the resulting measurements.

The variables in a study can be characterized by the type of values that can be assigned

to them. A discrete variable consists of separate, indivisible categories. For this type

of variable, there are no intermediate values between two adjacent categories. Consider

the values displayed when dice are rolled. Between neighboring values—for example,

five dots and six dots—no other values can ever be observed.

A discrete variable consists of separate, indivisible categories. No values can

exist between two neighboring categories.

Discrete variables are commonly restricted to whole, countable numbers—for

example, the number of children in a family or the number of students attending class.

If you observe class attendance from day to day, you may count 18 students one day

and 19 students the next day. However, it is impossible ever to observe a value between

18 and 19. A discrete variable may also consist of observations that differ qualitatively.

For example, people can be classified by gender (male or female), by occupation (nurse,

1.4

CO N ST R UC T S A N D O P E R AT I O N A L

D E F I N I T I O N S

D I S C R E T E A N D CO N T I N U O U S

VA R I A B L E S

D E F I N I T I O N

SECTION 1.4 / VARIABLES AND MEASUREMENT 2 1

teacher, lawyer, and so on), and college students can be classified by academic major

(art, biology, chemistry, and so on). In each case, the variable is discrete because it

consists of separate, indivisible categories.

On the other hand, many variables are not discrete. Variables such as time, height, and

weight are not limited to a fixed set of separate, indivisible categories. You can measure

time, for example, in hours, minutes, seconds, or fractions of seconds. These variables are

called continuous because they can be divided into an infinite number of fractional parts.

For a continuous variable, there are an infinite number of possible values that

fall between any two observed values. A continuous variable is divisible into an

infinite number of fractional parts.

Note that the terms continuous and discrete apply to the variables that are being

measured and not to the scores that are obtained from the measurement. For example,

people’s heights can be measured by simply classifying individuals into three broad

categories: tall, average, and short. Note that there is no measurement category be-

tween tall and average. Thus, it may appear that we are measuring a discrete variable.

However, the underlying variable, height, is continuous. In this example, we chose to

limit the measurement scale to three categories. We could have decided to measure

height to the nearest inch, or the nearest half inch, and so on. The key to determin-

ing whether a variable is continuous or discrete is that a continuous variable can be

divided into any number of fractional parts. Height can be measured to the nearest

inch, the nearest 0.1 inch, or the nearest 0.01 inch. Similarly, a professor evaluating

students’ knowledge could use a pass/fail system that classifies students into two broad

categories. However, the professor could choose to use a 10-point quiz that divides

student knowledge into 11 categories corresponding to quiz scores from 0 to 10. Or the

professor could use a 100-point exam that potentially divides student knowledge into

101 categories from 0 to 100. Whenever you are free to choose the degree of precision

or the number of categories for measuring a variable, the variable must be continuous.

Measuring a continuous variable Any continuous variable, for example, weight,

can be pictured as a continuous line (Figure 1.8). Note that there are an infinite number

of possible points on the line without any gaps or separations between neighboring

points. For any two different points on the line, it is always possible to find a third value

that is between the two points.

D E F I N I T I O N

149

149.5

150

149.6 150.3

150.5

151 152

148.5

149

149.5

150

150.5

Real limits

151

151.5

152

152.5

FIGURE 1.8

When measuring weight

to the nearest whole

pound, 149.6 and 150.3

are assigned the value

of 150 (top). Any value

in the interval between

149.5 and 150.5 is given

the value of 150.

2 2 CHAPTER 1 INTRODUCTION TO STATISTICS

Two other factors apply to continuous variables:

1. When measuring a continuous variable, it should be very rare to obtain identical

measurements for two different individuals. Because a continuous variable

has an infinite number of possible values, it should be almost impossible for

two people to have exactly the same score. If the data show a substantial

number of tied scores, then you should suspect that the measurement procedure

is relatively crude or that the variable is not really continuous.

2. When measuring a continuous variable, each measurement category is actually

an interval that must be defined by boundaries. For example, two people who

both claim to weigh 150 pounds are probably not exactly the same weight.

However, they are both around 150 pounds. One person may actually weigh

149.6 and the other 150.3. Thus, a score of 150 is not a specific point on the

scale but instead is an interval (see Figure 1.8). To differentiate a score of

150 from a score of 149 or 151, we must set up boundaries on the scale of

measurement. These boundaries are called real limits and are positioned exactly

halfway between adjacent scores. Thus, a score of X � 150 pounds is actually

an interval bounded by a lower real limit of 149.5 at the bottom and an upper

real limit of 150.5 at the top. Any individual whose weight falls between these

real limits is assigned a score of X � 150.

Real limits are the boundaries of intervals for scores that are represented on a

continuous number line. The real limit separating two adjacent scores is located

exactly halfway between the scores. Each score has two real limits. The upper

real limit is at the top of the interval, and the lower real limit is at the bottom.

The concept of real limits applies to any measurement of a continuous variable, even

when the score categories are not whole numbers. For example, if you were measur-

ing time to the nearest tenth of a second, the measurement categories would be 31.0,

31.1, 31.2, and so on. Each of these categories represents an interval on the scale that

is bounded by real limits. For example, a score of X � 31.1 seconds indicates that the

actual measurement is in an interval bounded by a lower real limit of 31.05 and an

upper real limit of 31.15. Remember that the real limits are always halfway between

adjacent categories.

Later in this book, real limits are used for constructing graphs and for various calcu-

lations with continuous scales. For now, however, you should realize that real limits are

a necessity whenever you make measurements of a continuous variable.

It should be obvious by now that data collection requires that we make measurements of

our observations. Measurement involves assigning individuals or events to categories.

The categories can simply be names such as male/female or employed/unemployed,

or they can be numerical values such as 68 inches or 175 pounds. The set of catego-

ries makes up a scale of measurement, and the relationships between the categories

determine different types of scales. The distinctions among the scales are important

because they identify the limitations of certain types of measurements and because

certain statistical procedures are appropriate for scores that have been measured on

some scales but not on others. If you were interested in people’s heights, for example,

you could measure a group of individuals by simply classifying them into three cat-

egories: tall, medium, and short. However, this simple classification would not tell you

much about the actual heights of the individuals, and these measurements would not

give you enough information to calculate an average height for the group. Although

D E F I N I T I O N S

S CA L E S O F M E AS U R E M E N T

SECTION 1.4 / VARIABLES AND MEASUREMENT 2 3

the simple classification would be adequate for some purposes, you would need more

sophisticated measurements before you could answer more detailed questions. In this

section, we examine four different scales of measurement, beginning with the simplest

and moving to the most sophisticated.

The word nominal means “having to do with names.” Measurement on a nominal scale

involves classifying individuals into categories that have different names but are not

related to each other in any systematic way. For example, if you were measuring the

academic majors for a group of college students, the categories would be art, biology,

business, chemistry, and so on. Each student would be classified in one category accord-

ing to his or her major. The measurements from a nominal scale allow us to determine

whether two individuals are different, but they do not identify either the direction or the

size of the difference. If one student is an art major and another is a biology major, we can

say that they are different, but we cannot say that art is “more than” or “less than” biology

and we cannot specify how much difference there is between art and biology. Other

examples of nominal scales include classifying people by race, gender, or occupation.

A nominal scale consists of a set of categories that have different names.

Measurements on a nominal scale label and categorize observations, but do not

make any quantitative distinctions between observations.

Although the categories on a nominal scale are not quantitative values, they are oc-

casionally represented by numbers. For example, the rooms or offices in a building may

be identified by numbers. You should realize that the room numbers are simply names

and do not reflect any quantitative information. Room 109 is not necessarily bigger than

Room 100 and certainly not 9 points bigger. It also is fairly common to use numerical

values as a code for nominal categories when data are entered into computer programs.

For example, the data from a survey may code males with a 0 and females with a 1.

Again, the numerical values are simply names and do not represent any quantitative dif-

ference. The scales that follow do reflect an attempt to make quantitative distinctions.

The categories that make up an ordinal scale not only have different names (as in a

nominal scale) but also are organized in a fixed order corresponding to differences of

magnitude.

An ordinal scale consists of a set of categories that are organized in an ordered

sequence. Measurements on an ordinal scale rank observations in terms of size

or magnitude.

Often, an ordinal scale consists of a series of ranks (first, second, third, and so on)

like the order of finish in a horse race. Occasionally, the categories are identified by

verbal labels like small, medium, and large drink sizes at a fast-food restaurant. In

either case, the fact that the categories form an ordered sequence means that there is a

directional relationship between categories. With measurements from an ordinal scale,

you can determine whether two individuals are different and you can determine the

direction of difference. However, ordinal measurements do not allow you to determine

the size of the difference between two individuals. In a NASCAR race, for example,

the first-place car finished faster than the second-place car, but the ranks don’t tell you

how much faster. Other examples of ordinal scales include socioeconomic class (upper,

middle, lower) and T-shirt sizes (small, medium, large). In addition, ordinal scales are

T H E N O M I N A L S CA L E

D E F I N I T I O N

T H E O R D I N A L S CA L E

D E F I N I T I O N

2 4 CHAPTER 1 INTRODUCTION TO STATISTICS

often used to measure variables for which it is difficult to assign numerical scores. For

example, people can rank their food preferences but might have trouble explaining

“how much” they prefer chocolate ice cream to steak.

Both an interval scale and a ratio scale consist of a series of ordered categories (like an

ordinal scale) with the additional requirement that the categories form a series of intervals

that are all exactly the same size. Thus, the scale of measurement consists of a series

of equal intervals, such as inches on a ruler. Other examples of interval and ratio scales

are the measurement of time in seconds, weight in pounds, and temperature in degrees

Fahrenheit. Note that, in each case, one interval (1 inch, 1 second, 1 pound, 1 degree) is the

same size, no matter where it is located on the scale. The fact that the intervals are all the

same size makes it possible to determine both the size and the direction of the difference

between two measurements. For example, you know that a measurement of 80° Fahrenheit

is higher than a measure of 60°, and you know that it is exactly 20° higher.

The factor that differentiates an interval scale from a ratio scale is the nature of the

zero point. An interval scale has an arbitrary zero point. That is, the value 0 is assigned

to a particular location on the scale simply as a matter of convenience or reference. In

particular, a value of zero does not indicate a total absence of the variable being mea-

sured. For example, a temperature of 0 degrees Fahrenheit does not mean that there is

no temperature, and it does not prohibit the temperature from going even lower. Interval

scales with an arbitrary zero point are relatively rare. The two most common examples

are the Fahrenheit and Celsius temperature scales. Other examples include golf scores

(above and below par) and relative measures such as above and below average rainfall.

A ratio scale is anchored by a zero point that is not arbitrary but rather is a meaning-

ful value representing none (a complete absence) of the variable being measured. The

existence of an absolute, nonarbitrary zero point means that we can measure the abso-

lute amount of the variable; that is, we can measure the distance from 0. This makes

it possible to compare measurements in terms of ratios. For example, a gas tank with

10 gallons (10 more than 0) has twice as much gas as a tank with only 5 gallons

(5 more than 0). Also note that a completely empty tank has 0 gallons. With a ratio

scale, we can measure the direction and the size of the difference between two

measurements and we can describe the difference in terms of a ratio. Ratio scales

are quite common and include physical measures such as height and weight, as well

as variables such as reaction time or the number of errors on a test. The distinction

between an interval scale and a ratio scale is demonstrated in Example 1.2.

An interval scale consists of ordered categories that are all intervals of exactly

the same size. Equal differences between numbers on the scale reflect equal

differences in magnitude. However, the zero point on an interval scale is arbitrary

and does not indicate a zero amount of the variable being measured.

A ratio scale is an interval scale with the additional feature of an absolute zero

point. With a ratio scale, ratios of numbers do reflect ratios of magnitude.

A researcher obtains measurements of height for a group of 8-year-old boys. Initially,

the researcher simply records each child’s height in inches, obtaining values such as

44, 51, 49, and so on. These initial measurements constitute a ratio scale. A value of

zero represents no height (absolute zero). Also, it is possible to use these measurements

to form ratios. For example, a child who is 60 inches tall is one-and-a-half times taller

than a child who is 40 inches tall.

T H E I N T E R VA L A N D R AT I O S CA L E S

D E F I N I T I O N S

E X A M P L E 1 . 2

SECTION 1.4 / VARIABLES AND MEASUREMENT 2 5

Now suppose that the researcher converts the initial measurement into a new scale by

calculating the difference between each child’s actual height and the average height for

this age group. A child who is 1 inch taller than average now gets a score of 11; a child who is 4 inches taller than average gets a score of 14. Similarly, a child who is 2 inches shorter than average gets a score of –2. On this scale, a score of zero is a convenient

reference point corresponding to the average height. Because zero no longer indicates a

complete absence of height, the new scores constitute an interval scale of measurement.

Notice that original scores and the converted scores both involve measurement in

inches, and you can compute differences, or distances, on either scale. For example,

there is a 6-inch difference in height between two boys who measure 57 and 51 inches

tall on the first scale. Likewise, there is a 6-inch difference between two boys who

measure 19 and 13 on the second scale. However, you should also notice that ratio comparisons are not possible on the second scale. For example, a boy who measures

19 is not three times taller than a boy who measures 13.

For our purposes, scales of measurement are important because they influence the kind

of statistics that can and cannot be used. For example, if you measure IQ scores for a

group of students, it is possible to add the scores together and calculate a mean score for

the group. On the other hand, if you measure the academic major for each student, you

STAT I ST I C S A N D S CA L E S

O F M E AS U R E M E N T

1. A tax form asks people to identify their annual income, number of dependents, and

social security number. For each of these three variables, identify the scale of measure-

ment that probably is used and identify whether the variable is continuous or discrete.

2. An English professor uses letter grades (A, B, C, D, and F) to evaluate a set of student

essays. What kind of scale is being used to measure the quality of the essays?

3. The teacher in a communications class asks students to identify their favorite

reality television show. The different television shows make up a ______ scale

of measurement.

4. A researcher studies the factors that determine the number of children that couples

decide to have. The variable, number of children, is a ______________ (discrete/

continuous) variable.

5. a. When measuring height to the nearest inch, what are the real limits for a score

of 68 inches?

b. When measuring height to the nearest half inch, what are the real limits for a

score of 68 inches?

1. Annual income and number of dependents are measured on ratio scales, and income is a

continuous variable. Social security number is measured on a nominal scale and is a discrete

variable. The number of dependents is also discrete.

2. ordinal

3. nominal

4. discrete

5. a. 67.5 and 68.5

b. 67.75 and 68.25

L E A R N I N G C H E C K

ANSWERS

2 6 CHAPTER 1 INTRODUCTION TO STATISTICS

cannot compute the mean. (What is the mean of three psychology majors, an English

major, and two chemistry majors?) The vast majority of the statistical techniques pre-

sented in this book are designed for numerical scores from an interval or a ratio scale.

For most statistical applications, the distinction between an interval scale and a ratio

scale is not important because both scales produce numerical values that permit us to

compute differences between scores, to add scores, and to calculate mean scores. On

the other hand, measurements from nominal or ordinal scales are typically not numeri-

cal values and are not compatible with many basic arithmetic operations. Therefore,

alternative statistical techniques are necessary for data from nominal or ordinal scales

of measurement (for example, the median and the mode in Chapter 3, the Spearman

correlation in Chapter 14, and the chi-square tests in Chapter 15).

STATISTICAL NOTATION

The measurements obtained in research studies provide the data for statistical analysis.

Most statistical techniques use the same general mathematical operations, notation,

and basic arithmetic that you have learned during previous years of school. In case you

are unsure of your mathematical skills, there is a mathematics review section in

Appendix A at the back of this book. The appendix also includes a skills-assessment

exam (p. 550) to help you determine whether you need the basic mathematics review. In

this section, we introduce some of the specialized notation that is used for statistical cal-

culations. In later chapters, additional statistical notation is introduced as it is needed.

Measuring a variable in a research study typically yields a value or a score for each

individual. Raw scores are the original, unchanged scores obtained in the study. Scores

for a particular variable are represented by the letter X. For example, if performance in

your statistics course is measured by tests and you obtain a 35 on the first test, then we

could state that X � 35. A set of scores can be presented in a column that is headed by

X. For example, a list of quiz scores from your class might be presented as shown in

the margin (the single column on the left).

When two variables are measured for each individual, the data can be presented as two

lists labeled X and Y. For example, measurements of people’s height in inches (variable X)

and weight in pounds (variable Y) can be presented as shown in the double column in the

margin. Each pair X, Y represents the observations made of a single participant.

The letter N is used to specify how many scores are in a set. An uppercase letter N

identifies the number of scores in a population and a lowercase letter n identifies the

number of scores in a sample. Throughout the remainder of the book you will notice

that we often use notational differences to distinguish between samples and populations.

For the height and weight data in the preceding table, n � 7 for both variables. Note that

by using a lowercase letter n, we are indicating that these scores come from a sample.

Many of the computations required in statistics involve adding a set of scores. Because

this procedure is used so frequently, a special notation is used to refer to the sum of a set

of scores. The Greek letter sigma, or o, is used to stand for summation. The expression

oX means to add all the scores for variable X. The summation sign, o, can be read as “the

sum of.” Thus, oX is read “the sum of the scores.” For the following set of quiz scores,

10, 6, 7, 4

oX � 27 and N � 4.

1.5

S CO R E S

S U M M AT I O N N OTAT I O N

Score

X X Y

37 72 165

35 68 151

35 67 160

30 67 160

25 68 146

17 70 160

16 66 133

SECTION 1.5 / STATISTICAL NOTATION 2 7

To use summation notation correctly, keep in mind the following two points:

1. The summation sign, o, is always followed by a symbol or mathematical ex-

pression. The symbol or expression identifies exactly which values are to be

added. To compute oX, for example, the symbol following the summation sign

is X, and the task is to find the sum of the X values. On the other hand, to com-

pute o(X – 1)2, the summation sign is followed by a relatively complex math-

ematical expression, so your first task is to calculate all of the (X – 1)2 values

and then add the results.

2. The summation process is often included with several other mathematical

operations, such as multiplication or squaring. To obtain the correct answer,

it is essential that the different operations be done in the correct sequence.

Following is a list showing the correct order of operations for performing

mathematical operations. Most of this list should be familiar, but you should

note that we have inserted the summation process as the fourth operation in

the list.

Order of Mathematical Operations

1. Any calculation contained within parentheses is done first.

2. Squaring (or raising to other exponents) is done second.

3. Multiplying and/or dividing is done third. A series of multiplication and/or

division operations should be done in order from left to right.

4. Summation using the o notation is done next.

5. Finally, any other addition and/or subtraction is done.

The following examples demonstrate how summation notation is used in most of the

calculations and formulas we present in this book.

A set of four scores consists of values 3, 1, 7, and 4. We will compute oX, oX2, and

(oX)2 for these scores. To help demonstrate the calculations, we will use a computa-

tional table showing the original scores (the X values) in the first column. Additional

columns can then be added to show additional steps in the series of operations.

You should notice that the first three operations in the list (parentheses, squaring, and

multiplying) all create a new column of values. The last two operations, however,

produce a single value corresponding to the sum.

The table to the left shows the original scores (the X values) and the squared scores

(the X2 values) that are needed to compute oX2.

The first calculation, oX, does not include any parentheses, squaring, or multiplica-

tion, so we go directly to the summation operation. The X values are listed in the first

column of the table, and we simply add the values in this column:

oX � 3 1 1 1 7 1 4 � 15

To compute oX2, the correct order of operations is to square each score and then

find the sum of the squared values. The computational table shows the original scores

and the results obtained from squaring (the first step in the calculation). The second

step is to find the sum of the squared values, so we simply add the numbers in the

X2 column.

oX2 � 9 1 1 1 49 1 16 � 75

E X A M P L E 1 . 3

More information on the

order of operations for

mathematics is available in

the Math Review appendix,

page 551.

X X2

3 9

1 1

7 49

4 16

2 8 CHAPTER 1 INTRODUCTION TO STATISTICS

The final calculation, (oX)2, includes parentheses, so the first step is to perform the

calculation inside the parentheses. Thus, we first find oX and then square this sum.

Earlier, we computed oX � 15, so

(oX)2 � (15)2 � 225

Next, we use the same set of four scores from Example 1.3 and compute o(X – 1) and

o(X – 1)2. The following computational table will help demonstrate the calculations.

E X A M P L E 1 . 4

X (X 2 1) (X 2 1)2 The first column lists the

original scores. A second

column lists the (X – 1)

values, and a third column

shows the (X – 1)2 values.

3 2 4

1 0 0

7 6 36

4 3 9

To compute o(X – 1), the first step is to perform the operation inside the parentheses.

Thus, we begin by subtracting one point from each of the X values. The resulting values

are listed in the middle column of the table. The next step is to add the (X – 1) values.

o(X – 1) � 2 1 0 1 6 1 3 1 � 11

The calculation of o(X – 1)2 requires three steps. The first step (inside parentheses)

is to subtract 1 point from each X value. The results from this step are shown in the

middle column of the computational table. The second step is to square each of the

(X – 1) values. The results from this step are shown in the third column of the table.

The final step is to add the (X – 1)2 values to obtain

o(X – 1)2 � 4 1 0 1 36 1 9 � 49

Notice that this calculation requires squaring before adding. A common mistake is

to add the (X – 1) values and then square the total. Be careful!

In both of the preceding examples, and in many other situations, the summation opera-

tion is the last step in the calculation. According to the order of operations, parentheses,

exponents, and multiplication all come before summation. However, there are situations

in which extra addition and subtraction are completed after the summation. For this

example, use the same scores that appeared in the previous two examples, and compute

oX – 1.

With no parentheses, exponents, or multiplication, the first step is the summation.

Thus, we begin by computing oX. Earlier we found oX � 15. The next step is to sub-

tract one point from the total. For these data,

oX – 1 � 15 – 1 � 14

E X A M P L E 1 . 5

SECTION 1.5 / STATISTICAL NOTATION 2 9

For this example, each individual has two scores. The first score is identified as X, and

the second score is Y. With the help of the following computational table, we compute

oX, oY, and oXY.

To find oX, simply add the values in the X column.

oX � 3 1 1 1 7 1 4 � 15

Similarly, oY is the sum of the Y values.

oY � 5 1 3 1 4 1 2 � 14

To compute oXY, the first step is to multiply X by Y for each individual. The result-

ing products (XY values) are listed in the third column of the table. Finally, we add the

products to obtain

oXY � 15 1 3 1 28 1 8 � 54

E X A M P L E 1 . 6

Person X Y XY

A 3 5 15

B 1 3 3

C 7 4 28

D 4 2 8

1. Calculate each value requested for the following scores: 4, 3, 7, 1.

a. oX d. oX – 1

b. oX2 e. o(X – 1)

c. (oX)2 f. o(X – 1)2

2. Identify the first step in each of the following calculations.

a. oX2 c. o(X – 2)2

b. (oX)2

3. Use summation notation to express each of the following.

a. Subtract 2 points from each score and then add the resulting values.

b. Subtract 2 points from each score, square the resulting values, and then add the

squared numbers.

c. Add the scores and then square the total.

1. a. 15 d. 14

b. 75 e. 11

c. 225 f. 49

2. a. Square each score.

b. Add the scores.

c. Subtract 2 points from each score.

3. a. o(X – 2) c. (oX)2

b. o(X – 2)2

L E A R N I N G C H E C K

ANSWERS

3 0 CHAPTER 1 INTRODUCTION TO STATISTICS

SUMMARY

1. The term statistics is used to refer to methods for organizing, summarizing, and interpreting data.

2. Scientific questions usually concern a population, which is the entire set of individuals one wishes to study. Usually, populations are so large that it is impossible to examine every individual, so most research is conducted with samples. A sample is a group selected from a population, usually for purposes of a research study.

3. A characteristic that describes a sample is called a statistic, and a characteristic that describes a popula- tion is called a parameter. Although sample statistics are usually representative of corresponding population parameters, there is typically some discrepancy between a statistic and a parameter. The naturally occurring difference between a statistic and a parameter is called sampling error.

4. Statistical methods can be classified into two broad categories: descriptive statistics, which organize and summarize data, and inferential statistics, which use sample data to draw inferences about populations.

5. The correlational method examines relationships between variables by measuring two different variables for each individual. This method allows researchers to measure and describe relationships, but cannot produce a cause-and-effect explanation for the relationship.

6. The experimental method examines relationships between variables by manipulating an independent variable to create different treatment conditions and then measuring a dependent variable to obtain a group of scores in each condition. The groups of scores are then compared. A systematic difference between groups provides evidence that changing the independent variable from one condition to another also caused a change in the dependent variable. All other variables are controlled to prevent them from influencing the relationship. The intent of the experimental method is to demonstrate a cause-and-effect relationship between variables.

7. Nonexperimental studies also examine relationships between variables by comparing groups of scores,

but they do not have the rigor of true experiments and cannot produce cause-and-effect explanations. Instead of manipulating a variable to create different groups, a nonexperimental study uses a preexisting participant characteristic (such as male/female) or the passage of time (before/after) to create the groups being compared.

8. A discrete variable consists of indivisible categories, often whole numbers that vary in countable steps. A continuous variable consists of categories that are infinitely divisible and each score corresponds to an interval on the scale. The boundaries that separate intervals are called real limits and are located exactly halfway between adjacent scores.

9. A measurement scale consists of a set of categories that are used to classify individuals. A nominal scale consists of categories that differ only in name and are not differentiated in terms of magnitude or direction. In an ordinal scale, the categories are differentiated in terms of direction, forming an ordered series. An interval scale consists of an ordered series of categories that are all equal-sized intervals. With an interval scale, it is possible to differentiate direc- tion and magnitude (or distance) between categories. Finally, a ratio scale is an interval scale for which the zero point indicates none of the variable being measured. With a ratio scale, ratios of measurements reflect ratios of magnitude.

10. The letter X is used to represent scores for a variable. If a second variable is used, Y represents its scores. The letter N is used as the symbol for the number of scores in a population; n is the symbol for the number of scores in a sample.

11. The Greek letter sigma (o) is used to stand for summation. Therefore, the expression oX is read “the sum of the scores.” Summation is a mathematical operation (like addition or multiplication) and must be performed in its proper place in the order of operations; summation occurs after operations in parentheses, exponents, and multiplication/division have been completed.

RESOURCES 31

KEY TERMS

statistics (5)

population (5)

sample (6)

variable (7)

datum (7)

raw score (7)

data (7)

data set (7)

parameter (7)

statistic (7)

descriptive statistics (8)

inferential statistics (8)

sampling error (8)

correlational method (12)

correlational research strategy (12)

experimental method (15)

confounded (15)

random assignment (15)

matching (15)

independent variable (16)

dependent variable (16)

control condition or control

group (16)

experimental condition or

experimental group (16)

nonequivalent groups study (17)

pre–post study (17)

quasi-independent variable (18)

construct or hypothetical

construct (20)

operational definition (20)

discrete variable (20)

continuous variable (21)

real limits (22)

lower real limit (22)

upper real limit (22)

nominal scale (23)

ordinal scale (23)

interval scale (24)

ratio scale (24)

sigma (26)

order of operations (27)

RESOURCES

Go to CengageBrain.com to access Psychology CourseMate, where you will find an

interactive eBook, glossaries, flashcards, quizzes, statistics workshops, and more.

If your professor has assigned Aplia:

1. Sign in to your account.

2. Complete the corresponding exercises as required by your professor.

3. When finished, click “Grade It Now” to see which areas you have mastered, which

areas need more work, and detailed explanations of every answer.

The Statistical Package for the Social Sciences, known as SPSS, is a computer program

that performs most of the statistical calculations that are presented in this book, and is

commonly available on college and university computer systems. Appendix D contains

a general introduction to SPSS. In the Resources section at the end of each chapter for

which SPSS is applicable, there are step-by-step instructions for using SPSS to perform

the statistical operations presented in the chapter.

3 2 CHAPTER 1 INTRODUCTION TO STATISTICS

FOCUS ON PROBLEM SOLVING

It may help to simplify summation notation if you observe that the summation sign is

always followed by a symbol or symbolic expression—for example, oX or o(X 1 3). This symbol specifies which values you are to add. If you use the symbol as a column

heading and list all the appropriate values in the column, your task is simply to add up

the numbers in the column. To find o(X 1 3) for example, start a column headed with (X 1 3) next to the column of Xs. List all the (X 1 3) values; then find the total for the column.

Often, summation notation is part of a relatively complex mathematical expres-

sion that requires several steps of calculation. The series of steps must be performed

according to the order of mathematical operations (see page 27). The best procedure

is to use a computational table that begins with the original X values listed in the first

column. Except for summation, each step in the calculation creates a new column

of values. For example, computing o(X 1 1)2 involves three steps and produces a computational table with three columns. The final step is to add the values in the

third column (see Example 1.4).

DEMONSTRATION 1.1

SUMMATION NOTATION

A set of scores consists of the following values:

7 3 9 5 4

For these scores, compute each of the following:

(oX)2

oX2

oX 1 5 o(X – 2)

Compute oX To compute oX, we simply add all of the scores in the group.

oX � 7 1 3 1 9 1 5 1 4 � 28

Compute (oX)2 The first step, inside the parentheses, is to compute oX. The second step

is to square the value for oX.

oX � 28 and (oX)2 � (28)2 � 784

Compute oX2 The first step is to square each score. The second step is to add the squared

scores. The computational table shows the scores and squared scores. To compute oX2 we

add the values in the X2 column.

oX2 � 49 1 9 1 81 1 25 1 16 � 180

X X2

7 49

3 9

9 81

5 25

4 16

PROBLEMS 33

Compute oX 1 5 The first step is to compute oX. The second step is to add 5 points to the total.

oX � 28 and oX 1 5 � 28 1 5 � 33

Compute o(X – 2) The first step, inside parentheses, is to subtract 2 points from each

score. The second step is to add the resulting values. The computational table shows the

scores and the (X – 2) values. To compute o(X – 2), add the values in the (X – 2) column

o(X – 2) � 5 1 1 1 7 1 3 1 2 � 18

X X – 2

7 5

3 1

9 7

5 3

4 2

PROBLEMS

*1. A researcher is investigating the effectiveness of a treatment for adolescent boys who are taking medi- cation for depression. A group of 30 boys is selected and half receive the new treatment in addition to their medication and the other half continue to take their medication without any treatment. For this study,

a. Identify the population. b. Identify the sample.

2. Define the terms population, sample, parameter, and statistic.

3. Statistical methods are classified into two major categories: descriptive and inferential. Describe the general purpose for the statistical methods in each category.

4. Define the concept of sampling error and explain why this phenomenon creates a problem to be ad- dressed by inferential statistics.

5. Describe the data for a correlational research study. Explain how these data are different from the data obtained in experimental and nonexperimental studies, which also evaluate relationships between two variables.

6. What is the goal for an experimental research study? Identify the two elements that are necessary for an experiment to achieve its goal.

7. Knight and Haslam (2010) found that office workers who had some input into the design of their office space were more productive and had higher well- being compared to workers for whom the office design was completely controlled by an office manager. For this study, identify the independent variable and the dependent variable.

8. Judge and Cable (2010) found that thin women had higher incomes than heavier women. Is this an exam- ple of an experimental or a nonexperimental study?

9. Two researchers are both interested in determining whether large doses of vitamin C can help prevent

the common cold. Each obtains a sample of n � 20 college students.

a. The first researcher interviews each student to determine whether they routinely take a vitamin C supplement. The researcher then records the number of colds each individual gets during the winter. Is this an experimental or a nonexperimental study? Explain your answer.

b. The second researcher separates the students into two roughly equivalent groups. The students in one group are given a daily multivitamin contain- ing a large amount of vitamin C, and the other group gets a multivitamin with no vitamin C. The researcher then records the number of colds each individual gets during the winter. Is this an experi- mental or a nonexperimental study? Explain your answer.

10. Weinstein, McDermott, and Roediger (2010) con- ducted an experiment to evaluate the effectiveness of different study strategies. One part of the study asked students to prepare for a test by reading a passage. In one condition, students generated and answered questions after reading the passage. In a second condition, students simply read the passage a second time. All students were then given a test on the passage material and the researchers recorded the number of correct answers.

a. Identify the dependent variable for this study. b. Is the dependent variable discrete or continuous? c. What scale of measurement (nominal, ordinal,

interval, or ratio) is used to measure the dependent variable?

11. A research study reports that alcohol consumption is significantly higher for students at a state university than for students at a religious college (Wells, 2010). Is this study an example of an experiment? Explain why or why not.

12. Oxytocin is a naturally occurring brain chemical that is nicknamed the “love hormone” because it seems to play a role in the formation of social relationships such as mating pairs and parent–child bonding. A recent study demonstrated that oxytocin appears to

*Solutions for odd-numbered problems are provided in Appendix C.

3 4 CHAPTER 1 INTRODUCTION TO STATISTICS

increase people’s tendency to trust others (Kosfeld, Heinrichs, Zak, Fischbacher, and Fehr, 2005). Using an investment game, the study demonstrated that people who inhaled oxytocin were more likely to give their money to a trustee compared to people who inhaled an inactive placebo. For this experimen- tal study, identify the independent variable and the dependent variable.

13. For each of the following, determine whether the variable being measured is discrete or continuous and explain your answer.

a. Social networking (number of daily minutes on Facebook)

b. Family size (number of siblings) c. Preference between digital or analog watch d. Number of correct answers on a statistics quiz

14. Four scales of measurement were introduced in this chapter: nominal, ordinal, interval, and ratio.

a. What additional information is obtained from measurements on an ordinal scale compared to measurements on a nominal scale?

b. What additional information is obtained from measurements on an interval scale compared to measurements on an ordinal scale?

c. What additional information is obtained from measurements on a ratio scale compared to measurements on an interval scale?

15. In an experiment examining the effects Tai Chi on arthritis pain, Callahan (2009) selected a large sample of individuals with doctor-diagnosed arthritis. Half of the participants immediately began a Tai Chi course and the other half (the control group) waited 8 weeks before beginning the program. At the end of 8 weeks, the individuals who had experienced Tai Chi had less arthritis pain that those who had not participated in the course.

a. Identify the independent variable for this study. b. What scale of measurement is used for the inde-

pendent variable? c. Identify the dependent variable for this study. d. What scale of measurement is used for the

dependent variable?

16. Explain why shyness is a hypothetical construct instead of a concrete variable. Describe how shyness might be measured and defined using an operational definition.

17. Ford and Torok (2008) found that motivational signs were effective in increasing physical activity on a college campus. Signs such as “Step up to a healthier lifestyle” and “An average person burns 10 calories a minute walking up the stairs” were posted by the elevators and stairs in a college building. Students and faculty increased their use of the stairs during

times that the signs were posted compared to times when there were no signs.

a. Identify the independent and dependent variables for this study.

b. What scale of measurement is used for the inde- pendent variable?

18. For the following scores, find the value of each expression:

a. oX

b. oX2

c. oX 1 1 d. o(X 1 1)

19. For the following set of scores, find the value of each expression:

a. oX2

b. (oX)2

c. o(X – 1) d. o(X – 1)2

20. For the following set of scores, find the value of each expression:

a. oX

b. oX2

c. o(X 1 3)

21. Two scores, X and Y, are recorded for each of n � 4 subjects. For these scores, find the value of each expression.

a. oX

b. oY

c. oXY

Subject X Y

A 3 4

B 0 7

C –1 5

D 2 2

22. Use summation notation to express each of the following calculations:

a. Add 1 point to each score, and then add the result- ing values.

PROBLEMS 35

b. Add 1 point to each score and square the result. Then add the squared values.

c. Add the scores and square the sum. Then subtract 3 points from the squared value.

23. For the set of scores at the right, find the value of each expression:

a. oX2

b. (oX)2

c. o(X – 3) d. o(X – 3)2

Aplia for Essentials of Statistics for the Behavioral Sciences

After reading, go to “Resources” at the end of this chapter for

an introduction on how to use Aplia’s homework and learning

resources.

Frequency Distributions

2.1 Introduction to Frequency Distributions

2.2 Frequency Distribution Tables

2.3 Frequency Distribution Graphs

2.4 The Shape of a Frequency Distribution

Summary

Focus on Problem Solving

Demonstration 2.1

Problems

C H A P T E R

2 Tools You Will Need The following items are considered essential background material for this chapter. If you doubt your knowledge of any of these items, you should review the appropriate chapter or section before proceeding.

• Proportions (math review, Appendix A) • Fractions • Decimals • Percentages

• Scales of measurement (Chapter 1): Nominal, ordinal, interval, and ratio

• Continuous and discrete variables (Chapter 1)

• Real limits (Chapter 1)

3 8 CHAPTER 2 FREQUENCY DISTRIBUTIONS

INTRODUCTION TO FREQUENCY DISTRIBUTIONS

The results from a research study usually consist of pages of numbers corresponding to

the measurements, or scores, collected during the study. The immediate problem for the

researcher is to organize the scores into some comprehensible form so that any patterns

in the data can be seen easily and communicated to others. This is the job of descrip-

tive statistics: to simplify the organization and presentation of data. One of the most

common procedures for organizing a set of data is to place the scores in a frequency

distribution.

A frequency distribution is an organized tabulation of the number of individuals

located in each category on the scale of measurement.

A frequency distribution takes a disorganized set of scores and places them in order

from highest to lowest, grouping together individuals who all have the same score. If

the highest score is X � 10, for example, the frequency distribution groups together all

the 10s, then all the 9s, then the 8s, and so on. Thus, a frequency distribution allows the

researcher to see “at a glance” the entire set of scores. It shows whether the scores are

generally high or low, whether they are concentrated in one area or spread out across

the entire scale, and generally provides an organized picture of the data. In addition

to providing a picture of the entire set of scores, a frequency distribution allows you

to see the location of any individual score relative to all of the other scores in the set.

A frequency distribution can be structured either as a table or as a graph, but in either

case, the distribution presents the same two elements:

1. The set of categories that make up the original measurement scale.

2. A record of the frequency, or number of individuals in each category.

Thus, a frequency distribution presents a picture of how the individual scores are

distributed on the measurement scale—hence the name frequency distribution.

FREQUENCY DISTRIBUTION TABLES

The simplest frequency distribution table presents the measurement scale by listing the

different measurement categories (X values) in a column from highest to lowest. Beside

each X value, we indicate the frequency, or the number of times that particular measure-

ment occurred in the data. It is customary to use an X as the column heading for the

scores and an f as the column heading for the frequencies. An example of a frequency

distribution table follows.

The following set of N � 20 scores was obtained from a 10-point statistics quiz. We

organize these scores by constructing a frequency distribution table. Scores:

8 9 8 7 10 9 6 4 9 8

7 8 10 9 8 6 9 7 8 9

1. The highest score is X � 10, and the lowest score is X � 4. Therefore, the first

column of the table lists the categories that make up the scale of measurement

2.1

D E F I N I T I O N

2.2

E X A M P L E 2 . 1

It is customary to list

categories from highest to

lowest, but this is an arbitrary

arrangement. Many computer

programs list categories from

lowest to highest.

SECTION 2.2 / FREQUENCY DISTRIBUTION TABLES 3 9

(X values) from 10 down to 4. Notice that all of the possible values are listed in

the table. For example, no one had a score of X � 5, but this value is included.

With an ordinal, interval, or ratio scale, the categories are listed in order (usually

highest to lowest). For a nominal scale, the categories can be listed in any order.

2. The frequency associated with each score is recorded in the second column.

For example, two people had scores of X � 10, so there is a 2 in the f column

beside X � 10.

Because the table organizes the scores, it is possible to see the general quiz results

very quickly. For example, there were only two perfect scores, but most of the class

had high grades (8s and 9s). With one exception (the score of X � 4), it appears that

the class has learned the material fairly well.

Notice that the X values in a frequency distribution table represent the scale

of measurement, not the actual set of scores. For example, the X column lists the value

10 only one time, but the frequency column indicates that there are actually two values

of X � 10. Also, the X column lists a value of X � 5, but the frequency column indicates

that no one actually had a score of X � 5.

You also should notice that the frequencies can be used to find the total number of

scores in the distribution. By adding up the frequencies, you obtain the total number

of individuals:

of � N

There may be times when you need to compute the sum of the scores, oX, or perform

other computations for a set of scores that has been organized into a frequency distribu-

tion table. To complete these calculations correctly, you must use all of the information

presented in the table. That is, it is essential to use the information in the f column as

well as that in the X column to obtain the full set of scores.

When it is necessary to perform calculations for scores that have been organized

into a frequency distribution table, the safest procedure is to take the individual scores

out of the table before you begin any computations. This process is demonstrated in the

following example.

Consider the frequency distribution table shown in the margin. The table shows that the

distribution has one 5, two 4s, three 3s, three 2s, and one 1, for a total of 10 scores. If

you simply list all 10 scores, you can safely proceed with calculations such as finding

oX or oX2. For example, to compute oX you must add all 10 scores:

oX � 5 1 4 1 4 1 3 1 3 1 3 1 2 1 2 1 2 1 1

For the distribution in this table, you should obtain oX � 29. Try it yourself. Similarly,

to compute oX2 you square each of the 10 scores and then add the squared values.

oX2 � 52 1 42 1 42 1 32 1 32 1 32 1 22 1 22 1 22 1 12

This time you should obtain oX2 � 97.

An alternative way to get oX from a frequency distribution table is to multi-

ply each X value by its frequency and then add these products. This sum may be

O B TA I N I N G  X F R O M A F R E Q U E N CY D I ST R I B U T I O N TA B L E

E X A M P L E 2 . 2

X f

10 2

9 5

8 7

7 3

6 2

5 0

4 1

X f

5 1

4 2

3 3

2 3

1 1

4 0 CHAPTER 2 FREQUENCY DISTRIBUTIONS

expressed in symbols as ofX. The computation is summarized as follows for the data

in Example 2.2:

X f fX

5 1 5 (the one 5 totals 5)

4 2 8 (the two 4s total 8)

3 3 9 (the three 3s total 9)

2 3 6 (the three 2s total 6)

1 1 1 (the one 1 totals 1)

oX � 29

Caution: Doing calculations

within the table works well

for oX but can lead to errors

for more complex formulas.

No matter which method you use to find oX, the important point is that you must

use the information given in the frequency column as well as the information in the

X column.

In addition to the two basic columns of a frequency distribution table, there are other

measures that describe the distribution of scores and can be incorporated into the table.

The two most common are proportion and percentage.

Proportion measures the fraction of the total group that is associated with each score.

In Example 2.2, there were two individuals with X � 4. Thus, 2 out of 10 people had

X � 4, so the proportion would be 2

10 � 0.20. In general, the proportion associated with

each score is

proportion � �p f

Because proportions describe the frequency (f) in relation to the total number (N),

they often are called relative frequencies. Although proportions can be expressed as

fractions (for example, 2

10 ), they more commonly appear as decimals. A column of

proportions, headed with a p, can be added to the basic frequency distribution table

(see Example 2.3).

In addition to using frequencies (f) and proportions (p), researchers often describe a

distribution of scores with percentages. For example, an instructor might describe the

results of an exam by saying that 15% of the class earned As, 23% earned Bs, and so

on. To compute the percentage associated with each score, you first find the proportion

(p) and then multiply by 100:

percentage (100) (100) 5 5p f

Percentages can be included in a frequency distribution table by adding a column

headed with % (see Example 2.3).

The frequency distribution table from Example 2.2 is repeated here. This time we have

added columns showing the proportion (p) and the percentage (%) associated with each

score.

P R O P O R T I O N S A N D P E R C E N TAG E S

E X A M P L E 2 . 3

SECTION 2.2 / FREQUENCY DISTRIBUTION TABLES 4 1

When a set of data covers a wide range of values, it is unreasonable to list all of the

individual scores in a frequency distribution table. Consider, for example, a set of exam

scores that range from a low of X 5 41 to a high of X 5 96. These scores cover a range

of more than 50 points.

If we were to list all of the individual scores from X 5 96 down to X 5 41, it would

take 56 rows to complete the frequency distribution table. Although this would orga-

nize the data, the table would be long and cumbersome. Remember: The purpose for

constructing a table is to obtain a relatively simple, organized picture of the data. This

can be accomplished by grouping the scores into intervals and then listing the intervals

in the table instead of listing each individual score. For example, we could construct

a table showing the number of students who had scores in the 90s, the number with

scores in the 80s, and so on. The result is called a grouped frequency distribution table

because we are presenting groups of scores rather than individual values. The groups,

or intervals, are called class intervals.

There are several guidelines that help guide you in the construction of a grouped

frequency distribution table. Note that these are simply guidelines, rather than absolute

requirements, but they do help to produce a simple, well-organized, and easily under-

stood table.

G R O U P E D F R E Q U E N CY D I ST R I B U T I O N TA B L E S

X f p 5 f/N % 5 p(100)

5 1 1/10 5 0.10 10%

4 2 2/10 5 0.20 20%

3 3 3/10 5 0.30 30%

2 3 3/10 5 0.30 30%

1 1 1/10 5 0.10 10%

X f

5 1

4 1

3 4

2 2

1 1

1. Construct a frequency distribution table for the following set of scores.

Scores: 3, 2, 3, 2, 4, 1, 3, 3, 5

2. Find each of the following values for the sample in the following frequency

distribution table.

a. n

b. oX

c. oX2

2. a. n 5 10 b. oX 5 28 c. oX2 5 92 (square then add all 10 scores)

L E A R N I N G C H E C K

ANSWERS

X f

5 1

4 2

3 2

2 4

1 1

When the scores are whole

numbers, the total number

of rows for a regular table

can be obtained by find-

ing the difference between

the highest and the lowest

scores and adding 1:

rows 5 highest – lowest 1 1

4 2 CHAPTER 2 FREQUENCY DISTRIBUTIONS

Guideline 1 The grouped frequency distribution table should have about 10 class inter-

vals. If a table has many more than 10 intervals, it becomes cumbersome and defeats the

purpose of a frequency distribution table. On the other hand, if you have too few intervals,

you begin to lose information about the distribution of the scores. At the extreme, with only

one interval, the table would not tell you anything about how the scores are distributed.

Remember that the purpose of a frequency distribution is to help a researcher see the data.

With too few or too many intervals, the table will not provide a clear picture. You should

note that 10 intervals is a general guide. If you are constructing a table on a blackboard, for

example, you probably want only 5 or 6 intervals. If the table is to be printed in a scientific

report, you may want 12 or 15 intervals. In each case, your goal is to present a table that is

relatively easy to see and understand.

Guideline 2 The width of each interval should be a relatively simple number. For

example, 2, 5, 10, or 20 would be a good choice for the interval width. Notice that it is

easy to count by 5s or 10s. These numbers are easy to understand and make it possible

for someone to see quickly how you have divided the range of scores.

Guideline 3 The bottom score in each class interval should be a multiple of the width.

If you are using a width of 10 points, for example, the intervals should start with 10,

20, 30, 40, and so on. Again, this makes it easier for someone to understand how the

table has been constructed.

Guideline 4 All intervals should be the same width. They should cover the range of

scores completely with no gaps and no overlaps, so that any particular score belongs in

exactly one interval.

The application of these rules is demonstrated in Example 2.4.

An instructor has obtained the set of N 5 25 exam scores shown here. To help organize

these scores, we will place them in a frequency distribution table. The scores are:

82 75 88 93 53 84 87 58 72 94 69 84 61

91 64 87 84 70 76 89 75 80 73 78 60

The first step is to determine the range of scores. For these data, the smallest score

is X 5 53 and the largest score is X 5 94, so a total of 42 rows would be needed for a

table that lists each individual score. Because 42 rows would not provide a simple table,

we have to group the scores into class intervals.

The best method for finding a good interval width is a systematic trial-and-error ap-

proach that uses guidelines 1 and 2 simultaneously. The goal is to find an interval width

that is an easy number and produces a table with around 10 intervals. For this example,

the scores cover a range of 42 points, so we will try several different interval widths to

see how many intervals are needed to cover the range. For example, if each interval is

2 points wide, it would take 21 intervals to cover a range of 42 points. This is too many,

so we move on to an interval width of 5 or 10 points. The following table shows how

many intervals would be needed for these possible widths:

E X A M P L E 2 . 4

Remember, when the

scores are whole numbers,

the number of rows is

determined by

highest – lowest 1 1

Width Number of Intervals Needed to Cover a Range of 42 Points

2 21 (too many)

5 9 (OK)

10 5 (too few)

SECTION 2.2 / FREQUENCY DISTRIBUTION TABLES 4 3

Note that the computed number of intervals is just an estimate of the actual number

needed for the final table. Because the bottom interval usually extends below the lowest

score and the top interval extends above the highest score, you may need slightly more

than the computed number of intervals. Also notice that an interval width of 5 points

will produce a table with about 10 intervals, which is exactly what we want.

The next step is to identify the actual intervals. The lowest score for these data is

X 5 53, so the lowest interval should contain this value. Because the interval should

have a multiple of 5 as its bottom score, the interval should begin at 50. The interval

has a width of 5, so it should contain 5 values: 50, 51, 52, 53, and 54. Thus, the bottom

interval is 50–54. The next interval would start at 55 and go to 59. Note that this interval

also has a bottom score that is a multiple of 5, and contains exactly 5 scores (55, 56,

57, 58, and 59). The complete frequency distribution table showing all of the class

intervals is presented in Table 2.1.

Once the class intervals are listed, you complete the table by adding a column of

frequencies. The values in the frequency column indicate the number of individu-

als who have scores located in that class interval. For this example, there were three

students with scores in the 60–64 interval, so the frequency for this class interval is

f 5 3 (see Table 2.1). The basic table can be extended by adding columns showing the

proportion and percentage associated with each class interval.

Finally, you should note that, after the scores have been placed in a grouped table,

you lose information about the specific value for any individual score. For example,

Table 2.1 shows that one person had a score between 65 and 69, but the table does

not identify the exact value for the score. In general, the wider the class intervals

are, the more information is lost. In Table 2.1, the interval width is 5 points, and the

table shows that there are three people with scores in the lower 60s and one person

with a score in the upper 60s. This information would be lost if the interval width

were increased to 10 points. With an interval width of 10, all of the 60s would be

grouped together into one interval labeled 60–69. The table would show a frequency

of four people in the 60–69 interval, but it would not tell whether the scores were

in the upper 60s or the lower 60s.

Recall from Chapter 1 that a continuous variable has an infinite number of possible

values and can be represented by a number line that is continuous and contains an infi-

nite number of points. However, when a continuous variable is measured, the resulting

measurements correspond to intervals on the number line rather than single points. If

you are measuring time in seconds, for example, a score of X 5 8 seconds actually

represents an interval bounded by the real limits 7.5 seconds and 8.5 seconds. Thus,

R E A L L I M I T S A N D F R E Q U E N CY

D I ST R I B U T I O N S

X f

90–94 3

85–89 4

80–84 5

75–79 4

70–74 3

65–69 1

60–64 3

55–59 1

50–54 1

TABLE 2.1

This grouped frequency distribu-

tion table shows the data from

Example 2.4. The original scores

range from a high of X 5 94

to a low of X 5 53. This range

has been divided into 9 inter-

vals with each interval exactly

5 points wide. The frequency

column ( f ) lists the number of

individuals with scores in each

of the class intervals.

4 4 CHAPTER 2 FREQUENCY DISTRIBUTIONS

a frequency distribution table showing a frequency of f 5 3 individuals all assigned a

score of X 5 8 does not mean that all three individuals had exactly the same measure-

ment. Instead, you should realize that the three measurements are simply located in the

same interval between 7.5 and 8.5.

The concept of real limits also applies to the class intervals of a grouped frequency

distribution table. For example, a class interval of 40–49 contains scores from X 5 40

to X 5 49. These values are called the apparent limits of the interval because it

appears that they form the upper and lower boundaries for the class interval. If you are

measuring a continuous variable, however, a score of X 5 40 is actually an interval

from 39.5 to 40.5. Similarly, X 5 49 is an interval from 48.5 to 49.5. Therefore, the

real limits of the interval are 39.5 (the lower real limit) and 49.5 (the upper real limit).

Notice that the next higher class interval is 50–59, which has a lower real limit of

49.5. Thus, the two intervals meet at the real limit 49.5, so there are no gaps in the

scale. You also should notice that the width of each class interval becomes easier

to understand when you consider the real limits of an interval. For example, the interval

50–59 has real limits of 49.5 and 59.5. The distance between these two real limits is

10 points, which is the width of the interval.

1. Place the following scores in a grouped frequency distribution table using an

interval width of 10 points.

Scores: 39 41 37 16 44 20 34 39 42

24 51 22 35 18 46 53 19 26

2. If the scores in the previous question were placed in a grouped table with an

interval width of 5 points, what are the apparent limits and the real limits for the

bottom interval?

3. Using only the frequency distribution table you constructed for Exercise 1, how

many individuals had a score of X 5 53?

X f

50–59 2

40–49 4

30–39 5

20–29 4

10–19 3

2. The apparent limits are 15–19 and the real limits are 14.5–19.5.

3. After a set of scores has been summarized in a grouped table, you cannot determine the

frequency for any specific score. There is no way to determine how many individuals had

X 5 53 from the table alone. (You can say that at most two people had X 5 53.)

L E A R N I N G C H E C K

ANSWERS

FREQUENCY DISTRIBUTION GRAPHS

A frequency distribution graph is basically a picture of the information available in

a frequency distribution table. We consider several different types of graphs, but all

start with two perpendicular lines called axes. The horizontal line is the X-axis, or the

abscissa (ab-SIS-uh). The vertical line is the Y-axis, or the ordinate. The measurement

2.3

SECTION 2.3 / FREQUENCY DISTRIBUTION GRAPHS 4 5

scale (set of X values) is listed along the X-axis with values increasing from left to right.

The frequencies are listed on the Y-axis with values increasing from bottom to top. As

a general rule, the point where the two axes intersect should have a value of zero for

both the scores and the frequencies. A final general rule is that the graph should be

constructed so that its height (Y-axis) is approximately two-thirds to three-quarters of

its length (X-axis). Violating these guidelines can result in graphs that give a misleading

picture of the data (see Box 2.1).

When the data consist of numerical scores that have been measured on an interval or a

ratio scale, there are two options for constructing a frequency distribution graph. The

two types of graphs are called histograms and polygons.

Histograms To construct a histogram, you first list the numerical scores (the catego-

ries of measurement) along the X-axis. Then you draw a bar above each X value so that

a. The height of the bar corresponds to the frequency for that category.

b. For continuous variables, the width of the bar extends to the real limits of the

category. For discrete variables, each bar extends exactly half the distance to the

adjacent category on each side.

For both continuous and discrete variables, each bar in a histogram extends to the

midpoint between adjacent categories. As a result, adjacent bars touch and there are no

spaces or gaps between bars. An example of a histogram is shown in Figure 2.1.

When data have been grouped into class intervals, you can construct a frequency

distribution histogram by drawing a bar above each interval so that the width of the bar

extends exactly half the distance to the adjacent category on each side. This process is

demonstrated in Figure 2.2.

For the two histograms shown in Figures 2.1 and 2.2, notice that the values on both

the vertical and horizontal axes are clearly marked and that both axes are labeled. Also

note that, whenever possible, the units of measurement are specified; for example,

Figure 2.2 shows a distribution of heights measured in inches. Finally, notice that the

horizontal axis in Figure 2.2 does not list all of the possible heights starting from zero

and going up to 48 inches. Instead, the graph clearly shows a break between zero and

30, indicating that some scores have been omitted.

A modified histogram A slight modification to the traditional histogram produces a

very easy to draw and simple to understand sketch of a frequency distribution. Instead

G R A P H S F O R I N T E R VA L O R R AT I O DATA

Quiz scores (number correct)

F re

q u

e n

c y

3 4

5 4 3 2 1

2 3 4 2 1

X fFIGURE 2.1

An example of a frequency

distribution histogram. The

same set of quiz scores is

presented in a frequency

distribution table and in a

histogram.

4 6 CHAPTER 2 FREQUENCY DISTRIBUTIONS

of drawing a bar above each score, the modification consists of drawing a stack of

blocks. Each block represents one individual, so the number of blocks above each score

corresponds to the frequency for that score. An example is shown in Figure 2.3.

Note that the number of blocks in each stack makes it very easy to see the abso-

lute frequency for each category. In addition, it is easy to see the exact difference in

frequency from one category to another. In Figure 2.3, for example, there are exactly

two more people with scores of X 5 2 than with scores of X 5 1. Because the frequencies

are clearly displayed by the number of blocks, this type of display eliminates the need for

a vertical line (the Y-axis) showing frequencies. In general, this kind of graph provides a

simple and concrete picture of the distribution for a sample of scores. Note that we often

use this kind of graph to show sample data throughout the rest of the book. You should

also note, however, that this kind of display simply provides a sketch of the distribution

and is not a substitute for an accurately drawn histogram with two labeled axes.

Polygons The second option for graphing a distribution of numerical scores from an

interval or a ratio scale of measurement is called a polygon. To construct a polygon,

you begin by listing the numerical scores (the categories of measurement) along the

X-axis. Then,

a. A dot is centered above each score so that the vertical position of the dot

corresponds to the frequency for the category.

b. A continuous line is drawn from dot to dot to connect the series of dots.

c. The graph is completed by drawing a line down to the X-axis (zero frequency)

at each end of the range of scores. The final lines are usually drawn so that they

reach the X-axis at a point that is one category below the lowest score on the

Children’s heights (in inches)

F re

q u

e n

c y

44–45 42–43 40–41 38–39 36–37 34–35 32–33 30–31

1 2 4 6 2 3 4 230 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

FIGURE 2.2

An example of a fre-

quency distribution

histogram for grouped

data. The same set of

children’s heights is

presented in a frequency

distribution table and in

a histogram.

1 2 3 4 5 6 7 x

FIGURE 2.3

A frequency distribution

in which each individual

is represented by a block

placed directly above the

individual’s score. For

example, three people had

scores of X 5 2.

SECTION 2.3 / FREQUENCY DISTRIBUTION GRAPHS 4 7

left side and one category above the highest score on the right side. An example

of a polygon is shown in Figure 2.4.

A polygon also can be used with data that have been grouped into class intervals.

For a grouped distribution, you position each dot directly above the midpoint of the

class interval. The midpoint can be found by averaging the highest and the lowest

scores in the interval. For example, a class interval that is listed as 20–29 would have

a midpoint of 24.5.

midpoint 5 1

5 5 . 20 29

2 24 5

An example of a frequency distribution polygon with grouped data is shown in

Figure 2.5.

6 5 4 3 2 1

1 2 2 4 2 1

X f

1 2 3 4 5 6 7

Scores

F re

q u

e n

c y

FIGURE 2.4

An example of a

frequency distribution

polygon. The same set

of data is presented in

a frequency distribution

table and in a polygon.

10 1 2 3 4 5 6 7 8 9 11 12 13 14 Scores

F re

q u

e n

c y

X f

12–13 10–11

4 5 3 3 2

8–9 6–7 4–5

FIGURE 2.5

An example of a fre-

quency distribution

polygon for grouped data.

The same set of data is

presented in a grouped

frequency distribution

table and in a polygon.

4 8 CHAPTER 2 FREQUENCY DISTRIBUTIONS

When the scores are measured on a nominal or ordinal scale (usually non-numerical

values), the frequency distribution can be displayed in a bar graph.

Bar graphs A bar graph is essentially the same as a histogram, except that spaces are

left between adjacent bars. For a nominal scale, the space between bars emphasizes that

the scale consists of separate, distinct categories. For ordinal scales, separate bars are

used because you cannot assume that the categories are all the same size.

To construct a bar graph, list the categories of measurement along the X-axis and

then draw a bar above each category so that the height of the bar equals the frequency

for the category. An example of a bar graph is shown in Figure 2.6.

When you can obtain an exact frequency for each score in a population, you can

construct frequency distribution graphs that are exactly the same as the histograms,

polygons, and bar graphs that are typically used for samples. For example, if a popu-

lation is defined as a specific group of N 5 50 people, we could easily determine how

many have IQs of X 5 110. However, if we are interested in the entire population

of adults in the United States, it would be impossible to obtain an exact count of the

number of people with an IQ of 110. Although it is still possible to construct graphs

showing frequency distributions for extremely large populations, the graphs usually

involve two special features: relative frequencies and smooth curves.

Relative frequencies Although you usually cannot find the absolute frequency

for each score in a population, you very often can obtain relative frequencies. For

example, you may not know exactly how many fish are in the lake, but after years of

fishing you do know that there are twice as many bluegill as there are bass. You can

represent these relative frequencies in a bar graph by making the bar above bluegill

two times taller than the bar above bass (Figure 2.7). Notice that the graph does not

show the absolute number of fish. Instead, it shows the relative number of bluegill

and bass.

Smooth curves When a population consists of numerical scores from an interval or a

ratio scale, it is customary to draw the distribution with a smooth curve instead of the jag-

ged, step-wise shapes that occur with histograms and polygons. The smooth curve indi-

cates that you are not connecting a series of dots (real frequencies) but instead are showing

G R A P H S F O R N O M I N A L O R O R D I N A L DATA

G R A P H S F O R P O P U L AT I O N

D I ST R I B U T I O N S

Personality type

A B C

F re

q u

e n

c y

FIGURE 2.6

A bar graph showing the

distribution of personality

types in a sample of

college students. Because

personality type is a

discrete variable measured

on a nominal scale, the

graph is drawn with space

between the bars.

SECTION 2.3 / FREQUENCY DISTRIBUTION GRAPHS 4 9

the relative changes that occur from one score to the next. One commonly occurring

population distribution is the normal curve. The word normal refers to a specific shape

that can be precisely defined by an equation. Less precisely, we can describe a nor-

mal distribution as being symmetrical, with the greatest frequency in the middle and

relatively smaller frequencies as you move toward either extreme. A good example of

a normal distribution is the population distribution for IQ scores shown in Figure 2.8.

Because normal-shaped distributions occur commonly and because this shape is math-

ematically guaranteed in certain situations, we give it extensive attention throughout

this book.

In the future, we will be referring to distributions of scores. Whenever the term

distribution appears, you should conjure up an image of a frequency distribution graph.

The graph provides a picture showing exactly where the individual scores are located.

To make this concept more concrete, you might find it useful to think of the graph as

showing a pile of individuals just like we showed a pile of blocks in Figure 2.3. For the

population of IQ scores shown in Figure 2.8, the pile is highest at an IQ score around

100 because most people have average IQs. There are only a few individuals piled up

at an IQ of 130; it must be lonely at the top.

R e

la ti v e

f re

q u

e n

c y

Type of fish

Bass Bluegill

FIGURE 2.7

A frequency distribu-

tion showing the relative

frequency for two types

of fish. Notice that the

exact number of fish is

not reported; the graph

simply says that there are

twice as many bluegill as

there are bass.

R e

la ti v e

f re

q u

e n

c y

IQ scores

70 85 100 115 130

FIGURE 2.8

The population distribution

of IQ scores: an example

of a normal distribution.

5 0 CHAPTER 2 FREQUENCY DISTRIBUTIONS

BOX

2.1 THE USE AND MISUSE OF GRAPHS

Although graphs are intended to provide an accurate

picture of a set of data, they can be used to exagger-

ate or misrepresent a set of scores. These misrep-

resentations generally result from failing to follow

the basic rules for graph construction. The following

example demonstrates how the same set of data can

be presented in two entirely different ways by

manipulating the structure of a graph.

For the past several years, the city has kept records

of the number of homicides. The data are summarized

as follows:

Year Number of Homicides

2007 42

2008 44

2009 47

2010 49

These data are shown in two different graphs in

Figure 2.9. In the first graph, we have exaggerated the

height and started numbering the Y-axis at 40 rather

than at zero. As a result, the graph seems to indicate a

rapid rise in the number of homicides over the 4-year

period. In the second graph, we have stretched out

the X-axis and used zero as the starting point for the

Y-axis. The result is a graph that shows little change

in the homicide rate over the 4-year period.

Which graph is correct? The answer is that neither

one is very good. Remember that the purpose of a

graph is to provide an accurate display of the data.

The first graph in Figure 2.9 exaggerates the differ-

ences between years, and the second graph conceals

the differences. Some compromise is needed. Also

note that in some cases a graph may not be the best

way to display information. For these data, for

example, showing the numbers in a table would be

better than either graph.

07 08 09 10

Year

N u

m b

e r

o f

h o

m ic

id e

2007

Year

N u

m b

e r

o f

h o

m ic

id e

2008 2009 2010

FIGURE 2.9

Two graphs showing the number of homicides in a

city over a 4-year period. Both graphs show exactly

the same data. However, the first graph gives the

appearance that the homicide rate is high and rising

rapidly. The second graph gives the impression that

the homicide rate is low and has not changed over

the 4-year period.

THE SHAPE OF A FREQUENCY DISTRIBUTION

Rather than drawing a complete frequency distribution graph, researchers often sim-

ply describe a distribution by listing its characteristics. There are three characteristics

that completely describe any distribution: shape, central tendency, and variability. In

simple terms, central tendency measures where the center of the distribution is located.

Variability tells whether the scores are spread over a wide range or are clustered together.

2.4

SECTION 2.4 / THE SHAPE OF A FREQUENCY DISTRIBUTION 5 1

Central tendency and variability will be covered in detail in Chapters 3 and 4. Technically,

the shape of a distribution is defined by an equation that prescribes the exact relationship

between each X and Y value on the graph. However, we rely on a few less-precise terms

that serve to describe the shape of most distributions.

Nearly all distributions can be classified as being either symmetrical or skewed.

In a symmetrical distribution, it is possible to draw a vertical line through

the middle so that one side of the distribution is a mirror image of the other

(Figure 2.10).

In a skewed distribution, the scores tend to pile up toward one end of the scale

and taper off gradually at the other end (see Figure 2.10).

The section where the scores taper off is called the tail of the distribution.

A skewed distribution with the tail on the right-hand side is positively skewed

because the tail points toward the positive (above-zero) end of the X-axis. If the

tail points to the left, the distribution is negatively skewed (see Figure 2.10).

For a very difficult exam, most scores tend to be low, with only a few individuals

earning high scores. This produces a positively skewed distribution. Similarly, a very

easy exam tends to produce a negatively skewed distribution, with most of the students

earning high scores and only a few with low values.

D E F I N I T I O N S

Symmetrical distributions

Skewed distributions

Positive skew Negative skew

FIGURE 2.10

Examples of different

shapes for distributions.

1. Sketch a frequency distribution histogram and a frequency distribution polygon for

the data in the following table:

X f

5 4

4 6

3 3

2 1

1 1

L E A R N I N G C H E C K

5 2 CHAPTER 2 FREQUENCY DISTRIBUTIONS

Exercise 1: histogram Exercise 1: polygon

F re

q u

e n

c y 6

5 4 3 2 1

F re

q u

e n

c y 6

5 4 3 2 1

1 2 3 4 5 6 1 2 3 4 5 6

FIGURE 2.11

Answers to Learning

Check Exercise 1.

2. Describe the shape of the distribution in Exercise 1.

3. A researcher records the gender and academic major for each student at a college

basketball game. If the distribution of majors is shown in a frequency distribution

graph, what type of graph should be used?

4. If the results from a research study are presented in a frequency distribution histo-

gram, would it also be appropriate to show the same results in a polygon? Explain

your answer.

5. A college reports that the youngest registered student is 17 years old, the major-

ity of students are between 18 and 25, and only 10% of the registered students are

older than 30. What is the shape of the distribution of ages for registered students?

1. The graphs are shown in Figure 2.11.

2. The distribution is negatively skewed.

3. A bar graph is used for nominal data.

4. Yes. Histograms and polygons are both used for data from interval or ratio scales.

5. It is positively skewed with most of the distribution around 17–25 and a few scores scattered

at 30 and higher.

ANSWERS

SUMMARY

1. The goal of descriptive statistics is to simplify the organization and presentation of data. One descriptive technique is to place the data in a frequency distribu- tion table or graph that shows exactly how many indi- viduals (or scores) are located in each category on the scale of measurement.

2. A frequency distribution table lists the categories that make up the scale of measurement (the X values) in one column. Beside each X value, in a second column, is the frequency of, or number of individuals in, that category. The table may include a proportion column showing the relative frequency for each category:

proportion 5 5p f

The table may include a percentage column showing the percentage associated with each X value:

percentage (100) (100)5 5p f

3. It is recommended that a frequency distribution table have a maximum of 15 rows to keep it simple. If the scores cover a range that is wider than this suggested maximum, it is customary to divide the range into sections called class intervals. These intervals are then listed in the frequency distribution table along with the frequency, or number of individ- uals with scores in each interval. The result is called a grouped frequency distribution. The guidelines for

RESOURCES

Go to CengageBrain.com to access Psychology CourseMate, where you will find an

interactive eBook, glossaries, flashcards, quizzes, statistics workshops, and more.

RESOURCES 5 3

constructing a grouped frequency distribution table are as follows:

a. There should be about 10 intervals. b. The width of each interval should be a simple num-

ber (e.g., 2, 5, or 10). c. The bottom score in each interval should be a mul-

tiple of the width. d. All intervals should be the same width, and they

should cover the range of scores with no gaps.

4. A frequency distribution graph lists scores on the hori- zontal axis and frequencies on the vertical axis. The type of graph used to display a distribution depends on the scale of measurement used. For interval or ratio scales, you should use a histogram or a polygon. For a histogram, a bar is drawn above each score so that

the height of the bar corresponds to the frequency. Each bar extends to the real limits of the score, so that adjacent bars touch. For a polygon, a dot is placed above the midpoint of each score or class interval so that the height of the dot corresponds to the frequency; then lines are drawn to connect the dots. Bar graphs are used with nominal or ordinal scales. Bar graphs are similar to histograms except that gaps are left between adjacent bars.

5. Shape is one of the basic characteristics used to de- scribe a distribution of scores. Most distributions can be classified as either symmetrical or skewed. A skewed distribution with the tail on the right is said to be positively skewed. If it has the tail on the left, it is negatively skewed.

KEY TERMS

frequency distribution (38)

range (41)

grouped frequency distribution (41)

class interval (41)

apparent limits (44)

axes (44)

histogram (45)

polygon (46)

bar graph (48)

relative frequency (48)

distribution of scores (49)

symmetrical distribution (51)

tail(s) of a distribution (51)

positively skewed distribution (51)

negatively skewed distribution (51)

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General instructions for using SPSS are presented in Appendix D. Following are detailed

instructions for using SPSS to produce Frequency Distribution Tables or Graphs.

5 4 CHAPTER 2 FREQUENCY DISTRIBUTIONS

Frequency Distribution Tables

Data Entry

Enter all the scores in one column of the data editor, probably VAR00001.

Data Analysis

1. Click Analyze on the tool bar, select Descriptive Statistics, and click on

Frequencies.

2. Highlight the column label for the set of scores (VAR00001) in the left box and

click the arrow to move it into the Variable box.

3. Be sure that the option to Display Frequency Table is selected.

4. Click OK.

SPSS Output

The frequency distribution table lists the score values in a column from smallest to largest,

with the percentage and cumulative percentage also listed for each score. Score values that

do not occur (zero frequencies) are not included in the table, and the program does not

group scores into class intervals (all values are listed).

Frequency Distribution Histograms or Bar Graphs

Data Entry

Enter all the scores in one column of the data editor, probably VAR00001.

Data Analysis

1. Click Analyze on the tool bar, select Descriptive Statistics, and click on

Frequencies.

2. Highlight the column label for the set of scores (VAR00001) in the left box and

click the arrow to move it into the Variable box.

3. Click Charts.

4. Select either Bar Graphs or Histogram.

5. Click Continue.

6. Click OK.

SPSS Output

SPSS displays a frequency distribution table and a graph. Note that SPSS often produces a

histogram that groups the scores in unpredictable intervals. A bar graph usually produces a

clearer picture of the actual frequency associated with each score.

FOCUS ON PROBLEM SOLVING

1. The reason for constructing frequency distributions is to put a disorganized set

of raw data into a comprehensible, organized format. Because several different

types of frequency distribution tables and graphs are available, one problem

is deciding which type to use. Tables have the advantage of being easier to

construct, but graphs generally give a better picture of the data and are easier to

understand.

DEMONSTRATION 2.1 5 5

To help you decide which type of frequency distribution is best, consider the

following points:

a. What is the range of scores? With a wide range, you need to group the scores

into class intervals.

b. What is the scale of measurement? With an interval or a ratio scale, you can

use a polygon or a histogram. With a nominal or an ordinal scale, you must

use a bar graph.

2. When using a grouped frequency distribution table, a common mistake is to

calculate the interval width by using the highest and lowest values that define

each interval. For example, some students are tricked into thinking that an

interval identified as 20–24 is only 4 points wide. To determine the correct

interval width, you can:

a. Count the individual scores in the interval. For this example, the scores are 20,

21, 22, 23, and 24, for a total of 5 values. Thus, the interval width is 5 points.

b. Use the real limits to determine the real width of the interval. For example, an

interval identified as 20–24 has a lower real limit of 19.5 and an upper real limit

of 24.5 (halfway to the next score). Using the real limits, the interval width is

24.5 – 19.5 5 5 points

DEMONSTRATION 2.1

A GROUPED FREQUENCY DISTRIBUTION TABLE

For the following set of N 5 20 scores, construct a grouped frequency distribution

table using an interval width of 5 points. The scores are:

14 8 27 16 10 22 9 13 16 12

10 9 15 17 6 14 11 18 14 11

Set up the class intervals.

The largest score in this distribution is X 5 27, and the lowest is X 5 6. Therefore,

a frequency distribution table for these data would have 22 rows and would be too

large. A grouped frequency distribution table would be better. We have asked

specifically for an interval width of 5 points, and the resulting table has five rows.

25–29

20–24

15–19

10–14

5–9

Remember that the interval width is determined by the real limits of the interval.

For example, the class interval 25–29 has an upper real limit of 29.5 and a lower

real limit of 24.5. The difference between these two values is the width of the

interval—namely, 5.

S T E P 1

5 6 CHAPTER 2 FREQUENCY DISTRIBUTIONS

Determine the frequencies for each interval.

Examine the scores, and count how many fall into the class interval of 25–29. Cross out

each score that you have already counted. Record the frequency for this class interval.

Now repeat this process for the remaining intervals. The result is the following table:

X f

25–29 1 (the score X 5 27)

20–24 1 (X 5 22)

15–19 5 (the scores X 5 16, 16, 15, 17, and 18)

10–14 9 (X 5 14, 10, 13, 12, 10, 14, 11, 14, and 11)

5–9 4 (X 5 8, 9, 9, and 6)

S T E P 2

PROBLEMS

1. Place the following set of n 5 20 scores in a frequency distribution table.

6 2 2 1 3 2 4 7 1 2

5 3 1 6 2 6 3 3 7 2

2. Construct a frequency distribution table for the fol- lowing set of scores. Include columns for proportion and percentage in your tables.

Scores: 5 7 8 4 7 9 6 6 5 3

9 6 4 7 7 8 6 7 8 5

3. Find each value requested for the distribution of scores in the following table.

a. n

b. oX

c. oX2

X f

5 2

4 3

3 5

2 1

1 1

4. Find each value requested for the distribution of scores in the following table.

a. n

b. oX

c. oX2

X f

5 1

4 2

3 3

2 5

1 3

5. For the following scores, the smallest value is X 5 17 and the largest value is X 5 53. Place the scores in a grouped frequency distribution table

a. using an interval width of 5 points. b. using an interval width of 10 points.

44 19 23 17 25 47 32 26

25 30 18 24 49 51 24 19

43 27 34 18 52 18 36 25

6. The following scores are the ages for a random sample of n 5 30 drivers who were issued speeding tickets in New York during 2008. Determine the best interval width and place the scores in a grouped fre- quency distribution table. From looking at your table, does it appear that tickets are issued equally across age groups?

17 30 45 20 39 53 28 19

24 21 34 38 22 29 64

22 44 36 16 56 20 23 58

32 25 28 22 51 26 43

7. For each of the following samples, determine the interval width that is most appropriate for a grouped frequency distribution and identify the approximate number of intervals needed to cover the range of scores.

a. Sample scores range from X 5 8 to X 5 41. b. Sample scores range from X 5 16 to X 5 33. c. Sample scores range from X 5 26 to X 5 98.

8. What information can you obtain about the scores in a regular frequency distribution table that is not available from a grouped table?

9. Describe the difference in appearance between a bar graph and a histogram and identify the circumstances in which each type of graph is used.

10. For the following set of scores:

8 5 9 6 8 7 4 10 6 7

9 7 9 9 5 8 8 6 7 10

PROBLEMS 5 7

a. Construct a frequency distribution table to orga- nize the scores.

b. Draw a frequency distribution histogram for these data.

11. Sketch a histogram and a polygon showing the distri- bution of scores presented in the following table:

X f

7 1

6 1

5 3

4 6

3 4

2 1

12. Sketch a histogram showing the distribution of scores shown in the following table:

X f

45–49 4

40–44 6

35–39 10

30–34 5

25–29 3

20–24 2

13. A survey given to a sample of college students con- tained questions about the following variables. For each variable, identify the kind of graph that should be used to display the distribution of scores (histo- gram, polygon, or bar graph).

a. number of brothers and sisters b. birth-order position among siblings (oldest 5 1st) c. gender (male/female) d. favorite television show during the previous year

14. Each year the college gives away T-shirts to new students during freshman orientation. The students are allowed to pick the shirt sizes that they want. To determine how many of each size shirt they should order, college officials look at the distribution from last year. The following table shows the distribution of shirt sizes selected last year.

Size f

S 27

M 48

L 136

XL 120

XXL 39

a. What kind of graph would be appropriate for showing this distribution?

b. Sketch the frequency distribution graph.

15. Gaucher, Friesen, and Kay (2011) found that masculine-themed words (such as competitive, inde- pendent, analyze, strong) are commonly used in job recruitment materials, especially for job advertise- ments in male-dominated areas. In a similar study, a researcher counted the number of masculine-themed words in job advertisements for job areas, and ob- tained the following data.

Area Number of Masculine Words

Plumber 14

Electrician 12

Security guard 17

Bookkeeper 9

Nurse 6

Early-childhood

educator

Determine what kind of graph would be appropri- ate for showing this distribution and sketch the fre- quency distribution graph.

16. Find each of the following values for the distribution shown in the following polygon.

a. n

b. oX

c. oX2

17. For the following set of scores:

Scores: 5 8 5 7 6 6 5 7 4 6

6 9 5 5 4 6 7 5 7 5

a. Place the scores in a frequency distribution table. b. Identify the shape of the distribution.

18. Place the following scores in a frequency distribution table. Based on the frequencies, what is the shape of the distribution?

13 14 12 15 15 14 15 11 13 14

11 13 15 12 14 14 10 14 13 15

19. For the following set of scores:

8 6 7 5 4 10 8 9 5 7 2 9

9 10 7 8 8 7 4 6 3 8 9 6

1 2 3 4 5 6 X

5 8 CHAPTER 2 FREQUENCY DISTRIBUTIONS

a. Construct a frequency distribution table. b. Sketch a histogram showing the distribution. c. Describe the distribution using the following

characteristics: (1) What is the shape of the distribution? (2) What score best identifies the center (average)

for the distribution? (3) Are the scores clustered together, or are they

spread out across the scale?

20. Fowler and Christakis (2008) report that personal happiness tends to be associated with having a social network including many other happy friends. To test this claim, a researcher obtains a sample of n 5 16 adults who claim to be happy people and a similar sample of n 5 16 adults who describe themselves as neutral or unhappy. Each individual is then asked to identify the number of their close friends whom they consider to be happy people. The scores are as follows:

Happy: 8 7 4 10 6 6 8 9 8 8

7 5 6 9 8 9

Unhappy: 5 8 4 6 6 7 9 6 2 8

5 6 4 7 5 6

Sketch a polygon showing the frequency distribu- tion for the happy people. In the same graph, sketch a polygon for the unhappy people. (Use two differ- ent colors, or use a solid line for one polygon and a dashed line for the other.) Does one group seem to have more happy friends?

21. Recent research suggests that the amount of time that parents spend talking about numbers can have a big effect on the mathematical development of their children (Levine, Suriyakham, Rowe, Huttenlocher, & Gunderson, 2010). In the study, the researchers visited the children’s homes between the ages of 14 and 30 months and recorded the amount of “number talk” they heard from the children’s parents. The researchers then tested the children’s knowledge of the meaning of numbers at 46 months. The following data are similar to the results obtained in the study.

Children’s Knowledge-of-Numbers Scores for Two Groups of Parents

Low Number-Talk Parents

High Number-Talk Parents

2, 1, 2, 3, 4 3, 4, 5, 4, 5

3, 3, 2, 2, 1 4, 2, 3, 5, 4

5, 3, 4, 1, 2 5, 3, 4, 5, 4

Sketch a polygon showing the frequency distribution for children with low number-talk parents. In the same graph, sketch a polygon showing the scores for the children with high number-talk parents. (Use two different colors or use a solid line for one polygon and a dashed line for the other.) Does it appear that there is a difference between the two groups?

C H A P T E R

3 Measures of Central Tendency

3.1 Defining Central Tendency

3.2 The Mean

3.3 The Median

3.4 The Mode

3.5 Selecting a Measure of Central Tendency

3.6 Central Tendency and the Shape of the Distribution

Summary

Focus on Problem Solving

Demonstration 3.1

Problems

Aplia for Essentials of Statistics for the Behavioral Sciences

After reading, go to “Resources” at the end of this chapter for

an introduction on how to use Aplia’s homework and learning

resources.

Tools You Will Need The following items are considered essential background material for this chapter. If you doubt your knowledge of any of these items, you should review the appropriate chapter or section before proceeding.

• Summation notation (Chapter 1) • Frequency distributions (Chapter 2)

6 0 CHAPTER 3 MEASURES OF CENTRAL TENDENCY

DEFINING CENTRAL TENDENCY

The general purpose of descriptive statistical methods is to organize and summarize

a set of scores. Perhaps the most common method for summarizing and describing a

distribution is to find a single value that defines the average score and can serve as a

representative for the entire distribution. In statistics, the concept of an average, or rep-

resentative, score is called central tendency. The goal in measuring central tendency

is to describe a distribution of scores by determining a single value that identifies

the center of the distribution. Ideally, this central value is the score that is the best

representative value for all of the individuals in the distribution.

Central tendency is a statistical measure that attempts to determine the single

value, usually located in the center of a distribution, that is most typical or most

representative of the entire set of scores.

In everyday language, central tendency attempts to identify the “average,” or “typi-

cal,” individual. This average value can then be used to provide a simple description of

an entire population or a sample. In addition to describing an entire distribution, mea-

sures of central tendency are also useful for making comparisons between groups of

individuals or between sets of figures. For example, weather data indicate that for Seattle,

Washington, the average yearly temperature is 53° and the average annual precipitation is

34 inches. By comparison, the average temperature in Phoenix, Arizona, is 71° and the

average precipitation is 7.4 inches. The point of these examples is to demonstrate the great

advantage of being able to describe a large set of data with a single, representative number.

Central tendency characterizes what is typical for a large population and, in doing so, makes

large amounts of data more digestible. Statisticians sometimes use the expression number

crunching to illustrate this aspect of data description. That is, we take a distribution consist-

ing of many scores and “crunch” them down to a single value that describes them all.

Unfortunately, there is no single, standard procedure for determining central ten-

dency. The problem is that no single measure produces a central, representative value

in every situation. The three distributions shown in Figure 3.1 should help demonstrate

this fact. Before we discuss the three distributions, take a moment to look at the figure

and try to identify the center or the most representative score for each distribution.

1. The first distribution [Figure 3.1(a)] is symmetrical, with the scores forming

a distinct pile centered around X 5 5. For this type of distribution, it is easy to

identify the center, and most people would agree that the value X 5 5 is an

appropriate measure of central tendency.

2. In the second distribution [Figure 3.1(b)], however, problems begin to appear.

Now the scores form a negatively skewed distribution, piling up at the high

end of the scale around X 5 8, but tapering off to the left all the way down to

X 5 1. Where is the center in this case? Some people might select X 5 8 as

the center because more individuals had this score than any other single value.

However, X 5 8 is clearly not in the middle of the distribution. In fact, the

majority of the scores (10 out of 16) have values less than 8, so it seems

reasonable that the center should be defined by a value that is less than 8.

3. Now consider the third distribution [Figure 3.1(c)]. Again, the distribution is

symmetrical, but now there are two distinct piles of scores. Because the distri-

bution is symmetrical with X 5 5 as the midpoint, you may choose X 5 5 as

the center. However, none of the scores is located at X 5 5 (or even close), so

this value is not particularly good as a representative score. On the other hand,

3.1

D E F I N I T I O N

SECTION 3.2 / THE MEAN 6 1

because there are two separate piles of scores with one group centered at X 5 2

and the other centered at X 5 8, it is tempting to say that this distribution has

two centers. But can one distribution have two centers?

Clearly, there can be problems defining the center of a distribution. Occasionally, you

will find a nice, neat distribution like the one shown in Figure 3.1(a), for which every-

one agrees on the center. But you should realize that other distributions are possible

and that there may be different opinions concerning the definition of the center. To deal

with these problems, statisticians have developed three different methods for measuring

central tendency: the mean, the median, and the mode. They are computed differently

and have different characteristics. To decide which of the three measures is best for

any particular distribution, you should keep in mind that the general purpose of central

tendency is to find the single most representative score. Each of the three measures we

present has been developed to work best in a specific situation. We examine this issue

in more detail after we introduce the three measures.

THE MEAN

The mean, also known as the arithmetic average, is computed by adding all the scores

in the distribution and dividing by the number of scores. The mean for a population is

identified by the Greek letter mu, m (pronounced “mew”), and the mean for a sample is identified by M or

– X (read “x-bar”).

The convention in many statistics textbooks is to use – X to represent the mean for a

sample. However, in manuscripts and in published research reports the letter M is the

standard notation for a sample mean. Because you will encounter the letter M when

reading research reports and because you should use the letter M when writing research

reports, we have decided to use the same notation in this text. Keep in mind that the – X

notation is still appropriate for identifying a sample mean, and you may find it used on

occasion, especially in textbooks.

T H R E E M E AS U R E S O F C E N T R A L T E N D E N CY

3.2

1 2 3 4 5 6 7 8 9 X

(b)

(c)

(a)

FIGURE 3.1

Three distributions demonstrating the

difficulty of defining central tendency.

In each case, try to locate the “center”

of the distribution.

6 2 CHAPTER 3 MEASURES OF CENTRAL TENDENCY

The mean for a distribution is the sum of the scores divided by the number of

scores.

The formula for the population mean is

m 5 X

N (3.1)

First, add all of the scores in the population, and then divide by N. For a sample, the

computation is exactly the same, but the formula for the sample mean uses symbols

(M and n) that signify sample values:

Sample mean 5 5 M X

 (3.2)

In general, we use Greek letters to identify characteristics of a population (param-

eters) and letters of our own alphabet to stand for sample values (statistics). If a mean

is identified with the symbol M, you should realize that we are dealing with a sample.

Also note that the equation for the sample mean uses a lowercase n as the symbol for

the number of scores in the sample.

For a population of N 5 4 scores,

3 7 4 6

the mean is

m 

5 5 5 X

20 4

Although the procedure of adding the scores and dividing by the number of scores pro-

vides a useful definition of the mean, there are two alternative definitions that may give

you a better understanding of this important measure of central tendency.

Dividing the total equally The first alternative is to think of the mean as the amount

each individual receives when the total (oX) is divided equally among all of the in-

dividuals (N) in the distribution. This somewhat socialistic viewpoint is particularly

useful in problems for which you know the mean and must find the total. Consider the

following example.

A group of n 5 6 boys buys a box of baseball cards at a garage sale and discovers that the box contains a total of 180 cards. If the boys divide the cards equally among

themselves, how many cards will each boy get? You should recognize that this problem

represents the standard procedure for computing the mean. Specifically, the total (oX)

is divided by the number (n) to produce the mean, 180

6 5 30 cards for each boy.

The previous example demonstrates that it is possible to define the mean as the

amount that each individual gets when the total is distributed equally. This new definition

can be useful for some problems involving the mean. Consider the following example.

D E F I N I T I O N

E X A M P L E 3 . 1

A LT E R N AT I V E D E F I N I T I O N S

F O R T H E M E A N

E X A M P L E 3 . 2

SECTION 3.2 / THE MEAN 6 3

Now suppose that the 6 boys from Example 3.2 decide to sell their baseball cards on

eBay. If they make an average of M 5 $5 per boy, what is the total amount of money

for the whole group? Although you do not know exactly how much money each boy

has, the new definition of the mean tells you that if they pool their money together and

then distribute the total equally, each boy will get $5. For each of n 5 6 boys to get

$5, the total must be 6($5) 5 $30. To check this answer, use the formula for the mean:

M X

n 5 5 5

 $ $

6 5

The mean as a balance point The second alternative definition of the mean describes

the mean as a balance point for the distribution. Consider a population consisting of

N 5 5 scores (1, 2, 6, 6, 10). For this population, oX 5 25 and m 5 5 25

5 5 . Figure 3.2

shows this population drawn as a histogram, with each score represented as a box that

is sitting on a seesaw. If the seesaw is positioned so that it pivots at a point equal to the

mean, then it will be balanced and will rest level.

The reason that the seesaw is balanced over the mean becomes clear when we mea-

sure the distance of each box (score) from the mean:

Score Distance from the Mean

X 5 1 4 points below the mean

X 5 2 3 points below the mean

X 5 6 1 point above the mean

X 5 10 5 points above the mean

Notice that the mean balances the distances. That is, the total distance below the

mean is the same as the total distance above the mean:

below the mean: 4 1 3 5 7 points

above the mean: 1 1 1 1 5 5 7 points

Because the mean serves as a balance point, the value of the mean is always located

somewhere between the highest score and the lowest score; that is, the mean can never

be outside the range of scores. If the lowest score in a distribution is X 5 8 and the

E X A M P L E 3 . 3

1 32 4 65 7 98 10

�

FIGURE 3.2

The frequency distribution

shown as a seesaw balanced

at the mean.

Based on Weinberg, G. A., Schumaker, J. A., & Oltman, D. (1981). Statistics: An Intuitive Approach (p. 14). Belmont, CA: Wadsworth.

6 4 CHAPTER 3 MEASURES OF CENTRAL TENDENCY

highest is X 5 15, then the mean must be between 8 and 15. If you calculate a value

that is outside this range, then you have made an error.

The image of a seesaw with the mean at the balance point is also useful for deter-

mining how a distribution is affected if a new score is added or if an existing score is

removed. For the distribution in Figure 3.2, for example, what would happen to the

mean (balance point) if a new score were added at X 5 10?

Often it is necessary to combine two sets of scores and then find the overall mean for

the combined group. Suppose that we begin with two separate samples. The first sample

has n 5 12 scores and a mean of M 5 6. The second sample has n 5 8 and M 5 7. If

the two samples are combined, what is the mean for the total group?

To calculate the overall mean, we need two values:

1. the overall sum of the scores for the combined group (oX), and

2. the total number of scores in the combined group (n).

The total number of scores in the combined group can be found easily by adding the

number of scores in the first sample (n 1 ) and the number in the second sample (n

2 ). In

this case, there are 12 1 8 5 20 scores in the combined group. Similarly, the overall

sum for the combined group can be found by adding the sum for the first sample (oX 1 )

and the sum for the second sample (oX 2 ). With these two values, we can compute the

mean using the basic equation

overall mean (overall sum for the combi

5 5M X nned group)

(total number in the combined group)n

5 1 

X X

n n

1 2

To find the sum of the scores for each sample, remember that the mean can be de-

fined as the amount each person receives when the total (oX) is distributed equally. The

first sample has n 5 12 and M 5 6. (Expressed in dollars instead of scores, this sample

has n 5 12 people and each person gets $6 when the total is divided equally.) For each

of 12 people to get M 5 6, the total must be oX 5 12 3 6 5 72. In the same way, the

second sample has n 5 8 and M 5 7 so the total must be oX 5 8 3 7 5 56. Using

these values, we obtain an overall mean of

overall mean 5 5 1

5 1

5M n n

 X X 1 2

1 2

72 56

12 8

1288

20 6 45 .

The following table summarizes the calculations.

First Sample Second Sample Combined Sample

n 5 12 n 5 8 n 5 20 (12 1 8)

oX 5 72 oX 5 56 oX 5 128 (72 1 56)

M 5 6 M 5 7 M 5 6.4

Note that the overall mean is not halfway between the original two sample means.

Because the samples are not the same size, one makes a larger contribution to the total

group and, therefore, carries more weight in determining the overall mean. For this rea-

son, the overall mean we have calculated is called the weighted mean. In this example,

the overall mean of M 5 6.4 is closer to the value of M 5 6 (the larger sample) than it

is to M 5 7 (the smaller sample).

T H E W E I G H T E D M E A N

SECTION 3.2 / THE MEAN 6 5

When a set of scores has been organized in a frequency distribution table, the calculation of

the mean is usually easier if you first remove the individual scores from the table. Table 3.1

shows a distribution of scores organized in a frequency distribution table. To compute the

mean for this distribution you must be careful to use both the X values in the first column

and the frequencies in the second column. The values in the table show that the distribution

consists of one 10, two 9s, four 8s, and one 6, for a total of n 5 8 scores. Remember that

you can determine the number of scores by adding the frequencies, n 5 of. To find the sum

of the scores, you must be careful to add all eight scores:

CO M P U T I N G T H E M E A N F R O M A F R E Q U E N CY D I ST R I B U T I O N TA B L E

1. Find the mean for the following sample of n 5 5 scores: 1, 8, 7, 5, 9.

2. A sample of n 5 6 scores has a mean of M 5 8. What is the value of oX for this

sample?

3. One sample has n 5 5 scores with a mean of M 5 4. A second sample has n 5 3

scores with a mean of M 5 10. If the two samples are combined, what is the mean

for the combined sample?

4. A sample of n 5 6 scores has a mean of M 5 40. One new score is added to the

sample and the new mean is found to be M 5 35. What can you conclude about

the value of the new score?

a. It must be greater 40.

b. It must be less than 40.

5. Find the values for n, oX, and M for the sample that is summarized in the follow-

ing frequency distribution table.

X f

5 1

4 2

3 3

2 5

1 1

L E A R N I N G C H E C K

Quiz Score (X) f fX

10 1 10

9 2 18

8 4 32

7 0 0

6 1 6

TABLE 3.1

Statistics quiz scores for a sam-

ple of n 5 8 students.

oX 5 10 1 9 1 9 1 8 1 8 1 8 1 8 1 6 5 66

Note that you can also find the sum of the scores by computing ofX as we demon-

strated in Chapter 2 (pp. 39–40). Once you have found oX and n, you compute the mean

as usual. For these data,

M X

n 5 5 5

 66

8 8 25.

6 6 CHAPTER 3 MEASURES OF CENTRAL TENDENCY

The mean has many characteristics that will be important in future discussions. In

general, these characteristics result from the fact that every score in the distribution

contributes to the value of the mean. Specifically, every score adds to the total (oX)

and every score contributes one point to the number of scores (n). These two values

(oX and n) determine the value of the mean. Also note that any factor that influences a

population mean will have exactly the same influence on a sample mean. Therefore, the

characteristics of the sample mean are the same as the characteristics of the population

mean. We now discuss four of the more important characteristics of the mean.

Changing a score Changing the value of any score changes the mean. For example,

a sample of quiz scores for a psychology lab section consists of 9, 8, 7, 5, and 1. Note

that the sample consists of n 5 5 scores with oX 5 30. The mean for this sample is

M X

n 5 5 5

 30

5 6 00.

Now suppose that the score of X 5 1 is changed to X 5 8. Note that we have added

7 points to this individual’s score, which also adds 7 points to the total (oX). After

changing the score, the new distribution consists of

9 8 7 5 8

There are still n 5 5 scores, but now the total is oX 5 37. Thus, the new mean is

M X

n 5 5 5

 37

5 7 40.

Notice that changing a single score in the sample has produced a new mean. You

should recognize that changing any score also changes the value of oX (the sum of the

scores), and, thus, always changes the value of the mean.

Introducing a new score or removing a score Adding a new score to a distribution,

or removing an existing score, usually changes the mean. The exception is when the

new score (or the removed score) is exactly equal to the mean. It is easy to visualize

the effect of adding or removing a score if you remember that the mean is defined as the

balance point for the distribution. Figure 3.3 shows a distribution of scores represented

as boxes on a seesaw that is balanced at the mean, µ 5 7. Imagine what would happen if

we added a new score (a new box) at X 5 10. Clearly, the seesaw would tip to the right

and we would need to move the pivot point (the mean) to the right to restore balance.

Now imagine what would happen if we removed the score (the box) at X 5 9. This

time the seesaw would tip to the left and, once again, we would need to change the

mean to restore balance.

Finally, consider what would happen if we added a new score of X 5 7, exactly

equal to the mean. It should be clear that the seesaw would not tilt in either direction,

so the mean would stay in exactly the same place. Also note that if we removed the new

C H A R AC T E R I ST I C S O F T H E M E A N

1. oX 5 30 and M 5 6

2. oX 5 48

3. The combined sample has n 5 8 scores that total oX 5 50. The mean is M 5 6.25.

4. b

5. For this sample n 5 12, oX 5 33, and M 5 33/12 5 2.75.

ANSWERS

SECTION 3.2 / THE MEAN 6 7

score at X 5 7, the seesaw would remain balanced and the mean would not change. In

general, adding a new score or removing an existing score causes the mean to change

unless that score is located exactly at the mean.

The following example demonstrates exactly how the new mean is computed when

a new score is added to an existing sample.

Adding a score (or removing a score) has the same effect on the mean whether the

original set of scores is a sample or a population. To demonstrate the calculation of

the new mean, we will use the set of scores that is shown in Figure 3.3. This time,

however, we will treat the scores as a sample with n 5 5 and M 5 7. Note that this

sample must have oX 5 35. What happens to the mean if a new score of X 5 13 is

added to the sample?

To find the new sample mean, we must determine how the values for n and oX are

changed by a new score. We begin with the original sample and then consider the effect

of adding the new score. The original sample had n 5 5 scores, so adding one new score

produces n 5 6. Similarly, the original sample had oX 5 35. Adding a score of X 5 13

increases the sum by 13 points, producing a new sum of oX 5 35 1 13 5 48. Finally,

the new mean is computed using the new values for n and oX.

M X

n 5 5 5

 48

6 8

The entire process can be summarized as follows:

Original Sample New Sample,

Adding X 5 13

n 5 5 n 5 6

X 5 35 X 5 48

M 5 35

5 5 7 M 5 48

6 5 8

Adding or subtracting a constant from each score If a constant value is added to

every score in a distribution, the same constant is added to the mean. Similarly, if you

subtract a constant from every score, the same constant is subtracted from the mean.

To demonstrate this characteristic of the mean, we use a study that confirms what

you already suspected to be true—alcohol consumption increases the attractiveness

of opposite-sex individuals (Jones, Jones, Thomas, & Piper, 2003). Participants in

the study were shown photographs of male and female faces and asked to rate the

attractiveness of each face. Table 3.2 shows results for a sample of n 5 6 male par-

ticipants who are rating a specific female face. The first column shows the ratings

E X A M P L E 3 . 4

2 3 54 6 7 8 9 10 11 12

FIGURE 3.3

A distribution of N 5 5 scores that is

balanced with a mean of µ 5 7.

6 8 CHAPTER 3 MEASURES OF CENTRAL TENDENCY

when the participants are sober. Note that the total for this column is oX 5 17 for a

sample of n 5 6 participants, so the mean is M 5 17

6 5 2.83. Now suppose that the

effect of alcohol is to add a constant amount (2 points) to each individual’s rating

score. The resulting scores, after moderate alcohol consumption, are shown in the

second column of the table. For these scores, the total is oX 5 29, so the mean is

M 5 29

6 5 4.83. Adding 2 points to each rating score has also added 2 points to the

mean, from M 5 2.83 to M 5 4.83. (It is important to note that treatment effects are

usually not as simple as adding or subtracting a constant amount. Nonetheless, the

concept of adding a constant to every score is important and will be addressed in later

chapters when we are using statistics to evaluate mean differences.)

Multiplying or dividing each score by a constant If every score in a distribution

is multiplied by (or divided by) a constant value, the mean changes in the same way.

Multiplying (or dividing) each score by a constant value is a common method for

changing the unit of measurement. To change a set of measurements from minutes

to seconds, for example, you multiply by 60; to change from inches to feet, you divide

by 12. One common task for researchers is converting measurements into metric units to

conform to international standards. For example, publication guidelines of the American

Psychological Association call for metric equivalents to be reported in parentheses when

most nonmetric units are used. Table 3.3 shows how a sample of n 5 5 scores measured in

inches would be transformed to a set of scores measured in centimeters. (Note that 1 inch

equals 2.54 centimeters.) The first column shows the original scores that total oX 5 50,

with M 5 10 inches. In the second column, each of the original scores has been multiplied

by 2.54 (to convert from inches to centimeters) and the resulting values total oX 5 127,

with M 5 25.4. Multiplying each score by 2.54 has also caused the mean to be multiplied

by 2.54. You should realize, however, that although the numerical values for the individual

scores and the sample mean have changed, the actual measurements have not changed.

TABLE 3.2

Attractiveness ratings of a

female face for a sample of

n 5 6 males.

Participant Sober Moderate Alcohol

A 4 6

B 2 4

C 3 5

D 3 5

E 2 4

F 3 5

oX 5 17 oX 5 29

M 5 2.83 M 5 4.83

TABLE 3.3

Measurements converted from

inches to centimeters.

Original Measurement in Inches

Conversion to Centimeters

(Multiply by 2.54)

10 25.40

9 22.86

12 30.48

8 20.32

11 27.94

oX 5 50 oX 5 127.00

M 510 M 5 25.40

SECTION 3.3 / THE MEDIAN 6 9

THE MEDIAN

The second measure of central tendency we consider is called the median. The goal

of the median is to locate the midpoint of the distribution. Unlike the mean, there are

no specific symbols or notation to identify the median. Instead, the median is simply

identified by the word median. In addition, the definition and the computations for the

median are identical for a sample and for a population.

If the scores in a distribution are listed in order from smallest to largest, the median

is the midpoint of the list. More specifically, the median is the point on the mea-

surement scale below which 50% of the scores in the distribution are located.

Defining the median as the midpoint of a distribution means that the scores are divided

into two equal-sized groups. We are not locating the midpoint between the highest and

lowest X values. To find the median, list the scores in order from smallest to largest.

Begin with the smallest score and count the scores as you move up the list. The median

is the first point you reach that is greater than 50% of the scores in the distribution.

The median can be equal to a score in the list or it can be a point between two scores.

Notice that the median is not algebraically defined (there is no equation for computing

the median), which means that there is a degree of subjectivity in determining the exact

value. However, the following two examples demonstrate the process of finding the

median for most distributions.

This example demonstrates the calculation of the median when n is an odd number.

With an odd number of scores, you list the scores in order (lowest to highest), and the

median is the middle score in the list. Consider the following set of N 5 5 scores, which

have been listed in order:

3 5 8 10 11

3.3

D E F I N I T I O N

F I N D I N G T H E M E D I A N F O R M O ST

D I ST R I B U T I O N S

E X A M P L E 3 . 5

1. Adding a new score to a distribution always changes the mean. (True or false?)

2. Changing the value of a score in a distribution always changes the mean. (True or

false?)

3. A population has a mean of µ 5 40.

a. If 5 points were added to every score, what would be the value for the new mean?

b. If every score were multiplied by 3, what would be the value for the new mean?

4. A sample of n 5 4 scores has a mean of 9. If one person with a score of X 5 3 is

removed from the sample, what is the value for the new sample mean?

1. False. If the score is equal to the mean, it does not change the mean.

2. True.

3. a. The new mean would be 45.

b. The new mean would be 120.

4. The original sample has n 5 4 and oX 5 36. The new sample has n 5 3 scores that total

oX 5 33. The new mean is M 5 11.

L E A R N I N G C H E C K

ANSWERS

7 0 CHAPTER 3 MEASURES OF CENTRAL TENDENCY

The middle score is X 5 8, so the median is equal to 8. Using the counting method,

with N 5 5 scores, the 50% point would be 2 1

2 scores. Starting with the smallest scores,

we must count the 3, the 5, and the 8 before we reach the target of at least 50%. Again,

for this distribution, the median is the middle score, X 5 8.

This example demonstrates the calculation of the median when n is an even number.

With an even number of scores in the distribution, you list the scores in order (lowest

to highest) and then locate the median by finding the average of the middle two scores.

Consider the following population:

1 1 4 5 7 8

Now we select the middle pair of scores (4 and 5), add them together, and divide

by 2:

median 5 1

5 5 4 5

2 4 5.

Using the counting procedure, with N 5 6 scores, the 50% point is 3 scores. Starting

with the smallest scores, we must count the first 1, the second 1, and the 4 before we

reach the target of at least 50%. Again, the median for this distribution is 4.5, which is

the first point on the scale beyond X 5 4. For this distribution, exactly 3 scores (50%)

are located below 4.5. Note: If there is a gap between the middle two scores, the conven-

tion is to define the median as the midpoint between the two scores. For example, if the

middle two scores are X 5 4 and X 5 6, the median would be defined as 5.

The simple technique of listing and counting scores is sufficient to determine the

median for most distributions and is always appropriate for discrete variables. Notice

that this technique always produces a median that either is a whole number or is half-

way between two whole numbers. With a continuous variable, however, it is possible to

divide a distribution precisely in half so that exactly 50% of the distribution is located

below (and above) a specific point. The procedure for locating the precise median is

discussed in the following section.

Recall from Chapter 1 that a continuous variable consists of categories that can be split

into an infinite number of fractional parts. For example, time can be measured in seconds,

tenths of a second, hundredths of a second, and so on. When the scores in a distribution are

measurements of a continuous variable, it is possible to split one of the categories into frac-

tional parts and find the median by locating the precise point that separates the bottom 50%

of the distribution from the top 50%. The following example demonstrates this process.

For this example, we will find the precise median for the following sample of n 5 8

scores: 1, 2, 3, 4, 4, 4, 4, 6

The frequency distribution for this sample is shown in Figure 3.4(a). With an even

number of scores, you normally would compute the average of the middle two scores to

find the median. This process produces a median of X 5 4. For a discrete variable, X 5 4

is the correct value for the median. Recall from Chapter 1 that a discrete variable consists

of indivisible categories, such as the number of children in a family. Some families have

4 children and some have 5, but none have 4.31 children. For a discrete variable, the

category X 5 4 cannot be divided and the whole number 4 is the median.

E X A M P L E 3 . 6

F I N D I N G T H E P R E C I S E M E D I A N F O R A

CO N T I N U O U S VA R I A B L E

E X A M P L E 3 . 7

SECTION 3.3 / THE MEDIAN 7 1

However, if you look at the distribution histogram, the value X 5 4 does not appear

to be the exact midpoint. The problem comes from the tendency to interpret a score of X

5 4 as meaning exactly 4.00. However, if the scores are measurements of a continuous

variable, then the score X 5 4 actually corresponds to an interval from 3.5 to 4.5, and

the median corresponds to a point within this interval.

To find the precise median, we first observe that the distribution contains n 5 8

scores represented by 8 boxes in the graph. The median is the point that has exactly

4 boxes (50%) on each side. Starting at the left-hand side and moving up the scale of

measurement, we accumulate a total of 3 boxes when we reach a value of 3.5 on the

X-axis [see Figure 3.4(a)]. What is needed is 1 more box to reach the goal of 4 boxes

(50%). The problem is that the next interval contains four boxes. The solution is to

take a fraction of each box so that the fractions combine to give you one box. For this

example, if we take 1

4 of each box, the four quarters will combine to make one whole

box. This solution is shown in Figure 3.4(b). The fraction is determined by the number

of boxes needed to reach 50% and the number of boxes in the interval.

fraction number needed to reach 50%

number i 5

nn the interval

For this example, we needed 1 out of the 4 boxes in the interval, so the fraction

is 1 4

. To obtain one-fourth of each box, the median is the point that is located exactly

one-fourth of the way into the interval. The interval for X 5 4 extends from 3.5 to 4.5.

The interval width is 1 point, so one-fourth of the interval corresponds to 0.25 points.

Starting at the bottom of the interval and moving up 0.25 points produces a value of

3.50 1 0.25 5 3.75. This is the median, with exactly 50% of the distribution (4 boxes)

on each side.

Remember, finding the precise midpoint by dividing scores into fractional parts is

sensible for a continuous variable, however; it is not appropriate for a discrete variable.

For example, a median time of 3.75 seconds is reasonable, but a median family size of

3.75 children is not.

1 2

3 4 5 6 7 X

1 4

3 4

Median = 3.75

F re

q u

e n

c y

1 2

(a) (b)

3 4 5 6 7 X

FIGURE 3.4

A distribution with several scores clustered at the median. The median for this distribution is positioned

so that each of the four boxes above X 5 4 is divided into two sections, with 1

4 of each box below the

median (to the left) and 3 4 of each box above the median (to the right). As a result, there are exactly four

boxes, 50% of the distribution, on each side of the median.

7 2 CHAPTER 3 MEASURES OF CENTRAL TENDENCY

Earlier, we defined the mean as the “balance point” for a distribution because the

distances above the mean must have the same total as the distances below the mean.

You should notice, however, that the concept of a balance point focuses on distances

rather than scores. In particular, it is possible to have a distribution in which the vast

majority of the scores are located on one side of the mean. Figure 3.5 shows a dis-

tribution of N 5 6 scores in which 5 out of 6 scores have values less than the mean.

In this figure, the total of the distances above the mean is 8 points and the total of

the distances below the mean is 8 points. Thus, the mean is located in the middle

of the distribution if you use the concept of distance to define the middle. However,

you should realize that the mean is not necessarily located at the exact center of the

group of scores.

The median, on the other hand, defines the middle of the distribution in terms

of scores. In particular, the median is located so that half of the scores are on one

side and half are on the other side. For the distribution in Figure 3.5, for example,

the median is located at X 5 2.5, with exactly 3 scores above this value and

exactly 3 scores below. Thus, it is possible to claim that the median is located in the

middle of the distribution, provided that the term middle is defined by the number

of scores.

In summary, the mean and the median are both methods for defining and measur-

ing central tendency. Although they both define the middle of the distribution, they use

different definitions of the term middle.

THE MEDIAN, THE MEAN, AND THE MIDDLE

1 2

F re

q u

e n

c y 3

3 4

µ 5 4

5 6 7 8 9 10 11 12 13 X

Median = 2.5FIGURE 3.5

A population of N 5 6 scores with a

mean of µ 5 4. Notice that the mean

does not necessarily divide the scores

into two equal groups. In this example,

5 out of the 6 scores have values less

than the mean.

1. Find the median for each distribution of scores:

a. 3, 4, 6, 7, 9, 10, 11

b. 8, 10, 11, 12, 14, 15

2. If you have a score of 52 on an 80-point exam, then you definitely scored above

the median. (True or false?)

3. The following is a distribution of measurements for a continuous variable. Find the

precise median that divides the distribution exactly in half.

Scores: 1, 2, 2, 3, 4, 4, 4, 4, 4, 5

1. a. The median is X 5 7. b. The median is X 5 11.5.

2. False. The value of the median would depend on where all of the scores are located.

3. The median is 3.70 (one-fifth of the way into the interval from 3.5 to 4.5).

L E A R N I N G C H E C K

ANSWERS

SECTION 3.4 / THE MODE 7 3

THE MODE

The final measure of central tendency that we consider is called the mode. In its com-

mon usage, the word mode means “the customary fashion” or “a popular style.” The

statistical definition is similar in that the mode is the most common observation among

a group of scores.

In a frequency distribution, the mode is the score or category that has the greatest

frequency.

As with the median, there are no symbols or special notation used to identify the

mode or to differentiate between a sample mode and a population mode. In addition,

the definition of the mode is the same for a population and for a sample distribution.

The mode is a useful measure of central tendency because it can be used to deter-

mine the typical or average value for any scale of measurement, including a nominal

scale (see Chapter 1). Consider, for example, the data shown in Table 3.4. These data

were obtained by asking a sample of 100 students to name their favorite restaurants in

town. The result is a sample of n 5 100 scores with each score corresponding to the

restaurant that the student named.

For these data, the mode is Luigi’s, the restaurant (score) that was named most fre-

quently as a favorite place. Although we can identify a modal response for these data, you

3.4

D E F I N I T I O N

TABLE 3.4

Favorite restaurants named by a

sample of n 5 100 students.

Caution: The mode is a score

or category, not a frequency.

For this example, the mode is

Luigi’s, not f 5 42.

Restaurant f

College Grill 5

George & Harry’s 16

Luigi’s 42

Oasis Diner 18

Roxbury Inn 7

Sutter’s Mill 12

should notice that it would be impossible to compute a mean or a median. For example,

you cannot add the scores to determine a mean (How much is 5 College Grills plus 42

Luigi’s?). Also, it is impossible to list the scores in order because the restaurants do not

form any natural order. For example, the College Grill is not “more than” or “less than”

the Oasis Diner, they are simply two different restaurants. Thus, it is impossible to obtain

the median by finding the midpoint of the list. In general, the mode is the only measure of

central tendency that can be used with data from a nominal scale of measurement.

The mode also can be useful because it is the only measure of central tendency that

corresponds to an actual score in the data; by definition, the mode is the most frequently

occurring score. The mean and the median, on the other hand, are both calculated val-

ues and often produce an answer that does not equal any score in the distribution. For

example, in Figure 3.5 on the previous page we presented a distribution with a mean of

4 and a median of 2.5. Note that none of the scores is equal to 4 and none of the scores

is equal to 2.5. However, the mode for this distribution is X 5 2 and there are three

individuals who actually have scores of X 5 2.

In a frequency distribution graph, the greatest frequency appears as the tallest part

of the figure. To find the mode, you simply identify the score located directly beneath

the highest point in the distribution.

7 4 CHAPTER 3 MEASURES OF CENTRAL TENDENCY

Although a distribution has only one mean and only one median, it is possible

to have more than one mode. Specifically, it is possible to have two or more scores

that have the same highest frequency. In a frequency distribution graph, the different

modes correspond to distinct, equally high peaks. A distribution with two modes is

said to be bimodal, and a distribution with more than two modes is called multimodal.

Occasionally, a distribution with several equally high points is said to have no mode.

Incidentally, a bimodal distribution is often an indication that two separate and dis-

tinct groups of individuals exist within the same population (or sample). For example,

if you measured height for each person in a set of 100 college students, the resulting

distribution would probably have two modes, one corresponding primarily to the males

in the group and one corresponding primarily to the females.

Technically, the mode is the score with the absolute highest frequency. However, the

term mode is often used more casually to refer to scores with relatively high frequencies—

that is, scores that correspond to peaks in a distribution even though the peaks are not the

absolute highest points. For example, Athos, et al. (2007) asked people to identify the

pitch for both pure tones and piano tones. Participants were presented with a series of

tones and had to name the note corresponding to each tone. Nearly half the participants

(44%) had extraordinary pitch-naming ability (absolute pitch), and were able to identify

most of the tones correctly. Most of the other participants performed around chance level,

apparently guessing the pitch names randomly. Figure 3.6 shows a distribution of scores

that is consistent with the results of the study. There are two distinct peaks in the distribu-

tion, one located at X 5 2 (chance performance) and the other located at X 5 10 (perfect

performance). Each of these values is a mode in the distribution. Note, however, that the

two modes do not have identical frequencies. Eight people scored at X 5 2 and only seven

had scores of X 5 10. Nonetheless, both of these points are called modes. When two

modes have unequal frequencies, researchers occasionally differentiate the two values by

calling the taller peak the major mode, and the shorter one the minor mode.

SELECTING A MEASURE OF CENTRAL TENDENCY

Deciding which measure of central tendency is best to use depends on several factors.

Before we discuss these factors, however, note that the mean is usually the preferred

measure of central tendency whenever the data consist of numerical scores. Because

the mean uses every score in the distribution, it typically produces a good representa-

tive value. Remember that the goal of central tendency is to find the single value that

best represents the entire distribution. Besides being a good representative, the mean

has the added advantage of being closely related to variance and standard deviation,

3.5

Tone Identification Score (number correct out of 10)

F re

q u

e n

c y

1 2 3 4

5 6 7 8 9

1 2 3 4 6 7 8 90

FIGURE 3.6

A frequency distribution for tone

identification scores. An example of a

bimodal distribution.

SECTION 3.5 / SELECTING A MEASURE OF CENTRAL TENDENCY 7 5

the most common measures of variability (see Chapter 4). This relationship makes the

mean a valuable measure for purposes of inferential statistics. For these reasons, and

others, the mean generally is considered to be the best of the three measures of central

tendency. But there are specific situations in which it is impossible to compute a mean

or in which the mean is not particularly representative. It is in these situations that the

mode and the median are used.

We consider four situations in which the median serves as a valuable alternative to the

mean. In the first three cases, the data consist of numerical values (interval or ratio

scales) for which you would normally compute the mean. However, each case also

involves a special problem so that either it is impossible to compute the mean, or the

calculation of the mean produces a value that is not central or not representative. The

fourth situation involves measuring central tendency for ordinal data.

Extreme scores or skewed distributions When a distribution has a few extreme

scores, scores that are very different in value from most of the others, then the mean

may not be a good representative of the majority of the distribution. The problem comes

from the fact that one or two extreme values can have a large influence and cause the

mean to be displaced. In this situation, the fact that the mean uses all of the scores

equally can be a disadvantage. Consider, for example, the distribution of n 5 10 scores

in Figure 3.7. For this sample, the mean is

M X

n 5 5 5

 203

10 20 3.

Notice that the mean is not very representative of any score in this distribution.

Although most of the scores are clustered between 10 and 13, the extreme score of

X 5 100 inflates the value of oX and distorts the mean.

The median, on the other hand, usually is not affected by extreme scores. For this

sample, n 5 10, so there should be five scores on either side of the median. The median

is 11.50. Notice that this is a very representative value. Also note that the median would

W H E N TO U S E T H E M E D I A N

1. During the month of October, an instructor recorded the number of absences for

each student in a class of n 5 20 and obtained the following distribution.

Number of Absences f

5 1

4 2

3 7

2 5

1 3

0 2

a. Using the mean, what is the average number of absences for the class?

b. Using the median, what is the average number of absences for the class?

c. Using the mode, what is the average number of absences for the class?

a. The mean is 47

20 5 2.35.

b. The median is 2.5.

c. The mode is 3.

L E A R N I N G C H E C K

ANSWERS

7 6 CHAPTER 3 MEASURES OF CENTRAL TENDENCY

be unchanged even if the extreme score were 1000 instead of only 100. Because it is

relatively unaffected by extreme scores, the median commonly is used when reporting

the average value for a skewed distribution. For example, the distribution of personal

incomes is very skewed, with a small segment of the population earning incomes that

are astronomical. These extreme values distort the mean, so that it is not very represen-

tative of the salaries that most of us earn. The median is the preferred measure of central

tendency when extreme scores exist.

Undetermined values Occasionally, you encounter a situation in which an individual

has an unknown or undetermined score. This often occurs when you are measuring the

number of errors or the amount of time required for an individual to complete a task.

For example, suppose that preschool children are asked to assemble a wooden puzzle

as quickly as possible. The experimenter records how long (in minutes) it takes each

child to arrange all of the pieces to complete the puzzle. Table 3.5 presents results for

a sample of n 5 6 children.

Notice that one child never completed the puzzle. After an hour, this child still

showed no sign of solving the puzzle, so the experimenter stopped him or her. This par-

ticipant has an undetermined score. (There are two important points to be noted. First,

the experimenter should not throw out this individual’s score. The whole purpose for

using a sample is to gain a picture of the population, and this child tells us that part of

the population cannot solve the puzzle. Second, this child should not be given a score of

F re

q u

e n

c y

10 1001514131211

FIGURE 3.7

Frequency distribution of er-

rors committed before reach-

ing learning criterion.

Notice that the graph shows

two breaks in the X-axis.

Rather than listing all of the

scores from 0 to 100, the

graph jumps directly to the

first score, which is X 5

10, and then jumps directly

from X 5 15 to X 5 100.

The breaks shown in the

X-axis are the conventional

way of notifying the reader

that some values have been

omitted.

TABLE 3.5

Number of minutes needed to

assemble a wooden puzzle.

Child Time (Min.)

1 8

2 11

3 12

4 13

5 17

6 Never finished

SECTION 3.5 / SELECTING A MEASURE OF CENTRAL TENDENCY 7 7

X 5 60 minutes. Even though the experimenter stopped the individual after 1 hour, the

child did not finish the puzzle. The score that is recorded is the amount of time needed

to finish. For this individual, we do not know how long this is.)

It is impossible to compute the mean for these data because of the undetermined

value. We cannot calculate the oX part of the formula for the mean. However, it is possi-

ble to determine the median. For these data, the median is 12.5. Three scores are below

the median, and three scores (including the undetermined value) are above the median.

Open-ended distributions A distribution is said to be open-ended when there is no

upper limit (or lower limit) for one of the categories. The table in the margin provides

an example of an open-ended distribution, showing the number of pizzas eaten during

a 1-month period for a sample of n 5 20 high school students. The top category in

this distribution shows that three of the students consumed “5 or more” pizzas. This is

an open-ended category. Notice that it is impossible to compute a mean for these data

because you cannot find oX (the total number of pizzas for all 20 students). However,

you can find the median. Listing the 20 scores in order produces X 5 1 and X 5 2 as

the middle two scores. For these data, the median is 1.5.

Ordinal scale Many researchers believe that it is not appropriate to use the mean to

describe central tendency for ordinal data. When scores are measured on an ordinal scale,

the median is always appropriate and is usually the preferred measure of central tendency.

You should recall that ordinal measurements allow you to determine direction (greater

than or less than) but do not allow you to determine distance. The median is compatible

with this type of measurement because it is defined by direction: half of the scores are above

the median and half are below the median. The mean, on the other hand, defines central

tendency in terms of distance. Remember that the mean is the balance point for the distribu-

tion, so that the distances above the mean are exactly balanced by the distances below the

mean. Because the mean is defined in terms of distances, and because ordinal scales do not

measure distance, it is not appropriate to compute a mean for scores from an ordinal scale.

We consider three situations in which the mode is commonly used as an alternative to

the mean, or is used in conjunction with the mean to describe central tendency.

Nominal scales The primary advantage of the mode is that it can be used to measure

and describe central tendency for data that are measured on a nominal scale. Recall that

the categories that make up a nominal scale are differentiated only by name. Because

nominal scales do not measure quantity (distance or direction), it is impossible to com-

pute a mean or a median for data from a nominal scale. Therefore, the mode is the only

option for describing central tendency for nominal data.

Discrete variables Recall that discrete variables are those that exist only in whole,

indivisible categories. Often, discrete variables are numerical values, such as the

number of children in a family or the number of rooms in a house. When these

variables produce numerical scores, it is possible to calculate means. In this situ-

ation, the calculated means are usually fractional values that cannot actually exist.

For example, computing means generates results such as “the average family has

2.4 children and a house with 5.33 rooms.” On the other hand, the mode always

identifies the most typical case and, therefore, it produces more sensible measures

of central tendency. Using the mode, our conclusion would be “the typical, or modal,

family has 2 children and a house with 5 rooms.” In many situations, especially with

discrete variables, people are more comfortable using the realistic, whole-number

values produced by the mode.

W H E N TO U S E T H E M O D E

Number of Pizzas (X) f

5 or more 3

4 2

3 2

2 3

1 6

0 4

7 8 CHAPTER 3 MEASURES OF CENTRAL TENDENCY

Describing shape Because the mode requires little or no calculation, it is often in-

cluded as a supplementary measure along with the mean or median as a no-cost extra.

The value of the mode (or modes) in this situation is that it gives an indication of the

shape of the distribution as well as a measure of central tendency. Remember that the

mode identifies the location of the peak (or peaks) in the frequency distribution graph.

For example, if you are told that a set of exam scores has a mean of 72 and a mode of

80, you should have a better picture of the distribution than would be available from the

mean alone (see Section 3.6).

IN THE LITERATURE REPORTING MEASURES OF CENTRAL TENDENCY

Measures of central tendency are commonly used in the behavioral sciences to sum-

marize and describe the results of a research study. For example, a researcher may

report the sample means from two different treatments or the median score for a large

sample. These values may be reported in verbal descriptions of the results, in tables,

or in graphs.

In reporting results, many behavioral science journals use guidelines adopted by

the American Psychological Association (APA), as outlined in the Publication Manual

of the American Psychological Association (2010). We refer to the APA manual from

time to time in describing how data and research results are reported in the scientific

literature. The APA style uses the letter M as the symbol for the sample mean. Thus, a

study might state:

The treatment group showed fewer errors (M 5 2.56) on the task than the control

group (M 5 11.76).

When there are many means to report, tables with headings provide an organized and

more easily understood presentation. Table 3.6 illustrates this point.

The median can be reported using the abbreviation Mdn, as in “Mdn 5 8.5 errors,”

or it can simply be reported in narrative text, as follows:

The median number of errors for the treatment group was 8.5, compared to a

median of 13 for the control group.

There is no special symbol or convention for reporting the mode. If mentioned at all,

the mode is usually just reported in narrative text.

PRESENTING MEANS AND MEDIANS IN GRAPHS

Graphs also can be used to report and compare measures of central tendency. Usually,

graphs are used to display values obtained for sample means, but occasionally sample

medians are reported in graphs (modes are rarely, if ever, shown in a graph). The value

of a graph is that it allows several means (or medians) to be shown simultaneously, so

it is possible to make quick comparisons between groups or treatment conditions. When

preparing a graph, it is customary to list the different groups or treatment conditions

on the horizontal axis. Typically, these are the different values that make up the

TABLE 3.6

The mean number of errors

made on the task for treatment

and control groups, divided by

gender.

Treatment Control

Females 1.45 8.36

Males 3.83 14.77

SECTION 3.5 / SELECTING A MEASURE OF CENTRAL TENDENCY 7 9

independent variable or the quasi-independent variable. Values for the dependent

variable (the scores) are listed on the vertical axis. The means (or medians) are then

displayed using a line graph, a histogram, or a bar graph, depending on the scale of

measurement used for the independent variable.

Figure 3.8 shows an example of a line graph displaying the relationship be-

tween drug dose (the independent variable) and food consumption (the dependent

variable). In this study, there were five different drug doses (treatment conditions)

and they are listed along the horizontal axis. The five means appear as points in

the graph. To construct this graph, a point was placed above each treatment con-

dition so that the vertical position of the point corresponds to the mean score for

the treatment condition. The points are then connected with straight lines. A line

graph is used when the values on the horizontal axis are measured on an interval or

a ratio scale. An alternative to the line graph is a histogram. For this example, the

histogram would show a bar above each drug dose so that the height of each bar

corresponds to the mean food consumption for that group, with no space between

adjacent bars.

Figure 3.9 shows a bar graph displaying the median selling price for single-family

homes in different regions of the United States. Bar graphs are used to present means

M e

a n

f o

o d

c o

n su

m p

ti o

n 20

1 2 3

Drug dose

FIGURE 3.8

The relationship between

an independent variable

(drug dose) and a

dependent variable (food

consumption). Because

drug dose is a continuous

variable, a continuous line

is used to connect the

different dose levels.

300

250

200

150

100

Northeast

Median new

house cost

(in $1000’s)

South

Region of the United States

Midwest West

FIGURE 3.9

Median cost of a new,

single-family home by

region.

8 0 CHAPTER 3 MEASURES OF CENTRAL TENDENCY

(or medians) when the groups or treatments shown on the horizontal axis are measured

on a nominal or an ordinal scale. To construct a bar graph, you simply draw a bar

directly above each group or treatment so that the height of the bar corresponds to the

mean (or median) for that group or treatment. For a bar graph, a space is left between

adjacent bars to indicate that the scale of measurement is nominal or ordinal.

When constructing graphs of any type, you should recall the basic rules that we

introduced in Chapter 2:

1. The height of a graph should be approximately two-thirds to three-quarters of

its length.

2. Normally, you start numbering both the X-axis and the Y-axis with zero at the

point where the two axes intersect. However, when a value of zero is part of the

data, it is common to move the zero point away from the intersection so that the

graph does not overlap the axes (see Figure 3.8).

Following these rules helps to produce a graph that provides an accurate presentation of

the information in a set of data. Although it is possible to construct graphs that distort

the results of a study (see Box 2.1), researchers have an ethical responsibility to present

an honest and accurate report of their research results.

CENTRAL TENDENCY AND THE SHAPE OF THE DISTRIBUTION

We have identified three different measures of central tendency, and often a researcher

calculates all three for a single set of data. Because the mean, the median, and the mode

are all trying to measure the same thing, it is reasonable to expect that these three val-

ues should be related. In fact, there are some consistent and predictable relationships

among the three measures of central tendency. Specifically, there are situations in which

all three measures have exactly the same value. On the other hand, there are situations

in which the three measures are guaranteed to be different. In part, the relationships

among the mean, median, and mode are determined by the shape of the distribution.

We consider two general types of distributions.

For a symmetrical distribution, the right-hand side of the graph is a mirror image of

the left-hand side. If a distribution is perfectly symmetrical, the median is exactly

at the center because exactly half of the area in the graph is on either side of the

center. The mean also is exactly at the center of a perfectly symmetrical distribution

because each score on the left side of the distribution is balanced by a correspond-

ing score (the mirror image) on the right side. As a result, the mean (the balance

point) is located at the center of the distribution. Thus, for a perfectly symmetrical

distribution, the mean and the median are the same (Figure 3.10). If a distribution is

roughly symmetrical, but not perfect, the mean and median are close together in the

center of the distribution.

If a symmetrical distribution has only one mode, then it is also in the center of the

distribution. Thus, for a perfectly symmetrical distribution with one mode, all three

measures of central tendency—the mean, the median, and the mode—have the same

value. For a roughly symmetrical distribution, the three measures are clustered together

in the center of the distribution. On the other hand, a bimodal distribution that is sym-

metrical [see Figure 3.10(b)] has the mean and median together in the center with

the modes on each side. A rectangular distribution [see Figure 3.10(c)] has no mode

3.6

SY M M E T R I CA L D I ST R I B U T I O N S

SECTION 3.6 / CENTRAL TENDENCY AND THE SHAPE OF THE DISTRIBUTION 8 1

because all X values occur with the same frequency. Still, the mean and the median are

in the center of the distribution.

In skewed distributions, especially distributions for continuous variables, there is a

strong tendency for the mean, median, and mode to be located in predictably differ-

ent positions. Figure 3.11(a), for example, shows a positively skewed distribution with

the peak (highest frequency) on the left-hand side. This is the position of the mode.

However, it should be clear that the vertical line drawn at the mode does not divide the

distribution into two equal parts. To have exactly 50% of the distribution on each side,

the median must be located to the right of the mode. Finally, the mean is located to

the right of the median because it is the measure of central tendency that is influenced

most by the extreme scores in the tail and, therefore, tends to be displaced toward the

tail of the distribution. Thus, in a positively skewed distribution, the typical order of the

three measures of central tendency from smallest to largest (left to right) is the mode,

the median, and the mean.

Negatively skewed distributions are lopsided in the opposite direction, with the scores

piling up on the right-hand side and the tail tapering off to the left. The grades on an easy

exam, for example, tend to form a negatively skewed distribution [see Figure 3.11(b)]. For a

distribution with negative skew, the mode is on the right-hand side (with the peak), whereas

the mean is displaced toward the left by the extreme scores in the tail. As before, the median

is usually located between the mean and the mode. Therefore, in a negatively skewed dis-

tribution, the most probable order for the three measures of central tendency from smallest

value to largest value (left to right), is the mean, the median, and the mode.

SKEWED DISTRIBUTIONS

The positions of the mean,

median, and mode are not as

consistently predictable in

distributions of discrete vari-

ables (see Von Hippel, 2005).

Mean Median Mode

Mean Median

Mode Mode Mean Median

No mode

FIGURE 3.10

Measures of central

tendency for three sym-

metrical distributions:

normal, bimodal, and

rectangular.

FIGURE 3.11

Measures of central tendency for skewed distributions.

Mode Median

Mean

Mean Median

Mode

F re

q u

e n

c y

F re

q u

e n

c y

(b)(a)

8 2 CHAPTER 3 MEASURES OF CENTRAL TENDENCY

SUMMARY

1. The goal of central tendency is to determine the single value that identifies the center of the distribution and best represents the entire set of scores. The three stan- dard measures of central tendency are the mode, the median, and the mean.

2. The mean is the arithmetic average. It is computed by adding all of the scores and then dividing by the number of scores. Conceptually, the mean is obtained by dividing the total (oX) equally among the number of individuals (N or n). The mean can also be defined as the balance point for the distribution. The distances above the mean are exactly balanced by the distances below the mean. Although the calculation is the same for a population or a sample mean, a population mean is identified by the symbol m, and a sample mean is identified by M. In most situations with numerical scores from an interval or a ratio scale, the mean is the preferred measure of central tendency.

3. Changing any score in the distribution causes the mean to be changed. When a constant value is added to (or subtracted from) every score in a distribution, the same constant value is added to (or subtracted from) the mean. If every score is multiplied by a constant, the mean is multiplied by the same constant.

4. The median is the midpoint of a distribution of scores. The median is the preferred measure of central ten- dency when a distribution has a few extreme scores that displace the value of the mean. The median also is used for open-ended distributions or when there are undetermined (infinite) scores that make it impos- sible to compute a mean. Finally, the median is the preferred measure of central tendency for data from an ordinal scale.

5. The mode is the most frequently occurring score in a distribution. It is easily located by finding the peak in a frequency distribution graph. For data measured on a nominal scale, the mode is the appropriate measure of central tendency. It is possible for a distribution to have more than one mode.

6. For symmetrical distributions, the mean is equal to the median. If there is only one mode, then it has the same value, too.

7. For skewed distributions, the mode is located toward the side where the scores pile up, and the mean tends to be pulled toward the extreme scores in the tail. The median is usually located between these two values.

1. Which measure of central tendency is most affected if one extremely large score is

added to a distribution? (mean, median, mode)

2. Why is it usually considered inappropriate to compute a mean for scores measured

on an ordinal scale?

3. In a perfectly symmetrical distribution, the mean, the median, and the mode will

all have the same value. (True or false?)

4. A distribution with a mean of 70 and a median of 75 is probably positively

skewed. (True or false?)

1. mean

2. The definition of the mean is based on distances (the mean balances the distances) and

ordinal scales do not measure distance.

3. False, if the distribution is bimodal.

4. False. The mean is displaced toward the tail on the left-hand side.

L E A R N I N G C H E C K

ANSWERS

RESOURCES 83

KEY TERMS

central tendency (60) mode (73) line graph (79)

population mean (m) (62) bimodal (74) symmetrical distribution (80)

sample mean (M) (62) multimodal (74) skewed distribution (81)

weighted mean (64) major mode (74) positively skewed (81)

median (69) minor mode (74) negatively skewed (81)

RESOURCES

Go to CengageBrain.com to access Psychology CourseMate, where you will find an interactive eBook, glossaries, flashcards, quizzes, statistics workshops, and more.

If your professor has assigned Aplia:

1. Sign in to your account.

2. Complete the corresponding exercises as required by your professor.

3. When finished, click “Grade It Now” to see which areas you have mastered, which

areas need more work, and detailed explanations of every answer.

General instructions for using SPSS are presented in Appendix D. Following are

detailed instructions for using SPSS to compute the Mean and oX for a set of scores.

Data Entry

Enter all of the scores in one column of the data editor, probably VAR00001.

Data Analysis

1. Click Analyze on the tool bar, select Descriptive Statistics, and click on

Descriptives.

2. Highlight the column label for the set of scores (VAR00001) in the left box and

click the arrow to move it into the Variable box.

3. If you want oX as well as the mean, click on the Options box, select Sum, then

click Continue.

4. Click OK.

SPSS Output

SPSS produces a summary table listing the number of scores (N), the maximum and

minimum scores, the sum of the scores (if you selected this option), the mean, and

the standard deviation. Note: The standard deviation is a measure of variability that is

presented in Chapter 4.

8 4 CHAPTER 3 MEASURES OF CENTRAL TENDENCY

FOCUS ON PROBLEM SOLVING

1. Although the three measures of central tendency appear to be very simple to

calculate, there is always a chance for errors. The most common sources of

error are listed next.

a. Many students find it very difficult to compute the mean for data presented in

a frequency distribution table. They tend to ignore the frequencies in the table

and simply average the score values listed in the X column. You must use the

frequencies and the scores! Remember that the number of scores is found

by N 5 of, and the sum of all N scores is found by ofX. For the distribution

shown in the margin, the mean is 24

10 5 2.40.

b. The median is the midpoint of the distribution of scores, not the midpoint of

the scale of measurement. For a 100-point test, for example, many students

incorrectly assume that the median must be X 5 50. To find the median, you

must have the complete set of individual scores. The median separates the

individuals into two equal-sized groups.

c. The most common error with the mode is for students to report the highest

frequency in a distribution rather than the score with the highest frequency.

Remember that the purpose of central tendency is to find the most represen-

tative score. For the distribution in the margin, the mode is X 5 3, not f 5 4.

DEMONSTRATION 3.1

COMPUTING MEASURES OF CENTRAL TENDENCY

For the following sample, find the mean, the median, and the mode. The scores are:

5 6 9 11 5 11 8 14 2 11

Compute the mean The calculation of the mean requires two pieces of information;

the sum of the scores, oX; and the number of scores, n. For this sample, n 5 10 and

oX 5 5 1 6 1 9 1 111 5 1 11 1 8 1 14 1 2 1 11 5 82

Therefore, the sample mean is

M X

n .5 5 5

 82

10 8 2

Find the median To find the median, first list the scores in order from smallest to

largest. With an even number of scores, the median is the average of the middle two

scores in the list. Listed in order, the scores are:

2 5 5 6 8 9 11 11 11 14

The middle two scores are 8 and 9, and the median is 8.5.

Find the mode For this sample, X 5 11 is the score that occurs most frequently. The

mode is X 5 11.

X f

4 1

3 4

2 3

1 2

PROBLEMS 85

PROBLEMS

1. Why is it necessary to have more than one method for measuring central tendency?

2. Find the mean, median, and mode for the following sample of scores:

5 4 5 2 7 1 3 5

3. Find the mean, median, and mode for the following sample of scores:

3 6 7 3 9 8 3 7 5

4. Find the mean, median, and mode for the scores in the following frequency distribution table:

X f

6 1

5 2

4 2

3 2

2 2

1 5

5. Find the mean, median, and mode for the scores in the following frequency distribution table:

X f

8 1

7 1

6 2

5 5

4 2

3 2

6. For the following sample:

a. Assume that the scores are measurements of a continuous variable and find the median by locating the precise midpoint of the distribution.

b. Assume that the scores are measurements of a discrete variable and find the median.

Scores: 1 2 3 3 3 4

7. A population of N 5 15 scores has oX 5 120. What is the population mean?

8. A sample of n 5 8 scores has a mean of M 5 12. What is the value of oX for this sample?

9. A population with a mean of m 5 8 has oX 5 40. How many scores are in the population?

10. A sample of n 5 7 scores has a mean of M 5 9. If one new person with a score of X 5 1 is added to the sample, what is the value for the new mean?

11. A sample of n 5 6 scores has a mean of M 5 13. If one person with a score of X 5 3 is removed from the sample, what is the value for the new mean?

12. A sample of n 5 15 scores has a mean of M 5 6. One person with a score of X 5 22 is added to the sample. What is the value for the new sample mean?

13. A sample of n 5 10 scores has a mean of M 5 9. One person with a score of X 5 0 is removed from the sample. What is the value for the new sample mean?

14. A population of N 5 15 scores has a mean of m 5 8. One score in the population is changed from X 5 20 to X 5 5. What is the value for the new population mean?

15. A sample of n 5 7 scores has a mean of M 5 16. One score in the sample is changed from X 5 6 to X 5 20. What is the value for the new sample mean?

16. A sample of n 5 7 scores has a mean of M 5 5. After one new score is added to the sample, the new mean is found to be M 5 6. What is the value of the new score? (Hint: Compare the values for oX before and after the score was added.)

17. A population of N 5 8 scores has a mean of m 5 16. After one score is removed from the population, the new mean is found to be m 5 15. What is the value of the score that was removed? (Hint: Compare the values for oX before and after the score was removed.)

18. A sample of n 5 9 scores has a mean of M 5 13. After one score is added to the sample, the mean is found to be M 5 12. What is the value of the score that was added?

19. A sample of n 5 9 scores has a mean of M 5 20. One of the scores is changed and the new mean is found to be M 5 22. If the changed score was originally X 5 7, what is its new value?

20. One sample of n 5 12 scores has a mean of M 5 7 and a second sample of n 5 8 scores has a mean of M 5 12. If the two samples are combined, what is the mean for the combined sample?

21. One sample has a mean of M 5 8 and a second sample has a mean of M 5 16. The two samples are combined into a single set of scores.

a. What is the mean for the combined set if both of the original samples have n 5 4 scores?

b. What is the mean for the combined set if the first sample has n 5 3 and the second sample has n 5 5?

8 6 CHAPTER 3 MEASURES OF CENTRAL TENDENCY

c. What is the mean for the combined set if the first sample has n 5 5 and the second sample has n 5 3?

22. One sample has a mean of M 5 5 and a second sample has a mean of M 5 10. The two samples are combined into a single set of scores.

a. What is the mean for the combined set if both of the original samples have n 5 5 scores?

b. What is the mean for the combined set if the first sample has n 5 4 scores and the second sample has n 5 6?

c. What is the mean for the combined set if the first sample has n 5 6 scores and the second sample has n 5 4?

23. Explain why the mean is often not a good measure of central tendency for a skewed distribution.

24. A researcher conducts a study comparing two dif- ferent treatments with a sample of n 5 16 partici- pants in each treatment. The study produced the following data:

Treatment 1: 6 7 11 4 19 17 2 5

9 13 6 23 11 4 6 1

Treatment 2: 10 9 6 6 1 11 8 6 3

2 11 1 12 7 10 9

a. Calculate the mean for each treatment. Based on the two means, which treatment produces the higher scores?

b. Calculate the median for each treatment. Based on the two medians, which treatment produces the higher scores?

c. Calculate the mode for each treatment. Based on the two modes, which treatment produces the higher scores?

25. Schmidt (1994) conducted a series of experiments examining the effects of humor on memory. In one study, participants were shown a list of sentences, of which half were humorous and half were non- humorous. A humorous example is, “If at first you don’t succeed, you are probably not related to the boss.” Other participants would see a nonhumorous version of this sentence, such as “People who are related to the boss often succeed the very first time.”

Schmidt then measured the number of each type of sentence recalled by each participant. The follow- ing scores are similar to the results obtained in the study.

Number of Sentences Recalled

Humorous Sentences Nonhumorous Sentences

4 5 2 4 5 2 4 2

6 7 6 6 2 3 1 6

2 5 4 3 3 2 3 3

1 3 5 5 4 1 5 3

Calculate the mean number of sentences recalled for each of the two conditions. Do the data suggest that humor helps memory?

26. Stephens, Atkins, and Kingston (2009) conducted a research study demonstrating that swearing can help reduce pain. In the study, each participant was asked to plunge a hand into icy water and keep it there as long as the pain would allow. In one condi- tion, the participants repeatedly yelled their favorite curse words while their hands were in the water. In the other condition the participants repeated a neu- tral word. Data similar to the results obtained in the study are shown in the following table. Calculate the mean number of seconds that the participants could tolerate the pain for each of the two treatment condi- tions. Does it appear that swearing helped with pain tolerance?

Amount of Time (in seconds)

Participant Swear Words Neutral Words

1 94 59

2 70 61

3 52 47

4 83 60

5 46 35

6 117 92

7 69 53

8 39 30

9 51 56

10 73 61

27. Earlier in this chapter (p. 67), we mentioned a research study demonstrating that alcohol consumption in- creases attractiveness ratings for members of the oppo- site sex (Jones, Jones, Thomas, & Piper, 2003). In the actual study, college-age participants were recruited from bars and restaurants near campus and asked to participate in a “market research” study. During the introductory conversation, they were asked to report their alcohol consumption for the day and were told that moderate consumption would not prevent them from taking part in the study. Participants were then shown a series of photographs of male and female faces and asked to rate the attractiveness of each face

PROBLEMS 87

on a 1–7 scale. The following data duplicate the general pattern of results obtained in the study. The two sets of scores are attractiveness ratings for one female obtained from two groups of males: those who had no alcohol and those with moderate alcohol con- sumption. Calculate the mean for each group. Does it appear from these data that alcohol has an effect on judgments of attractiveness?

Group 1 No Alcohol

Group 2 Moderate Alcohol

3 4 5 1 2 5 3 5 2 4

4 2 3 4 4 6 5 6 5 4

5 6 3 4 3 7 5 6 5 6

C H A P T E R

4 Measures of Variability

4.1 Defining Variability

4.2 The Range

4.3 Standard Deviation and Variance for a Population

4.4 Standard Deviation and Variance for a Sample

4.5 More About Variance and Standard Deviation

Summary

Focus on Problem Solving

Demonstration 4.1

Problems

Aplia for Essentials of Statistics for the Behavioral Sciences

After reading, go to “Resources” at the end of this chapter for

an introduction on how to use Aplia’s homework and learning

resources.

• Summation notation (Chapter 1) • Central tendency (Chapter 3) • Mean • Median

9 0 CHAPTER 4 MEASURES FOR VARIABILITY

DEFINING VARIABILITY

The term variability has much the same meaning in statistics as it has in everyday

language; to say that things are variable means that they are not all the same. In sta-

tistics, our goal is to measure the amount of variability for a particular set of scores:

a distribution. In simple terms, if the scores in a distribution are all the same, then

there is no variability. If there are small differences between scores, then the vari-

ability is small, and if there are large differences between scores, then the variability

is large.

Variability provides a quantitative measure of the differences between scores in

a distribution and describes the degree to which the scores are spread out or clus-

tered together.

Figure 4.1 shows two distributions of familiar values for the population of adult

males: Part (a) shows the distribution of men’s heights (in inches), and part (b) shows

the distribution of men’s weights (in pounds). Notice that the two distributions differ in

terms of central tendency. The mean height is 70 inches (5 feet, 10 inches) and the mean

weight is 170 pounds. In addition, notice that the distributions differ in terms of vari-

ability. For example, most heights are clustered close together, within 5 or 6 inches of

the mean. On the other hand, weights are spread over a much wider range. In the weight

distribution it is not unusual to find individuals who are located more than 30 pounds

away from the mean, and it would not be surprising to find two individuals whose

weights differ by more than 30 or 40 pounds. The purpose for measuring variability

is to obtain an objective measure of how the scores are spread out in a distribution. In

general, a good measure of variability serves two purposes:

1. Variability describes the distribution. Specifically, it tells whether the scores

are clustered close together or are spread out over a large distance. Usually,

variability is defined in terms of distance. It tells how much distance to expect

between one score and another, or how much distance to expect between an

individual score and the mean. For example, we know that the heights for most

adult males are clustered close together, within 5 or 6 inches of the average.

Although more extreme heights exist, they are relatively rare.

4.1

D E F I N I T I O N

FIGURE 4.1

Population distribution of adult heights and adult weights.

X 76 82706458

Adult weights (in pounds)

200 230170140110 Adult heights

(in inches)

(a) (b)

SECTION 4.2 / THE RANGE 9 1

2. Variability measures how well an individual score (or group of scores) rep-

resents the entire distribution. This aspect of variability is very important for

inferential statistics, in which relatively small samples are used to answer ques-

tions about populations. For example, suppose that you selected one person to

represent the entire population. Because most adult males have heights that are

within a few inches of the population average (the distances are small), there

is a very good chance that you would select someone whose height is within

6 inches of the population mean. On the other hand, the scores are much more

spread out (greater distances) in the distribution of weights. In this case, you

probably would not obtain someone whose weight was within 6 pounds of the

population mean. Thus, variability provides information about how much error

to expect if you are using a sample to represent a population.

In this chapter, we consider three different measures of variability: the range, stan-

dard deviation, and the variance. Of these three, the standard deviation and the related

measure of variance are by far the most important.

THE RANGE

The range is the distance covered by the scores in a distribution, from the smallest score

to the largest score. When the scores are measurements of a continuous variable, the

range can be defined as the difference between the upper real limit (URL) for the largest

score (Xmax) and the lower real limit (LRL) for the smallest score (Xmin).

range 5 URL for Xmax 2 LRL for Xmin

If the scores have values from 1 to 5, for example, the range is 5.5 2 0.5 5 5 points. When the scores are whole numbers, this definition of the range is also a measure of the

number of measurement categories. If every individual is classified as either 1, 2, 3, 4,

or 5, then there are five measurement categories and the range is 5 points.

Defining the range as the number of measurement categories also works for discrete

variables that are measured with numerical scores. For example, if you are measuring

the number of children in a family and the data produce values from 0 to 4, then there

are five measurement categories (0, 1, 2, 3, and 4) and the range is 5 points. By this

definition, when the scores are all whole numbers, the range can be obtained by

X max

2 X min

1 1.

A commonly used alternative definition of the range simply measures the distance

between the largest score (X max

) and the smallest score (X min

), without any reference to

real limits.

range 5 Xmax 2 Xmax

By this definition, scores having values from 1 to 5 cover a range of only 4 points.

Many computer programs, such as SPSS, use this definition. For discrete variables,

which do not have real limits, this definition is often considered more appropriate. Also,

this definition works well for variables with precisely defined upper and lower boundar-

ies. For example, if you are measuring proportions of an object, like pieces of a pizza,

you can obtain values such as 1 8

, 1 4

, 1 2 , 3

4 , and so on. Expressed as decimal values, the

proportions range from 0 to 1. You can never have a value less than 0 (none of the pizza)

and you can never have a value greater than 1 (all of the pizza). Thus, the complete set

4.2

Continuous and discrete

variables were discussed in

Chapter 1 on pages 20-22.

9 2 CHAPTER 4 MEASURES FOR VARIABILITY

of proportions is bounded by 0 at one end and by 1 at the other. As a result, the propor-

tions cover a range of 1 point.

Using either definition, the range is probably the most obvious way to describe how

spread out the scores are—simply find the distance between the maximum and the

minimum scores. The problem with using the range as a measure of variability is that it

is completely determined by the two extreme values and ignores the other scores in the

distribution. Thus, a distribution with one unusually large (or small) score has a large

range even if the other scores are all clustered close together.

Because the range does not consider all of the scores in the distribution, it often does not

give an accurate description of the variability for the entire distribution. For this reason, the

range is considered to be a crude and unreliable measure of variability. Therefore, in most

situations, it does not matter which definition you use to determine the range.

STANDARD DEVIATION AND VARIANCE FOR A POPULATION

The standard deviation is the most commonly used and the most important measure

of variability. Standard deviation uses the mean of the distribution as a reference point

and measures variability by considering the distance between each score and the mean.

In simple terms, the standard deviation provides a measure of the standard, or aver-

age, distance from the mean, and describes whether the scores are clustered closely

around the mean or are widely scattered. The fundamental definition of the standard

deviation is the same for both samples and populations, but the calculations differ

slightly. We look first at the standard deviation as it is computed for a population, and

then turn our attention to samples in Section 4.4.

Although the concept of standard deviation is straightforward, the actual equations

appear complex. Therefore, we begin by looking at the logic that leads to these equa-

tions. If you remember that our goal is to measure the standard, or typical, distance from

the mean, then this logic and the equations that follow should be easier to remember.

Step 1 The first step in finding the standard distance from the mean is to determine

the deviation, or distance from the mean, for each individual score. By definition, the

deviation for each score is the difference between the score and the mean.

Deviation is distance from the mean:

deviation score 5 X 2 m

For a distribution of scores with m 5 50, if your score is X 5 53, then your deviation score is

X 2 m 5 53 2 50 5 3

If your score is X 5 45, then your deviation score is

X 2 m 5 45 2 50 5 25

Notice that there are two parts to a deviation score: the sign (1 or 2) and the num- ber. The sign tells the direction from the mean—that is, whether the score is located

above (1) or below (2) the mean. The number gives the actual distance from the mean.

4.3

T H E P R O C E S S O F CO M P U T I N G STA N DA R D

D E V I AT I O N

D E F I N I T I O N

A deviation score is often

represented by a lowercase

letter x.

SECTION 4.3 / STANDARD DEVIATION AND VARIANCE FOR A POPULATION 9 3

For example, a deviation score of 26 corresponds to a score that is below the mean by

a distance of 6 points.

Step 2 Because our goal is to compute a measure of the standard distance from the

mean, the obvious next step is to calculate the mean of the deviation scores. To compute

this mean, you first add up the deviation scores and then divide by N. This process is

demonstrated in the following example.

We start with the following set of N 5 4 scores. These scores add up to oX 5 12, so the mean is m 5 12

4 5 3. For each score, we have computed the deviation.

X X 2 m

8 15

1 22

3 0

0 23

0 5 o(X 2 m)

Note that the deviation scores add up to zero. This should not be surprising if you

remember that the mean serves as a balance point for the distribution. The total of the

distances above the mean is exactly equal to the total of the distances below the mean

(see page 63). Thus, the total for the positive deviations is exactly equal to the total

for the negative deviations, and the complete set of deviations always adds up to zero.

Because the sum of the deviations is always zero, the mean of the deviations is also

zero and is of no value as a measure of variability. Specifically, it is zero if the scores

are closely clustered and it is zero if the scores are widely scattered. You should note,

however, that the constant value of zero can be useful in other ways. Whenever you are

working with deviation scores, you can check your calculations by making sure that the

deviation scores add up to zero.

Step 3 The average of the deviation scores does not work as a measure of variability

because it is always zero. Clearly, this problem results from the positive and negative values

canceling each other out. The solution is to get rid of the signs (1 and 2). The standard procedure for accomplishing this is to square each deviation score. Using the squared

values, you then compute the mean squared deviation, which is called population variance.

Population variance equals the mean squared deviation. Variance is the average

squared distance from the mean.

Note that the process of squaring deviation scores does more than simply get rid

of plus and minus signs. It results in a measure of variability based on squared dis-

tances. Although variance is valuable for some of the inferential statistical methods

covered later, the concept of squared distance is not an intuitive or easy to understand

descriptive measure. For example, it is not particularly useful to know that the squared

distance from New York City to Boston is 26,244 miles squared. The squared value

becomes meaningful, however, if you take the square root. For example, the distance

from New York City to Boston is 26,244 5 162 miles. Therefore, we continue the process with one more step.

E X A M P L E 4 . 1

D E F I N I T I O N

9 4 CHAPTER 4 MEASURES FOR VARIABILITY

Step 4 Remember that our goal is to compute a measure of the standard distance

from the mean. Variance, which measures the average squared distance from the mean,

is not exactly what we want. The final step simply takes the square root of the variance

to obtain the standard deviation, which measures the standard distance from the mean.

Standard deviation is the square root of the variance and provides a measure of

the standard, or average, distance from the mean.

Standard deviation variance=

Figure 4.2 shows the overall process of computing variance and standard deviation.

Remember that our goal is to measure variability by finding the standard distance from

the mean. However, we cannot simply calculate the average of the distances because

this value will always be zero. Therefore, we begin by squaring each distance, then we

find the average of the squared distances, and finally we take the square root to obtain

a measure of the standard distance. Technically, the standard deviation is the square

root of the average squared deviation. Conceptually, however, the standard deviation

provides a measure of the average distance from the mean.

Because the standard deviation and variance are defined in terms of distance from the

mean, these measures of variability are used only with numerical scores that are obtained

from measurements on an interval or a ratio scale. Recall from Chapter 1 (page 24) that

these two scales are the only ones that provide information about distance; nominal and

ordinal scales do not. Also, recall from Chapter 3 (page 77) that it is inappropriate to

compute the mean for ordinal data and impossible to compute the mean for nominal data.

Because the mean is a critical component in the calculation of standard deviation and

D E F I N I T I O N

Square each deviation

DEAD END This value is always zero.

Take the square root of the variance. This is the standard deviation (the standard distance

from the mean).

Find the average of the squared deviations.

This is the variance.

Find the deviation (distance from the mean)

for each score.

Add the deviations and compute the average.

FIGURE 4.2

The calculation of variance

and standard deviation.

SECTION 4.3 / STANDARD DEVIATION AND VARIANCE FOR A POPULATION 9 5

variance, the same restrictions that apply to the mean also apply to these two measures of

variability. Specifically, the mean, the standard deviation, and the variance should be used

only with numerical scores from interval or ordinal scales of measurement.

Although we still have not presented any formulas for variance or standard devia-

tion, you should be able to compute these two statistical values from their definitions.

The following example demonstrates this process.

We will calculate the variance and standard deviation for the following population of

N 5 5 scores:

1 9 5 8 7

Remember that the purpose of standard deviation is to measure the standard distance

from the mean, so we begin by computing the population mean. These five scores add up

to oX 5 30 so the mean is m 5 30 5

5 6. Next, we find the deviation (distance from the mean) for each score and then square the deviations. Using the population mean m 5 6, these calculations are shown in the following table.

Score X Deviation

X 2 m

Squared Deviation (X 2 m)2

1 25 25

9 3 9

5 21 1

8 2 4

7 1 1

40 5 the sum of the squared deviations

For this set of N 5 5 scores, the squared deviations add up to 40. The mean of the squared deviations, the variance, is

40 5

5 8, and the standard deviation is 8 5 2.83.

You should note that a standard deviation of 2.83 is a sensible answer for this distri-

bution. The five scores in the population are shown in a histogram in Figure 4.3 so that

you can see the distances more clearly. Note that the scores closest to the mean are only

1 point away. Also, the score farthest from the mean is 5 points away. For this distribution,

E X A M P L E 4 . 2

81 2 4 53 6 9 107 X

µ = 6

F re

q u

e n

c y

5 1

FIGURE 4.3

A frequency distribution

histogram for a population

of N 5 5 scores. The mean for this population is µ 5 6. The smallest distance from

the mean is 1 point, and the

largest distance is 5 points.

The standard distance (or

standard deviation) should be

between 1 and 5 points.

9 6 CHAPTER 4 MEASURES FOR VARIABILITY

the largest distance from the mean is 5 points and the smallest distance is 1 point. Thus,

the standard distance should be somewhere between 1 and 5. By looking at a distribution

in this way, you should be able to make a rough estimate of the standard deviation. In this

case, the standard deviation should be between 1 and 5, probably around 3 points. The

value we calculated for the standard deviation is in excellent agreement with this estimate.

Making a quick estimate of the standard deviation can help you avoid errors in calcula-

tion. For example, if you calculated the standard deviation for the scores in Figure 4.3 and

obtained a value of 12, you should realize immediately that you have made an error. If the

biggest deviation is only 5 points, then it is impossible for the standard deviation to be 12.

1. Briefly explain what is measured by the standard deviation and what is measured

by the variance.

2. The deviation scores are calculated for each individual in a population of N 5 4. The first three individuals have deviations of 12, 14, and 21. What is the deviation for the fourth individual?

3. What is the standard deviation for the following set of N 5 5 scores: 10, 10, 10, 10, and 10? (Note: You should be able to answer this question directly from the

definition of standard deviation, without doing any calculations.)

4. Calculate the variance for the following population of N 5 5 scores: 4, 0, 7, 1, 3.

1. Standard deviation measures the standard distance from the mean, and variance measures

the average squared distance from the mean.

2. The deviation scores for the entire set must add up to zero. The first three deviations add to

15 so the fourth deviation must be 25.

3. Because there is no variability (the scores are all the same), the standard deviation is zero.

4. For these scores, the sum of the squared deviations is 30 and the variance is 30 5

5 6.

L E A R N I N G C H E C K

ANSWERS

The concepts of standard deviation and variance are the same for both samples and

populations. However, the details of the calculations differ slightly, depending on

whether you have data from a sample or from a complete population. We first consider

the formulas for populations and then look at samples in Section 4.4.

Recall that variance is defined as the mean of the squared deviations. This mean is

computed in exactly the same way you compute any mean: First find the sum, and then

divide by the number of scores.

variance mean squared deviation sum of squa

5 5 rred deviations

number of scores

The sum of squared deviations (SS) The value in the numerator of this equation, the

sum of the squared deviations, is a basic component of variability, and we focus on it.

To simplify things, it is identified by the notation SS (for sum of squared deviations),

and it generally is referred to as the sum of squares.

SS, or sum of squares, is the sum of the squared deviation scores.

You need to know two formulas to compute SS. These formulas are algebraically

equivalent (they always produce the same answer), but they look different and are used

in different situations.

F O R M U L AS F O R P O P U L AT I O N VA R I A N C E

A N D STA N DA R D D E V I AT I O N

D E F I N I T I O N

SECTION 4.3 / STANDARD DEVIATION AND VARIANCE FOR A POPULATION 9 7

The Definitional Formula for SS The first of these formulas is called the defini-

tional formula because the symbols in the formula literally define the process of adding

up the squared deviations:

Definitional Formula: SS 5 o(X – m)2 (4.1)

To find the sum of the squared deviations, the formula instructs you to perform the

following sequence of calculations:

1. Find each deviation score (X 2 m).

2. Square each deviation score (X 2 m)2.

3. Add the squared deviations.

The result is SS, the sum of the squared deviations. Note that this is the process we

used (without a formula) to compute the sum of the squared deviations in Example 4.2.

The following example demonstrates the formula.

We compute SS for the following set of N 5 4 scores. These scores have a sum of oX 5 8, so the mean is m 5

4 5 2. The following table shows the deviation and the

squared deviation for each score. The sum of the squared deviation is SS 5 22.

Score X Deviation

X 2 m

Squared Deviation (X 2 m)2

1 21 1 oX 5 8

0 22 4 m 5 2

6 14 16 o(X 2 m)2 5 22

1 21 1

The Computational Formula for SS Although the definitional formula is the

most direct method for computing SS, it can be awkward to use. In particular, when

the mean is not a whole number, the deviations all contain decimals or fractions,

and the calculations become difficult. In addition, calculations with decimal values

introduce the opportunity for rounding error, which can make the result less accu-

rate. For these reasons, an alternative formula has been developed for computing SS.

The alternative, known as the computational formula, performs calculations with the

scores (not the deviations) and therefore minimizes the complications of decimals

and fractions.

Computational formula: SS X X

N =

( ) 

 2

2 (4.2)

The first part of this formula directs you to square each score and then add the

squared values, oX2. In the second part of the formula, you find the sum of the scores, oX, then square this total and divide the result by N. Finally, subtract the second part from the first. The use of this formula is shown in Example 4.4 with the same scores

that we used to demonstrate the definitional formula.

E X A M P L E 4 . 3

9 8 CHAPTER 4 MEASURES FOR VARIABILITY

The computational formula can be used to calculate SS for the same set of N 5 4 scores we used in Example 4.3. Note that the formula requires the calculation of two sums:

first, compute oX, and then square each score and compute oX2. These calculations are shown in the following table. The two sums are used in the formula to compute SS.

X X2

1 1

0 0

6 36

1 1

SS X X

N 5 2

5 2

 

38 8

38 64

38 16

( )

oX 5 8 oX2 5 38

Note that the two formulas produce exactly the same value for SS. Although the for-

mulas look different, they are in fact equivalent. The definitional formula provides the

most direct representation of the concept of SS; however, this formula can be awkward

to use, especially if the mean includes a fraction or decimal value. If you have a small

group of scores and the mean is a whole number, then the definitional formula is fine;

otherwise the computational formula is usually easier to use.

With the definition and calculation of SS behind you, the equations for variance and

standard deviation become relatively simple. Remember that variance is defined as the

mean squared deviation. The mean is the sum of the squared deviations divided by N,

so the equation for the population variance is

variance 5 SS

Standard deviation is the square root of variance, so the equation for the population

standard deviation is

standard deviation 5 SS

There is one final bit of notation before we work completely through an example

computing SS, variance, and standard deviation. Like the mean (m), variance and standard deviation are parameters of a population and are identified by Greek letters. To identify the

standard deviation, we use the Greek letter sigma (the Greek letter s, standing for standard

deviation). The capital letter sigma (o) has been used already, so we now use the lowercase sigma, s, as the symbol for the population standard deviation. To emphasize the relation- ship between standard deviation and variance, we use s2 as the symbol for population variance (standard deviation is the square root of the variance). Thus,

population standard deviation 5 s 5 s 5 2 SS

N (4.3)

population variance 5 s 5 2 SS

N (4.4)

E X A M P L E 4 . 4

F I N A L F O R M U L AS A N D N OTAT I O N

In the same way that sum

of squares, or SS, is used to

refer to the sum of squared

deviations, the term mean

square, or MS, is often used

to refer to variance, which is

the mean squared deviation.

SECTION 4.4 / STANDARD DEVIATION AND VARIANCE FOR A SAMPLE 9 9

Earlier, in Examples 4.3 and 4.4, we computed the sum of squared deviations for a popu-

lation of N 5 4 scores (1, 0, 6, 1) and obtained SS 5 22. For this population, the variance is

s 5 5 5 2 22

4 5 50

N .

and the standard deviation is s 5 55 50 2 345. .

1. Find the sum of the squared deviations, SS, for each of the following populations.

Note that the definitional formula works well for one population but the computa-

tional formula is better for the other.

Population 1: 3 1 5 1

Population 2: 6 4 2 0 9 3

2. a. Sketch a histogram showing the frequency distribution for the following popu-

lation of N 5 6 scores: 12, 0, 1, 7, 4, 6. Locate the mean in your sketch, and estimate the value of the standard deviation.

b. Calculate SS, variance, and the standard deviation for these scores. How well

does your estimate compare with the actual standard deviation?

1. For population 1, the mean is not a whole number (M 5 2.5) and the computational formula is better and produces SS 5 11. The mean is a whole number (M 5 4) and definitional formula works well for population 2, which has SS 5 50.

2. a. Your sketch should show a mean of m 5 5. The scores closest to the mean are X 5 4 and X 5 6, both of which are only 1 point away. The score farthest from the mean is X 5 12, which is 7 points away. The standard deviation should have a value between 1 and 7, prob-

ably around 4 points.

b. For these scores, SS 5 96, the variance is 96

6 5 16, and the standard deviation is s 5 4.

L E A R N I N G C H E C K

ANSWERS

STANDARD DEVIATION AND VARIANCE FOR A SAMPLE

The goal of inferential statistics is to use the limited information from samples to

draw general conclusions about populations. The basic assumption of this process

is that samples should be representative of the populations from which they come.

This assumption poses a special problem for variability because samples consistently

tend to be less variable than their populations. An example of this general tendency

is shown in Figure 4.4. Notice that a few extreme scores in the population tend to

make the population variability relatively large. However, these extreme values are

unlikely to be obtained when you are selecting a sample, which means that the sample

variability is relatively small. The fact that a sample tends to be less variable than its

population means that sample variability gives a biased estimate of population vari-

ability. This bias is in the direction of underestimating the population value rather

than being right on the mark. (The concept of a biased statistic is discussed in more

detail in Section 4.5.)

Fortunately, the bias in sample variability is consistent and predictable, which

means it can be corrected. For example, if the speedometer in your car consistently

shows speeds that are 5 mph slower than you are actually going, it does not mean that

4.4

1 0 0 CHAPTER 4 MEASURES FOR VARIABILITY

the speedometer is useless. It simply means that you must make an adjustment to the

speedometer reading to get an accurate speed. In the same way, we make an adjust-

ment in the calculation of sample variance. The purpose of the adjustment is to make

the resulting value for sample variance an accurate and unbiased representative of the

population variance.

The calculations of variance and standard deviation for a sample follow the same

steps that were used to find population variance and standard deviation. First, calculate

the sum of squared deviations (SS). Second, calculate the variance. Third, find the

square root of the variance, which is the standard deviation.

The sum of squared deviations for a sample Except for minor changes in notation,

calculating the sum of the squared deviations, SS, is the same for a sample as it is for

a population. The changes in notation involve using M for the sample mean instead of

m, and using n (instead of N) for the number of scores. For example, the definitional formula for SS for a sample is

Definitional formula: SS X M= ( ) 2 2 (4.5)

Note that the sample formula has exactly the same structure as the population for-

mula (Equation 4.1 on p. 97) and instructs you to find the sum of the squared deviations

using the following sequence of three steps:

1. Find the deviation from the mean for each score: deviation 5 X 2 M

2. Square each deviation: squared deviation 5 (X 2 M)2

3. Add the squared deviations: SS 5 o(X 2 M)2

Population variability

Population distribution

SampleX X X X XX XX

Sample variability

X X

FIGURE 4.4

The population of adult heights forms a

normal distribution. If you select a sample

from this population, you are most likely

to obtain individuals who are near aver-

age in height. As a result, the scores in the

sample will be less variable (spread out)

than the scores in the population.

A sample statistic is said to

be biased if, on average, it

consistently overestimates or

underestimates the

corresponding population

parameter.

SECTION 4.4 / STANDARD DEVIATION AND VARIANCE FOR A SAMPLE 1 0 1

The value of SS also can be obtained using a computational formula. Except for one

minor difference in notation (using n in place of N), the computational formula for SS

is the same for a sample as it was for a population (see Equation 4.2). Using sample

notation, this formula is:

Computational formula: SS X X

n =

( ) 

 2

2 (4.6)

Formulas for sample variance and sample standard deviation Again, calcu-

lating SS for a sample is exactly the same as for a population, except for minor

changes in notation. After you compute SS, however, it becomes critical to differen-

tiate between samples and populations. To correct for the bias in sample variability,

it is necessary to make an adjustment in the formulas for sample variance and

standard deviation. With this in mind, sample variance (identified by the symbol s2)

is defined as

sample variance 5 5 2

s SS

1 (4.7)

Sample standard deviation (identified by the symbol s) is simply the square root of

the variance.

sample standard deviation 5 5 5 2

s s SS

1 (4.8)

Notice that the sample formulas divide by n 2 1, unlike the population formulas, which divide by N (see Equations 4.3 and 4.4). This is the adjustment that is necessary

to correct for the bias in sample variability. The effect of the adjustment is to increase

the value that you obtain. Dividing by a smaller number (n 2 1 instead of n) produces a larger result and makes sample variance an accurate and unbiased estimator of popu-

lation variance. The following example demonstrates the calculation of variance and

standard deviation for a sample.

We have selected a sample of n 5 8 scores from a population. The scores are 4, 6, 5, 11, 7, 9, 7, 3. The frequency distribution histogram for this sample is shown in

Figure 4.5. Before we begin any calculations, you should be able to look at the sample

distribution and make a preliminary estimate of the outcome. Remember that standard

deviation measures the standard distance from the mean. For this sample the mean is

M 5 52

8 5 6.5. The scores closest to the mean are X 5 6 and X 5 7, both of which

are exactly 0.5 points away. The score farthest from the mean is X 5 2, which is 4.5 points away. With the smallest distance from the mean equal to 0.5 and the larg-

est distance equal to 4.5, we should obtain a standard distance somewhere between

0.5 and 4.5, probably around 2.5.

We begin the calculations by finding the value of SS for this sample. Because

the mean is not a whole number (M 5 6.5), the computational formula is easier to use. The scores, and the squared scores, needed for this formula are shown in the

following table.

E X A M P L E 4 . 5

Remember, sample variabil-

ity tends to underestimate

population variability unless

some correction is made.

1 0 2 CHAPTER 4 MEASURES FOR VARIABILITY

Scores X Squared Scores X2

4 16

6 36

5 25

11 121

7 49

9 81

7 49

3 9

oX 5 52 oX2 5 386

Using the two sums,

SS X X

n 5 2 5 2

5 2

 

2 2

386 52

386 338

( ) ( )

The sum of squared deviations for this sample is SS 5 48. Continuing the calculations,

sample variance 5 5 2

5 2

5s SS

8 1 6 86.

Finally, the standard deviation is

s 5 s 2 5 6 86. 5 2.62

Note that the value we obtained is in excellent agreement with our preliminary pre-

diction (see Figure 4.5).

Remember that the formulas for sample variance and standard deviation were con-

structed so that the sample variability would provide a good estimate of population

variability. For this reason, the sample variance is often called estimated population

1 2 3 4 5 6 7 8 9 10 11 X

= 6.5M

4.5

3 1

FIGURE 4.5

The frequency distribution

histogram for a sample

of n 5 8 scores. The sample mean is M 5 6.5. The smallest distance

from the mean is

0.5 points, and the largest

distance from the mean is

4.5 points. The standard

distance (standard devia-

tion) should be between

0.5 and 4.5 points, or

about 2.5.

SECTION 4.4 / STANDARD DEVIATION AND VARIANCE FOR A SAMPLE 1 0 3

variance, and the sample standard deviation is called estimated population standard

deviation. When you have only a sample to work with, the variance and standard de-

viation for the sample provide the best possible estimates of the population variability.

Although the concept of a deviation score and the calculation of SS are almost exactly

the same for samples and populations, the minor differences in notation are really very

important. Specifically, with a population, you find the deviation for each score by mea-

suring its distance from the population mean, m. With a sample, on the other hand, the value of m is unknown and you must measure distances from the sample mean. Because the value of the sample mean varies from one sample to another, you must first compute

the sample mean before you can begin to compute deviations. However, calculating

the value of M places a restriction on the variability of the scores in the sample. This

restriction is demonstrated in the following example.

Suppose we select a sample of n 5 3 scores and compute a mean of M 5 5. The first two scores in the sample have no restrictions; they are independent of each other and

they can have any values. For this demonstration, we assume that we obtained X 5 2 for the first score and X 5 9 for the second. At this point, however, the third score in the sample is restricted.

X A sample of n 5 3 scores with a mean of M 5 5.

— ← What is the third score?

For this example, the third score must be X 5 4. The reason that the third score is restricted to X 5 4 is that the sample has a mean of M 5 5. For n 5 3 scores to have a mean of 5, the scores must have a total of oX 5 15. Because the first two scores add up to 11 (9 1 2), the third score must be X 5 4.

In Example 4.6, the first two out of three scores were free to have any values, but

the final score was dependent on the values chosen for the first two. In general, with a

sample of n scores, the first n 2 1 scores are free to vary, but the final score is restricted. As a result, the sample is said to have n 2 1 degrees of freedom.

For a sample of n scores, the degrees of freedom, or df, for the sample variance

are defined as df 5 n 2 1. The degrees of freedom determine the number of scores in the sample that are independent and free to vary.

The n 2 1 degrees of freedom for a sample is the same n 2 1 that is used in the for- mulas for sample variance and standard deviation. Remember that variance is defined

as the mean squared deviation. As always, this mean is computed by finding the sum

and dividing by the number of scores:

mean sum

number 5

To calculate sample variance (mean squared deviation), we find the sum of the

squared deviations (SS) and divide by the number of scores that are free to vary. This

number is n 2 1 5 df. Thus, the formula for sample variance is

SA M P L E VA R I A B I L I T Y A N D D E G R E E S O F

F R E E D O M

E X A M P L E 4 . 6

D E F I N I T I O N

1 0 4 CHAPTER 4 MEASURES FOR VARIABILITY

s 2

5 sum of squared deviations

number of scoress free to vary 5 5

n 1

Later in this book, we use the concept of degrees of freedom in other situations. For

now, remember that knowing the sample mean places a restriction on sample variability.

Only n 2 1 of the scores are free to vary; df 5 n 2 1.

1. a. Sketch a histogram showing the frequency distribution for the following sample

of n 5 5 scores: 3, 1, 9, 4, 3. Locate the mean in your sketch, and estimate the value of the sample standard deviation.

b. Calculate SS, variance, and standard deviation for this sample. How well does

your estimate from part a compare with the real standard deviation?

2. For the following set of scores: 1, 5, 7, 3, 4

a. Assume that this is a population of N 5 5 scores and compute SS and variance for the population.

b. Assume that this is a sample of n 5 5 scores and compute SS and variance for the sample.

3. Explain why the formula for sample variance divides SS by n 2 1 instead of dividing by n.

1. a. Your graph should show a sample mean of M 5 4. The score farthest from the mean is X 5 9 (which is 5 points away), and the closest score is X 5 3 (which is 1 point away). You should estimate the standard deviation to be between 1 and 5 points, probably

around 3 points.

b. For this sample, SS 5 36; the sample variance is 36

4 5 9; the sample standard deviation is

9 5 3.

2. a. SS 5 20 and the population variance is 20

5 5 4.

b. SS 5 20 and the sample variance is 20

4 5 5.

3. Without some correction, sample variability consistently underestimates the population vari-

ability. Dividing by a smaller number (n 2 1 instead of n) increases the value of the sample variance and makes it an unbiased estimate of the population variance.

L E A R N I N G C H E C K

ANSWERS

MORE ABOUT VARIANCE AND STANDARD DEVIATION

In frequency distribution graphs, we identify the position of the mean by drawing a ver-

tical line and labeling it with m or M. Because the standard deviation measures distance from the mean, it is represented by a line or an arrow drawn from the mean outward

for a distance equal to the standard deviation and labeled with a s or an s. Figure 4.6(a) shows an example of a population distribution with a mean of m 5 80 and a standard deviation of s 5 8, and Figure 4.6(b) shows the frequency distribution for a sample with a mean of M 5 16 and a standard deviation of s 5 2. For rough sketches, you can identify the mean with a vertical line in the middle of the distribution. The standard

deviation line should extend approximately halfway from the mean to the most extreme

score. [Note: In Figure 4.6(a) we show the standard deviation as a line to the right of the

4.5

P R E S E N T I N G T H E M E A N A N D

STA N DA R D D E V I AT I O N I N A F R E Q U E N CY

D I ST R I B U T I O N G R A P H

SECTION 4.5 / MORE ABOUT VARIANCE AND STANDARD DEVIATION 1 0 5

mean. You should realize that we could have drawn the line pointing to the left, or we

could have drawn two lines (or arrows), with one pointing to the right and one pointing

to the left, as in Figure 4.6(b). In each case, the goal is to show the standard distance

from the mean.]

Earlier we noted that sample variability tends to underestimate the variability in the cor-

responding population. To correct for this problem we adjusted the formula for sample

variance by dividing by n 2 1 instead of dividing by n. The result of the adjustment is that sample variance provides a much more accurate representation of the population

variance. Specifically, dividing by n 2 1 produces a sample variance that provides an unbiased estimate of the corresponding population variance. This does not mean that

each individual sample variance is exactly equal to its population variance. In fact,

some sample variances overestimate the population value and some underestimate it.

However, the average of all the sample variances produces an accurate estimate of the

population variance. This is the idea behind the concept of an unbiased statistic.

A sample statistic is unbiased if the average value of the statistic is equal to the

population parameter. (The average value of the statistic is obtained from all the

possible samples for a specific sample size, n.)

A sample statistic is biased if the average value of the statistic either underesti-

mates or overestimates the corresponding population parameter.

The following example demonstrates the concept of biased and unbiased statistics.

We begin with a population that consists of exactly N 5 6 scores: 0, 0, 3, 3, 9, 9. With a few calculations you should be able to verify that this population has a mean of m 5 4 and a variance of s2 5 14.

Next, we select samples of n 5 2 scores from this population. In fact, we obtain every single possible sample with n 5 2. The complete set of samples is listed in Table 4.1. Notice that the samples are listed systematically to ensure that every possible sample is

included. We begin by listing all the samples that have X 5 0 as the first score, then all

SA M P L E VA R I A N C E AS A N U N B I AS E D

STAT I ST I C

D E F I N I T I O N S

E X A M P L E 4 . 7

13 14

(b)

σ 5 8

µ 5 80

(a)

15 16 17 18 19

M 5 16

s 5 2 s 5 2

We have structured this

example to mimic “sampling

with replacement,” which is

covered in Chapter 6.

FIGURE 4.6

Showing means and standard deviations in frequency distribution graphs. (a) A population

distribution with a mean of µ 5 80 and a standard deviation of s 5 8. (b) A sample with a mean of M 5 16 and a standard deviation of s 5 2.

1 0 6 CHAPTER 4 MEASURES FOR VARIABILITY

the samples with X 5 3 as the first score, and so on. Notice that the table shows a total of 9 samples.

Finally, we have computed the mean and the variance for each sample. Note that

the sample variance has been computed two different ways. First, we examine what

happens if the sample variance is computed as the mean squared deviation (SS divided

by n) without any correction for bias. Second, we examine the correct sample variance

for which SS is divided by n 2 1 to produce an unbiased measure of variance. You should verify our calculations by computing one or two of the values for yourself. The

complete set of sample means and sample variances is presented in Table 4.1.

First, consider the column of biased sample variances, which were calculated by

dividing by n. These 9 sample variances add up to a total of 63, which produces an

average value of 63

9 5 7. The original population variance, however, is s2 5 14. Note

that the average of the sample variances is not equal to the population variance. If the

sample variance is computed by dividing by n, the resulting values do not produce an

accurate estimate of the population variance. On average, these sample variances under-

estimate the population variance and, therefore, are biased statistics.

Next, consider the column of sample variances that are computed using n 2 1. Although the population has a variance of s2 5 14, you should notice that none of the samples has a variance exactly equal to 14. However, if you consider the complete set of sample vari-

ances, you will find that the 9 values add up to a total of 126, which produces an average

value of 126

9 5 14. Thus, the average of the sample variances is exactly equal to the original

population variance. On average, the sample variance (computed using n 2 1) produces an accurate, unbiased estimate of the population variance.

Finally, direct your attention to the column of sample means. For this example, the

original population has a mean of m 5 4. Although none of the samples has a mean exactly equal to 4, if you consider the complete set of sample means, you will find that

the 9 sample means add up to a total of 36, so the average of the sample means is 36

9 5 4.

Note that the average of the sample means is exactly equal to the population mean. Again,

this is what is meant by the concept of an unbiased statistic. On average, the sample values

provide an accurate representation of the population. In this example, the average of the

9 sample means is exactly equal to the population mean.

In summary, both the sample mean and the sample variance (using n 2 1) are examples of unbiased statistics. This fact makes the sample mean and sample variance

extremely valuable for use as inferential statistics. Although no individual sample is

likely to have a mean and variance exactly equal to the population values, both the

TABLE 4.1

The set of all the possible

samples for n 5 2 selected from the population described in

Example 4.7. The mean is com-

puted for each sample, and the

variance is computed two dif-

ferent ways: (1) dividing by n,

which is incorrect and produces

a biased statistic; and (2) divid-

ing by n 2 1, which is correct and produces an unbiased sta-

tistic.

Sample Statistics

Sample First

Score Second Score

Mean M

Biased Variance (Using n)

Unbiased Variance

(Using n 2 1)

1 0 0 0.00 0.00 0.00

2 0 3 1.50 2.25 4.50

3 0 9 4.50 20.25 40.50

4 3 0 1.50 2.25 4.50

5 3 3 3.00 0.00 0.00

6 3 9 6.00 9.00 18.00

7 9 0 4.50 20.25 40.50

8 9 3 6.00 9.00 18.00

9 9 9 9.00 0.00 0.00

Totals 36.00 63.00 126.00

SECTION 4.5 / MORE ABOUT VARIANCE AND STANDARD DEVIATION 1 0 7

sample mean and the sample variance, on average, do provide accurate estimates of the

corresponding population values.

Because standard deviation requires extensive calculations, there is a tendency to get lost

in the arithmetic and forget what standard deviation is and why it is important. Standard

deviation is primarily a descriptive measure; it describes how variable, or how spread out,

the scores are in a distribution. Behavioral scientists must deal with the variability that

comes from studying people and animals. People are not all the same; they have different

attitudes, opinions, talents, IQs, and personalities. Although we can calculate the average

value for any of these variables, it is equally important to describe the variability. Standard

deviation describes variability by measuring distance from the mean. In any distribu-

tion, some individuals are close to the mean, and others are relatively far from the mean.

Standard deviation provides a measure of the typical, or standard, distance from the mean.

Describing an entire distribution Rather than listing all of the individual scores in a dis-

tribution, research reports typically summarize the data by reporting only the mean and the

standard deviation. When you are given these two descriptive statistics, however, you should

be able to visualize the entire set of data. For example, consider a sample with a mean of

M 5 36 and a standard deviation of s 5 4. Although there are several different ways to picture the data, one simple technique is to imagine (or sketch) a histogram in which each

score is represented by a box in the graph. For this sample, the data can be pictured as a pile

of boxes (scores) with the center of the pile located at a value of M 5 36. The individual scores, or boxes, are scattered on both sides of the mean with some of the boxes relatively

close to the mean and some farther away. As a rule of thumb, roughly 70% of the scores in

a distribution are located within a distance of one standard deviation from the mean, and

almost all of the scores (roughly 95%) are within two standard deviations of the mean. In

this example, the standard distance from the mean is s 5 4 points, so your image should have most of the boxes within 4 points of the mean, and nearly all of the boxes within

8 points. One possibility for the resulting image is shown in Figure 4.7.

Describing the location of individual scores Notice that Figure 4.7 not only shows

the mean and the standard deviation, but also uses these two values to reconstruct the

underlying scale of measurement (the X values along the horizontal line). The scale of

STA N DA R D D E V I AT I O N A N D D E S C R I P T I V E

STAT I ST I C S

28 30 32 34 36 38 40 42 44 46

s 5 4 s 5 4

M 5 36

FIGURE 4.7

A sample of n 5 20 scores with a mean of M 5 36 and a standard deviation of s 5 4.

1 0 8 CHAPTER 4 MEASURES FOR VARIABILITY

measurement helps to complete the picture of the entire distribution and relate each

individual score to the rest of the group. In this example, you should realize that a score

of X 5 34 is located near the center of the distribution, only slightly below the mean. On the other hand, a score of X 5 45 is an extremely high score, located far out in the right-hand tail of the distribution.

Notice that the relative position of a score depends in part on the size of the standard

deviation. In Figure 4.6 (p. 105), for example, we show a population distribution with a

mean of m 5 80 and a standard deviation of s 5 8, and a sample distribution with a mean of M 5 16 and a standard deviation of s 5 2. In the population distribution, a score that is 4 points above the mean is slightly above average but is certainly not an extreme value. In

the sample distribution, however, a score that is 4 points above the mean is an extremely

high score. In each case, the relative position of the score depends on the size of the stan-

dard deviation. For the population, a deviation of 4 points from the mean is relatively small,

corresponding to only half of the standard deviation. For the sample, on the other hand, a

4-point deviation is very large, twice the size of the standard deviation.

The general point of this discussion is that the mean and standard deviation are not

simply abstract concepts or mathematical equations. Instead, these two values should

be concrete and meaningful, especially in the context of a set of scores. The mean and

standard deviation are central concepts for most of the statistics that are presented in

the following chapters. A good understanding of these two statistics will help you with

the more complex procedures that follow (see Box 4.1).

Occasionally a set of scores is transformed by adding a constant to each score or by mul-

tiplying each score by a constant value. This happens, for example, when exposure to a

treatment adds a fixed amount to each participant’s score or when you want to change

the unit of measurement (to convert from minutes to seconds, multiply each score by 60).

What happens to the standard deviation when the scores are transformed in this manner?

The easiest way to determine the effect of a transformation is to remember that the stan-

dard deviation is a measure of distance. If you select any two scores and see what happens

to the distance between them, you also find out what happens to the standard deviation.

1. Adding a constant to each score does not change the standard deviation If you

begin with a distribution that has a mean of m 5 40 and a standard deviation of s 5 10, what

T R A N S F O R M AT I O N S O F S CA L E

BOX

4.1 AN ANALOGY FOR THE MEAN AND THE STANDARD DEVIATION

Although the basic concepts of the mean and the stan-

dard deviation are not overly complex, the following

analogy often helps students gain a more complete

understanding of these two statistical measures.

In our local community, the site for a new high

school was selected because it provides a central

location. An alternative site on the western edge of

the community was considered, but this site was re-

jected because it would require extensive busing for

students living on the east side. In this example, the

location of the high school is analogous to the con-

cept of the mean; just as the high school is located

in the center of the community, the mean is located

in the center of the distribution of scores. For each

student in the community, it is possible to mea-

sure the distance between home and the new high

school. Some students live only a few blocks from

the new school and others live as much as 3 miles

away. The average distance that a student must

travel to school was calculated to be 0.80 miles. The

average distance from the school is analogous to the

concept of the standard deviation; that is, the stan-

dard deviation measures the standard distance from

an individual score to the mean.

SECTION 4.5 / MORE ABOUT VARIANCE AND STANDARD DEVIATION 1 0 9

happens to s if you add 5 points to every score? Consider any two scores in this distribution: Suppose, for example, that these are exam scores and that you had a score of X 5 41 and your friend had X 5 43. The distance between these two scores is 43 2 41 5 2 points. After add- ing the constant, 5 points, to each score, your score would be X 5 46, and your friend would have X 5 48. The distance between scores is still 2 points. Adding a constant to every score does not affect any of the distances and, therefore, does not change the standard deviation.

This fact can be seen clearly if you imagine a frequency distribution graph. If, for example,

you add 5 points to each score, then every score in the graph is moved 5 points to the right.

The result is that the entire distribution is shifted to a new position 5 points up the scale.

Note that the mean moves along with the scores and is increased by 5 points. However, the

variability does not change because each of the deviation scores (X 2 m) does not change.

2. Multiplying each score by a constant causes the standard deviation to be multi-

plied by the same constant Consider the same distribution of exam scores we looked

at earlier. If m 5 40 and s 5 10, what would happen to s if each score were multiplied by 2? Again, we look at two scores, X 5 41 and X 5 43, with a distance between them equal to 2 points. After the scores have been multiplied by 2, these two scores become

X 5 82 and X 5 86. Now the distance between scores is 4 points, twice the original distance. Multiplying each score causes each distance to be multiplied, so the standard

deviation also is multiplied by the same amount.

IN THE LITERATURE

REPORTING THE STANDARD DEVIATION

In reporting the results of a study, the researcher often provides descriptive informa-

tion for both central tendency and variability. The dependent variables in psychology

research are often numerical values obtained from measurements on interval or ratio

scales. With numerical scores, the most common descriptive statistics are the mean

(central tendency) and the standard deviation (variability), which are usually reported

together. In many journals, especially those following APA style, the symbol SD is

used for the sample standard deviation. For example, the results might state:

Children who viewed the violent cartoon displayed more aggressive responses

(M 5 12.45, SD 5 3.7) than those who viewed the control cartoon (M 5 4.22, SD 5 1.04).

When reporting the descriptive measures for several groups, the findings may be

summarized in a table. Table 4.2 illustrates the results of hypothetical data.

Sometimes the table also indicates the sample size, n, for each group. You should

remember that the purpose of the table is to present the data in an organized, concise,

and accurate manner.

TABLE 4.2

The number of aggressive

responses in male and female

children after viewing cartoons.

Type of Cartoon

Violent Control

Males M 5 15.72 M 5 6.94

SD 5 4.43 SD 5 2.26

Females M 5 3.47 M 5 2.61

SD 5 1.12 SD 5 0.98

1 1 0 CHAPTER 4 MEASURES FOR VARIABILITY

In very general terms, the goal of inferential statistics is to detect meaningful and sig-

nificant patterns in research results. The basic question is whether the patterns observed

in the sample data reflect corresponding patterns that exist in the population, or are

simply random fluctuations that occur by chance. Variability plays an important role

in the inferential process because the variability in the data influences how easy it is to

see patterns. In general, low variability means that existing patterns can be seen clearly,

whereas high variability tends to obscure any patterns that might exist. The following

example provides a simple demonstration of how variance can influence the perception

of patterns.

In most research studies the goal is to compare means for two (or more) sets of data.

For example:

Is the mean level of depression lower after therapy than it was before therapy?

Is the mean attitude score for men different from the mean score for women?

Is the mean reading achievement score higher for students in a special program than

for students in regular classrooms?

In each of these situations, the goal is to find a clear difference between two means

that would demonstrate a significant, meaningful pattern in the results. Variability plays

an important role in determining whether a clear pattern exists. Consider the follow-

ing data representing hypothetical results from two experiments, each comparing two

treatment conditions. For both experiments, your task is to determine whether there

appears to be any consistent difference between the scores in treatment 1 and the scores

in treatment 2.

Experiment A

Treatment 1 Treatment 2

35 39

34 40

36 41

35 40

VA R I A N C E A N D I N F E R E N T I A L

STAT I ST I C S

E X A M P L E 4 . 8

Experiment B

Treatment 1 Treatment 2

31 46

15 21

57 61

37 32

For each experiment, the data have been constructed so that there is a 5-point mean

difference between the two treatments: On average, the scores in treatment 2 are

5 points higher than the scores in treatment 1. The 5-point difference is relatively easy

to see in experiment A, where the variability is low, but the same 5-point difference is

difficult to see in experiment B, where the variability is large. Again, high variability

tends to obscure any patterns in the data. This general fact is perhaps even more con-

vincing when the data are presented in a graph. Figure 4.8 shows the two sets of data

from experiments A and B. Notice that the results from experiment A clearly show the

5-point difference between treatments. One group of scores piles up around 35 and

the second group piles up around 40. On the other hand, the scores from experiment B

[Figure 4.8(b)] seem to be mixed together randomly with no clear difference between

the two treatments.

SECTION 4.5 / MORE ABOUT VARIANCE AND STANDARD DEVIATION 1 1 1

In the context of inferential statistics, the variance that exists in a set of sample

data is often classified as error variance. This term is used to indicate that the sample

variance represents unexplained and uncontrolled differences between scores. As the

error variance increases, it becomes more difficult to see any systematic differences or

patterns that might exist in the data. An analogy is to think of variance as the static that

appears on a radio station or a cell phone when you enter an area of poor reception. In

general, variance makes it difficult to get a clear signal from the data. High variance can

make it difficult or impossible to see a mean difference between two sets of scores, or

to see any other meaningful patterns in the results from a research study.

Sample 1

Data from Experiment A

34 35 36

33 37 39 40 4138 42 X

Sample 2

Sample 1

Data from Experiment B

10 20 X

Sample 2

30 40 50 60

M 5 35 M 5 35

M 5 40M 5 40

(b)(a)

FIGURE 4.8

Graphs showing the results from two experiments. In experiment A, the variability is small and it is easy to see the

5-point mean difference between the two treatments. In experiment B, however, the 5-point mean difference between

treatments is obscured by the large variability.

1. Explain the difference between a biased and an unbiased statistic.

2. In a population with a mean of m 5 50 and a standard deviation of s 5 10, would a score of X 5 58 be considered an extreme value (far out in the tail of the distri- bution)? What if the standard deviation were s 5 3?

3. A population has a mean of m 5 70 and a standard deviation of s 5 5.

a. If 10 points were added to every score in the population, what would be the

new values for the population mean and standard deviation?

b. If every score in the population were multiplied by 2, what would be the new

values for the population mean and standard deviation?

1. If a statistic is biased, it means that the average value of the statistic does not accurately represent

the corresponding population parameter. Instead, the average value of the statistic either overes-

timates or underestimates the parameter. If a statistic is unbiased, it means that the average value

of the statistic is an accurate representation of the corresponding population parameter.

2. With s 5 10, a score of X 5 58 would be located in the central section of the distribution (within one standard deviation). With s 5 3, a score of X 5 58 would be an extreme value, located more than two standard deviations above the mean.

3. a. The new mean would be m 5 80 but the standard deviation would still be s 5 5.

b. The new mean would be m 5 140 and the new standard deviation would be s 5 10.

L E A R N I N G C H E C K

ANSWERS

1 1 2 CHAPTER 4 MEASURES FOR VARIABILITY

SUMMARY

1. The purpose of variability is to measure and describe the degree to which the scores in a distribution are spread out or clustered together. There are three basic measures of variability: the range, the variance, and the standard deviation.

The range is the distance covered by the set of scores, from the smallest score to the largest score. The range is completely determined by the two extreme scores and is considered to be a relatively crude measure of variability.

Standard deviation and variance are the most commonly used measures of variability. Both of these measures are based on the idea that each score can be described in terms of its deviation, or distance, from the mean. The variance is the mean of the squared deviations. The standard deviation is the square root of the variance and provides a measure of the standard distance from the mean.

2. To calculate variance or standard deviation, you first need to find the sum of the squared deviations, SS. Except for minor changes in notation, the calculation of SS is identical for samples and populations. There are two methods for calculating SS:

I. By definition, you can find SS using the following steps: a. Find the deviation (X 2 m) for each score. b. Square each deviation. c. Add the squared deviations.

This process can be summarized in a formula as follows:

Definitional Formula: SS 5 o(X 2 m)2

II. The sum of the squared deviations can also be found using a computational formula, which is especially useful when the mean is not a whole number:

Computational formula: SS X X

N 5 2

 2

( )

3. Variance is the mean squared deviation and is obtained by finding the sum of the squared deviations and then

dividing by the number of scores. For a population, variance is

s 5 2 SS

For a sample, only n 2 1 of the scores are free to vary (degrees of freedom or df 5 n 2 1), so sample variance is

s SS

1 5

2 5

Using n 2 1 in the sample formula makes the sample variance an accurate and unbiased estimate of the

population variance.

4. Standard deviation is the square root of the variance. For a population, this is

s 5 SS

Sample standard deviation is

s SS

df 5

2 5

5. Adding a constant value to every score in a distribution does not change the standard deviation. Multiplying every score by a constant, however, causes the stan- dard deviation to be multiplied by the same constant.

RESOURCES 1 1 3

KEY TERMS

variability (90)

range (91)

deviation (92)

deviation score (92)

sum of squares (SS) (96)

mean squared deviation (93)

population variance (s2) (93)

population standard deviation (s) (94)

sample variance (s2) (101)

sample standard deviation (s) (101)

degrees of freedom (df) (103)

unbiased statistic (105)

biased statistic (105)

error variance (111)

If your professor has assigned Aplia:

1. Sign in to your account.

2. Complete the corresponding exercises as required by your professor.

3. When finished, click “Grade It Now” to see which areas you have mastered, which

areas need more work, and detailed explanations of every answer.

General instructions for using SPSS are presented in Appendix D. Following are

detailed instructions for using SPSS to compute the Range, Standard Deviation, and

Variance for a sample of scores.

Data Entry

Enter all of the scores in one column of the data editor, probably VAR00001.

Data Analysis

1. Click Analyze on the tool bar, select Descriptive Statistics, and click on

Descriptives.

2. Highlight the column label for the set of scores (VAR00001) in the left box and

click the arrow to move it into the Variable box.

3. If you want the variance and/or the range reported along with the standard devia-

tion, click on the Options box, select Variance and/or Range, then click Continue.

4. Click OK.

RESOURCES

Go to CengageBrain.com to access Psychology CourseMate, where you will find an

interactive eBook, glossaries, flashcards, quizzes, statistics workshops, and more.

1 1 4 CHAPTER 4 MEASURES FOR VARIABILITY

SPSS Output

We used SPSS to find the variance and standard deviation for the sample of n 5 8 scores from Example 4.5 (p. 101), and the SPSS output is shown in Figure 4.9. The

summary table lists the number of scores, the maximum and minimum scores, the

mean, the range, the standard deviation, and the variance. Note that the range and

variance are included because these values were selected using the Options box dur-

ing data analysis. Caution: SPSS computes the sample standard deviation and sample

variance using n 2 1. If your scores are intended to be a population, you can multiply the sample standard deviation by the square root of (n 2 1)/n to obtain the population standard deviation.

Note: You can also obtain the mean and standard deviation for a sample if you

use SPSS to display the scores in a frequency distribution histogram (see the SPSS

section at the end of Chapter 2). The mean and standard deviation are displayed

beside the graph.

FOCUS ON PROBLEM SOLVING

1. The purpose of variability is to provide a measure of how spread out the scores

in a distribution are. Usually this is described by the standard deviation. Because

the calculations are relatively complicated, it is wise to make a preliminary

estimate of the standard deviation before you begin. Remember that standard

deviation provides a measure of the typical, or standard, distance from the mean.

Therefore, the standard deviation must have a value somewhere between the

largest and the smallest deviation scores. As a rule of thumb, the standard devia-

tion should be about one-fourth of the range.

2. Rather than trying to memorize all of the formulas for SS, variance, and stan-

dard deviation, you should focus on the definitions of these values and the logic

that relates them to each other:

SS is the sum of squared deviations.

Variance is the mean squared deviation.

Standard deviation is the square root of variance.

The only formula you should need to memorize is the computational formula for SS.

3. A common error is to use n – 1 in the computational formula for SS when you

have scores from a sample. Remember that the SS formula always uses n (or N).

After you compute SS for a sample, you must correct for the sample bias by using

n – 1 in the formulas for variance and standard deviation.

VAR00001

Valid N (listwise)

8 8.00 3.00 11.00 6.5000 2.61861 6.85714

N Range Minimum Maximum Mean Std. Deviation Variance

FIGURE 4.9

The SPSS summary table showing descriptive statistics for the sample of n 5 8 scores from Example 4.5.

PROBLEMS 1 1 5

DEMONSTRATION 4.1

COMPUTING MEASURES OF VARIABILITY

For the following sample data, compute the variance and standard deviation. The

scores are:

10 7 6 10 6 15

Compute SS, the sum of squared deviations

We use the computational formula. For this sample, n 5 6 and

oX 5 10 1 7 1 6 1 10 1 6 1 15 5 54

oX2 5 102 1 72 1 62 1 102 1 62 1 152 5 546

SS X X

N 5 2 5 2

 2

2 2

546 54

( ) ( )

5 546 2 486

5 60

Compute the sample variance

For sample variance, SS is divided by the degrees of freedom, df 5 n – 1.

s SS

5 125

2 5 5

Compute the sample standard deviation

Standard deviation is simply the square root of the variance.

s 5 512 3 46.

S T E P 1

S T E P 2

S T E P 3

PROBLEMS

1. In words, explain what is measured by each of the following:

a. SS

b. Variance c. Standard deviation

2. Can SS ever have a value less than zero? Explain your answer.

3. Is it possible to obtain a negative value for the variance or the standard deviation?

4. What does it mean for a sample to have a standard deviation of zero? Describe the scores in such a sample.

5. Explain why the formulas for sample variance and population variance are different.

6. A population has a mean of m 5 80 and a standard deviation of s 5 20.

a. Would a score of X 5 70 be considered an extreme value (out in the tail) in this sample?

b. If the standard deviation were s 5 5, would a score of X 5 70 be considered an extreme value?

1 1 6 CHAPTER 4 MEASURES FOR VARIABILITY

7. On an exam with a mean of M 5 78, you obtain a score of X 5 84.

a. Would you prefer a standard deviation of s 5 2 or s 5 10? (Hint: Sketch each distribution and find the location of your score.)

b. If your score were X 5 72, would you prefer s 5 2 or s 5 10? Explain your answer.

8. Calculate the mean and SS (sum of squared devia- tions) for each of the following samples. Based on the value for the mean, you should be able to decide which SS formula is better to use.

Sample A: 1 4 8 5

Sample B: 3 0 9 4

9. For the following population of N 5 6 scores:

3 1 4 3 3 4

a. Sketch a histogram showing the population distribution.

b. Locate the value of the population mean in your sketch, and make an estimate of the standard deviation (as done in Example 4.2).

c. Compute SS, variance, and standard deviation for the population. (How well does your estimate compare with the actual value of s?)

10. For the following sample of n 5 7 scores:

8 6 5 2 6 3 5

a. Sketch a histogram showing the sample distribution. b. Locate the value of the sample mean in your

sketch, and make an estimate of the standard deviation (as done in Example 4.5).

c. Compute SS, variance, and standard deviation for the sample. (How well does your estimate compare with the actual value of s?)

11. For the following population of N 5 6 scores:

11 0 2 9 9 5

a. Calculate the range and the standard deviation. (Use either definition for the range.)

b. Add 2 points to each score and compute the range and standard deviation again. Describe how adding a constant to each score influences measures of variability.

12. The range is completely determined by the two extreme scores in a distribution. The standard devia- tion, on the other hand, uses every score.

a. Compute the range (choose either definition) and the standard deviation for the following sample of n 5 5 scores. Note that there are three scores clustered around the mean in the center of the distribution, and two extreme values.

Scores: 0 6 7 8 14

b. Now we break up the cluster in the center of the distribution by moving two of the central scores out to the extremes. Once again compute the range and the standard deviation.

New scores: 0 0 7 14 14

c. According to the range, how do the two distribu- tions compare in variability? How do they com- pare according to the standard deviation?

13. A population has a mean of m 5 30 and a standard deviation of s 5 5.

a. If 5 points were added to every score in the popu- lation, what would be the new values for the mean and standard deviation?

b. If every score in the population were multiplied by 3, what would be the new values for the mean and standard deviation?

14. a. After 3 points have been added to every score in a sample, the mean is found to be M 5 83 and the standard deviation is s 5 8. What were the values for the mean and standard deviation for the original sample?

b. After every score in a sample has been multiplied by 4, the mean is found to be M 5 48 and the standard deviation is s 5 12. What were the values for the mean and standard deviation for the original sample?

15. For the following sample of n 5 4 scores: 82, 88, 82, and 86:

a. Simplify the arithmetic by first subtracting 80 points from each score to obtain a new sample of 2, 8, 2, and 6. Then, compute the mean and standard deviation for the new sample.

b. Using the values you obtained in part a, what are the values for the mean and standard deviation for the original sample?

16. For the following sample of n 5 8 scores: 0, 1, 1

2 , 0, 3, 1

2 , 0, and 1:

a. Simplify the arithmetic by first multiplying each score by 2 to obtain a new sample of 0, 2, 1, 0, 6, 1, 0, and 2. Then, compute the mean and standard deviation for the new sample.

b. Using the values you obtained in part a, what are the values for the mean and standard deviation for the original sample?

17. For the data in the following sample:

8 1 5 1 5

a. Find the mean and the standard deviation. b. Now change the score of X 5 8 to X 5 18, and

find the new mean and standard deviation. c. Describe how one extreme score influences the

mean and standard deviation.

PROBLEMS 1 1 7

18. Calculate SS, variance, and standard deviation for the following sample of n 5 4 scores: 7, 4, 2, 1. (Note: The computational formula for SS works well with these scores.)

19. Calculate SS, variance, and standard deviation for the following population of N 5 8 scores: 0, 0, 5, 0, 3, 0, 0, 4. (Note: The computational formula for SS works well with these scores.)

20. Calculate SS, variance, and standard deviation for the following population of N 5 6 scores: 1, 6, 10, 9, 4, 6. (Note: The definitional formula for SS works well with these scores.)

21. Calculate SS, variance, and standard deviation for the following sample of n 5 5 scores: 10, 4, 8, 5, 8. (Note: The definitional formula for SS works well with these scores.)

22. In an extensive study involving thousands of British children, Arden and Plomin (2006) found significantly higher variance in the intelligence scores for males than for females. Following are hypothetical data, similar to the results obtained in the study. Note that the scores are not regular IQ scores but have been standardized so that the entire sample has a mean of M 5 10 and a standard deviation of s 5 2.

a. Calculate the mean and the standard deviation for the sample of n 5 8 females and for the sample of n 5 8 males.

b. Based on the means and the standard deviations, describe the differences in intelligence scores for males and females.

Female Male

9 8

11 10

10 11

13 12

8 6

9 10

11 14

9 9

23. Within a population, the differences that exist from one person to another are often called diversity. Researchers comparing cognitive skills for younger

adults and older adults, typically find greater differences (greater diversity) in the older popula- tion (Morse, 1993). Following are typical data showing problem-solving scores for two groups of participants.

Older Adults (average age 72)

Younger Adults (average age 31)

9 4 7 3 8 7 9 6 7 8

6 2 8 4 5 6 7 6 6 8

7 5 2 6 6 9 7 8 6 9

a. Compute the mean, the variance, and the standard deviation for each group.

b. Is one group of scores noticeably more variable (more diverse) than the other?

24. In the previous problem we noted that the differ- ences in cognitive skills tend to be bigger among older people than among younger people. These differences are often called diversity. Similarly, the differences in performance from trial to trial for the same person are often called consistency. Research in this area suggests that consistency of perfor- mance seems to decline as people age. A study by Wegesin and Stern (2004) found lower consistency (more variability) in the memory performance scores for older women than for younger women. The following data represent memory scores obtained for two women, one older and one younger, over a series of memory trials.

a. Calculate the variance of the scores for each woman.

b. Are the scores for the younger woman more con- sistent (less variable)?

Younger Older

8 7

6 5

6 8

7 5

8 7

7 6

8 8

8 5

1. Familiarity with statistical terminology and notation (Chapter 1).

2. The ability to organize a set of scores in a frequency distribution table or a frequency distribution graph (Chapter 2).

3. The ability to summarize and describe a distribution of scores by computing a measure of central tendency (Chapter 3).

4. The ability to summarize and describe a distribution of scores by computing a measure of variability (Chapter 4).

The general goal of descriptive statistics is to simplify a set

of data by organizing or summarizing a large set of scores.

A frequency distribution table or graph organizes the entire

set of scores so that it is possible to see the complete distri-

bution all at once. Measures of central tendency describe the

distribution by identifying its center. They also summarize

the distribution by condensing all of the individual scores

into one value that represents the entire group. Measures of

variability describe whether the scores in a distribution are

widely scattered or closely clustered. Variability also pro-

vides an indication of how accurately a measure of central

tendency represents the entire group. Of the basic skills presented in this part, the most com-

monly used are calculating the mean and standard deviation for a sample of numerical scores. The following exercises should provide an opportunity to use and reinforce these statistical skills.

REVIEW EXERCISES

1. a. What is the general goal for descriptive statistics? b. How is the goal served by putting scores in a fre-

quency distribution? c. How is the goal served by computing a measure of

central tendency? d. How is the goal served by computing a measure of

variability?

2. In Example 1.1, we proposed an experiment in which one group of students prepared for an exam by studying text on printed pages and a second group studied the same text on a computer screen. The goal is to determine whether one study technique leads to better exam scores than the other. This same experiment was actually con- ducted by Ackerman and Goldsmith (2011). In one part of the experiment, the students were allowed to control the amount of time that they spent studying. Data similar to the results obtained from this condition are presented in the following table. The data are the test scores ob- tained from the students in the two conditions.

Number Correct on a 20-Question Multiple-Choice Exam

Text Studied on Printed Pages

Text Studied on a Computer Screen

16 17 18 16 14 17

18 19 15 13 18 17

19 14 18 10 17 12

18 18 19 16 15 18

14 17 17 15 19 16

20 15 19 14 12 15

16 17 13 11 13 17

a. Sketch a polygon showing the distribution of test scores for the students who read printed pages. On the same graph, sketch a polygon for the students who read the computer screen. (Use two different colors or use a dashed line for one group and a solid line for the other.) Based on the appearance of your graph, describe the differences between the two groups.

b. Calculate the mean test score for each sample. Does the mean difference support your description from part a?

c. Calculate the variance and standard deviation for each sample. Based on the measures of variability, is one group of scores more widely scattered than the other?

REVIEW

By completing this part, you should understand and be able to perform basic descriptive statisti-

cal procedures. These include:

1 1 9

P A R T I

P A R T

121

Chapter 5 z-Scores: Location of Scores and Standardized Distributions 123

Chapter 6 Probability 149

Chapter 7 Probability and Samples: The Distribution of Sample Means 175

Chapter 8 Introduction to Hypothesis Testing 203

Y ou should recall from Chapter 1 that statistical methods are

classified into two general categories: descriptive statistics,

which attempt to organize and summarize data, and infer-

ential statistics, which use the limited information from samples

to answer general questions about populations. In most research

situations, both kinds of statistics are used to gain a complete

understanding of the research results. In Part I of this book we in-

troduced the techniques of descriptive statistics. We now are ready

to turn our attention to inferential statistics.

Before we proceed with inferential statistics, however, it is

necessary to present some additional information about samples.

We know that it is possible to obtain hundreds or even thousands of

different samples from the same population. We need to determine

how all the different samples are related to each other and how

individual samples are related to the population from which they

were obtained. Finally, we need a system for designating which

samples are representative of their populations and which are not.

In the next four chapters we develop the concepts and skills

that form the foundation for inferential statistics. In general, these

chapters establish formal, quantitative relationships between sam-

ples and populations and introduce a standardized procedure for

determining whether the data from a sample justify a conclusion

about the population. After we have developed this foundation,

we can begin inferential statistics. That is, we can begin to look at

statistical techniques that use the sample data obtained in research

studies as the basis for answering questions about populations.

Foundations of Inferential Statistics

Aplia for Essentials of Statistics for the Behavioral Sciences

After reading, go to “Resources” at the end of this chapter for

an introduction on how to use Aplia’s homework and learning

resources.

z-Scores: Location of Scores and Standardized Distributions 5.1 Introduction to z-Scores

5.2 z-Scores and Location in a Distribution

5.3 Using z-Scores to Standardize a Distribution

5.4 Other Standardized Distributions Based on z-Scores

5.5 Computing z-Scores for a Sample

5.6 Looking Ahead to Inferential Statistics

Summary

Focus on Problem Solving

Demonstrations 5.1 and 5.2

Problems

C H A P T E R

5 Tools You Will Need The following items are considered essential background material for this chapter. If you doubt your knowledge of any of these items, you should review the appropriate chapter or section before proceeding.

• The mean (Chapter 3) • The standard deviation (Chapter 4) • Basic algebra (math review,

Appendix A)

1 2 4 CHAPTER 5 z-SCORES: LOCATION OF SCORES AND STANDARDIZED DISTRIBUTIONS

INTRODUCTION TO z-SCORES

In the previous two chapters, we introduced the concepts of the mean and the standard

deviation as methods for describing an entire distribution of scores. Now we shift at-

tention to the individual scores within a distribution. In this chapter, we introduce a sta-

tistical technique that uses the mean and the standard deviation to transform each score

(X value) into a z-score, or a standard score. The purpose of z-scores, or standard

scores, is to identify and describe the exact location of each score in a distribution.

The following example demonstrates why z-scores are useful and introduces the

general concept of transforming X values into z-scores.

Suppose you received a score of X 5 76 on a statistics exam. How did you do? It should be clear that you need more information to predict your grade. Your score of

X 5 76 could be one of the best scores in the class, or it might be the lowest score in the distribution. To find the location of your score, you must have information about the

other scores in the distribution. It would be useful, for example, to know the mean for

the class. If the mean were µ 5 70, you would be in a much better position than if the mean were µ 5 85. Obviously, your position relative to the rest of the class depends on the mean. However, the mean by itself is not sufficient to tell you the exact location of

your score. Suppose you know that the mean for the statistics exam is µ 5 70 and your score is X 5 76. At this point, you know that your score is 6 points above the mean, but you still do not know exactly where it is located. Six points may be a relatively big

distance and you may have one of the highest scores in the class, or 6 points may be

a relatively small distance and you may be only slightly above the average. Figure 5.1

shows two possible distributions of exam scores. Both distributions have a mean of

µ 5 70, but for one distribution, the standard deviation is s 5 3, and for the other, s 5 12. The location of X 5 76 is highlighted in each of the two distributions. When the

standard deviation is s 5 3, your score of X 5 76 is in the extreme right-hand tail, the

highest score in the distribution. However, in the other distribution, where s 5 12, your

score is only slightly above average. Thus, the relative location of your score within the

distribution depends on the standard deviation as well as the mean.

5.1

E X A M P L E 5 . 1

X = 76

7370

σ = 3

X = 76

8270

σ = 12

FIGURE 5.1

Two distributions of exam scores. For both distributions, µ 5 70, but for one distribution, s 5 3, and for the other,

s 5 12. The relative position of X 5 76 is very different for the two distributions.

(a) (b)

SECTION 5.2 / z-SCORES AND LOCATION IN A DISTRIBUTION 1 2 5

The purpose of the preceding example is to demonstrate that a score by itself does

not necessarily provide much information about its position within a distribution. These

original, unchanged scores that are the direct result of measurement are called raw

scores. To make raw scores more meaningful, they are often transformed into new val-

ues that contain more information. This transformation is one purpose for z-scores. In

particular, we transform X values into z-scores so that the resulting z-scores tell exactly

where the original scores are located.

A second purpose for z-scores is to standardize an entire distribution. A common ex-

ample of a standardized distribution is the distribution of IQ scores. Although there are

several different tests for measuring IQ, the tests usually are standardized so that they

have a mean of 100 and a standard deviation of 15. Because all the different tests are

standardized, it is possible to understand and compare IQ scores even though they come

from different tests. For example, we all understand that an IQ score of 95 is a little

below average, no matter which IQ test was used. Similarly, an IQ of 145 is extremely

high, no matter which IQ test was used. In general terms, the process of standardizing

takes different distributions and makes them equivalent. The advantage of this process

is that it is possible to compare distributions even though they may have been quite

different before standardization.

In summary, the process of transforming X values into z-scores serves two useful

purposes:

1. Each z-score tells the exact location of the original X value within the distribution.

2. The z-scores form a standardized distribution that can be directly compared to

other distributions that also have been transformed into z-scores.

Each of these purposes is discussed in the following sections.

z-SCORES AND LOCATION IN A DISTRIBUTION

One of the primary purposes of a z-score is to describe the exact location of a score

within a distribution. The z-score accomplishes this goal by transforming each X value

into a signed number (1 or –) so that

1. The sign tells whether the score is located above (1) or below (–) the mean, and

2. The number tells the distance between the score and the mean in terms of the

number of standard deviations.

In Figure 5.1(a), for example, we show a distribution with a mean of µ 5 70 and s 5 3. In this distribution, a score of X 5 76 would be transformed into z 5 12.00. The

z value indicates that the score is located above the mean (1) by a distance equal to

2 standard deviations (6 points). For the distribution in Figure 5.1(b), µ 5 70 and

s 5 12. In this distribution, a score of X 5 76 would be transformed into z 5 1 1 2 or

z 5 10.50. In this case, the score is above the mean (1) by a distance of only 1 2 of a

standard deviation (6 points).

A z-score specifies the precise location of each X value within a distribution. The sign

of the z-score (1 or –) signifies whether the score is above the mean (positive) or

below the mean (negative). The numerical value of the z-score specifies the distance

from the mean by counting the number of standard deviations between X and µ.

5.2

D E F I N I T I O N

1 2 6 CHAPTER 5 z-SCORES: LOCATION OF SCORES AND STANDARDIZED DISTRIBUTIONS

Notice that a z-score always consists of two parts: a sign (1 or –) and a magnitude. Both parts are necessary to describe completely where a raw score is located within a

distribution.

Figure 5.2 shows a population distribution with various positions identified by their

z-score values. Notice that all z-scores above the mean are positive and all z-scores below

the mean are negative. The sign of a z-score tells you immediately whether the score is

located above or below the mean. Also, note that a z-score of z 5 11.00 corresponds to a position exactly 1 standard deviation above the mean. A z-score of z 5 12.00 is always located exactly 2 standard deviations above the mean. The numerical value of the z-score

tells you the number of standard deviations it is from the mean. Finally, you should

notice that Figure 5.2 does not give any specific values for the population mean or the

standard deviation. The locations identified by z-scores are the same for all distributions,

no matter what mean or standard deviation the distributions may have.

Whenever you are working

with z-scores, you should

imagine or draw a picture

similar to Figure 5.2. Although

you should realize that not all

distributions are normal, we

use the normal shape as an

example when showing

z-scores for populations.

1. Identify the z-score value corresponding to each of the following locations in a

distribution.

a. Below the mean by 2 standard deviations.

b. Above the mean by 1 2

standard deviation.

c. Below the mean by 1.50 standard deviations.

2. Describe the location in the distribution for each of the following z-scores. (For

example, z 5 11.00 is located above the mean by 1 standard deviation.)

a. z 5 21.50 b. z 5 0.25 c. z 5 22.50 d. z 5 0.50

3. For a population with µ 5 30 and s 5 8, find the z-score for each of the following scores:

a. X 5 32 b. X 5 26 c. X 5 42

4. For a population with µ 5 50 and s 5 12, find the X value corresponding to each

of the following z-scores:

a. z 5 20.25 b. z 5 2.00 c. z 5 0.50

1. a. z 5 22.00 b. z 5 10.50 c. z 5 21.50

2. a. Below the mean by 1 1 2

standard deviations.

b. Above the mean by 1 4

standard deviation.

c. Below the mean by 2 1 2

standard deviations.

d. Above the mean by 1 2

standard deviation.

3. a. z 5 10.25 b. z 5 20.50 c. z 5 11.50

4. a. X 5 47 b. X 5 74 c. X 5 56

L E A R N I N G C H E C K

ANSWERS

The z-score definition is adequate for transforming back and forth from X values to

z-scores as long as the arithmetic is easy to do in your head. For more complicated

values, it is best to have an equation to help structure the calculations. Fortunately,

the relationship between X values and z-scores is easily expressed in a formula. The

formula for transforming scores into z-scores is

z 5 2m

X (5.1)

T H E Z- S CO R E F O R M U L A

SECTION 5.2 / z-SCORES AND LOCATION IN A DISTRIBUTION 1 2 7

The numerator of the equation, X – µ, is a deviation score (Chapter 4, page 92),

which measures the distance in points between X and µ and indicates whether X is

located above or below the mean. The deviation score is then divided by s because we

want the z-score to measure distance in terms of standard deviation units. The formula

performs exactly the same arithmetic that is used with the z-score definition, and it

provides a structured equation to organize the calculations when the numbers are more

difficult. The following examples demonstrate the use of the z-score formula.

A distribution of scores has a mean of µ 5 100 and a standard deviation of s 5 10.

What z-score corresponds to a score of X 5 130 in this distribution?

According to the definition, the z-score has a value of 13 because the score is located

above the mean by exactly 3 standard deviations. Using the z-score formula, we obtain

z X

5 2m

s 5

2 5 5

130 100

10 3 00.

The formula produces exactly the same result that is obtained using the z-score

definition.

A distribution of scores has a mean of µ 5 86 and a standard deviation of s 5 7. What

z-score corresponds to a score of X 5 95 in this distribution?

Note that this problem is not particularly easy, especially if you try to use the z-score

definition and perform the calculations in your head. However, the z-score formula

organizes the numbers and allows you to finish the final arithmetic with a calculator.

Using the formula, we obtain

z X

5 2m

s 5

2 5 5

95 86

7 1 29.

According to the formula, a score of X 5 95 corresponds to z 5 1.29. The z-score

indicates a location that is above the mean (positive) by slightly more than 1 standard

deviation.

E X A M P L E 5 . 2

E X A M P L E 5 . 3

z +1 +2

–1–2

FIGURE 5.2

The relationship between

z-score values and locations

in a population distribution.

1 2 8 CHAPTER 5 z-SCORES: LOCATION OF SCORES AND STANDARDIZED DISTRIBUTIONS

When you use the z-score formula, it can be useful to pay attention to the definition

of a z-score as well. For example, we used the formula in Example 5.3 to calculate the

z-score corresponding to X 5 95, and obtained z 5 1.29. Using the z-score definition, we note that X 5 95 is located above the mean by 9 points, which is slightly more than one standard deviation (s 5 7). Therefore, the z-score should be positive and have a

value slightly greater than 1.00. In this case, the answer predicted by the definition is in

perfect agreement with the calculation. However, if the calculations produce a different

value, for example z 5 0.78, you should realize that this answer is not consistent with

the definition of a z-score. In this case, an error has been made and you should double

check the calculations.

Although the z-score equation (Equation 5.1) works well for transforming X val-

ues into z-scores, it can be awkward when you are trying to work in the opposite

direction and change z-scores back into X values. In general, it is easier to use the

definition of a z-score, rather than a formula, when you are changing z-scores into

X values. Remember, the z-score describes exactly where the score is located by

identifying the direction and distance from the mean. It is possible, however, to

express this definition as a formula, and we use a sample problem to demonstrate

how the formula can be created.

For a distribution with a mean of µ 5 60 and s 5 5, what X value corresponds to a

z-score of z 5 23.00?

To solve this problem, we use the z-score definition and carefully monitor the step-

by-step process. The value of the z-score indicates that X is located below the mean

by a distance equal to 3 standard deviations. Thus, the first step in the calculation is to

determine the distance corresponding to 3 standard deviations. For this problem, the

standard deviation is s 5 5 points, so 3 standard deviations is 3(5) 5 15 points. The

next step is to find the value of X that is located below the mean by 15 points. With a

mean of µ 5 60, the score is

X 5 µ 2 15 5 60 2 15 5 45

The two steps can be combined to form a single formula:

X 5 µ 1 zs (5.2)

In the formula, the value of zs is the deviation of X and determines both the direction

and the size of the distance from the mean. In this problem, zs 5 (–3)(5) 5 –15, or

15 points below the mean. Equation 5.2 simply combines the mean and the deviation

from the mean to determine the exact value of X.

Finally, you should realize that Equation 5.1 and Equation 5.2 are actually two dif-

ferent versions of the same equation. If you begin with either formula and use algebra

to shuffle the terms around, you soon end up with the other formula. We leave this as

an exercise for those who want to try it.

In most cases, we simply transform scores (X values) into z-scores, or change

z-scores back into X values. However, you should realize that a z-score establishes

a relationship between the score, the mean, and the standard deviation. This rela-

tionship can be used to answer a variety of different questions about scores and the

distributions in which they are located. The following examples demonstrate some

possibilities.

D E T E R M I N I N G A R AW S CO R E

( X ) F R O M A z - S CO R E

OT H E R R E L AT I O N S H I P S B E T W E E N z , X , µ ,

A N D 

SECTION 5.2 / z-SCORES AND LOCATION IN A DISTRIBUTION 1 2 9

In a population with a mean of µ 5 65, a score of X 5 59 corresponds to z 5 22.00. What is the standard deviation for the population?

To answer the question, we begin with the z-score value. A z-score of 22.00

indicates that the corresponding score is located below the mean by a distance of

2 standard deviations. You also can determine that the score (X 5 59) is located below the mean (µ 5 65) by a distance of 6 points. Thus, 2 standard deviations correspond to a distance of 6 points, which means that 1 standard deviation must

be s 5 3 points.

In a population with a standard deviation of s 5 6, a score of X 5 33 corresponds to

z 5 11.50. What is the mean for the population?

Again, we begin with the z-score value. In this case, a z-score of 11.50 indicates

that the score is located above the mean by a distance corresponding to 1.50 standard

deviations. With a standard deviation of s 5 6, this distance is (1.50)(6) 5 9 points.

Thus, the score is located 6 points above the mean. The score is X 5 33, so the mean

must be µ 5 24.

Many students find problems like those in Examples 5.4 and 5.5 easier to under-

stand if they draw a picture showing all of the information presented in the problem.

For the problem in Example 5.4, the picture would begin with a distribution that has

a mean of µ 5 65 (we use a normal distribution, which is shown in Figure 5.3). The

value of the standard deviation is unknown, but you can add arrows to the sketch

pointing outward from the mean for a distance corresponding to 1 standard devia-

tion. Finally, use standard deviation arrows to identify the location of z 5 22.00

(2 standard deviations below the mean) and add X 5 59 at that location. All of these

factors are shown in Figure 5.3. In the figure, it is easy to see that X 5 59 is located

6 points below the mean, and that the 6-point distance corresponds to exactly

2 standard deviations. Again, if 2 standard deviations equal 6 points, then 1 standard

deviation must be s 5 3 points.

A slight variation on Examples 5.4 and 5.5 is demonstrated in the following example.

This time you must use the z-score information to find both the population mean and the

standard deviation.

E X A M P L E 5 . 4

E X A M P L E 5 . 5

6 points

FIGURE 5.3

A visual presentation

of the question in

Example 5.4. If 2 standard

deviations correspond to a

6-point distance, then one

standard deviation must

equal 3 points.

1 3 0 CHAPTER 5 z-SCORES: LOCATION OF SCORES AND STANDARDIZED DISTRIBUTIONS

In a population distribution, a score of X 5 54 corresponds to z 5 12.00 and a score of X 5 42 corresponds to z 5 21.00. What are the values for the mean and the standard deviation for the population?

The key to solving this kind of problem is to focus on the distance between the

two scores. By subtraction, the distance between X 5 54 and X 5 42 is 12 points. The two z-scores also provide information about the distance between the scores.

A z-score of 12.00 indicates that X 5 52 is above the mean by 2.00 standard deviations, and z 5 21.00 indicates that X 5 44 is below the mean by 1.00 standard deviation. Thus, the total distance between the scores is equal to 2.00 1 1.00 5 3 standard deviations. At this point, you know that the distance between the scores

is 12 points and the distance is equal to 3 standard deviations. If a distance of

12 points is equal to 3 standard deviations, then the standard deviation must be

s 5 4. Now, you can pick either score and use the standard deviation to find the

value of the mean. For example, X 5 42 is below the mean by 1.00 standard devia-

tion. One standard deviation is 4 points, so the mean must be m 5 46. Thus, the

population has m 5 46 and s 5 4.

Once again, solving problems like the one is Example 5.6 can be easier if you sketch

a picture showing the information. Figure 5.4 shows a generic distribution (similar to

the one in Figure 5.2), for which the mean and the standard deviation are shown with

lines and arrows but are not assigned specific values. Next, we locate the two scores in

the distribution; X 5 42 is placed one standard deviation below the mean (z 5 21.00)

and X 5 54 is located two standard deviations above the mean (z 5 12.00). At this

point, you can see that the distance between the scores is 12 points, and that this

12-point distance corresponds to 3 standard deviations. Therefore, one standard devia-

tion is s 5 4. With a standard deviation of 4 points, you can see that X 5 42 is below the

mean by 4 points, so the mean must be µ 5 46. Again, we conclude that the population

has µ 5 46 and s 5 4.

E X A M P L E 5 . 6

1. For a distribution with µ 5 40 and s 5 12, find the z-score for each of the following

scores.

a. X 5 36 b. X 5 46 c. X 5 56

2. For a distribution with µ 5 40 and s 5 12, find the X value corresponding to each

of the following z-scores.

a. z 5 1.50 b. z 5 21.25 c. z 5 1

3. In a distribution with µ 5 50, a score of X 5 42 corresponds to z 5 22.00. What

is the standard deviation for this distribution?

4. In a distribution with s 5 12, a score of X 5 56 corresponds to z 5 20.25. What

is the mean for this distribution?

1. a. z 5 20.33 (or 2 1

3 ) b. z 5 0.50 c. z 5 1.33 ( 11 1

3 )

2. a. X 5 58 b. X 5 25 c. X 5 44

3. s 5 4

4. µ 5 59

L E A R N I N G C H E C K

ANSWERS

SECTION 5.3 / USING z-SCORES TO STANDARDIZE A DISTRIBUTION 1 3 1

54µ42 X

1 2–1–2 0

12 points

FIGURE 5.4

A visual presentation

of the question in

Example 5.6. The

12-point distance from

42 to 54 corresponds to

3 standard deviations.

Therefore, the standard

deviation must be s 5 4.

Also, the score X 5 42 is

below the mean by one

standard deviation, so the

mean must be µ 5 46.

USING z-SCORES TO STANDARDIZE A DISTRIBUTION

It is possible to transform every X value in a distribution into a corresponding z-score. The

result of this process is that the entire distribution of X values is transformed into a dis-

tribution of z-scores (Figure 5.5). The new distribution of z-scores has characteristics that

make the z-score transformation a very useful tool. Specifically, if every X value is trans-

formed into a z-score, then the distribution of z-scores will have the following properties:

1. Shape The distribution of z-scores will have exactly the same shape as the original

distribution of scores. If the original distribution is negatively skewed, for example,

then the z-score distribution will also be negatively skewed. If the original distribution

is normal, the distribution of z-scores will also be normal. Transforming raw scores into

5.3

Transform X to z

Population of scores (X values)

110 1201009080 µ

σ � 10

Population of z-scores (z values)

+1 +20−1−2 µ

σ � 1

FIGURE 5.5

An entire population of scores is transformed into z-scores. The transformation does not change the shape of the popula-

tion, but the mean is transformed into a value of 0 and the standard deviation is transformed to a value of 1.

1 3 2 CHAPTER 5 z-SCORES: LOCATION OF SCORES AND STANDARDIZED DISTRIBUTIONS

z-scores does not change anyone’s position in the distribution. For example, any raw

score that is above the mean by 1 standard deviation will be transformed to a z-score of

11.00, which is still above the mean by 1 standard deviation. Transforming a distribu- tion from X values to z values does not move scores from one position to another; the

procedure simply relabels each score (see Figure 5.5). Because each individual score

stays in its same position within the distribution, the overall shape of the distribution

does not change.

2. The mean The z-score distribution will always have a mean of zero. In Figure 5.5,

the original distribution of X values has a mean of µ 5 100. When this value, X 5 100, is transformed into a z-score, the result is

z X

5 2m

s 5

2 5

100 100

10 0

Thus, the original population mean is transformed into a value of zero in the z-score

distribution. The fact that the z-score distribution has a mean of zero makes the mean

a convenient reference point. Recall from the definition of z-scores that all positive

z-scores are above the mean and all negative z-scores are below the mean. In other

words, for z-scores, µ 5 0.

3. The standard deviation The distribution of z-scores will always have a standard

deviation of 1. In Figure 5.5, the original distribution of X values has µ 5 100 and

s 5 10. In this distribution, a value of X 5 110 is above the mean by exactly 10 points

or 1 standard deviation. When X 5 110 is transformed, it becomes z 5 11.00, which

is above the mean by exactly 1 point in the z-score distribution. Thus, the standard

deviation corresponds to a 10-point distance in the X distribution and is transformed

into a 1-point distance in the z-score distribution. The advantage of having a standard

deviation of 1 is that the numerical value of a z-score is exactly the same as the number

of standard deviations from the mean. For example, a z-score of z 5 1.50 is exactly

1.50 standard deviations from the mean.

In Figure 5.5, we showed the z-score transformation as a process that changed a

distribution of X values into a new distribution of z-scores. In fact, there is no need to

create a whole new distribution. Instead, you can think of the z-score transformation as

simply relabeling the values along the X-axis. That is, after a z-score transformation,

you still have the same distribution, but now each individual is labeled with a z-score

instead of an X value. Figure 5.6 demonstrates this concept with a single distribution

that has two sets of labels: the X values along one line and the corresponding z-scores

along another line. Notice that the mean for the distribution of z-scores is zero and the

standard deviation is 1.

When any distribution (with any mean or standard deviation) is transformed into

z-scores, the resulting distribution will always have a mean of µ 5 0 and a standard

deviation of s 5 1. Because all z-score distributions have the same mean and the same

standard deviation, the z-score distribution is called a standardized distribution.

A standardized distribution is composed of scores that have been transformed

to create predetermined values for µ and s. Standardized distributions are used to

make dissimilar distributions comparable.

A z-score distribution is an example of a standardized distribution with µ 5 0 and

s 5 1. That is, when any distribution (with any mean or standard deviation) is trans-

formed into z-scores, the transformed distribution will always have µ 5 0 and s 5 1.

D E F I N I T I O N

SECTION 5.3 / USING z-SCORES TO STANDARDIZE A DISTRIBUTION 1 3 3

Although the basic characteristics of a z-score distribution have been explained logi-

cally, the following example provides a concrete demonstration that a z-score transfor-

mation creates a new distribution with a mean of zero, a standard deviation of 1, and

the same shape as the original population.

We begin with a population of N 5 6 scores consisting of the following values: 0, 6, 5, 2, 3, 2. This population has a mean of m 5

18 6

5 3 and a standard deviation of s 5 2 (check the calculations for yourself).

Each of the X values in the original population is then transformed into a z-score as

summarized in the following table.

X 5 0 Below the mean by 1 1 2

standard deviations z 5 21.50

X 5 6 Above the mean by 1 1 2

standard deviations z 5 11.50

X 5 5 Above the mean by 1 standard deviation z 5 11.00

X 5 2 Below the mean by 1 2

standard deviation z 5 20.50

X 5 3 Exactly equal to the mean—zero deviation z 5 0

X 5 2 Below the mean by 1 2

standard deviation z 5 20.50

The frequency distribution for the original population of X values is shown in

Figure 5.7(a) and the corresponding distribution for the z-scores is shown in

Figure 5.7(b). A simple comparison of the two distributions demonstrates the results of

a z-score transformation.

1. The two distributions have exactly the same shape. Each individual has exactly

the same relative position in the X distribution and in the z-score distribution.

2. After the transformation to z-scores, the mean of the distribution becomes

µ 5 0. For these z-scores values, N 5 6 and oz 5 –1.50 1 1.50 1 1.00 1 20.50 1 0 1 20.50 5 0. Thus, the mean for the z-scores is µ 5

o z

N 5

0 6 5 0.

D E M O N ST R AT I O N O F A z - S CO R E

T R A N S F O R M AT I O N

E X A M P L E 5 . 7

z 0

100 110 1209080

−1−2 +1 +2

FIGURE 5.6

Following a z-score trans-

formation, the X-axis is

relabeled in z-score units.

The distance that is

equivalent to 1 standard

deviation on the X-axis

(s 5 10 points in this

example) corresponds to

1 point on the z-score

scale.

1 3 4 CHAPTER 5 z-SCORES: LOCATION OF SCORES AND STANDARDIZED DISTRIBUTIONS

Note that the individual with a score of X 5 3 is located exactly at the mean in the X distribution and this individual is transformed into z 5 0, exactly at the mean in the z-distribution.

3. After the transformation, the standard deviation becomes s 5 1. For these

z-scores, oz 5 0 and

oz2 5 (21.50)2 1 (1.50)2 1 (1.00)2 1 (20.50)2 1 (0)2 1 (20.50)2

5 2.25 1 2.25 1 1.00 1 0.25 1 0 1 0.25

5 6.00

Using the computational formula for SS, substituting z in place of X, we obtain

SS N

5 2 5 2 5o o

z z2 ( ) ( )

2 2

6 0

6 6 00.

For these z-scores, the variance is s 5 5 5 2 SS

6 1.00 and the standard deviation is

s 5 51.00 1.00

Note that the individual with X 5 5 is located above the mean by 2 points, which

is exactly one standard deviation in the X distribution. After transformation, this indi-

vidual has a z-score that is located above the mean by 1 point, which is exactly one

standard deviation in the z-score distribution.

One advantage of standardizing distributions is that it makes it possible to compare dif-

ferent scores or different individuals even though they come from completely different

distributions. Normally, if two scores come from different distributions, it is impossible

U S I N G z - S CO R E S TO M A K E CO M PA R I S O N S

F re

q u

e n

c y

1 2 3

X 4

5 6

–1.5

F re

q u

e n

c y

–1.0 –0.5 0

z +0.5

+1.0 +1.5

FIGURE 5.7

Transforming a distribu-

tion of raw scores

(a) into z-scores (b) will

not change the shape of

the distribution.

(a)

(b)

SECTION 5.3 / USING z-SCORES TO STANDARDIZE A DISTRIBUTION 1 3 5

to make any direct comparison between them. Suppose, for example, Dave received a

score of X 5 60 on a psychology exam and a score of X 5 56 on a biology test. For which course should Dave expect the better grade?

Because the scores come from two different distributions, you cannot make any direct

comparison. Without additional information, it is even impossible to determine whether

Dave is above or below the mean in either distribution. Before you can begin to make

comparisons, you must know the values for the mean and standard deviation for each dis-

tribution. Suppose the psychology scores had µ 5 50 and s 5 10, and the biology scores had µ 5 48 and s 5 4. With this new information, you could sketch the two distributions,

locate Dave’s score in each distribution, and compare the two locations.

Instead of drawing the two distributions to compare Dave’s two scores, we simply

can compute the corresponding z-scores to determine the two locations. For psychology,

Dave’s z-score is

z 5 2m

s 5

2 5 51

X 60 50

10 1.0

For biology, Dave’s z-score is

z 5 2

5 51 56 48

4 2.0

Note that Dave’s z-score for biology is 12.0, which means that his test score is

2 standard deviations above the class mean. On the other hand, his z-score is 11.0 for

psychology, or 1 standard deviation above the mean. In terms of relative class standing,

Dave is doing much better in the biology class.

Notice that we cannot compare Dave’s two exam scores (X 5 60 and X 5 56)

because the scores come from different distributions with different means and standard

deviations. However, we can compare the two z-scores because all distributions of

z-scores have the same mean (µ 5 0) and the same standard deviation (s 5 1).

Be sure to use the µ and

s values for the distribution

to which X belongs.

1. A normal-shaped distribution with µ 5 40 and s 5 8 is transformed into z-scores.

Describe the shape, the mean, and the standard deviation for the resulting distribution

of z-scores.

2. What is the advantage of having a mean of µ 5 0 for a distribution of z-scores?

3. A distribution of English exam scores has µ 5 70 and s 5 4. A distribution of

history exam scores has µ 5 60 and s 5 20. For which exam would a score of

X 5 78 have a higher standing? Explain your answer.

4. A distribution of English exam scores has µ 5 50 and s 5 12. A distribution of

history exam scores has µ 5 58 and s 5 4. For which exam would a score of

X 5 62 have a higher standing? Explain your answer.

1. The z-score distribution would be normal with a mean of 0 and a standard deviation of 1.

2. With a mean of zero, all positive scores are above the mean and all negative scores are

below the mean.

3. For the English exam, X 5 78 corresponds to z 5 2.00, which is a higher standing than

z 5 0.90 for the history exam.

4. The score X 5 62 corresponds to z 5 11.00 in both distributions. The score has exactly the

same standing for both exams.

L E A R N I N G C H E C K

ANSWERS

1 3 6 CHAPTER 5 z-SCORES: LOCATION OF SCORES AND STANDARDIZED DISTRIBUTIONS

OTHER STANDARDIZED DISTRIBUTIONS BASED ON z-SCORES

Although z-score distributions have distinct advantages, many people find them cum-

bersome because they contain negative values and decimals. For this reason, it is com-

mon to standardize a distribution by transforming the scores into a new distribution

with a predetermined mean and standard deviation that are whole round numbers.

The goal is to create a new (standardized) distribution that has “simple” values for the

mean and standard deviation but does not change any individual’s location within the

distribution. Standardized scores of this type are frequently used in psychological or

educational testing. For example, raw scores of the Scholastic Aptitude Test (SAT) are

transformed to a standardized distribution that has µ 5 500 and s 5 100. For intelli- gence tests, raw scores are frequently converted to standard scores that have a mean of

100 and a standard deviation of 15. Because most IQ tests are standardized so that they

have the same mean and standard deviation, it is possible to compare IQ scores even

though they may come from different tests.

The procedure for standardizing a distribution to create new values for µ and s

involves two steps:

1. The original raw scores are transformed into z-scores.

2. The z-scores are then transformed into new X values so that the specific µ and s

are attained.

This procedure ensures that each individual has exactly the same z-score location in

the new distribution as in the original distribution. The following example demonstrates

the standardization procedure.

A distribution of exam scores has a mean of µ 5 57 with s 5 14. The instructor would

like to simplify the distribution by transforming all scores into a new, standardized

distribution with µ 5 50 and s 5 10. To demonstrate this process, we consider what

happens to two specific students: Maria, who has a score of X 5 64 in the original

distribution; and Joe, whose original score is X 5 43.

Step 1 Transform each of the original scores into z-scores. Maria started with

X 5 64, so her z-score is

z X

5 2m

s 5

2 51

64 57

14 0.5

For Joe, X 5 43, and his z-score is

z X

5 2m

s 5

2 52

43 57

14 1.0

Remember: The values of µ and s are for the distribution from which X was taken.

Step 2 Change each z-score into an X value in the new standardized distribution that

has a mean of µ 5 50 and a standard deviation of s 5 10.

Maria’s z-score, z 5 10.50, indicates that she is located above the mean by 1 2

standard deviation. In the new, standardized distribution, this location corresponds

to X 5 55 (above the mean by 5 points).

5.4

T R A N S F O R M I N G z - S CO R E S TO

A D I ST R I B U T I O N W I T H A P R E D E T E R M I N E D

m A N D 

E X A M P L E 5 . 8

SECTION 5.4 / OTHER STANDARDIZED DISTRIBUTIONS BASED ON z-SCORES 1 3 7

Joe’s z-score, z 5 –1.00, indicates that he is located below the mean by exactly 1 standard deviation. In the new distribution, this location corresponds to X 5 40 (below the mean by 10 points).

The results of this two-step transformation process are summarized in Table 5.1.

Note that Joe, for example, has exactly the same z-score (z 5 –1.00) in both the original distribution and the new standardized distribution. This means that Joe’s position rela-

tive to the other students in the class has not changed.

Figure 5.8 provides another demonstration of the concept that standardizing a distribu-

tion does not change the individual positions within the distribution. The figure shows the

original exam scores from Example 5.7, with a mean of µ 5 57 and a standard deviation of s 5 14. In the original distribution, Joe is located at a score of X 5 43. In addition to

the original scores, we have included a second scale showing the z-score value for each

location in the distribution. In terms of z-scores, Joe is located at a value of z 5 –1.00.

Finally, we have added a third scale showing the standardized scores, for which the mean

is µ 5 50 and the standard deviation is s 5 10. For the standardized scores, Joe is located

at X 5 40. Note that Joe is always in the same place in the distribution. The only thing

that changes is the number that is assigned to Joe: For the original scores, Joe is at 43; for

the z-scores, Joe is at –1.00; and for the standardized scores, Joe is at 40.

TABLE 5.1

A demonstration of how two

individual scores are changed

when a distribution is standard-

ized. See Example 5.8.

Original Scores µ 5 57 and s 5 14

z-Score Location

Standardized Scores µ 5 50 and s 5 10

Maria X 5 64 → z 5 10.50 → X 5 55

Joe X 5 43 → z 5 21.00 → X 5 40

�2

�1 �1 �2

Joe

�� Original scores (� � 57 and � � 14)

�� z-Scores (� � 0 and � � 1)

�� Standardized scores (� � 50 and � � 10)

FIGURE 5.8

The distribution of exam scores from Example 5.7. The original distribution was standardized to

produce a new distribution with µ 5 50 and s 5 10. Note that each individual is identified by an

original score, a z-score, and a new, standardized score. For example, Joe has an original score of

43, a z-score of –100, and a standardized score of 40.

1 3 8 CHAPTER 5 z-SCORES: LOCATION OF SCORES AND STANDARDIZED DISTRIBUTIONS

COMPUTING z-SCORES FOR A SAMPLE

Although z-scores have been presented in the context of a population, the same prin-

ciples can be used to compute z-scores within a sample. The definition of a z-score is

the same for a sample as for a population, provided that you use the sample mean and

the sample standard deviation to specify each z-score location. Thus, for a sample, each

X value is transformed into a z-score so that

1. The sign of the z-score indicates whether the X value is above (1) or below (–) the sample mean, and

2. The numerical value of the z-score identifies the distance from the sample mean

by measuring the number of sample standard deviations between the score (X)

and the sample mean (M).

Expressed as a formula, each X value in a sample can be transformed into a z-score

as follows:

z X M

s 5

(5.3)

Similarly, each z-score can be transformed back into an X value, as follows:

X 5 M 1 zs (5.4)

In a sample with a mean of M 5 40 and a standard deviation of s 5 10, what is the z-score corresponding to X 5 35 and what is the X value corresponding to z 5 12.00?

The score, X 5 35, is located below the mean by 5 points, which is exactly half of the standard deviation. Therefore, the corresponding z-score is z 5 20.50. The z-score, z 5 12.00, corresponds to a location above the mean by 2 standard deviations. With a

5.5

E X A M P L E 5 . 9

1. A population of scores has µ 5 73 and s 5 8. If the distribution is standardized to create a new distribution with µ 5 100 and s 5 20, what are the new values for

each of the following scores from the original distribution?

a. X 5 65 b. X 5 71 c. X 5 81 d. X 5 83

2. A population with a mean of µ 5 44 and a standard deviation of s 5 6 is

standardized to create a new distribution with µ 5 50 and s 5 10.

a. What is the new standardized value for a score of X 5 47 from the original

distribution?

b. One individual has a new standardized score of X 5 65. What was this person’s

score in the original distribution?

1. a. z 5 21.00, X 5 80 b. z 5 20.25, X 5 95

c. z 5 1.00, X 5 120 d. z 5 1.25, X 5 125

2. a. X 5 47 corresponds to z 5 10.50 in the original distribution. In the new distribution, the

corresponding score is X 5 55.

b. In the new distribution, X 5 65 corresponds to z 5 11.50. The corresponding score in

the original distribution is X 5 53.

L E A R N I N G C H E C K

ANSWERS

See the population equations

(5.1 and 5.2) on pages 126

and 128 for comparison.

SECTION 5.5 / COMPUTING z-SCORES FOR A SAMPLE 1 3 9

standard deviation of s 5 10, this is a distance of 20 points. The score that is located 20 points above the mean is X 5 60. Note that it is possible to find these answers using either the z-score definition or one of the equations (5.3 or 5.4).

If all the scores in a sample are transformed into z-scores, the result is a sample of

z-scores. The transformed distribution of z-scores will have the same properties that

exist when a population of X values is transformed into z-scores. Specifically,

1. The sample of z-scores will have the same shape as the original sample of

scores.

2. The sample of z-scores will have a mean of M z 5 0.

3. The sample of z-scores will have a standard deviation of s z 5 1.

Note that the set of z-scores is still considered to be a sample (just like the set of

X values) and the sample formulas must be used to compute variance and standard

deviation. The following example demonstrates the process of transforming the scores

from a sample into z-scores.

We begin with a sample of n 5 5 scores: 0, 2, 4, 4, 5. With a few simple calculations, you should be able to verify that the sample mean is M 5 3, the sample variance is s2 5 4, and the sample standard deviation is s 5 2. Using the sample mean and sample standard deviation, we can convert each X value into a z-score. For example, X 5 5 is located above the mean by 2 points. Thus, X 5 5 is above the mean by exactly 1 standard deviation and has a z-score of z 5 11.00. The z-scores for the entire sample are shown in the following table.

X z

0 21.50

2 20.50

4 10.50

5 11.00

Again, a few simple calculations demonstrate that the sum of the z-score values is

oz 5 0, so the mean is M z 5 0.

Because the mean is zero, each z-score value is its own deviation from the mean.

Therefore, the sum of the squared z-scores is also the sum of the squared deviations.

For this sample of z-scores,

SS 5 oz2 5 (–1.50)2 1 (20.50)2 1 (10.50)2 1 (0.50)2 1 (11.00)2

5 2.25 1 0.25 1 0.25 1 0.25 1 1.00

5 4.00

STA N DA R D I Z I N G A SA M P L E

D I ST R I B U T I O N

E X A M P L E 5 . 1 0

1 4 0 CHAPTER 5 z-SCORES: LOCATION OF SCORES AND STANDARDIZED DISTRIBUTIONS

The variance for the sample of z-scores is

s SS

n z

4 1.005

2 5 5

Finally, the standard deviation for the sample of z-scores is s z 5 1 00. 5 1.00. As

always, the distribution of z-scores has a mean of 0 and a standard deviation of 1.

LOOKING AHEAD TO INFERENTIAL STATISTICS

Recall that inferential statistics are techniques that use the information from samples to

answer questions about populations. In later chapters, we use inferential statistics to help

interpret the results from research studies. A typical research study begins with a question

about how a treatment will affect the individuals in a population. Because it is usually

impossible to study an entire population, the researcher selects a sample and administers

the treatment to the individuals in the sample. This general research situation is shown

in Figure 5.9. To evaluate the effect of the treatment, the researcher simply compares the

treated sample with the original population. If the individuals in the sample are noticeably

different from the individuals in the original population, the researcher has evidence that

the treatment has had an effect. On the other hand, if the sample is not noticeably different

from the original population, it would appear that the treatment has had no effect.

Notice that the interpretation of the research results depends on whether the sample is

noticeably different from the population. One technique for deciding whether a sample

is noticeably different is to use z-scores. For example, an individual with a z-score near

0 is located in the center of the population and would be considered to be a fairly typical

or representative individual. However, an individual with an extreme z-score, beyond

5.6

Original population

(Without treatment)

Sample Treated sample

T r e a t

m e n t

FIGURE 5.9

A diagram of a research

study. The goal of the

study is to evaluate the

effect of a treatment. A

sample is selected from

the population, and the

treatment is administered

to the sample. If, after

treatment, the individuals

in the sample are

noticeably different

from the individuals in

the original population,

then we have evidence

that the treatment does

have an effect.

Notice that the set of z-scores

is considered to be a sample

and the variance is computed

using the sample formula

with df 5 n – 1.

SECTION 5.6 / LOOKING AHEAD TO INFERENTIAL STATISTICS 1 4 1

12.00 or 22.00 for example, would be considered noticeably different from most of the individuals in the population. Thus, we can use z-scores to help decide whether the treat-

ment has caused a change. Specifically, if the individuals who receive the treatment in

a research study tend to have extreme z-scores, we can conclude that the treatment does

appear to have an effect. The following example demonstrates this process.

A researcher is evaluating the effect of a new growth hormone. It is known that regular

adult rats weigh an average of µ 5 400 grams. The weights vary from rat to rat, and the distribution of weights is normal with a standard deviation of s 5 20 grams. The popu-

lation distribution is shown in Figure 5.10. The researcher selects one newborn rat and

injects the rat with the growth hormone. When the rat reaches maturity, it is weighed to

determine whether there is any evidence that the hormone has an effect.

First, assume that the hormone-injected rat weighs X 5 418 grams. Although this is

more than the average nontreated rat (µ 5 400 grams), is it convincing evidence that the

hormone has an effect? If you look at the distribution in Figure 5.10, you should realize

that a rat weighing 418 grams is not noticeably different from the regular rats that did

not receive any hormone injection. Specifically, our injected rat would be located near

the center of the distribution for regular rats with a z-score of

z X

5 2m

s 5

2 5 5

418 400

20 0.90

Because the injected rat still looks the same as a regular, nontreated rat, the conclu-

sion is that the hormone does not appear to have an effect.

Now, assume that our injected rat weighs X 5 450 grams. In the distribution of

regular rats (see Figure 5.10), this animal would have a z-score of

z X

5 2m

s 5

2 5 5

450 400

20 2.50

E X A M P L E 5 . 1 1

X 5 450

m 5 400 440420380360

z 0

Population of

nontreated rats

Representative individuals (z near 0)

Extreme individuals

(z beyond 12.00)

Extreme individuals

(z beyond 22.00)

1.00 2.0021.0022.00

X 5 418

FIGURE 5.10

The distribution of weights

for the population of adult

rats. Note that individuals

with z-scores near 0 are

typical or representative.

However, individuals with

z-scores beyond 12.00

or –2.00 are extreme and

noticeably different from

most of the others in the

distribution.

1 4 2 CHAPTER 5 z-SCORES: LOCATION OF SCORES AND STANDARDIZED DISTRIBUTIONS

In this case, the hormone-injected rat is substantially bigger than most ordinary rats,

and it would be reasonable to conclude that the hormone does have an effect on weight.

In the preceding example, we used z-scores to help interpret the results obtained

from a sample. Specifically, if the individuals who receive the treatment in a research

study have extreme z-scores compared to those who do not receive the treatment, we

can conclude that the treatment does appear to have an effect. The example, however,

used an arbitrary definition to determine which z-score values are noticeably differ-

ent. Although it is reasonable to describe individuals with z-scores near 0 as “highly

representative” of the population, and individuals with z-scores beyond 62.00 as “extreme,” you should realize that these z-score boundaries were not determined

by any mathematical rule. In the following chapter we introduce probability, which

gives us a rationale for deciding exactly where to set the boundaries.

1. For a sample with a mean of M 5 40 and a standard deviation of s 5 12, find the z-score corresponding to each of the following X values.

X 5 43 X 5 58 X 5 49

X 5 34 X 5 28 X 5 16

2. For a sample with a mean of M 5 80 and a standard deviation of s 5 20, find the X value corresponding to each of the following z-scores.

z 5 21.00 z 5 20.50 z 5 20.20

z 5 1.50 z 5 0.80 z 5 1.40

3. For a sample with a mean of M 5 85, a score of X 5 80 corresponds to z 5 20.50. What is the standard deviation for the sample?

4. For a sample with a standard deviation of s 5 12, a score of X 5 83 corresponds to z 5 0.50. What is the mean for the sample?

5. A sample has a mean of M 5 30 and a standard deviation of s 5 8.

a. Would a score of X 5 36 be considered a central score or an extreme score in the sample?

b. If the standard deviation were s 5 2, would X 5 36 be central or extreme?

1. z 5 0.25 z 5 1.50 z 5 0.75

z 5 20.50 z 5 21.00 z 5 22.00

2. X 5 60 X 5 70 X 5 76

X 5 110 X 5 96 X 5 108

3. s 5 10

4. M 5 77

5. a. X 5 36 is a central score corresponding to z 5 0.75.

b. X 5 36 would be an extreme score corresponding to z 5 3.00.

L E A R N I N G C H E C K

ANSWERS

RESOURCES 1 4 3

RESOURCES

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SUMMARY

1. Each X value can be transformed into a z-score that specifies the exact location of X within the distribu- tion. The sign of the z-score indicates whether the lo- cation is above the mean (positive) or below the mean (negative). The numerical value of the z-score specifies the number of standard deviations between X and µ.

2. The z-score formula is used to transform X values into z-scores. For a population:

z 5 2m

For a sample:

z X M

s 5

3. To transform z-scores back into X values, it usually is easier to use the z-score definition rather than a formula. However, the z-score formula can be trans- formed into a new equation.

For a population: X 5 µ 1 zs

For a sample: X 5 M 1 zs

4. When an entire distribution of X values is transformed into z-scores, the result is a distribution of z-scores.

The z-score distribution will have the same shape as the distribution of raw scores, and it always will have a mean of 0 and a standard deviation of 1.

5. One method for comparing scores from different distributions is to standardize the distributions with a z-score transformation. The distributions will then be comparable because they will have the same parameters (µ 5 0, s 5 1). In practice, it is necessary to transform only those raw scores that are being compared.

6. A distribution also can be standardized by converting the original X values into z-scores and then convert- ing the z-scores into a new distribution of scores with predetermined values for the mean and the standard deviation.

7. In inferential statistics, z-scores provide an objective method for determining how well a specific score represents its population. A z-score near 0 indicates that the score is close to the population mean and, therefore, is representative. A z-score beyond 12.00 (or 22.00) indicates that the score is extreme and is noticeably different from the other scores in the distribution.

KEY TERMS

raw score (125)

z-score (125)

deviation score (127)

z-score transformation (131)

standardized distribution (132)

standardized score (137)

1 4 4 CHAPTER 5 z-SCORES: LOCATION OF SCORES AND STANDARDIZED DISTRIBUTIONS

General instructions for using SPSS are presented in Appendix D. Following are

detailed instructions for using SPSS to Transform X Values into z-Scores for a

Sample.

Data Entry

Enter all of the scores in one column of the data editor, probably VAR00001.

Data Analysis

1. Click Analyze on the tool bar, select Descriptive Statistics, and click on

Descriptives.

2. Highlight the column label for the set of scores (VAR0001) in the left box and click

the arrow to move it into the Variable box.

3. Click the box to Save standardized values as variables at the bottom of the

Descriptives screen.

4. Click OK.

SPSS Output

The program produces the usual output display listing the number of scores (N), the

maximum and minimum scores, the mean, and the standard deviation. However, if you

go back to the Data Editor (use the tool bar at the bottom of the screen), you can see

that SPSS has produced a new column showing the z-score corresponding to each of

the original X values.

Caution: The SPSS program computes the z-scores using the sample standard devia-

tion instead of the population standard deviation. If your set of scores is intended to be

a population, SPSS does not produce the correct z-score values. You can convert the

SPSS values into population z-scores by multiplying each z-score value by the square

root of n

n( 21) .

FOCUS ON PROBLEM SOLVING

1. When you are converting an X value to a z-score (or vice versa), do not rely

entirely on the formula. You can avoid careless mistakes if you use the defini-

tion of a z-score (sign and numerical value) to make a preliminary estimate

of the answer before you begin computations. For example, a z-score of

z 5 20.85 identifies a score located below the mean by almost 1 standard deviation. When computing the X value for this z-score, be sure that your

answer is smaller than the mean, and check that the distance between X and

µ is slightly less than the standard deviation.

2. When comparing scores from distributions that have different standard devia-

tions, it is important to be sure that you use the correct values for µ and s in

the z-score formula. Use the values for the distribution from which the score

was taken.

3. Remember that a z-score specifies a relative position within the context of a

specific distribution. A z-score is a relative value, not an absolute value. For

DEMONSTRATION 5.2 1 4 5

example, a z-score of z 5 22.0 does not necessarily suggest a very low raw score—it simply means that the raw score is among the lowest within that

specific group.

DEMONSTRATION 5.1

TRANSFORMING X VALUES INTO z-SCORES

A distribution of scores has a mean of µ 5 60 with s 5 12. Find the z-score for X 5 75.

Determine the sign of the z-score.

First, determine whether X is above or below the mean. This determines the sign of the

z-score. For this demonstration, X is larger than (above) µ, so the z-score is positive.

Convert the distance between X and µ into standard deviation units.

For X 5 75 and µ 5 60, the distance between X and µ is 15 points. With s 5 12

points, this distance corresponds to 15 12

5 1.25 standard deviations.

Combine the sign from step 1 with the numerical value from step 2.

The score is above the mean (1) by a distance of 1.25 standard deviations. Thus,

z 5 11.25.

Confirm the answer using the z-score formula.

For this example, X 5 75, µ 5 60, and s 5 12.

z X

5 2m

s 5

2 5

1 51

75 60

12 1.25

DEMONSTRATION 5.2

CONVERTING z-SCORES TO X VALUES

For a population with µ 5 60 and s 5 12, what is the X value corresponding to

z 5 20.50?

Locate X in relation to the mean.

A z-score of 20.50 indicates a location below the mean by half of a standard deviation.

Convert the distance from standard deviation units to points.

With s 5 12, half of a standard deviation is 6 points.

Identify the X value.

The value we want is located below the mean by 6 points. The mean is µ 5 60, so the

score must be X 5 54.

S T E P 1

S T E P 2

S T E P 3

S T E P 4

S T E P 1

S T E P 2

S T E P 3

1 4 6 CHAPTER 5 z-SCORES: LOCATION OF SCORES AND STANDARDIZED DISTRIBUTIONS

PROBLEMS

1. What information is provided by the sign (1/–) of a z-score? What information is provided by the numerical value of the z-score?

2. A distribution has a standard deviation of s 5 10. Find the z-score for each of the following locations in the distribution.

a. Above the mean by 5 points. b. Above the mean by 2 points. c. Below the mean by 20 points. d. Below the mean by 15 points.

3. For a distribution with a standard deviation of s 5 20, describe the location of each of the following z-scores in terms of its position relative to the mean. For example, z 5 11.00 is a location that is 20 points above the mean.

a. z 5 12.00 b. z 5 10.50 c. z 5 21.00 d. z 5 20.25

4. For a population with µ 5 80 and s 5 10, a. Find the z-score for each of the following X values.

(Note: You should be able to find these values using the definition of a z-score. You should not need to use a formula or do any serious calculations.)

X 5 75 X 5 100 X 5 60

X 5 95 X 5 50 X 5 85 b. Find the score (X value) that corresponds to each

of the following z-scores. (Again, you should not need a formula or any serious calculations.)

z 5 1.00 z 5 0.20 z 5 1.50

z 5 20.50 z 5 22.00 z 5 21.50

5. For a population with µ 5 40 and s 5 11, find the z-score for each of the following X values. (Note: You probably will need to use a formula and a calculator to find these values.)

X 5 45 X 5 52 X 5 41 X 5 30 X 5 25 X 5 38

6. For a population with a mean of µ 5 100 and a standard deviation of s 5 20,

a. Find the z-score for each of the following X values.

X 5 108 X 5 115 X 5 130

X 5 90 X 5 88 X 5 95 b. Find the score (X value) that corresponds to each

of the following z-scores.

z 5 20.40 z 5 20.50 z 5 1.80

z 5 0.75 z 5 1.50 z 5 21.25

7. A population has a mean of µ 5 60 and a standard deviation of s 5 12.

a. For this population, find the z-score for each of the following X values.

X 5 69 X 5 84 X 5 63

X 5 54 X 5 48 X 5 45

b. For the same population, find the score (X value) that corresponds to each of the following z-scores.

z 5 0.50 z 5 1.50 z 5 22.50 z 5 20.25 z 5 20.50 z 5 1.25

8. A sample has a mean of M 5 30 and a standard deviation of s 5 8. Find the z-score for each of the following X values from this sample.

X 5 32 X 5 34 X 5 36

X 5 28 X 5 20 X 5 18

9. A sample has a mean of M 5 25 and a standard deviation of s 5 5. For this sample, find the X value corresponding to each of the following z-scores.

z 5 0.40 z 5 1.20 z 5 2.00

z 5 20.80 z 5 20.60 z 5 21.40

10. Find the z-score corresponding to a score of X 5 45 for each of the following distributions.

a. µ 5 40 and s 5 20 b. µ 5 40 and s 5 10 c. µ 5 40 and s 5 5 d. µ 5 40 and s 5 2

11. Find the X value corresponding to z 5 0.25 for each of the following distributions.

a. µ 5 40 and s 5 4 b. µ 5 40 and s 5 8 c. µ 5 40 and s 5 16 d. µ 5 40 and s 5 32

12. A score that is 6 points below the mean corresponds to a z-score of z 5 22.00. What is the population standard deviation?

13. A score that is 9 points above the mean corresponds to a z-score of z 5 1.50. What is the population standard deviation?

14. For a population with a standard deviation of s 5 12, a score of X 5 44 corresponds to z 5 20.50. What is the population mean?

15. For a sample with a standard deviation of s 5 8, a score of X 5 65 corresponds to z 5 1.50. What is the sample mean?

16. For a sample with a mean of M 5 51, a score of X 5 59 corresponds to z 5 2.00. What is the sample standard deviation?

PROBLEMS 1 4 7

17. For a population with a mean of µ 5 70, a score of X 5 64 corresponds to z 5 21.50. What is the population standard deviation?

18. In a population distribution, a score of X 5 28 cor- responds to z 5 21.00 and a score of X 5 34 cor- responds to z 5 20.50. Find the mean and standard deviation for the population. (Hint: Sketch the distri- bution and locate the two scores on your sketch.)

19. In a sample distribution, X 5 56 corresponds to z 5 1.00, and X 5 47 corresponds to z 5 20.50. Find the mean and standard deviation for the sample.

20. For each of the following populations, would a score of X 5 50 be considered a central score (near the middle of the distribution) or an extreme score (far out in the tail of the distribution)?

a. µ 5 45 and s 5 10 b. µ 5 45 and s 5 2 c. µ 5 90 and s 5 20 d. µ 5 60 and s 5 20

21. A distribution of exam scores has a mean of µ 5 78. a. If your score is X 5 70, which standard deviation

would give you a better grade: s 5 4 or s 5 8? b. If your score is X 5 80, which standard deviation

would give you a better grade: s 5 4 or s 5 8?

22. For each of the following, identify the exam score that should lead to the better grade. In each case, explain your answer.

a. A score of X 5 74 on an exam with M 5 82 and s 5 8; or a score of X 5 40 on an exam with µ 5 50 and s 5 20.

b. A score of X 5 51 on an exam with µ 5 45 and s 5 2; or a score of X 5 90 on an exam with µ 5 70 and s 5 20.

c. A score of X 5 62 on an exam with µ 5 50 and s 5 8; or a score of X 5 23 on an exam with µ 5 20 and s 5 2.

23. A distribution with a mean of µ 5 38 and a standard deviation of s 5 5 is transformed into a standardized distribution with µ 5 50 and s 5 10. Find the new, standardized score for each of the following values from the original population.

a. X 5 39 b. X 5 43 c. X 5 35 d. X 5 28

24. A distribution with a mean of µ 5 76 and a standard deviation of s 5 12 is transformed into a standard- ized distribution with µ 5 100 and s 5 20. Find the new, standardized score for each of the following values from the original population.

a. X 5 61 b. X 5 70 c. X 5 85 d. X 5 94

25. A population consists of the following N 5 5 scores: 0, 6, 4, 3, and 12.

a. Compute µ and s for the population. b. Find the z-score for each score in the population. c. Transform the original population into a new

population of N 5 5 scores with a mean of µ 5 100 and a standard deviation of s 5 20.

26. A sample consists of the following n 5 7 scores: 5, 0, 4, 5, 1, 2, and 4.

a. Compute the mean and standard deviation for the sample.

b. Find the z-score for each score in the sample. c. Transform the original sample into a new sample

with a mean of M 5 50 and s 5 10.

Aplia for Essentials of Statistics for the Behavioral Sciences

After reading, go to “Resources” at the end of this chapter for

an introduction on how to use Aplia’s homework and learning

resources.

Probability

6.1 Introduction to Probability

6.2 Probability and the Normal Distribution

6.3 Probabilities and Proportions for Scores from a Normal Distribution

6.4 Looking Ahead to Inferential Statistics

Summary

Focus on Problem Solving

Demonstration 6.1

Problems

C H A P T E R

6 Tools You Will Need The following items are considered essential background material for this chapter. If you doubt your knowledge of any of these items, you should review the appropriate chapter or section before proceeding.

• Proportions (math review, Appendix A)

• Fractions • Decimals • Percentages • Basic algebra (math review,

Appendix A) • z-Scores (Chapter 5)

1 5 0 CHAPTER 6 PROBABILITY

INTRODUCTION TO PROBABILITY

In Chapter 1, we introduced the idea that research studies begin with a general question

about an entire population, but the actual research is conducted using a sample. In this

situation, the role of inferential statistics is to use the sample data as the basis for an-

swering questions about the population. To accomplish this goal, inferential procedures

are typically built around the concept of probability. Specifically, the relationships be-

tween samples and populations are usually defined in terms of probability.

Suppose, for example, that you are selecting a single marble from a jar that contains

50 black and 50 white marbles. (In this example, the jar of marbles is the population

and the single marble to be selected is the sample.) Although you cannot guarantee the

exact outcome of your sample, it is possible to talk about the potential outcomes in

terms of probabilities. In this case, you have a 50-50 chance of getting either color. Now

consider another jar (population) that has 90 black and only 10 white marbles. Again,

you cannot predict the exact outcome of a sample, but now you know that the sample

probably will be a black marble. By knowing the makeup of a population, we can de-

termine the probability of obtaining specific samples. In this way, probability gives us

a connection between populations and samples, and this connection is the foundation

for the inferential statistics that are presented in the chapters that follow.

You may have noticed that the preceding examples begin with a population and

then use probability to describe the samples that could be obtained. This is exactly

backward from what we want to do with inferential statistics. Remember that the

goal of inferential statistics is to begin with a sample and then answer a general

question about the population. We reach this goal in a two-stage process. In the

first stage, we develop probability as a bridge from populations to samples. This

stage involves identifying the types of samples that probably would be obtained

from a specific population. Once this bridge is established, we simply reverse the

probability rules to allow us to move from samples to populations (Figure 6.1). The

process of reversing the probability relationship can be demonstrated by considering

again the two jars of marbles we looked at earlier. (Jar 1 has 50 black and 50 white

marbles; jar 2 has 90 black and only 10 white marbles.) This time, suppose you are

blindfolded when the sample is selected, so you do not know which jar is being used.

6.1

SamplePopulation

INFERENTIAL STATISTICS

PROBABILITY

FIGURE 6.1

The role of probability

in inferential statistics.

Probability is used to pre-

dict what kind of samples

are likely to be obtained

from a population. Thus,

probability establishes

a connection between

samples and populations.

Inferential statistics rely

on this connection when

they use sample data as the

basis for making conclu-

sions about populations.

SECTION 6.1 / INTRODUCTION TO PROBABILITY 1 5 1

Your task is to look at the sample that you obtain and then decide which jar is most

likely. If you select a sample of n 5 4 marbles and all are black, which jar would you choose? It should be clear that it would be relatively unlikely (low probability) to

obtain this sample from jar 1; in four draws, you almost certainly would get at least 1

white marble. On the other hand, this sample would have a high probability of com-

ing from jar 2, where nearly all of the marbles are black. Your decision, therefore, is

that the sample probably came from jar 2. Note that you now are using the sample

to make an inference about the population.

Probability is a huge topic that extends far beyond the limits of introductory statistics,

and we do not attempt to examine it all here. Instead, we concentrate on the few con-

cepts and definitions that are needed for an introduction to inferential statistics. We

begin with a relatively simple definition of probability.

For a situation in which several different outcomes are possible, the probability

for any specific outcome is defined as a fraction or a proportion of all the possible

outcomes. If the possible outcomes are identified as A, B, C, D, and so on, then

probability of number of outcomes classif

A 5 iied as

total number of possible outcomes

For example, if you are selecting a card from a complete deck, there are 52 possible

outcomes. The probability of selecting the king of hearts is p 5 1 52

. The probability of

selecting an ace is p 5 4 52

because there are 4 aces in the deck.

To simplify the discussion of probability, we use a notation system that eliminates

a lot of the words. The probability of a specific outcome is expressed with a p (for

probability) followed by the specific outcome in parentheses. For example, the prob-

ability of selecting a king from a deck of cards is written as p(king). The probability

of obtaining heads for a coin toss is written as p(heads).

Note that probability is defined as a proportion, or a part of the whole. This defini-

tion makes it possible to restate any probability problem as a proportion problem. For

example, the probability problem “What is the probability of selecting a king from

a deck of cards?” can be restated as “What proportion of the whole deck consists of

kings?” In each case, the answer is 4 52

, or “4 out of 52.” This translation from probability

to proportion may seem trivial now, but it is a great aid when the probability problems

become more complex. In most situations, we are concerned with the probability of

obtaining a particular sample from a population. The terminology of sample and popu-

lation do change the basic definition of probability. For example, the whole deck of

cards can be considered as a population, and the single card we select is the sample.

Probability values The definition we are using identifies probability as a fraction or

a proportion. If you work directly from this definition, the probability values you obtain

are expressed as fractions. For example, if you are selecting a card at random,

p(spade) 13

4 5 5

Or, if you are tossing a coin,

p(heads) 1

2 5

D E F I N I N G P R O BA B I L I T Y

D E F I N I T I O N

1 5 2 CHAPTER 6 PROBABILITY

You should be aware that these fractions can be expressed equally well as either

decimals or percentages:

5 5 5

4 0.25 25%

2 0.50 50%

By convention, probability values most often are expressed as decimal values. But

you should realize that any of these three forms is acceptable.

You also should note that all of the possible probability values are contained in a

limited range. At one extreme, when an event never occurs, the probability is zero, or

0%. At the other extreme, when an event always occurs, the probability is 1, or 100%.

Thus, all probability values are contained in a range from 0 to 1. For example, suppose

that you have a jar containing 10 white marbles. The probability of randomly selecting

a black marble is

p(black) 0

10 05 5

The probability of selecting a white marble is

p(white) 10

10 15 5

For the preceding definition of probability to be accurate, it is necessary that the out-

comes be obtained by a process called random sampling.

A simple random sample requires that each individual in the population has an

equal chance of being selected.

A second requirement, necessary for many statistical applications, states that if more

than one individual is being selected, the probabilities must stay constant from one se-

lection to the next. Adding this second requirement produces what technically is called

independent random sampling. The term independent refers to the fact that the probability

of selecting any particular individual is independent of those individuals who have already

been selected for the sample. For example, the probability that you will be selected is

constant and does not change even when other individuals are selected before you are.

Because independent random sampling is a fundamental requirement for many sta-

tistical applications, we always assume that this is the sampling method being used. To

simplify discussion, we typically omit the word “independent” and simply refer to this

sampling technique as random sampling.

Random sampling requires that each individual has an equal chance of being

selected and that the probability of being selected stays constant from one selec-

tion to the next if more than one individual is selected. A sample produced by

this technique is known as a random sample.

Each of the two requirements for random sampling has some interesting conse-

quences. The first ensures that there is no bias in the selection process. For a population

R A N D O M SA M P L I N G

D E F I N I T I O N

D E F I N I T I O N S

If you are unsure of how to

convert from fractions to

decimals or percentages, you

should review the section

on proportions in the math

review, Appendix A.

SECTION 6.1 / INTRODUCTION TO PROBABILITY 1 5 3

with N individuals, each individual must have the same probability, p 5 1 N

, of being

selected. This means, for example, that you would not get a random sample of people

in your city by selecting names from a yacht-club membership list. Similarly, you

would not get a random sample of college students by selecting individuals from your

psychology classes. You also should note that the first requirement of random sampling

prohibits you from applying the definition of probability to situations in which the pos-

sible outcomes are not equally likely. Consider, for example, the question of whether

you will win a million dollars in the lottery tomorrow. There are only two possible

alternatives.

1. You will win.

2. You will not win.

According to our simple definition, the probability of winning would be one out of

two, or p 5 1 2

. However, the two alternatives are not equally likely, so the simple defini-

tion of probability does not apply.

The second requirement also is more interesting than may be apparent at first glance.

Consider, for example, the selection of n 5 2 cards from a complete deck. For the first draw, the probability of obtaining the jack of diamonds is

p(jack of diamonds) 1

52 5

After selecting one card for the sample, you are ready to draw the second card. What

is the probability of obtaining the jack of diamonds this time? Assuming that you still

are holding the first card, there are two possibilities:

p(jack of diamonds) 1

51 if the first card w5 aas not the jack of diamonds

p(jack of diamonds) 5 0 if the first card was the jack of diamonds

In either case, the probability is different from its value for the first draw. This

contradicts the requirement for random sampling, which says that the probability

must stay constant. To keep the probabilities from changing from one selection to

the next, it is necessary to return each individual to the population before you make

the next selection. This process is called sampling with replacement. The second re-

quirement for random samples (constant probability) demands that you sample with

replacement.

(Note: We are using a definition of random sampling that requires equal chance

of selection and constant probabilities. This kind of sampling often is called random

sampling with replacement. Many of the statistics we encounter later are founded on

this kind of sampling. However, you should realize that other definitions exist for

the concept of random sampling. In particular, it is very common to define random

sampling without the requirement of constant probabilities—that is, random sampling

without replacement. In addition, there are many different sampling techniques that

are used when researchers are selecting individuals to participate in research studies.)

The situations in which we are concerned with probability usually involve a popula-

tion of scores that can be displayed in a frequency distribution graph. If you think

of the graph as representing the entire population, then different proportions of the

P R O BA B I L I T Y A N D F R E Q U E N CY

D I ST R I B U T I O N S

1 5 4 CHAPTER 6 PROBABILITY

graph represent different proportions of the population. Because probabilities and

proportions are equivalent, a particular proportion of the graph corresponds to a

particular probability in the population. Thus, whenever a population is presented in

a frequency distribution graph, it is possible to represent probabilities as proportions

of the graph. The relationship between graphs and probabilities is demonstrated in

the following example.

We use a very simple population that contains only N 5 10 scores with values 1, 1, 2, 3, 3, 4, 4, 4, 5, 6. This population is shown in the frequency distribution graph in

Figure 6.2. If you take a random sample of n 5 1 score from this population, what is the probability of obtaining an individual with a score greater than 4? In probability

notation,

p(X . 4) 5 ?

Using the definition of probability, there are 2 scores that meet this criterion out

of the total group of N 5 10 scores, so the answer would be p 5 2

10 . This answer can

be obtained directly from the frequency distribution graph if you recall that prob-

ability and proportion measure the same thing. Looking at the graph (see Figure 6.2),

what proportion of the population consists of scores greater than 4? The answer is

the shaded part of the distribution—that is, 2 squares out of the total of 10 squares in

the distribution. Notice that we now are defining probability as a proportion of area

in the frequency distribution graph. This provides a very concrete and graphic way of

representing probability.

Using the same population once again, what is the probability of selecting an

individual with a score less than 5? In symbols,

p(X , 5) 5 ?

Going directly to the distribution in Figure 6.2, we now want to know what part

of the graph is not shaded. The unshaded portion consists of 8 out of the 10 blocks

(eight-tenths of the area of the graph), so the answer is p 5 8 10

E X A M P L E 6 . 1

F re

q u

e n

c y

2 3 4 5 6 7 8 X

FIGURE 6.2

A frequency distribution

histogram for a population

that consists of N 5 10

scores. The shaded part

of the figure indicates

the portion of the whole

population that corre-

sponds to scores greater

than X 5 4. The shaded

portion is two-tenths

(p 5 2

10 ) of the whole

distribution.

SECTION 6.2 / PROBABILITY AND THE NORMAL DISTRIBUTION 1 5 5

PROBABILITY AND THE NORMAL DISTRIBUTION

The normal distribution was first introduced in Chapter 2 as an example of a commonly

occurring shape for population distributions. An example of a normal distribution is

shown in Figure 6.3.

Note that the normal distribution is symmetrical, with the highest frequency in

the middle and frequencies tapering off as you move toward either extreme. Although

the exact shape for the normal distribution is defined by an equation (see Figure 6.3), the

normal shape can also be described by the proportions of area contained in each section

of the distribution. Statisticians often identify sections of a normal distribution by using

z-scores. Figure 6.4 shows a normal distribution with several sections marked in z-score

units. You should recall that z-scores identify locations in a distribution in terms of standard

deviations from the mean. For example, z 5 11 is 1 standard deviation above the mean, z 5 12 is 2 standard deviations above the mean, and so on. Figure 6.4 shows the percent- age of scores that fall in each of these sections for a normal distribution. For example, the

section between the mean (z 5 0) and the point that is 1 standard deviation above the mean (z 5 1) contains 34.13% of the scores. Similarly, 13.59% of the scores are located in the section between 1 and 2 standard deviations above the mean. In this way it is possible to

define a normal distribution in terms of its proportions; that is, a distribution is normal if

and only if it has all the right proportions.

There are two additional points to be made about the distribution shown in

Figure 6.4. First, you should realize that the sections on the left side of the distribution

6.2

1. A survey of the students in a psychology class revealed that there were 19 females

and 8 males. Of the 19 females, only 4 had no brothers or sisters, and 3 of the

males were also the only child in the household. If a student is randomly selected

from this class,

a. What is the probability of obtaining a male?

b. What is the probability of selecting a student who has at least one brother or sister?

c. What is the probability of selecting a female who has no siblings?

2. A jar contains 10 red marbles and 30 blue marbles.

a. If you randomly select 1 marble from the jar, what is the probability of

obtaining a red marble?

b. If you take a random sample of n 5 3 marbles from the jar and the first two marbles are both blue, what is the probability that the third marble will be red?

3. Suppose that you are going to select a random sample of n 5 1 score from the distribution in Figure 6.2. Find the following probabilities:

a. p(X . 2) b. p(X . 5) c. p(X , 3)

1. a. p 5 8 27

b. p 5 20 27

c. p 5 4 27

2. a. p 5 10 40 5 0.25

b. p 5 10 40

5 0.25. Remember that random sampling requires sampling with replacement.

3. a. p 5 7

10 5 0.70 b. p 5

1 10

5 0.10 c. p 5 3

10 5 0.30

L E A R N I N G C H E C K

ANSWERS

1 5 6 CHAPTER 6 PROBABILITY

have exactly the same areas as the corresponding sections on the right side because the

normal distribution is symmetrical. Second, because the locations in the distribution are

identified by z-scores, the percentages shown in the figure apply to any normal distribu-

tion regardless of the values for the mean and the standard deviation. Remember: When

any distribution is transformed into z-scores, the mean becomes zero and the standard

deviation becomes one.

Because the normal distribution is a good model for many naturally occurring

distributions and because this shape is guaranteed in some circumstances (as we see in

Chapter 7), we devote considerable attention to this particular distribution. The process

of answering probability questions about a normal distribution is introduced in the

following example.

µ X

FIGURE 6.3

The normal distribution.

The exact shape of the nor-

mal distribution is specified

by an equation relating to

each X value (score) with

each Y value (frequency).

The equation is

Y e X

5 s

2 2m s1

2 2 2

 2

( ) /

(π and e are mathemati-

cal constants.) In simpler

terms, the normal distribu-

tion is symmetrical with a

single mode in the middle.

The frequency tapers off as

you move farther from the

middle in either direction.

–2 –1 0

+1 +2

34.13%

13.59%

2.28%

FIGURE 6.4

The normal distribution

following a z-score

transformation.

SECTION 6.2 / PROBABILITY AND THE NORMAL DISTRIBUTION 1 5 7

The population distribution of SAT scores is normal with a mean of m 5 500 and a standard deviation of s � 100. Given this information about the population and the

known proportions for a normal distribution (see Figure 6.4), we can determine the

probabilities associated with specific samples. For example, what is the probability

of randomly selecting an individual from this population who has an SAT score

greater than 700?

Restating this question in probability notation, we get

p(X . 700) 5 ?

We follow a step-by-step process to find the answer to this question.

1. First, the probability question is translated into a proportion question: Out

of all possible SAT scores, what proportion corresponds to scores greater

than 700?

2. The set of “all possible SAT scores” is simply the population distribution. This

population is shown in Figure 6.5. The mean is m 5 500, so the score X 5 700

is to the right of the mean. Because we are interested in all scores greater than

700, we shade in the area to the right of 700. This area represents the propor-

tion we are trying to determine.

3. Identify the exact position of X 5 700 by computing a z-score. For this

example,

z X

5 2m

s 5

700 500

100

200

100 2 00

2 5 5 .

That is, an SAT score of X 5 700 is exactly 2 standard deviations above the

mean and corresponds to a z-score of z 5 12.00. We have also located this

z-score in Figure 6.5.

4. The proportion we are trying to determine may now be expressed in terms of its

z-score:

p(z . 2.00) 5 ?

E X A M P L E 6 . 2

X µ = 500 X = 700

σ =100

0 z

2.00

FIGURE 6.5

The distribution of SAT

scores described in

Example 6.2.

1 5 8 CHAPTER 6 PROBABILITY

According to the proportions shown in Figure 6.4, all normal distributions,

regardless of the values for µ and s, have 2.28% of the scores in the tail beyond

z � 12.00. Thus, for the population of SAT scores,

p(X . 700) 5 p(z . 12.00) 5 2.28%

Before we attempt any more probability questions, we must introduce a more useful

tool than the graph of the normal distribution shown in Figure 6.4. The graph shows

proportions for only a few selected z-score values. A more complete listing of z-scores

and proportions is provided in the unit normal table. This table lists proportions of the

normal distribution for a full range of possible z-score values.

The complete unit normal table is provided in Appendix B Table B.1, and part of the

table is reproduced in Figure 6.6. Notice that the table is structured in a four-column

format. The first column (A) lists z-score values corresponding to different locations in a

T H E U N I T N O R M A L TA B L E

Mean z

(A) z

(B) Proportion

in body

mean and z

Proportion in ta il

0.00 0.01 0.02 0.03 0.20 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29 0.30 0.31 0.32 0.33 0.34

.5000

.5040

.5080

.5120 0.5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 .6179 .6217 .6255 .6293 .6331

.5000 .0000 .0040 .0080 .0120 .0120 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141 .1179 .1217 .1255 .1293 .1331

.4960

.4920

.4880 0.4207 .4168 .4129 .4090 .4052 .4013 .3974 .3936 .3897 .3859 .3821 .3783 .3745 .3707 .3669

Mean z

FIGURE 6.6

A portion of the unit normal table. This table lists proportions of the normal distribution

corresponding to each z-score value. Column A of the table lists z-scores. Column B lists the

proportion in the body of the normal distribution up to the z-score value. Column C lists the

proportion of the normal distribution that is located in the tail of the distribution beyond

the z-score value. Column D lists the proportion between the mean and the z-score value.

SECTION 6.2 / PROBABILITY AND THE NORMAL DISTRIBUTION 1 5 9

normal distribution. If you imagine a vertical line drawn through a normal distribution,

then the exact location of the line can be described by one of the z-score values listed in

column A. You should also realize that a vertical line separates the distribution into two

sections: a larger section called the body and a smaller section called the tail. Columns B

and C in the table identify the proportion of the distribution in each of the two sections.

Column B presents the proportion in the body (the larger portion), and column C presents

the proportion in the tail. Finally, we have added a fourth column, column D, that identi-

fies the proportion of the distribution that is located between the mean and the z-score.

We use the distribution in Figure 6.7(a) to demonstrate how the unit normal table

can be used to find specific proportions of a normal distribution. The figure shows a

normal distribution with a vertical line drawn at z 5 10.25. Using the portion of the table shown in Figure 6.6, find the row in the table that contains z 5 0.25 in column A. Reading across the row, you should find that the line drawn at z 5 1 0.25 separates the distribution into two sections with the larger section containing 0.5987 (59.87%) of the

distribution and the smaller section containing 0.4013 (40.13%) of the distribution. Also,

there is exactly 0.0987 (9.87%) of the distribution between the mean and z 5 10.25. To make full use of the unit normal table, there are a few facts to keep in mind:

1. The body always corresponds to the larger part of the distribution whether it

is on the right-hand side or the left-hand side. Similarly, the tail is always the

smaller section whether it is on the right or the left.

2. Because the normal distribution is symmetrical, the proportions on the right-

hand side are exactly the same as the corresponding proportions on the left-hand

side. Earlier, for example, we used the unit normal table to obtain proportions

for z 5 10.25. Figure 6.7(b) shows the same proportions for z 5 20.25. For a negative z-score, however, notice that the tail of the distribution is on the left side

and the body is on the right. For a positive z-score [Figure 6.7(a)], the positions

are reversed. However, the proportions in each section are exactly the same, with

0.55987 in the body and 0.4013 in the tail. Once again, the table does not list

negative z-score values. To find proportions for negative z-scores, you must look

up the corresponding proportions for the positive value of z.

3. Although the z-score values change signs (1 and 2) from one side to the other, the proportions are always positive. Thus, column C in the table always lists the

proportion in the tail whether it is the right-hand tail or the left-hand tail.

0 �0.25

Tail 0.4013

Body 0.5987

0�0.25

Tail 0.4013

Body 0.5987

FIGURE 6.7

Proportions of a normal distribution corresponding to z 5 10.25 (a) and 20.25 (b).

(a) (b)

1 6 0 CHAPTER 6 PROBABILITY

The unit normal table lists relationships between z-score locations and proportions in a

normal distribution. For any z-score location, you can use the table to look up the cor-

responding proportions. Similarly, if you know the proportions, you can use the table

to find the specific z-score location. Because we have defined probability as equivalent

to proportion, you can also use the unit normal table to look up probabilities for normal

distributions. The following examples demonstrate a variety of different ways that the

unit normal table can be used.

Finding proportions or probabilities for specific z-score values For each of the

following examples, we begin with a specific z-score value and then use the unit normal

table to find probabilities or proportions associated with the z-score.

What proportion of the normal distribution corresponds to z-score values greater than

z 5 1.00? First, you should sketch the distribution and shade in the area you are try- ing to determine. This is shown in Figure 6.8(a). In this case, the shaded portion is the

tail of the distribution beyond z 5 1.00. To find this shaded area, you simply look for z 5 1.00 in column A to find the appropriate row in the unit normal table. Then scan across the row to column C (tail) to find the proportion. Using the table in Appendix B,

you should find that the answer is 0.1587.

You also should notice that this same problem could have been phrased as a prob-

ability question. Specifically, we could have asked, “For a normal distribution, what is

the probability of selecting a z-score value greater than z 5 11.00?” Again, the answer is p(z . 1.00) 5 0.1587 (or 15.87%).

For a normal distribution, what is the probability of selecting a z-score less than z 5 1.50?

In symbols, p(z , 1.50) 5 ? Our goal is to determine what proportion of the normal dis-

tribution corresponds to z-scores less than 1.50. A normal distribution is shown in Figure

6.8(b) and z 5 1.50 is marked in the distribution. Notice that we have shaded all of the

values to the left of (less than) z 5 1.50. This is the portion we are trying to find. Clearly

the shaded portion is more than 50%, so it corresponds to the body of the distribution.

Therefore, find z 5 1.50 in column A of the unit normal table and read across the row to

obtain the proportion from column B. The answer is p(z , 1.50) 5 0.9332 (or 93.32%).

Many problems require that you find proportions for negative z-scores. For example,

what proportion of the normal distribution is contained in the tail beyond z 5 20.50?

P R O BA B I L I T I E S , P R O P O R T I O N S , A N D

Z - S CO R E S

E X A M P L E 6 . 3 A

E X A M P L E 6 . 3 B

E X A M P L E 6 . 3 C

0 1.00 µ

0 1.50 µ

0−0.5 µ

FIGURE 6.8

The distribution for Example 6.3A26.3C.

(a) (b) (c)

SECTION 6.2 / PROBABILITY AND THE NORMAL DISTRIBUTION 1 6 1

That is, p(z � 20.50). This portion has been shaded in Figure 6.8(c). To answer ques-

tions with negative z-scores, simply remember that the normal distribution is symmetri-

cal with a z-score of zero at the mean, positive values to the right, and negative values to

the left. The proportion in the left tail beyond z 5 20.50 is identical to the proportion in the right tail beyond z 5 10.50. To find this proportion, look up z 5 0.50 in column A, and read across the row to find the proportion in column C (tail). You should get an

answer of 0.3085 (30.85%).

Finding the z-score location that corresponds to specific proportions The preced-

ing examples all involved using a z-score value in column A to look up proportions in

column B or C. You should realize, however, that the table also allows you to begin

with a known proportion and then look up the corresponding z-score. The following

examples demonstrate this process.

For a normal distribution, what z-score separates the top 10% from the remainder of the

distribution? To answer this question, we have sketched a normal distribution [Figure

6.9(a)] and drawn a vertical line that separates the highest 10% (approximately) from

the rest. The problem is to find the exact location of this line. For this distribution, we

know that the tail contains 0.1000 (10%) and the body contains 0.9000 (90%). To find

the z-score value, you simply locate the row in the unit normal table that has 0.1000 in

column C or 0.9000 in column B. For example, you can scan down the values in column

C (tail) until you find a proportion of 0.1000. Note that you probably will not find the

exact proportion, but you can use the closest value listed in the table. For this example,

a proportion of 0.1000 is not listed in column C but you can use 0.1003, which is listed.

Once you have found the correct proportion in the table, simply read across the row to

find the corresponding z-score value in column A.

For this example, the z-score that separates the extreme 10% in the tail is z 5 1.28. At this point you must be careful because the table does not differentiate between the

right-hand tail and the left-hand tail of the distribution. Specifically, the final answer

could be either z 5 11.28, which separates 10% in the right-hand tail, or z 5 21.28, which separates 10% in the left-hand tail. For this problem, we want the right-hand tail

(the highest 10%), so the z-score value is z 5 11.28.

For a normal distribution, what z-score values form the boundaries that separate the

middle 60% of the distribution from the rest of the scores?

Again, we have sketched a normal distribution [Figure 6.9(b)] and drawn vertical

lines so that roughly 60% of the distribution in the central section, with the remainder

E X A M P L E 6 . 4 A

E X A M P L E 6 . 4 B

Moving to the left on the

X-axis results in smaller

X values and smaller

z-scores. Thus, a z-score

of 23.00 reflects a smaller

value than a z-score of 21.

z 5 ? z 5 ?z 5 ?

10% (.1000)

90% (.9000) 60%

(.6000)

FIGURE 6.9

The distributions for

Examples 6.4A and 6.4B.

(a) (b)

1 6 2 CHAPTER 6 PROBABILITY

split equally between the two tails. The problem is to find the z-score values that define

the exact locations for the lines. To find the z-score values, we begin with the known

proportions: 0.6000 in the center and 0.4000 divided equally between the two tails.

Although these proportions can be used in several different ways, this example provides

an opportunity to demonstrate how column D in the table can be used to solve problems.

For this problem, the 0.6000 in the center can be divided in half with exactly 0.3000 to

the right of the mean and exactly 0.3000 to the left. Each of these sections corresponds

to the proportion listed in column D. Begin by scanning down column D, looking for

a value of 0.3000. Again, this exact proportion is not in the table, but the closest value

is 0.2995. Reading across the row to column A, you should find a z-score value of

z 5 0.84. Looking again at the sketch [see Figure 6.9(b)], the right-hand line is located at z 5 10.84 and the left-hand line is located at z 5 20.84.

You may have noticed that we sketched distributions for each of the preceding prob-

lems. As a general rule, you should always sketch a distribution, locate the mean with

a vertical line, and shade in the portion that you are trying to determine. Look at your

sketch. It will help you to determine which columns to use in the unit normal table. If you

make a habit of drawing sketches, you will avoid careless errors when using the table.

1. Find the proportion of a normal distribution that corresponds to each of the follow-

ing sections:

a. z � 0.25 b. z . 0.80 c. z , 21.50 d. z . 20.75

2. For a normal distribution, find the z-score location that divides the distribution as

follows:

a. Separate the top 20% from the rest.

b. Separate the top 60% from the rest.

c. Separate the middle 70% from the rest.

3. The tail will be on the right-hand side of a normal distribution for any positive

z-score. (True or false?)

1. a. p 5 0.5987 b. p 5 0.2119 c. p 5 0.0668 d. p 5 0.7734

2. a. z 5 0.84 b. z 5 20.25 c. z 5 21.04 and 11.04.

3. True

L E A R N I N G C H E C K

ANSWERS

PROBABILITIES AND PROPORTIONS FOR SCORES FROM A NORMAL DISTRIBUTION

In the preceding section, we used the unit normal table to find probabilities and propor-

tions corresponding to specific z-score values. In most situations, however, it is necessary

to find probabilities for specific X values. Consider the following example:

It is known that IQ scores form a normal distribution with m 5 100 and s 5 15. Given

this information, what is the probability of randomly selecting an individual with an

IQ score that is less than 120?

6.3

SECTION 6.3 / PROBABILITIES AND PROPORTIONS FOR SCORES FROM A NORMAL DISTRIBUTION 1 6 3

This problem is asking for a specific probability or proportion of a normal distribu-

tion. However, before we can look up the answer in the unit normal table, we must

first transform the IQ scores (X values) into z-scores. Thus, to solve this new kind of

probability problem, we must add one new step to the process. Specifically, to answer

probability questions about scores (X values) from a normal distribution, you must use

the following two-step procedure:

1. Transform the X values into z-scores.

2. Use the unit normal table to look up the proportions corresponding to the

z-score values.

This process is demonstrated in the following examples. Once again, we suggest

that you sketch the distribution and shade the portion you are trying to find to avoid

careless mistakes.

We now answer the probability question about IQ scores that we presented earlier.

Specifically, what is the probability of randomly selecting an individual with an IQ

score that is less than 120? Restated in terms of proportions, we want to find the propor-

tion of the IQ distribution that corresponds to scores less than 120. The distribution is

drawn in Figure 6.10, and the portion we want has been shaded.

The first step is to change the X values into z-scores. In particular, the score of

X 5 120 is changed to

z X

5 2m

s �

2 � �

120 100

15 1 33.

Thus, an IQ score of X � 120 corresponds to a z-score of z � 1.33, and IQ scores

less than 120 correspond to z-scores less than 1.33.

Next, look up the z-score value in the unit normal table. Because we want the propor-

tion of the distribution in the body to the left of X � 120 (see Figure 6.10), the answer

is in column B. First, locate z � 1.33 in column A, and then read across the row to find

p � 0.9082 in column B. Thus, the probability of randomly selecting an individual with

an IQ less than 120 is p � 0.9082. In symbols,

p(X � 120) 5 p(z � 1.33) 5 0.9082 (or 90.82%)

E X A M P L E 6 . 5

Caution: The unit normal

table can be used only with

normal-shaped distributions.

If a distribution is not normal,

transforming raw scores to

z-scores does not make it

normal.

� � 100

� � 15

120

1.330

FIGURE 6.10

The distribution of IQ

scores. The problem is

to find the probability or

proportion of the distri-

bution corresponding to

scores less than 120.

1 6 4 CHAPTER 6 PROBABILITY

Finally, notice that we phrased this question in terms of a probability. Specifically,

we asked, “What is the probability of selecting an individual with an IQ less than 120?”

However, the same question can be phrased in terms of a proportion: “What proportion

of all of the individuals in the population have IQ scores that are less than 120?” Both

versions ask exactly the same question and produce exactly the same answer. A third

alternative for presenting the same question is introduced in Box 6.1.

Finding proportions/probabilities located between two scores The next example

demonstrates the process of finding the probability of selecting a score that is located

between two specific values. Although these problems can be solved using the pro-

portions of columns B and C (body and tail), they are often easier to solve with the

proportions listed in column D.

The highway department conducted a study measuring driving speeds on a local section

of interstate highway. They found an average speed of m 5 58 miles per hour with a standard deviation of s 5 10. The distribution was approximately normal. Given this

information, what proportion of the cars are traveling between 55 and 65 miles per hour?

Using probability notation, we can express the problem as

p(55 � X � 65) 5 ?

The distribution of driving speeds is shown in Figure 6.11 with the appropriate area

shaded. The first step is to determine the z-score corresponding to the X value at each

end of the interval.

For :

For

X z X

5 5 2m

s 5

2 5

2 52

55 55 58

10 0 30

:: z X

5 2m

s 5

2 5 5

65 58

10 0 70.

E X A M P L E 6 . 6

BOX

6.1 PROBABILITIES, PROPORTIONS, AND PERCENTILE RANKS

working. In Example 6.5, the problem is presented as

“What is the probability of randomly selecting an

individual with an IQ of less than 120?” Exactly the

same question could be phrased as “What is the

percentile rank for an IQ score of 120?” In each case,

we are drawing a line at X 5 120 and looking for the

proportion of the distribution on the left-hand side of

the line. Similarly, Example 6.8 asks “How much time

do you have to spend commuting each day to be in the

highest 10% nationwide?” Because this score separates

the top 10% from the bottom 90%, the same question

could be rephrased as “What is the 90th percentile for

the distribution of commuting times?”

Thus far we have discussed parts of distributions in

terms of proportions and probabilities. However, there

is another set of terminology that deals with many of

the same concepts. Specifically, the percentile rank

for a specific score is defined as the percentage of the

individuals in the distribution who have scores that are

less than or equal to the specific score. For example,

if 70% of the individuals have scores of X 5 45 or

lower, then X 5 45 has a percentile rank of 70%.

When a score is referred to by its percentile rank, the

score is called a percentile. For example, a score with

a percentile rank of 70% is called the 70th percentile.

Using this terminology, it is possible to rephrase

some of the probability problems that we have been

SECTION 6.3 / PROBABILITIES AND PROPORTIONS FOR SCORES FROM A NORMAL DISTRIBUTION 1 6 5

Looking again at Figure 6.11, we see that the proportion we are seeking can be di-

vided into two sections: (1) the area left of the mean, and (2) the area right of the mean.

The first area is the proportion between the mean and z 5 20.30, and the second is the proportion between the mean and z 5 10.70. Using column D of the unit normal

table, these two proportions are 0.1179 and 0.2580. The total proportion is obtained by

adding these two sections:

p(55 � X � 65) 5 p(20.30 � z � 10.70) 5 0.1179 1 0.2580 5 0.3759

Using the same distribution of driving speeds from the previous example, what

proportion of cars are traveling between 65 and 75 miles per hour?

p(65 � X � 75) 5 ?

The distribution is shown in Figure 6.12 with the appropriate area shaded. Again, we

start by determining the z-score corresponding to each end of the interval.

For :

X z X

5 5 2m

s 5

2 5 5

65 65 58

10 0 70

zz X

5 2m

2 5= =

75 58

10 1 70.

E X A M P L E 6 . 7

� � 58

� � 10

6555 X

02.30 .70

FIGURE 6.11

The distribution for

Example 6.6.

� � 58

� � 10

65 75

0 .70 1.70

FIGURE 6.12

The distribution for

Example 6.7.

1 6 6 CHAPTER 6 PROBABILITY

There are several different ways to use the unit normal table to find the proportion

between these two z-scores. For this example, we use the proportions in the tail of the

distribution (column C). According to column C in the unit normal table, the proportion

in the tail beyond z 5 0.70 is p 5 0.2420. Note that this proportion includes the section that we want, but it also includes an extra, unwanted section located in the tail beyond

z 5 1.70. Locating z 5 1.70 in the table, and reading across the row to column C, we see that the unwanted section is p 5 0.0446. To obtain the correct answer, we subtract the unwanted portion from the total proportion in the tail beyond z 5 0.70.

p(65 � X � 75) 5 p(0.70 � z � 1.70) 5 0.2420 2 0.0446 5 0.1974

Finding scores corresponding to specific proportions or probabilities In the pre-

vious three examples, the problem was to find the proportion or probability correspond-

ing to specific X values. The two-step process for finding these proportions is shown

in Figure 6.13. Thus far, we have only considered examples that move in a clockwise

direction around the triangle shown in the figure; that is, we start with an X value that

is transformed into a z-score, and then we use the unit normal table to look up the

appropriate proportion. You should realize, however, that it is possible to reverse this

two-step process so that we move counterclockwise around the triangle. This reverse

process allows us to find the score (X value) corresponding to a specific proportion in

the distribution. Following the lines in Figure 6.13, we begin with a specific proportion,

use the unit normal table to look up the corresponding z-score, and then transform the

z-score into an X value. The following example demonstrates this process.

The U.S. Census Bureau (2005) reports that Americans spend an average of m 5 24.3 minutes

commuting to work each day. Assuming that the distribution of commuting times is normal

with a standard deviation of s 5 10 minutes, how much time do you have to spend com-

muting each day to be in the highest 10% nationwide? (An alternative form of the same

question is presented in Box 6.1.) The distribution is shown in Figure 6.14 with a portion

representing approximately 10% shaded in the right-hand tail.

In this problem, we begin with a proportion (10% or 0.10), and we are looking

for a score. According to the map in Figure 6.13, we can move from p (propor-

tion) to X (score) via z-scores. The first step is to use the unit normal table to

E X A M P L E 6 . 8

X z-score formula

z-score

Unit normal table

Proportions or

probabilities

FIGURE 6.13

Determining probabilities

or proportions for a normal

distribution is shown as

a two-step process with

z-scores as an intermediate

stop along the way.

Note that you cannot

move directly along the

dashed line between

X values and probabilities

or proportions. Instead,

you must follow the solid

lines around the corner.

SECTION 6.3 / PROBABILITIES AND PROPORTIONS FOR SCORES FROM A NORMAL DISTRIBUTION 1 6 7

find the z-score that corresponds to a proportion of 0.10 in the tail. First, scan the

values in column C to locate the row that has a proportion of 0.10 in the tail of the

distribution. Note that you will not find 0.1000 exactly, but locate the closest value

possible. In this case, the closest value is 0.1003. Reading across the row, we find

z 5 1.28 in column A. The next step is to determine whether the z-score is positive or negative. Remember

that the table does not specify the sign of the z-score. Looking at the distribution in

Figure 6.14, you should realize that the score we want is above the mean, so the z-score

is positive, z 5 11.28. The final step is to transform the z-score into an X value. By definition, a z-score

of 11.28 corresponds to a score that is located above the mean by 1.28 standard

deviations. One standard deviation is equal to 10 points (s 5 10), so 1.28 standard

deviations is

1.28s 5 1.28(10) 5 12.8 points

Thus, our score is located above the mean (µ 5 24.3) by a distance of 12.8 points.

Therefore,

X 5 24.3 1 12.8 5 37.1

The answer for our original question is that you must commute at least 37.1 minutes

a day to be in the top 10% of American commuters.

Again, the distribution of commuting time for American workers is normal with a mean

of m 5 24.3 minutes and a standard deviation of s 5 10 minutes. For this example,

we find the range of values that defines the middle 90% of the distribution. The entire

distribution is shown in Figure 6.15 with the middle portion shaded.

The 90% (0.9000) in the middle of the distribution can be split in half with 45%

(0.4500) on each side of the mean. Looking up 0.4500, in column D of the unit normal

table, you will find that the exact proportion is not listed. However, you will find 0.4495

and 0.4505, which are equally close. Technically, either value is acceptable, but we use

E X A M P L E 6 . 9

24.3

0 1.28

37.1

Highest 10%

FIGURE 6.14

The distribution of

commuting times for

American workers. The

problem is to find the score

that separates the highest

10% of commuting times

from the rest.

1 6 8 CHAPTER 6 PROBABILITY

0.4505 so that the total area in the middle is at least 90%. Reading across the row, you

should find a z-score of z 5 1.65 in column A. Thus, the z-score at the right boundary is z 5 11.65 and the z-score at the left boundary is z 5 21.65. In either case, a z-score of 1.65 indicates a location that is 1.65 standard deviations away from the mean. For

the distribution of commuting times, one standard deviation is s 5 10, so 1.65 standard

deviations is a distance of

1.65s 5 1.65(10) 5 16.5 points

Therefore, the score at the right-hand boundary is located above the mean by

16.5 points and corresponds to X 5 24.3 1 16.5 5 40.8. Similarly, the score at the

left-hand boundary is below the mean by 16.5 points and corresponds to X 5 24.3

2 16.5 5 7.8. The middle 90% of the distribution corresponds to values between

7.8 and 40.8. Thus, 90% of American commuters spend between 7.8 and 40.8 min-

utes commuting to work each day. Only 10% of commuters spend either more time

or less time.

µ = 24.3

0 1.65

40.8

–1.65

7.8

σ = 10

Middle 90%FIGURE 6.15

The distribution of

commuting times for

American workers. The

problem is to find the

middle 90% of the

distribution.

1. For a normal distribution with a mean of m 5 60 and a standard deviation of

s 5 12, find each probability value requested.

a. p(X . 66) b. p(X , 75) c. p(X , 57) d. p(48 , X , 72)

2. Scores on the Mathematics section of the SAT Reasoning Test form a normal

distribution with a mean of m 5 500 and a standard deviation of s 5 100.

a. If the state college only accepts students who score in the top 60% on this test,

what is the minimum score needed for admission?

b. What is the minimum score necessary to be in the top 10% of the distribution?

c. What scores form the boundaries for the middle 50% of the distribution?

3. What is the probability of selecting a score greater than 45 from a positively

skewed distribution with m 5 40 and s 5 10? (Be careful.)

L E A R N I N G C H E C K

SECTION 6.4 / LOOKING AHEAD TO INFERENTIAL STATISTICS 1 6 9

LOOKING AHEAD TO INFERENTIAL STATISTICS

Probability forms a direct link between samples and the populations from which they

come. As we noted at the beginning of this chapter, this link is the foundation for the

inferential statistics in future chapters. The following example provides a brief preview

of how probability is used in the context of inferential statistics.

We ended Chapter 5 with a demonstration of how inferential statistics are used

to help interpret the results of a research study. A general research situation was

shown in Figure 5.9 and is repeated here in Figure 6.16. The research begins with a

population that forms a normal distribution with a mean of µ 5 400 and a standard deviation of s 5 20. A sample is selected from the population and a treatment is

administered to the sample. The goal for the study is to evaluate the effect of the

treatment.

To determine whether the treatment has an effect, the researcher simply com-

pares the treated sample with the original population. If the individuals in the

sample have scores around 400 (the original population mean), then we must con-

clude that the treatment appears to have no effect. On the other hand, if the treated

individuals have scores that are noticeably different from 400, then the researcher

has evidence that the treatment does have an effect. Notice that the study is using

6.4

1. a. p 5 0.3085 b. p 5 0.8944 c. p 5 0.4013 d. p 5 0.6826

2. a. z 5 20.25; X 5 475

b. z 5 1.28; X 5 628

c. z 5 60.67; X 5433 and X 5567

3. You cannot obtain the answer. The unit normal table cannot be used to answer this question

because the distribution is not normal.

ANSWERS

Population

Normal � � 400 � � 20

Sample Treated sample

T r e a t

m e n t

FIGURE 6.16

A diagram of a research

study. A sample is selected

from the population and

receives a treatment.

The goal is to determine

whether the treatment has

an effect.

1 7 0 CHAPTER 6 PROBABILITY

a sample to help answer a question about a population; this is the essence of infer-

ential statistics.

The problem for the researcher is determining exactly what is meant by “notice-

ably different” from 400. If a treated individual has a score of X 5 415, is that enough to say that the treatment has an effect? What about X 5 420 or X 5 450? In Chapter 5, we suggested that z-scores provide one method for solving this prob-

lem. Specifically, we suggested that a z-score value beyond z 5 2.00 (or 22.00) was an extreme value and, therefore, noticeably different. However, the choice of

z 5 ±2.00 was purely arbitrary. Now we have another tool, probability, to help us

decide exactly where to set the boundaries. Specifically, we can determine exactly

which samples have a high probability of being selected and which are extremely

unlikely. For example, we can set the boundaries to separate the most likely 95%

of the samples in the middle of the distribution from the extremely unlikely 5% in

the tails.

Figure 6.17 shows the original population from our hypothetical research study.

Note that most of the scores are located close to µ 5 400. Also note that we have added

boundaries separating the middle 95% of the distribution from the extreme 5%, or

0.0500, in the two tails. Dividing the 0.0500 in half produces proportions of 0.0250 in

the right-hand tail and 0.0250 in the left-hand tail. Using column C of the unit normal

table, the z-score boundaries for the right and left tails are z 5 11.96 and z 5 21.96,

respectively.

The boundaries set at z 5 ±1.96 provide objective criteria for deciding whether

the treated sample is noticeably different from the original population. Specifically,

any sample that is beyond the ±1.96 boundaries is an extreme value and is extremely

unlikely to occur (p 5 0.05 or less) if the treatment has no effect. Therefore, this

kind of sample provides convincing evidence that the treatment really does have an

effect.

� � 400

z � 21.96 z � 11.96

Middle 95%

High probability values

(scores near � � 400)

indicating that the treatment

has no effect

Extreme 5%

Scores that are very unlikely to be obtained from the original population

and therefore provide evidence of a treatment effect

FIGURE 6.17

Using probability to

evaluate a treatment

effect. Values that are

extremely unlikely to be

obtained from the original

population are viewed as

evidence of a treatment

effect.

RESOURCES 1 7 1

SUMMARY

1. The probability of a particular event A is defined as a fraction or proportion:

p A A

( ) 5 number of outcomes classified as

totall number of possible outcomes

2. Our definition of probability is accurate only for random samples. There are two requirements that must be satisfied for a random sample:

a. Every individual in the population has an equal chance of being selected.

b. When more than one individual is being selected, the probabilities must stay constant. This means that there must be sampling with replacement.

3. All probability problems can be restated as proportion problems. The “probability of selecting a king from a deck of cards” is equivalent to the “proportion of the deck that consists of kings.” For frequency dis- tributions, probability questions can be answered by determining proportions of area. The “probability of selecting an individual with an IQ greater than 108” is

equivalent to the “proportion of the whole population that consists of IQs greater than 108.”

4. For normal distributions, probabilities (proportions) can be found in the unit normal table. The table pro- vides a listing of the proportions of a normal distribu- tion that correspond to each z-score value. With the table, it is possible to move between X values and probabilities using a two-step procedure:

a. The z-score formula (Chapter 5) allows you to transform X to z or to change z back to X.

b. The unit normal table allows you to look up the prob- ability (proportion) corresponding to each z-score or the z-score corresponding to each probability.

5. Percentiles and percentile ranks measure the relative standing of a score within a distribution (see Box 6.1). Percentile rank is the percentage of individuals with scores at or below a particular X value. A percentile is an X value that is identified by its rank. The percentile rank always corresponds to the proportion to the left of the score in question.

KEY TERMS

probability (151)

random sample (152)

independent random sample (152)

sampling with replacement (153)

unit normal table (158)

percentile rank (164)

percentile (164)

RESOURCES

Go to CengageBrain.com to access Psychology CourseMate, where you will find an

interactive eBook, glossaries, flashcards, quizzes, statistics workshops, and more.

If your professor has assigned Aplia:

1. Sign in to your account.

2. Complete the corresponding exercises as required by your professor.

3. When finished, click “Grade It Now” to see which areas you have mastered, which

areas need more work, and detailed explanations of every answer.

1 7 2 CHAPTER 6 PROBABILITY

The statistics computer package SPSS is not structured to compute probabilities.

However, the program does report probability values as part of the inferential statistics

that we examine later in this book. In the context of inferential statistics, the probabilities

are called significance levels, and they warn researchers about the probability of misinter-

preting their research results.

FOCUS ON PROBLEM SOLVING

1. We have defined probability as being equivalent to a proportion, which means

that you can restate every probability problem as a proportion problem.

This definition is particularly useful when you are working with frequency

distribution graphs in which the population is represented by the whole graph

and probabilities (proportions) are represented by portions of the graph. When

working problems with the normal distribution, you always should start with

a sketch of the distribution. You should shade the portion of the graph that

reflects the proportion you are looking for.

2. Remember that the unit normal table shows only positive z-scores in column A.

However, because the normal distribution is symmetrical, the proportions in the

table apply to both positive and negative z-score values.

3. A common error for students is to use negative values for proportions on the

left-hand side of the normal distribution. Proportions (or probabilities) are always

positive: 10% is 10% whether it is in the left or right tail of the distribution.

4. The proportions in the unit normal table are accurate only for normal distribu-

tions. If a distribution is not normal, you cannot use the table.

DEMONSTRATION 6.1

FINDING PROBABILITY FROM THE UNIT NORMAL TABLE

A population is normally distributed with a mean of µ 5 45 and a standard devia- tion of s 5 4. What is the probability of randomly selecting a score that is greater

than 43? In other words, what proportion of the distribution consists of scores

greater than 43?

Sketch the distribution. For this demonstration, the distribution is normal with µ 5 45

and s 5 4. The score of X 5 43 is lower than the mean and therefore is placed to the

left of the mean. The question asks for the proportion corresponding to scores greater

than 43, so shade in the area to the right of this score. Figure 6.18 shows the sketch.

Transform the X value to a z-score.

z X

5 2m

s 5

2 5

2 52

43 45

4 0 5.

S T E P 1

S T E P 2

PROBLEMS 1 7 3

Find the appropriate proportion in the unit normal table. Ignoring the negative size, locate z 5 20.50 in column A. In this case, the proportion we want corresponds to the body of the distribution and the value is found in column B. For this example,

p(X . 43) 5 p(z . 20.50) 5 0.6915

S T E P 3

= 4

FIGURE 6.18

A sketch of the distribution

for Demonstration 6.1.

PROBLEMS

1. A local hardware store has a “Savings Wheel” at the checkout. Customers get to spin the wheel and, when the wheel stops, a pointer indicates how much they will save. The wheel can stop in any one of 50 sections. Of the sections, 10 produce 0% off, 20 sections are for 10% off, 10 sections for 20%, 5 for 30%, 3 for 40%, 1 for 50%, and 1 for 100% off. Assuming that all 50 sections are equally likely,

a. What is the probability that a customer’s purchase will be free (100% off)?

b. What is the probability that a customer will get no savings from the wheel (0% off)?

c. What is the probability that a customer will get at least 20% off?

2. A psychology class consists of 14 males and 36 females. If the professor selects names from the class list using random sampling,

a. What is the probability that the first student selected will be a female?

b. If a random sample of n 5 3 students is selected and the first two are both females, what is the prob- ability that the third student selected will be a male?

3. What are the two requirements that must be satisfied for a random sample?

4. Draw a vertical line through a normal distribution for each of the following z-score locations. Determine whether the tail is on the right or left side of the line and find the proportion in the tail.

a. z 5 1.00 b. z 5 0.50 c. z 5 21.25 d. z 5 20.40

5. Draw a vertical line through a normal distribution for each of the following z-score locations. Determine

whether the body is on the right or left side of the line and find the proportion in the body.

a. z 5 2.50 b. z 5 0.80 c. z 5 20.50 d. z 5 20.77

6. Find each of the following probabilities for a normal distribution.

a. p(z . 1.25) b. p(z . 20.60) c. p(z , 0.70) d. p(z , 21.30)

7. What proportion of a normal distribution is located between each of the following z-score boundaries?

a. z 5 20.25 and z 5 10.25 b. z 5 20.67 and z 5 10.67 c. z 5 21.20 and z 5 11.20

8. Find each of the following probabilities for a normal distribution.

a. p(20.80 , z , 0.80) b. p(20.50 , z , 1.00) c. p(0.20 , z , 1.50) d. p(21.20 , z , 20.80)

9. Find the z-score location of a vertical line that sepa- rates a normal distribution as described in each of the following.

a. 5% in the tail on the left b. 30% in the tail on the right c. 65% in the body on the left d. 80% in the body on the right

10. Find the z-score boundaries that separate a normal distribution as described in each of the following.

a. The middle 30% from the 70% in the tails. b. The middle 40% from the 60% in the tails. c. The middle 50% from the 50% in the tails. d. The middle 60% from the 40% in the tails.

1 7 4 CHAPTER 6 PROBABILITY

11. A normal distribution has a mean of µ 5 70 and a standard deviation of s 5 8. For each of the follow- ing scores, indicate whether the tail is to the right or left of the score and find the proportion of the distri- bution located in the tail.

a. X 5 72 b. X 5 76 c. X 5 66 d. X 5 60

12. A normal distribution has a mean of µ 5 30 and a standard deviation of s 5 12. For each of the follow- ing scores, indicate whether the body is to the right or left of the score and find the proportion of the distribution located in the body.

a. X 5 33 b. X 5 18 c. X 5 24 d. X 5 39

13. For a normal distribution with a mean of µ 5 60 and a standard deviation of s 5 10, find the propor- tion of the population corresponding to each of the following.

a. Scores greater than 65. b. Scores less than 68. c. Scores between 50 and 70.

14. IQ test scores are standardized to produce a normal distribution with a mean of µ 5 100 and a standard deviation of s 515. Find the proportion of the population in each of the following IQ categories.

a. Genius or near genius: IQ greater than 140 b. Very superior intelligence: IQ between 120 and 140 c. Average or normal intelligence: IQ between

90 and 109

15. The distribution of SAT scores is normal with µ 5 500 and s 5 100.

a. What SAT score, X value, separates the top 15% of the distribution from the rest?

b. What SAT score, X value, separates the top 10% of the distribution from the rest?

c. What SAT score, X value, separates the top 2% of the distribution from the rest?

16. According to a recent report, people smile an average of µ 5 62 time per day. Assuming that the distribu- tion of smiles is approximately normal with a standard deviation of s 5 18, find each of the following values.

a. What proportion of people smile more than 80 times a day?

b. What proportion of people smile at least 50 times a day?

17. A recent newspaper article reported the results of a survey of well-educated suburban parents. The re- sponses to one question indicated that by age 2, chil- dren were watching an average of µ 5 60 minutes of

television each day. Assuming that the distribution of television-watching times is normal with a standard deviation of s 5 25 minutes, find each of the follow- ing proportions.

a. What proportion of 2-year-old children watch more than 90 minutes of television each day?

b. What proportion of 2-year-old children watch less than 20 minutes a day?

18. Information from the Department of Motor Vehicles indicates that the average age of licensed drivers is µ 5 45.7 years with a standard deviation of s 5 12.5 years. Assuming that the distribution of drivers’ ages is approximately normal,

a. What proportion of licensed drivers are older than 50 years old?

b. What proportion of licensed drivers are younger than 30 years old?

19. A consumer survey indicates that the average house- hold spends µ 5 $185 on groceries each week. The distribution of spending amounts is approximately normal with a standard deviation of s 5 $25. Based on this distribution,

a. What proportion of the population spends more than $200 per week on groceries?

b. What is the probability of randomly selecting a family that spends less than $150 per week on groceries?

c. How much money do you need to spend on groceries each week to be in the top 20% of the distribution?

20. A report in 2010 indicates that Americans between the ages of 8 and 18 spend an average of µ 5 7.5 hours per day using some sort of electronic device such as smart phones, computers, or tablets. Assume that the distribution of times is normal with a stan- dard deviation of s 5 2.5 hours and find the follow- ing values.

a. What is the probability of selecting an individual who uses electronic devices more than 12 hours a day?

b. What proportion of 8- to 18-year-old Americans spend between 5 and 10 hours per day using elec- tronic devices? In symbols, p(5 � X � 10) 5 ?

21. Rochester, New York, averages µ 5 21.9 inches of snow for the month of December. The distribution of snowfall amounts is approximately normal with a standard deviation of s 5 6.5 inches. This year, a local jewelry store is advertising a refund of 50% off of all purchases made in December, if Rochester finishes the month with more than 3 feet (36 inches) of total snowfall. What is the probability that the jewelry store will have to pay off on its promise?

Aplia for Essentials of Statistics for the Behavioral Sciences

After reading, go to “Resources” at the end of this chapter for

an introduction on how to use Aplia’s homework and learning

resources.

Probability and Samples: The Distribution of Sample Means 7.1 Samples and Populations

7.2 The Distribution of Sample Means

7.3 Probability and the Distribution of Sample Means

7.4 More About Standard Error

7.5 Looking Ahead to Inferential Statistics

Summary

Focus on Problem Solving

Demonstration 7.1

Problems

C H A P T E R

7 Tools You Will Need The following items are considered essential background material for this chapter. If you doubt your knowledge of any of these items, you should review the appropriate chapter or section before proceeding.

• Random sampling (Chapter 6) • Probability and the normal distribu-

tion (Chapter 6) • z-Scores (Chapter 5)

1 7 6 CHAPTER 7 PROBABILITY AND SAMPLES: THE DISTRIBUTION OF SAMPLE MEANS

SAMPLES AND POPULATIONS

The preceding two chapters presented the topics of z-scores and probability. Whenever a

score is selected from a population, you should be able to compute a z-score that describes

exactly where the score is located in the distribution. If the population is normal, you also

should be able to determine the probability value for obtaining any individual score. In a

normal distribution, for example, any score located in the tail of the distribution beyond

z 5 �2.00 is an extreme value, and a score this large has a probability of only p 5 0.0228. However, the z-scores and probabilities that we have considered so far are limited to

situations in which the sample consists of a single score. Most research studies involve

much larger samples, such as n 5 25 preschool children or n 5 100 American Idol contestants. In these situations, the sample mean, rather than a single score, is used

to answer questions about the population. In this chapter, we extend the concepts of

z-scores and probability to cover situations with larger samples. In particular, we intro-

duce a procedure for transforming a sample mean into a z-score. Thus, a researcher is

able to compute a z-score that describes an entire sample. As always, a z-score near zero

indicates a central, representative sample; a z-score beyond �2.00 or –2.00 indicates an

extreme sample. Thus, it is possible to describe how any specific sample is related to all

the other possible samples. In addition, we can use the z-scores to look up probabilities

for obtaining certain samples, no matter how many scores the sample contains.

In general, the difficulty of working with samples is that a sample provides an in-

complete picture of the population. Suppose, for example, a researcher randomly selects

a sample of n 5 25 students from the state college. Although the sample should be representative of the entire student population, there are almost certainly some segments

of the population that are not included in the sample. In addition, any statistics that are

computed for the sample are not identical to the corresponding parameters for the entire

population. For example, the average IQ for the sample of 25 students is not the same as

the overall mean IQ for the entire population. This difference, or error, between sample

statistics and the corresponding population parameters is called sampling error and was

illustrated in Figure 1.2 (p. 9).

Sampling error is the natural discrepancy, or amount of error, between a sample

statistic and its corresponding population parameter.

Furthermore, samples are variable; they are not all the same. If you take two separate

samples from the same population, the samples are different. They contain different

individuals, they have different scores, and they have different sample means. How

can you tell which sample gives the best description of the population? Can you even

predict how well a sample describes its population? What is the probability of select-

ing a sample with specific characteristics? These questions can be answered once we

establish the rules that relate samples and populations.

THE DISTRIBUTION OF SAMPLE MEANS

As noted, two separate samples probably are different even though they are taken from

the same population. The samples have different individuals, different scores, different

means, and so on. In most cases, it is possible to obtain thousands of different samples

from one population. With all these different samples coming from the same population,

it may seem hopeless to try to establish some simple rules for the relationships between

7.1

D E F I N I T I O N

7.2

SECTION 7.2 / THE DISTRIBUTION OF SAMPLE MEANS 1 7 7

samples and populations. Fortunately, however, the huge set of possible samples forms a

relatively simple and orderly pattern that makes it possible to predict the characteristics

of a sample with some accuracy. The ability to predict sample characteristics is based on

the distribution of sample means.

The distribution of sample means is the collection of sample means for all of

the possible random samples of a particular size (n) that can be obtained from a

population.

Notice that the distribution of sample means contains all of the possible samples. It is

necessary to have all of the possible values to compute probabilities. For example, if the

entire set contains exactly 100 samples, then the probability of obtaining any specific

sample is 1 out of 100: p 5 1

100 (Box 7.1).

Also, you should notice that the distribution of sample means is different from the

distributions that we have considered before. Until now we always have discussed

distributions of scores; now the values in the distribution are not scores, but statistics

(sample means). Because statistics are obtained from samples, a distribution of statis-

tics is referred to as a sampling distribution.

A sampling distribution is a distribution of statistics obtained by selecting all of

the possible samples of a specific size from a population.

Thus, the distribution of sample means is an example of a sampling distribution. In

fact, it often is called the sampling distribution of M.

If you actually wanted to construct the distribution of sample means, you would first

select a random sample of a specific size (n) from a population, calculate the sample

mean, and place the sample mean in a frequency distribution. Then you would select

another random sample with the same number of scores. Again, you would calculate

the sample mean and add it to your distribution. You would continue selecting samples

and calculating means, over and over, until you had the complete set of all the possible

D E F I N I T I O N

BOX

7.1 PROBABILITY AND THE DISTRIBUTION OF SAMPLE MEANS

I have a bad habit of losing playing cards. This habit

is compounded by the fact that I always save the old

deck in the hope that someday I will find the missing

cards. As a result, I have a drawer filled with partial

decks of playing cards. Suppose that I take one of

these almost-complete decks, shuffle the cards care-

fully, and then randomly select one card. What is the

probability that I will draw a king?

You should realize that it is impossible to answer

this probability question. To find the probability of

selecting a king, you must know how many cards

are in the deck and exactly which cards are miss-

ing. (It is crucial that you know whether any kings

are missing.) The point of this simple example is

that any probability question requires that you have

complete information about the population from

which the sample is being selected. In this case,

you must know all of the possible cards in the deck

before you can find the probability for selecting any

specific card.

In this chapter, we are examining probability and

sample means. To find the probability for any

specific sample mean, you first must know all of the

possible sample means. Therefore, we begin by

defining and describing the set of all possible sample

means that can be obtained from a particular popula-

tion. Once we have specified the complete set of all

possible sample means (i.e., the distribution of sample

means), we can find the probability of selecting any

specific sample mean.

1 7 8 CHAPTER 7 PROBABILITY AND SAMPLES: THE DISTRIBUTION OF SAMPLE MEANS

random samples. At this point, your frequency distribution would show the distribution

of sample means.

We demonstrate the process of constructing a distribution of sample means in

Example 7.1, but first we use common sense and a little logic to predict the general

characteristics of the distribution.

1. The sample means should pile up around the population mean. Samples are not

expected to be perfect but they are representative of the population. As a result,

most of the sample means should be relatively close to the population mean.

2. The pile of sample means should tend to form a normal-shaped distribution.

Logically, most of the samples should have means close to m, and it should be

relatively rare to find sample means that are substantially different from m. As a

result, the sample means should pile up in the center of the distribution (around m)

and the frequencies should taper off as the distance between M and m increases.

This describes a normal-shaped distribution.

3. In general, the larger the sample size, the closer the sample means should be to

the population mean, m. Logically, a large sample should be a better representa-

tive than a small sample. Thus, the sample means obtained with a large sample

size should cluster relatively close to the population mean; the means obtained

from small samples should be more widely scattered.

As you will see, each of these three commonsense characteristics is an accurate

description of the distribution of sample means. The following example demonstrates

the process of constructing the distribution of sample means by repeatedly selecting

samples from a population.

Consider a population that consists of only 4 scores: 2, 4, 6, 8. This population is pic-

tured in the frequency distribution histogram in Figure 7.1.

We are going to use this population as the basis for constructing the distribution of

sample means for n 5 2. Remember: This distribution is the collection of sample means

from all of the possible random samples of n 5 2 from this population. We begin by

looking at all of the possible samples. For this example, there are 16 different samples,

and they are all listed in Table 7.1. Notice that the samples are listed systematically.

First, we list all of the possible samples with X 5 2 as the first score, then all of the

possible samples with X 5 4 as the first score, and so on. In this way, we can be sure

that we have all of the possible random samples.

Next, we compute the mean, M, for each of the 16 samples (see the last column of

Table 7.1). The 16 means are then placed in a frequency distribution histogram in Figure 7.2.

E X A M P L E 7 . 1

0 1 2 3 4 5 6 7 8 9

F re

q u

e n

c y

Scores

FIGURE 7.1

Frequency distribution

histogram for a popula-

tion of 4 scores: 2, 4, 6, 8.

Remember that random

sampling requires sampling

with replacement.

SECTION 7.2 / THE DISTRIBUTION OF SAMPLE MEANS 1 7 9

This is the distribution of sample means. Note that the distribution in Figure 7.2 demon-

strates two of the characteristics that we predicted for the distribution of sample means.

1. The sample means pile up around the population mean. For this example, the

population mean is m 5 5, and the sample means are clustered around a value of

5. It should not surprise you that the sample means tend to approximate the popu-

lation mean. After all, samples are supposed to be representative of the population.

2. The distribution of sample means is roughly normal in shape. This is a character-

istic that is discussed in detail later and is extremely useful because we already

know a great deal about probabilities and the normal distribution (Chapter 6).

Finally, you should notice that we can use the distribution of sample means to an-

swer probability questions about sample means. For example, if you take a sample of

n 5 2 scores from the original population, what is the probability of obtaining a sample

mean greater than 7? In symbols,

p(M . 7) 5 ?

TABLE 7.1

All the possible samples of

n 5 2 scores that can be

obtained from the population

presented in Figure 7.1. Notice

that the table lists random

samples. This requires sampling

with replacement, so it is pos-

sible to select the same score

twice.

Scores Sample Mean

Sample First Second (M)

1 2 2 2

2 2 4 3

3 2 6 4

4 2 8 5

5 4 2 3

6 4 4 4

7 4 6 5

8 4 8 6

9 6 2 4

10 6 4 5

11 6 6 6

12 6 8 7

13 8 2 5

14 8 4 6

15 8 6 7

16 8 8 8

Remember that our goal

in this chapter is to answer

probability questions about

samples with n . 1.

0 1 2 3 4 5 6 7 8 9

F re

q u

e n

c y

Sample means

FIGURE 7.2

The distribution of sam-

ple means for n 5 2. The

distribution shows the

16 sample means from

Table 7.1.

1 8 0 CHAPTER 7 PROBABILITY AND SAMPLES: THE DISTRIBUTION OF SAMPLE MEANS

Because probability is equivalent to proportion, the probability question can be

restated as follows: Of all of the possible sample means, what proportion have values

greater than 7? In this form, the question is easily answered by looking at the distribu-

tion of sample means. All of the possible sample means are pictured (see Figure 7.2),

and only 1 out of the 16 means has a value greater than 7. The answer, therefore, is 1

out of 16, or p 5 1

16 .

Example 7.1 demonstrated the construction of the distribution of sample means for an

overly simplified situation with a very small population and samples that each contain

only n 5 2 scores. In more realistic circumstances, with larger populations and larger samples, the number of possible samples increases dramatically and it is virtually

impossible to actually obtain every possible random sample. Fortunately, it is possible

to determine exactly what the distribution of sample means looks like without taking

hundreds or thousands of samples. Specifically, a mathematical proposition known as

the central limit theorem provides a precise description of the distribution that would

be obtained if you selected every possible sample, calculated every sample mean, and

constructed the distribution of the sample mean. This important and useful theorem

serves as a cornerstone for much of inferential statistics. Following is the essence of

the theorem.

Central limit theorem: For any population with mean m and standard deviation

s, the distribution of sample means for sample size n will have a mean of m and a

standard deviation of s n and will approach a normal distribution as n approaches

infinity.

The value of this theorem comes from two simple facts. First, it describes the distri-

bution of sample means for any population, no matter what shape, mean, or standard

deviation. Second, the distribution of sample means “approaches” a normal distribution

very rapidly. By the time the sample size reaches n 5 30, the distribution is almost

perfectly normal.

Note that the central limit theorem describes the distribution of sample means by

identifying the three basic characteristics that describe any distribution: shape, central

tendency, and variability. We examine each of these.

It has been observed that the distribution of sample means tends to be a normal distribu-

tion. In fact, this distribution is almost perfectly normal if either of the following two

conditions is satisfied:

1. The population from which the samples are selected is a normal distribution.

2. The number of scores (n) in each sample is relatively large, around 30 or more.

(As n gets larger, the distribution of sample means more closely approximates a

normal distribution. When n . 30, the distribution is almost normal, regardless of the

shape of the original population.)

As we noted earlier, the fact that the distribution of sample means tends to be

normal is not surprising. Whenever you take a sample from a population, you expect

the sample mean to be near to the population mean. When you take lots of differ-

ent samples, you expect the sample means to “pile up” around m, resulting in a

normal-shaped distribution. You can see this tendency emerging (although it is not

yet normal) in Figure 7.2.

T H E C E N T R A L L I M I T T H E O R E M

T H E S H A P E O F T H E D I ST R I B U T I O N O F

SA M P L E M E A N S

SECTION 7.2 / THE DISTRIBUTION OF SAMPLE MEANS 1 8 1

In Example 7.1, the distribution of sample means is centered around the mean of the

population from which the samples were obtained. In fact, the average value of all of

the sample means is exactly equal to the value of the population mean. This fact should

be intuitively reasonable; the sample means are expected to be close to the population

mean, and they do tend to pile up around µ. The formal statement of this phenomenon

is that the mean of the distribution of sample means always is identical to the popula-

tion mean. This mean value is called the expected value of M. In commonsense terms,

a sample mean is “expected” to be near its population mean. When all of the possible

sample means are obtained, the average value is identical to m.

The fact that the average value of M is equal to µ was first introduced in Chapter 4

(p. 106) in the context of biased versus unbiased statistics. The sample mean is an

example of an unbiased statistic, which means that, on average, the sample statistic

produces a value that is exactly equal to the corresponding population parameter. In this

case, the average value of all of the sample means is exactly equal to m.

The mean of the distribution of sample means is equal to the mean of the popula-

tion of scores, m, and is called the expected value of M.

So far, we have considered the shape and the central tendency of the distribution of

sample means. To completely describe this distribution, we need one more character-

istic: variability. The value we will be working with is the standard deviation for the

distribution of sample means. This standard deviation is identified by the symbol s M

and is called the standard error of M.

When the standard deviation was first introduced in Chapter 4, we noted that

this measure of variability serves two general purposes. First, the standard deviation

describes the distribution by telling whether the individual scores are clustered close

together or scattered over a wide range. Second, the standard deviation measures

how well any individual score represents the population by providing a measure

of how much distance is reasonable to expect between a score and the population

mean. The standard error serves the same two purposes for the distribution of sample

means.

1. The standard error describes the distribution of sample means. It provides a

measure of how much difference is expected from one sample to another. When

the standard error is small, then all of the sample means are close together and

have similar values. If the standard error is large, then the sample means are

scattered over a wide range and there are big differences from one sample to

another.

2. Standard error measures how well an individual sample mean represents the

entire distribution. Specifically, it provides a measure of how much distance is

reasonable to expect between a sample mean and the overall mean for the dis-

tribution of sample means. However, because the overall mean is equal to m, the

standard error also provides a measure of how much distance to expect between

a sample mean (M) and the population mean (m).

Remember that a sample is not expected to provide a perfectly accurate reflection

of its population. Although a sample mean should be representative of the population

mean, there typically is some error between the sample and the population. The stan-

dard error measures exactly how much difference is expected on average between a

sample mean, M, and the population mean, m.

T H E M E A N O F T H E D I ST R I B U T I O N O F

SA M P L E M E A N S : T H E E X P E C T E D VA L U E O F M

D E F I N I T I O N

T H E STA N DA R D E R R O R O F M

1 8 2 CHAPTER 7 PROBABILITY AND SAMPLES: THE DISTRIBUTION OF SAMPLE MEANS

The standard deviation of the distribution of sample means, s M , is called the stan-

dard error of M. The standard error provides a measure of how much distance is

expected on average between a sample mean (M) and the population mean (m).

Once again, the symbol for the standard error is s M . The s indicates that this value is

a standard deviation, and the subscript M indicates that it is the standard deviation for the

distribution of sample means. Similarly, it is common to use the symbol m M to represent

the mean of the distribution of sample means. However, m M is always equal to µ and

our primary interest in inferential statistics is to compare sample means (M) with their

population means (m). Therefore, we simply use the symbol µ to refer to the mean of the

distribution of sample means.

The standard error is an extremely valuable measure because it specifies precisely

how well a sample mean estimates its population mean—that is, how much error you

should expect, on the average, between M and m. Remember that one basic reason for

taking samples is to use the sample data to answer questions about the population.

However, you do not expect a sample to provide a perfectly accurate picture of the

population. There always is some discrepancy, or error, between a sample statistic and

the corresponding population parameter. Now we are able to calculate exactly how

much error to expect. For any sample size (n), we can compute the standard error, which

measures the average distance between a sample mean and the population mean.

The magnitude of the standard error is determined by two factors: (1) the size of

the sample and (2) the standard deviation of the population from which the sample is

selected. We examine each of these factors.

The sample size Earlier we predicted, based on common sense, that the size of a sam-

ple should influence how accurately the sample represents its population. Specifically,

a large sample should be more accurate than a small sample. In general, as the sample

size increases, the error between the sample mean and the population mean should

decrease. This rule is also known as the law of large numbers.

The law of large numbers states that the larger the sample size (n), the more

probable it is that the sample mean is close to the population mean.

The population standard deviation As we noted earlier, there is an inverse relationship

between the sample size and the standard error: bigger samples have smaller error, and

smaller samples have bigger error. At the extreme, the smallest possible sample (and the

largest standard error) occurs when the sample consists of n 5 1 score. At this extreme,

each sample is a single score and the distribution of sample means is identical to the origi-

nal distribution of scores. In this case, the standard deviation for the distribution of sample

means, which is the standard error, is identical to the standard deviation for the distribution

of scores. In other words, when n 5 1, the standard error 5 s M is identical to the standard

deviation 5 s.

When n 5 1, s M 5 s (standard error 5 standard deviation).

You can think of the standard deviation as the “starting point” for standard error.

When n 5 1, the standard error and the standard deviation are the same: s M 5 s. As

sample size increases beyond n 5 1, the sample becomes a more accurate representa-

tive of the population, and the standard error decreases. The formula for standard error

expresses this relationship between standard deviation and sample size (n).

standard error 5 s 5 s

M n (7.1)

D E F I N I T I O N

This formula is contained in

the central limit theorem.

SECTION 7.2 / THE DISTRIBUTION OF SAMPLE MEANS 1 8 3

Note that the formula satisfies all of the requirements for the standard error.

Specifically,

a. As sample size (n) increases, the size of the standard error decreases. (Larger

samples are more accurate.)

b. When the sample consists of a single score (n 5 1), the standard error is the same as the standard deviation (s

M 5 s).

In Equation 7.1 and in most of the preceding discussion, we defined standard error

in terms of the population standard deviation. However, the population standard devia-

tion (s) and the population variance (s2) are directly related, and it is easy to substitute

variance into the equation for standard error. Using the simple equality s 5 s2 , the

equation for standard error can be rewritten as follows:

standard error 5 s 5 s

5 s

M n n n

2 2

(7.2)

Throughout the rest of this chapter (and in Chapter 8), we continue to define

standard error in terms of the standard deviation (Equation 7.1). However, in later

chapters (starting in Chapter 9), the formula based on variance (Equation 7.2) will

become more useful.

Figure 7.3 illustrates the general relationship between standard error and sample

size. (The calculations for the data points in Figure 7.3 are presented in Table 7.2.)

Again, the basic concept is that the larger a sample is, the more accurately it rep-

resents its population. Also note that the standard error decreases in relation to the

square root of the sample size. As a result, researchers can substantially reduce

error by increasing sample size up to around n 5 30. However, increasing sample

size beyond n 5 30 does not produce much additional improvement in how well the

sample represents the population.

Standard distance between a sample

mean and the population

mean

Standard Error (based on � � 10)

4 9 16 25 36 49 64 100

Number of scores in the sample (n)

9 8 7 6 5 4 3 2 1

FIGURE 7.3

The relationship between standard error and sample size. As the sample size is increased, there

is less error between the sample mean and the population mean.

1 8 4 CHAPTER 7 PROBABILITY AND SAMPLES: THE DISTRIBUTION OF SAMPLE MEANS

Before we move forward with our discussion of the distribution of sample means, we

pause for a moment to emphasize the idea that we are now dealing with three different

but interrelated distributions.

1. First, we have the original population of scores. This population contains the

scores for thousands or millions of individual people, and it has its own shape,

mean, and standard deviation. For example, the population of IQ scores consists

of millions of individual IQ scores that form a normal distribution with a mean

of m 5 100 and a standard deviation of s 5 15. An example of a population is

shown in Figure 7.4(a).

2. Next, we have a sample that is selected from the population. The sample

consists of a small set of scores for a few people who have been selected to

represent the entire population. For example, we could select a sample of

n 5 25 people and measure each individual’s IQ score. The 25 scores could

be organized in a frequency distribution and we could calculate the sample

mean and the sample standard deviation. Note that the sample also has its

own shape, mean, and standard deviation. An example of a sample is shown

in Figure 7.4(b).

3. The third distribution is the distribution of sample means. This is a theoretical

distribution consisting of the sample means obtained from all of the possible

random samples of a specific size. For example, the distribution of sample

means for samples of n 5 25 IQ scores would be normal with a mean

(expected value) of m 5 100 and a standard deviation (standard error) of

s M

5 15 25 5 3. This distribution, shown in Figure 7.4(c), also has its own shape,

mean, and standard deviation.

T H R E E D I F F E R E N T D I ST R I B U T I O N S

TABLE 7.2

Calculations for the points

shown in Figure 7.3. Again,

notice that the size of the stan-

dard error decreases as the size

of the sample increases.

Sample Size (n) Standard Error

1 s 5 M

1 5 10.00

4 s 5 M

4 5 5.00

9 s 5M 10

9 5 3.33

16 s 5 M

16 5 2.50

25 s 5 M

25 5 2.00

49 s 5M 10

49 5 1.43

64 s 5M 10

64 5 1.25

100 s 5M 10

100 5 1.00

SECTION 7.2 / THE DISTRIBUTION OF SAMPLE MEANS 1 8 5

µ 5 100

� 5 15

(a) Original population of IQ scores.

80 90 100 110 120 130

s 5 11.5

M 5 101.2

(b) A sample of n 5 25 IQ scores.

µ 5 100

�M 5 3

FIGURE 7.4

Three distributions. Part

(a) shows the population

of IQ scores. Part (b)

shows a sample of n 5 25 IQ scores. Part (c) shows

the distribution of sample

means for samples of

n 5 100 scores. Note that the sample mean from

part (b) is one of the thou-

sands of sample means in

the part (c) distribution.

Note that the scores for the sample [Figure 7.4(b)] were taken from the original

population [Figure 7.4(a)] and that the mean for the sample is one of the values con-

tained in the distribution of sample means [Figure 7.4(c)]. Thus, the three distributions

are all connected, but they are all distinct.

1 8 6 CHAPTER 7 PROBABILITY AND SAMPLES: THE DISTRIBUTION OF SAMPLE MEANS

L E A R N I N G C H E C K PROBABILITY AND THE DISTRIBUTION OF SAMPLE MEANS The primary use of the distribution of sample means is to find the probability associated

with any specific sample. Recall that probability is equivalent to proportion. Because

the distribution of sample means presents the entire set of all possible sample means,

we can use proportions of this distribution to determine probabilities. The following

example demonstrates this process.

The population of scores on the SAT forms a normal distribution with m 5 500 and

s 5 100. If you take a random sample of n 5 25 students, what is the probability that

the sample mean will be greater than M 5 540?

First, you can restate this probability question as a proportion question: Out of all of

the possible sample means, what proportion have values greater than 540? You know

about “all of the possible sample means”; this is the distribution of sample means. The

problem is to find a specific portion of this distribution.

Although we cannot construct the distribution of sample means by repeatedly taking

samples and calculating means (as in Example 7.1), we know exactly what the distribu-

tion looks like based on the information from the central limit theorem. Specifically, the

distribution of sample means has the following characteristics:

a. The distribution is normal because the population of SAT scores is normal.

b. The distribution has a mean of 500 because the population mean is m 5 500.

c. For n 5 25, the distribution has a standard error of s M 5 20:

s 5 s

5 5 5 M

100

5 20

7.3

E X A M P L E 7 . 2

1. A population has a mean of m 5 65 and a standard deviation of s 5 16.

a. Describe the distribution of sample means (shape, central tendency, and vari-

ability) for samples of size n 5 4 selected from this population.

b. Describe the distribution of sample means (shape, central tendency, and vari-

ability) for samples of size n 5 64 selected from this population.

2. Describe the relationship between the sample size and the standard error of M.

3. For a population with of m 5 40 and a standard deviation of s 5 8, the standard

error for a sample mean can never be larger than 8. (True or false.)

1. a The distribution of sample means has an expected value of m 5 65 and a standard error of

s M

5 16

4�� 5 8. The shape of the distribution is unknown because the sample is not large

enough to guarantee a normal distribution and the population shape is not known.

b. The distribution of sample means has an expected value of m 5 65 and a standard error

of s M

5 16

��64 5 2. The shape of the distribution is normal because the sample is large

enough to guarantee a normal distribution.

2. The standard error decreases as sample size increases.

3. True. If each sample has n 5 1 score, then the standard error is 8. For any other sample size,

the standard error is smaller than 8.

L E A R N I N G C H E C K

ANSWERS

Caution: Whenever you

have a probability question

about a sample mean, you

must use the distribution of

sample means.

SECTION 7.3 / PROBABILITY AND THE DISTRIBUTION OF SAMPLE MEANS 1 8 7

This distribution of sample means is shown in Figure 7.5.

We are interested in sample means greater than 540 (the shaded area in Figure

7.5), so the next step is to use a z-score to locate the exact position of M 5 540 in the distribution. The value 540 is located above the mean by 40 points, which is exactly

2 standard deviations (in this case, exactly 2 standard errors). Thus, the z-score for

M 5 540 is z 5 �2.00. Because this distribution of sample means is normal, you can use the unit normal

table to find the probability associated with z 5 �2.00. The table indicates that 0.0228 of the distribution is located in the tail of the distribution beyond z 5 �2.00. Our con- clusion is that it is very unlikely, p 5 0.0228 (2.28%), to obtain a random sample of n 5 25 students with an average SAT score greater than 540.

As demonstrated in Example 7.2, it is possible to use a z-score to describe the exact

location of any specific sample mean within the distribution of sample means. The

z-score tells exactly where the sample mean is located in relation to all of the other pos-

sible sample means that could have been obtained. As defined in Chapter 5, a z-score

identifies the location with a signed number so that

1. The sign tells whether the location is above (�) or below (–) the mean.

2. The number tells the distance between the location and the mean in terms of the

number of standard deviations.

However, we are now finding a location within the distribution of sample means.

Therefore, we must use the notation and terminology appropriate for this distribution.

First, we are finding the location for a sample mean (M) rather than a score (X). Second,

the standard deviation for the distribution of sample means is the standard error, s M .

With these changes, the z-score formula for locating a sample mean is

z M

5 2m

s (7.3)

A Z - S CO R E F O R SA M P L E M E A N S

500

0 21

540

M 20

FIGURE 7.5

The distribution of

sample means for n 5 25.

Samples were selected

from a normal population

with µ 5 500 and

s 5 100.

1 8 8 CHAPTER 7 PROBABILITY AND SAMPLES: THE DISTRIBUTION OF SAMPLE MEANS

Just as every score (X) has a z-score that describes its position in the distribution of

scores, every sample mean (M) has a z-score that describes its position in the distribu-

tion of sample means. When the distribution of sample means is normal, it is possible

to use z-scores and the unit normal table to find the probability associated with any

specific sample mean (as in Example 7.2). The following example demonstrates that it

also is possible to make quantitative predictions about the kinds of samples that should

be obtained from any population.

Once again, the distribution of SAT scores forms a normal distribution with a mean

of m 5 500 and a standard deviation of s 5 100. For this example, we are going to

determine what kind of sample mean is likely to be obtained as the average SAT score

for a random sample of n 5 25 students. Specifically, we determine the exact range of

values that is expected for the sample mean 80% of the time.

We begin with the distribution of sample means for n 5 25. As demonstrated in

Example 7.2, this distribution is normal with an expected value of m 5 500 and a stan-

dard error of s M 5 20 (Figure 7.6). Our goal is to find the range of values that make

up the middle 80% of the distribution. Because the distribution is normal, we can use

the unit normal table. First, the 80% in the middle is split in half, with 40% (0.4000)

on each side of the mean. Looking up 0.4000 in column D (the proportion between the

mean and z), we find a corresponding z-score of z 5 1.28. Thus, the z-score boundar-

ies for the middle 80% are z 5 �1.28 and z 5 –1.28. By definition, a z-score of 1.28 represents a location that is 1.28 standard deviations (or standard errors) from the

mean. With a standard error of 20 points, the distance from the mean is 1.28(20) 5 25.6 points. The mean is m 5 500, so a distance of 25.6 in both directions produces a

range of values from 474.4 to 525.6.

Thus, 80% of all the possible sample means are contained in a range between 474.4

and 525.6. If we select a sample of n 5 25 students, we can be 80% confident that the

mean SAT score for the sample will be in this range.

E X A M P L E 7 . 3

500

40% 40% 10%10%

0 +1.28−1.28

525.6474.4 µ

FIGURE 7.6

The middle 80% of the

distribution of sample

means for n 5 25.

Samples were selected

from a normal population

with µ 5 500 and

s 5 100.

Caution: When computing

z for a single score, use the

standard deviation, s. When

computing z for a sample

mean, you must use the stan-

dard error, s M (see Box 7.2).

SECTION 7.3 / PROBABILITY AND THE DISTRIBUTION OF SAMPLE MEANS 1 8 9

The point of Example 7.3 is that the distribution of sample means makes it possible

to predict the value that ought to be obtained for a sample mean. We know, for example,

that a sample of n 5 25 students ought to have a mean SAT score around 500. More specifically, we are 80% confident that the value of the sample mean will be between

474.4 and 525.6. The ability to predict sample means in this way is a valuable tool for

the inferential statistics that follow.

BOX

7.2 THE DIFFERENCE BETWEEN STANDARD DEVIATION AND STANDARD ERROR

standard error 5 s 5 s

M n

If you are working with a single score, then n 5 1,

and the standard error becomes

standard error standard deviatio5 s 5 s

5 s

5 s 5 M

n 1 nn

Thus, standard error always measures the standard

distance from the population mean for any sample

size, including n 5 1.

A constant source of confusion for many students is

the difference between standard deviation and standard

error. Remember that standard deviation measures the

standard distance between a score and the population

mean, X – m. If you are working with a distribution of

scores, the standard deviation is the appropriate mea-

sure of variability. Standard error, on the other hand,

measures the standard distance between a sample

mean and the population mean, M – m. Whenever you

have a question concerning a sample, the standard

error is the appropriate measure of variability.

If you still find the distinction confusing, there is

a simple solution. Namely, if you always use stan-

dard error, you always will be right. Consider the

formula for standard error:

1. A sample is selected from a population with a mean of m 5 90 and a standard

deviation of s 5 15. Find the z-score for the sample mean for each of the follow-

ing samples.

a. n 5 9 scores with M 5 100

b. n 5 25 scores with M 5 89

2. What is the probability of obtaining a sample mean greater than M 5 60 for a ran-

dom sample of n 5 16 scores selected from a normal population with a mean of

m 5 65 and a standard deviation of s 5 20?

3. What are the boundaries for the middle 50% of all possible random samples of

n 5 25 scores selected from a normal population with m 5 80 and s 5 10?

1. a. The standard error is s M 5 5, and z 5 2.00.

b. The standard error is s M 5 3, and z 5 –0.33.

2. The standard error is s M 5 5, and M 5 60 corresponds to z 5 –1.00, p(M . 60) 5

p(z . –1.00) 5 0.8413 (or 84.13%).

3. The standard error is s M 5 2, and the z-score boundaries for the middle 50% are z 5 –0.67

and z 5 �0.67. The boundaries are 78.66 and 81.34.

L E A R N I N G C H E C K

ANSWERS

1 9 0 CHAPTER 7 PROBABILITY AND SAMPLES: THE DISTRIBUTION OF SAMPLE MEANS

MORE ABOUT STANDARD ERROR

At the beginning of this chapter, we introduced the idea that it is possible to obtain

thousands of different samples from a single population. Each sample has its own in-

dividuals, its own scores, and its own sample mean. The distribution of sample means

provides a method for organizing all of the different sample means into a single picture.

Figure 7.7 shows a prototypical distribution of sample means. To emphasize the fact

that the distribution contains many different samples, we have constructed this figure

so that the distribution is made up of hundreds of small boxes, each box representing

a single sample mean. Also notice that the sample means tend to pile up around the

population mean (µ), forming a normal-shaped distribution as predicted by the central

limit theorem.

The distribution shown in Figure 7.7 provides a concrete example for reviewing the

general concepts of sampling error and standard error. Although the following points

may seem obvious, they are intended to provide you with a better understanding of

these two statistical concepts.

1. Sampling Error. The general concept of sampling error is that a sample typi-

cally does not provide a perfectly accurate representation of its population.

More specifically, there typically is some discrepancy (or error) between a

statistic computed for a sample and the corresponding parameter for the popula-

tion. As you look at Figure 7.7, notice that the individual sample means are not

exactly equal to the population mean. In fact, 50% of the samples have means

that are smaller than µ (the entire left-hand side of the distribution). Similarly,

50% of the samples produce means that overestimate the true population mean.

In general, there is some discrepancy, or sampling error, between the mean for

a sample and the mean for the population from which the sample was obtained.

2. Standard Error. Again looking at Figure 7.7, notice that most of the sample

means are relatively close to the population mean (those in the center of the distri-

bution). These samples provide a fairly accurate representation of the population.

7.4

M µ

FIGURE 7.7

An example of a typical

distribution of sample

means. Each of the small

boxes represents the mean

obtained for one sample.

SECTION 7.4 / MORE ABOUT STANDARD ERROR 1 9 1

On the other hand, some samples produce means that are out in the tails of the

distribution, relatively far from the population mean. These extreme sample means

do not accurately represent the population. For each individual sample, you can

measure the error (or distance) between the sample mean and the population mean.

For some samples, the error is relatively small, but for other samples, the error is

relatively large. The standard error provides a way to measure the “average,” or

standard, distance between a sample mean and the population mean.

Thus, the standard error provides a method for defining and measuring sampling

error. Knowing the standard error gives researchers a good indication of how accurately

their sample data represent the populations that they are studying. In most research

situations, for example, the population mean is unknown, and the researcher selects

a sample to help obtain information about the unknown population. Specifically, the

sample mean provides information about the value of the unknown population mean.

The sample mean is not expected to give a perfectly accurate representation of the

population mean; there will be some error, and the standard error tells exactly how much

error, on average, should exist between the sample mean and the unknown population

mean. The following example demonstrates the use of standard error and provides addi-

tional information about the relationship between standard error and standard deviation.

A recent survey of students at a local college included the following question: How

many minutes do you spend each day watching electronic video (e.g., online, TV, cell

phone, iPad, etc.). The average response was m 5 80 minutes, and the distribution of

viewing times was approximately normal with a standard deviation of s 5 20 minutes.

Next, we take a sample from this population and examine how accurately the sample

mean represents the population mean. More specifically, we will examine how sample

size affects accuracy by considering three different samples: one with n 5 1 student,

one with n 5 4 students, and one with n 5 100 students.

Figure 7.8 shows the distributions of sample means based on samples of n 5 1,

n 5 4, and n 5 100. Each distribution shows the collection of all possible sample means

that could be obtained for that particular sample size. Notice that all three sampling

E X A M P L E 7 . 4

Distribution of M for n 5 100

σ M

5 2

Distribution of M for n 5 4 σ

M 5 10

Distribution of M for n 5 1

σ M

5 σ 5 20

FIGURE 7.8

The distribution of sample means for random samples of size (a) n 5 1, (b) n 5 4, and

the size of the standard error decreases as the sample size increases.

(a) (b) (c)

1 9 2 CHAPTER 7 PROBABILITY AND SAMPLES: THE DISTRIBUTION OF SAMPLE MEANS

distributions are normal (because the original population is normal), and all three have

the same mean, m 5 80, which is the expected value of M. However, the three distribu-

tions differ greatly with respect to variability. We will consider each one separately.

The smallest sample size is n 5 1. When a sample consists of a single student, the

mean for the sample equals the score for the student, M 5 X. Thus, when n 5 1, the

distribution of sample means is identical to the original population of scores. In this

case, the standard error for the distribution of sample means is equal to the standard

deviation for the original population. Equation 7.1 confirms this observation.

s 5 s

5 5 M

1 20

When the sample consists of a single student, you expect, on average, a 20-point dif-

ference between the sample mean and the mean for the population. As we noted earlier,

the population standard deviation is the “starting point” for the standard error. With the

smallest possible sample, n 5 1, the standard error is equal to the standard deviation

[see Figure 7.8(a)].

As the sample size increases, however, the standard error gets smaller. For a sample

of n 5 4 students, the standard error is

s 5 s

5 5 5 M

2 10

That is, the typical (or standard) distance between M and m is 10 points. Figure 7.8(b)

illustrates this distribution. Notice that the sample means in this distribution approximate

the population mean more closely than in the previous distribution where n 5 1.

With a sample of n 5 100, the standard error is still smaller.

s 5 s

5 5 5 M

100

10 2

A sample of n 5 100 students should produce a sample mean that represents the popu-

lation much more accurately than a sample of n 5 4 or n 5 1. As shown in Figure 7.8(c),

there is very little error between M and µ when n 5 100. Specifically, you would expect,

on average, only a 2-point difference between the population mean and the sample mean.

In summary, this example illustrates that with the smallest possible sample (n 5 1),

the standard error and the population standard deviation are the same. When sample

size is increased, the standard error gets smaller, and the sample means tend to approxi-

mate µ more closely. Thus, standard error defines the relationship between sample size

and the accuracy with which M represents µ.

IN THE LITERATURE

REPORTING STANDARD ERROR

As we will see later, standard error plays a very important role in inferential

statistics. Because of its crucial role, the standard error for a sample mean, rather

than the sample standard deviation, is often reported in scientific papers. Scientific

journals vary in how they refer to the standard error, but frequently the symbols

SE and SEM (for standard error of the mean) are used. The standard error is reported in

SECTION 7.4 / MORE ABOUT STANDARD ERROR 1 9 3

TABLE 7.3

The mean self-consciousness

scores for participants who were

working in front of a video

camera and those who were not

(controls).

n Mean SE

Control 17 32.23 2.31

Camera 15 45.17 2.78

two ways. Much like the standard deviation, it may be reported in a table along with the

sample means (Table 7.3). Alternatively, the standard error may be reported in graphs.

Figure 7.9 illustrates the use of a bar graph to display information about the sample

mean and the standard error. In this experiment, two samples (groups A and B) are

given different treatments, and then the subjects’ scores on a dependent variable are

recorded. The mean for group A is M 5 15, and for group B, it is M 5 30. For both samples, the standard error of M is s

M 5 5. Note that the mean is represented by the

height of the bar, and the standard error is depicted by brackets at the top of each bar.

Each bracket extends 1 standard error above and 1 standard error below the sample

mean. Thus, the graph illustrates the mean for each group plus or minus 1 standard error

(M ± SE). When you glance at Figure 7.9, not only do you get a “picture” of the sample

means, but you also get an idea of how much error you should expect for those means.

Figure 7.10 shows how sample means and standard error are displayed in a line

graph. In this study, two samples representing different age groups are tested on a

Group A

M s

c o

re (

± S E

)

Group B

35FIGURE 7.9

The mean (± SE) score for

treatment groups A and B.

1 M n

u m

b e

r o

f m

is ta

k e

s (±

S E

)

2 3

Trials

Group A

Group B

FIGURE 7.10

The mean (± SE) number

of mistakes made for

groups A and B on

each trial.

1 9 4 CHAPTER 7 PROBABILITY AND SAMPLES: THE DISTRIBUTION OF SAMPLE MEANS

task for four trials. The number of errors committed on each trial is recorded for

all participants. The graph shows the mean (M) number of errors committed for

each group on each trial. The brackets show the size of the standard error for each sample

mean. Again, the brackets extend 1 standard error above and below the value of the mean.

1. If a sample is selected from a population with a mean of m 5 120 and a standard

deviation of s 5 20, then, on average, how much difference should there be be-

tween the sample mean and the population mean

a. for a sample of n 5 25 scores?

b. for a sample of n 5 100 scores?

2. Can the value of the standard error ever be larger than the value of the population

standard deviation? Explain your answer.

3. If a random sample is selected from a population with a standard deviation of s 5 40,

then how large a sample is needed to have a standard error of 2 points or less?

4. A sample of n 5 25 scores is selected from a population. If the sample mean has a

standard error of 4 points, then what is the population standard deviation?

1. a. s M 5 4 points

b. s M 5 2 points

2. No. The standard error is computed by dividing the standard deviation by the square root of

n. The standard error is always less than or equal to the standard deviation.

3. A sample of n 5 400 or larger.

4. s 5 20

L E A R N I N G C H E C K

ANSWERS

LOOKING AHEAD TO INFERENTIAL STATISTICS

Inferential statistics are methods that use sample data as the basis for drawing general con-

clusions about populations. However, we have noted that a sample is not expected to give

a perfectly accurate reflection of its population. In particular, there will be some error or

discrepancy between a sample statistic and the corresponding population parameter. In this

chapter, we have observed that a sample mean is not exactly equal to the population mean.

The standard error of M specifies how much difference is expected on average between the

mean for a sample and the mean for the population.

The natural differences that exist between samples and populations introduce a de-

gree of uncertainty and error into all inferential processes. Specifically, there is always

a margin of error that must be considered whenever a researcher uses a sample mean as

the basis for drawing a conclusion about a population mean. Remember that the sample

mean is not perfect. In the next seven chapters we introduce a variety of statistical meth-

ods that all use sample means to draw inferences about population means.

In each case, the distribution of sample means and the standard error are critical

elements in the inferential process. Before we begin this series of chapters, we pause

briefly to demonstrate how the distribution of sample means, along with z-scores and

probability, can help us use sample means to draw inferences about population means.

We ended Chapters 5 and 6 with a demonstration of how inferential statistics are used

to help interpret the results of a research study. A general research situation was shown

7.5

E X A M P L E 7 . 5

SECTION 7.5 / LOOKING AHEAD TO INFERENTIAL STATISTICS 1 9 5

in Figures 5.9 and 6.16, and is repeated here in Figure 7.11. The research begins with

a population that forms a normal distribution with a mean of µ 5 400 and a standard deviation of s 5 20. A sample is selected from the population and a treatment is admin- istered to the sample. The goal for the study is to evaluate the effect of the treatment.

In the previous two chapters, however, we were limited to using a sample of n 5 1 individual. Now, we can use any sample size and have selected n 5 25 for this example.

The psychologist makes a decision about the effect of the treatment by comparing

the treated sample with the original population. If the scores in the sample are notice-

ably different from the scores in the population, then the researcher has evidence that

the treatment has an effect. The problem is to determine exactly how much difference

is necessary before we can say that the sample is noticeably different.

The distribution of sample means and the standard error can help researchers make

this decision. In particular, the distribution of sample means can be used to show ex-

actly what would be expected for samples that do not receive any treatment. This allows

researchers to make a simple comparison between

a. The treated sample (from the research study)

b. Untreated samples (from the distribution of sample means)

If our treated sample is noticeably different from the untreated samples, then we have

evidence that the treatment has an effect. On the other hand, if our treated sample still

looks like one of the untreated samples, then we must conclude that the treatment does

not appear to have any effect.

We begin with the original population and consider the distribution of sample means

for all of the possible samples of n 5 25. The distribution of sample means has the fol- lowing characteristics:

1. It is a normal distribution, because the original population is normal.

2. It has an expected value of 400, because the population mean is m 5 400.

3. It has a standard error of s M = = = 20

5 4, because the population standard

deviation is s 5 20 and the sample size is n 5 25.

T r e a t

m e n t

Treated sample n = 25

Sample n = 25

Population

Normal

µ = 400 σ = 20

FIGURE 7.11

The structure of the

research study described

in Example 7.5. The

purpose of the study is

to determine whether the

treatment has an effect.

1 9 6 CHAPTER 7 PROBABILITY AND SAMPLES: THE DISTRIBUTION OF SAMPLE MEANS

The distribution of sample means is shown in Figure 7.12. Notice that an untreated

sample of n 5 25 should have a mean around m 5 400. To be more precise, we can use z-scores to determine the middle 95% of all the possible sample means. As dem-

onstrated in Chapter 6 (p. 170), the middle 95% of a normal distribution is located

between z-score boundaries of z 5 �1.96 and z 5 –1.96 (check the unit normal table). These z-score boundaries are shown in Figure 7.12. With a standard error of s 5 4 points, a z-score of 1.96 corresponds to a distance of 1.96(4) 5 7.84 points from the mean. Thus, the z-score boundaries of ±1.96 correspond to sample means of 392.16

and 407.84.

We have demonstrated that an untreated sample is almost guaranteed (95% prob-

ability) to have a sample mean between 392.16 and 407.84. If our sample has a mean

within this range, then we must conclude that our sample of is not noticeably different

from untreated samples. In this case, we conclude that the treatment does not appear

to have any effect.

On the other hand, if the mean for the treated sample is outside the 95% range, then

we can conclude that our sample is noticeably different from the samples that would be

obtained without any treatment. In this case, the research results provide evidence that

the treatment has an effect.

In Example 7.5 we used the distribution of sample means, together with z-

scores and probability, to provide a description of what is reasonable to expect for

an untreated sample. Then, we evaluated the effect of a treatment by determining

whether the treated sample was noticeably different from an untreated sample. This

procedure forms the foundation for the inferential technique known as hypothesis

testing, which is introduced in Chapter 8 and repeated throughout the remainder of

this book.

µ = 400392.16

−1.96 +1.96

407.84

σM = 4

FIGURE 7.12

The distribution of sam-

ple means for samples

of n 5 25 untreated rats (from Example 7.5).

SECTION 7.5 / LOOKING AHEAD TO INFERENTIAL STATISTICS 1 9 7

1. A population forms a normal distribution with a mean of m 5 80 and a standard

deviation of s 5 20.

a. If single score is selected from this population, how much distance would you

expect, on average, between the score and the population mean?

b. If a sample of n 5 100 scores is selected from this population, how much dis-

tance would you expect, on average, between the sample mean and the popula-

tion mean?

2. A population forms a normal shaped distribution with m 5 40 and s 5 8.

a. A sample of n 5 16 scores from this population has a mean of M 5 36. Would

you describe this as a relatively typical sample, or is the sample mean an

extreme value? Explain your answer.

b. If the sample from part a had n 5 4 scores, would it be considered typical or

extreme?

3. The SAT scores for the entering freshman class at a local college form a normal

distribution with a mean of m 5 530 and a standard deviation of s 5 80.

a. For a random sample of n 5 16 students, what range of values for the sample

mean would be expected 95% of the time?

b. What range of values would be expected 95% of the time if the sample size

were n 5 100?

4. A sample of n 516 individuals is selected from a normal population with a mean

of m 5 50 with s 5 12. After a treatment is administered to the sample, the sam-

ple mean is found to be M 5 57. Is this sample mean likely to occur if the treat-

ment has no effect? Specifically, is the sample mean within the range of values

that would be expected 95% of the time?

1. a. For a single score, the standard distance from the mean is the standard deviation, s 5 20.

b. For a sample of n 5 100 scores, the average distance between the sample mean and the

population mean is the standard error, s M

5 20 100

5 2.

2. a. With n 5 16, the standard error is 2, and the sample mean corresponds to z 5 22.00.

This is an extreme value.

b. With n 5 4, the standard error is 4, and the sample mean corresponds to z 5 21.00. This

is a relatively typical value.

3. a. With n 5 16, the standard error is s M 5 20 points. Using z 5 ±1.96, the 95% range

extends from 490.8 to 569.2.

b. With n 5 100, the standard error is only 8 points and the range extends from 514.32 to

545.68.

4. With n 5 16, the standard error is s M 5 3. If the treatment has no effect, then the population

mean is still m 5 50 and 95% of all the possible sample means should be within 1.96(3) 5

5.88 points of m 5 50. This is a range of values from 44.12 to 55.88. Our sample mean is

outside this range, so it is not the kind of sample that ought to be obtained if the treatment

has no effect.

L E A R N I N G C H E C K

ANSWERS

1 9 8 CHAPTER 7 PROBABILITY AND SAMPLES: THE DISTRIBUTION OF SAMPLE MEANS

SUMMARY

1. The distribution of sample means is defined as the set of Ms for all of the possible random samples for a spe- cific sample size (n) that can be obtained from a given population. According to the central limit theorem, the parameters of the distribution of sample means are as follows:

a. Shape. The distribution of sample means is normal if either one of the following two conditions is satisfied:

(1) The population from which the samples are selected is normal.

(2) The size of the samples is relatively large (n 5 30 or more).

b. Central Tendency. The mean of the distribution of sample means is identical to the mean of the popu- lation from which the samples are selected. The mean of the distribution of sample means is called the expected value of M.

c. Variability. The standard deviation of the distribu- tion of sample means is called the standard error of M and is defined by the formula

s 5 s

M M n n

or 2

Standard error measures the standard distance between a sample mean (M) and the population mean (m).

2. One of the most important concepts in this chapter is standard error. The standard error is the standard

deviation of the distribution of sample means. It mea- sures the standard distance between a sample mean (M) and the population mean (m). The standard error tells how much error to expect if you are using a sam- ple mean to represent a population mean.

3. The location of each M in the distribution of sample means can be specified by a z-score:

z M

5 2m

Because the distribution of sample means tends to be normal, we can use these z-scores and the unit normal table to find probabilities for specific sample means. In particular, we can identify which sample means are likely and which are very unlikely to be obtained from any given population. This ability to find probabilities for samples is the basis for the inferential statistics in the chapters ahead.

4. In general terms, the standard error measures how much discrepancy you should expect between a sample statistic and a population parameter. Statistical inference involves using sample statistics to make a general conclusion about a population parameter. Thus, standard error plays a crucial role in inferential statistics.

KEY TERMS

sampling error (176)

distribution of sample means (177)

sampling distribution (177)

central limit theorem (180)

expected value of M (181)

standard error of M (181)

law of large numbers (182)

RESOURCES

Go to CengageBrain.com to access Psychology CourseMate, where you will find an

interactive eBook, glossaries, flashcards, quizzes, statistics workshops, and more.

FOCUS ON PROBLEM SOLVING 1 9 9

If your professor has assigned Aplia:

1. Sign in to your account.

2. Complete the corresponding exercises as required by your professor.

3. When finished, click “Grade It Now” to see which areas you have mastered, which

areas need more work, and detailed explanations of every answer.

The statistical computer package SPSS is not structured to compute the standard error

or a z-score for a sample mean. In later chapters, however, we introduce new inferential

statistics that are included in SPSS. When these new statistics are computed, SPSS typi-

cally includes a report of standard error that describes how accurately, on average, the

sample represents its population.

FOCUS ON PROBLEM SOLVING

1. Whenever you are working probability questions about sample means,

you must use the distribution of sample means. Remember that every

probability question can be restated as a proportion question. Probabilities

for sample means are equivalent to proportions of the distribution of

sample means.

2. When computing probabilities for sample means, the most common error

is to use standard deviation (s) instead of standard error (s M ) in the z-score

formula. Standard deviation measures the typical deviation (or error) for a

single score. Standard error measures the typical deviation (or error) for a

sample. Remember: The larger the sample is, the more accurately the sample

represents the population. Thus, sample size (n) is a critical part of the stan-

dard error.

Standard error 5 s 5 s

M n

3. Although the distribution of sample means is often normal, it is not always

a normal distribution. Check the criteria to be certain that the distribution is

normal before you use the unit normal table to find probabilities (see item 1a

of the Summary). Remember that all probability problems with a normal dis-

tribution are easier to solve if you sketch the distribution and shade in the area

of interest.

2 0 0 CHAPTER 7 PROBABILITY AND SAMPLES: THE DISTRIBUTION OF SAMPLE MEANS

DEMONSTRATION 7.1

PROBABILITY AND THE DISTRIBUTION OF SAMPLE MEANS

A population forms a normal distribution with a mean of m 5 60 and a standard

deviation of s 5 12. For a sample of n 5 36 scores from this population, what is the

probability of obtaining a sample mean greater than 64?

p(M . 64) 5 ?

Rephrase the probability question as a proportion question. Out of all of the possible sample means for n 5 36, what proportion has values greater than 64? All of the possible

sample means is simply the distribution of sample means, which is normal, with a mean of

m 5 60 and a standard error of

s 5 s

5 5 5 M

6 2

The distribution is shown in Figure 7.13(a). Because the problem is asking for the pro-

portion greater than M 5 64, this portion of the distribution is shaded in Figure 7.13(b).

Compute the z-score for the sample mean. A sample mean of M 5 64 corresponds to a z-score of

z M

5 2m

s 5

2 5 5

64 60

2 2 00.

Therefore, p(M . 64) 5 p(z . 2.00)

Look up the proportion in the unit normal table. Find z 5 2.00 in column A and read across the row to find p 5 0.0228 in column C. This is the answer as shown in

Figure 7.13(c).

p(M . 64) 5 p(z . 2.00) 5 0.0228 (or 2.28%)

S T E P 1

S T E P 2

S T E P 3

64 64

Column C p = 0.0228

Column B p = 0.9772

µµ

σ M

= 2 σ M

= 2

M M M

FIGURE 7.13

Sketches of the distribution for Demonstration 7.1.

(a) (b) (c)

PROBLEMS 2 0 1

PROBLEMS

1. Describe the distribution of sample means (shape, expected value, and standard error) for samples of n 5 100 selected from a population with a mean of m 5 40 and a standard deviation of s 5 10.

2. A sample is selected from a population with a mean of m 5 40 and a standard deviation of s 5 8.

a. If the sample has n 5 4 scores, what is the expected value of M and the standard error of M?

b. If the sample has n 5 16 scores, what is the expected value of M and the standard error of M?

3. The distribution of sample means is not always a normal distribution. Under what circumstances is the distribution of sample means not normal?

4. A population has a standard deviation of s 5 24. a. On average, how much difference should exist be-

tween the population mean and the sample mean for n 5 4 scores randomly selected from the population?

b. On average, how much difference should exist for a sample of n 5 9 scores?

c. On average, how much difference should exist for a sample of n 5 16 scores?

5. For a population with a mean of m 5 70 and a standard deviation of s 5 20, how much error, on average, would you expect between the sample mean (M) and the popu- lation mean for each of the following sample sizes?

a. n 5 4 scores b. n 5 16 scores c. n 5 25 scores

6. For a population with a standard deviation of s 5 20, how large a sample is necessary to have a standard error that is:

a. less than or equal to 5 points? b. less than or equal to 2 points? c. less than or equal to 1 point?

7. For a population with s 5 12, how large a sample is necessary to have a standard error that is:

a. less than 4 points? b. less than 3 points? c. less than 2 point?

8. For a sample of n 5 25 scores, what is the value of the population standard deviation (s) necessary to produce each of the following a standard error values?

a. s M 5 10 points?

b. s M 5 5 points?

c. s M 5 2 points?

9. For a population with a mean of m 5 80 and a standard deviation of s 5 12, find the z-score corre- sponding to each of the following samples.

a. M 5 83 for a sample of n 5 4 scores b. M 5 83 for a sample of n 5 16 scores c. M 5 83 for a sample of n 5 36 scores

10. A sample of n 5 4 scores has a mean of M 5 75. Find the z-score for this sample:

a. If it was obtained from a population with m 5 80 and s 5 10.

b. If it was obtained from a population with m 5 80 and s 5 20.

c. If it was obtained from a population with m 5 80 and s 5 40.

11. A normal distribution has a mean of m 5 60 and a standard deviation of s 5 18. For each of the fol- lowing samples, compute the z-score for the sample mean and determine whether the sample mean is a typical, representative value or an extreme value for a sample of this size.

a. M 5 67 for n 5 4 scores b. M 5 67 for n 5 36 scores

12. A random sample is obtained from a normal popula- tion with a mean of m 5 95 and a standard deviation of s 5 40. The sample mean is M 5 86.

a. Is this a representative sample mean or an extreme value for a sample of n 5 16 scores?

b. Is this a representative sample mean or an extreme value for a sample of n 5 100 scores?

13. The population of IQ scores forms a normal distribu- tion with a mean of m 5 100 and a standard deviation of s 5 15. What is the probability of obtaining a sample mean greater than M 5 97,

a. for a random sample of n 5 9 people? b. for a random sample of n 5 25 people?

14. The scores on a standardized mathematics test for 8th-grade children in New York State form a normal distribution with a mean of m 5 70 and a standard deviation of s 5 10.

a. What proportion of the students in the state have scores less than X 5 75?

b. If samples of n 5 4 are selected from the popula- tion, what proportion of the samples will have means less than M 5 75?

c. If samples of n 5 25 are selected from the popu- lation, what proportion of the samples will have means less than M 5 75?

15. A normal distribution has a mean of m 5 54 and a standard deviation of s 5 6.

a. What is the probability of randomly selecting a score less than X 5 51?

b. What is the probability of selecting a sample of n 5 4 scores with a mean less than M 5 51?

c. What is the probability of selecting a sample of n 5 36 scores with a mean less than M 5 51?

16. A population of scores forms a normal distribution with a mean of m 5 80 and a standard deviation of s 5 10.

a. What proportion of the scores have values between 75 and 85?

b. For samples of n 5 4, what proportion of the samples will have means between 75 and 85?

2 0 2 CHAPTER 7 PROBABILITY AND SAMPLES: THE DISTRIBUTION OF SAMPLE MEANS

c. For samples of n 5 16, what proportion of the samples will have means between 75 and 85?

17. For random samples of size n 5 25 selected from a normal distribution with a mean of m 5 50 and a standard deviation of s 5 20, find each of the following:

a. The range of sample means that defines the middle 95% of the distribution of sample means.

b. The range of sample means that defines the middle 99% of the distribution of sample means.

18. The distribution ages for students at the state college is positively skewed with a mean of m 5 21.5 and a standard deviation of s 5 3.

a. What is the probability of selecting a random sam- ple of n 5 4 students with an average age greater than 23? (Careful: This is a trick question.)

b. What is the probability of selecting a random sample of n 5 36 students with an average age greater than 23?

c. For a sample of n 5 36 students, what is the prob- ability that the average age is between 21 and 22?

19. At the end of the spring semester, the Dean of Students sent a survey to the entire freshman class. One question asked the students how much weight they had gained or lost since the beginning of the school year. The average was a gain of m 5 9 pounds with a standard deviation of s 5 6. The distribution of scores was approximately normal. A sample of n 5 4 students is selected and the average weight change is computed for the sample.

a. What is the probability that the sample mean will be greater than M 5 10 pounds? In symbols, what is p(M . 10)?

b. Of all of the possible samples, what proportion will show an average weight loss? In symbols, what is p(M , 0)?

c. What is the probability that the sample mean will be a gain of between M 5 9 and M 5 12 pounds? In symbols, what is p(9 , M , 12)?

20. Jumbo shrimp are those that require 10 to 15 shrimp to make a pound. Suppose that the number of jumbo shrimp in a 1-pound bag averages m 5 12.5 with a standard deviation of s 5 1, and forms a normal distribution. What is the probability of randomly picking a sample of n 5 25 1-pound bags that average more than M 5 13 shrimp per bag?

21. The average age for licensed drivers in the county is m 5 40.3 years with a standard deviation of s 5 13.2 years.

a. A researcher obtained a random sample of n 5 16 parking tickets and computed an average age of M 5 38.9 years for the drivers. Compute the z-score for the sample mean and find the probability

of obtaining an average age this young or younger for a random sample of licensed drivers. Is it reasonable to conclude that this set of n 5 16 people is a representative sample of licensed drivers?

b. The same researcher obtained a random sample of n 5 36 speeding tickets and computed an average age of M 5 36.2 years for the drivers. Compute the z-score for the sample mean and find the probability of obtaining an average age this young or younger for a random sample of licensed drivers. Is it reasonable to conclude that this set of n 5 36 people is a representative sample of licensed drivers?

22. Callahan (2009) conducted a study to evaluate the effectiveness of physical exercise programs for in- dividuals with chronic arthritis. Participants with doctor-diagnosed arthritis either received a Tai Chi course immediately or were placed in a control group to begin the course 8 weeks later. At the end of the 8-week period, self-reports of pain were obtained for both groups. Data similar to the results obtained in the study are shown in the following table.

Self-Reported Level of Pain

Mean SE

Tai Chi course 3.7 1.2

No Tai Chi course 7.6 1.7

a. Construct a bar graph that incorporates all of the information in the table.

b. Looking at your graph, do you think that partici- pation in the Tai Chi course reduces arthritis pain?

23. Xu and Garcia (2008) conducted a research study demonstrating that 8-month-old infants appear to recognize which samples are likely to be obtained from a population and which are not. In the study, the infants watched as a sample of n 5 5 ping-pong balls was selected from a large box. In one condition, the sample consisted of 1 red ball and 4 white balls. After the sample was selected, the front panel of the box was removed to reveal the contents. In the ex- pected condition, the box contained primarily white balls like the sample, and the infants looked at it for an average of M 5 7.5 seconds. In the unexpected condition, the box had primarily red balls, unlike the sample, and the infants looked at it for M 5 9.9 seconds. The researchers interpreted the results as demonstrating that the infants found the unexpected result surprising and, therefore, more interesting than the expected result. Assuming that the standard error for both means is s

M 5 1 second, draw a bar graph

showing the two sample means using brackets to show the size of the standard error for each mean.

Aplia for Essentials of Statistics for the Behavioral Sciences

After reading, go to “Resources” at the end of this chapter for

an introduction on how to use Aplia’s homework and learning

resources.

Introduction to Hypothesis Testing

8.1 The Logic of Hypothesis Testing

8.2 Uncertainty and Errors in Hypothesis Testing

8.3 More About Hypothesis Tests

8.4 Directional (One-Tailed) Hypothesis Tests

8.5 Concerns About Hypothesis Testing: Measuring Effect Size

8.6 Statistical Power

Summary

Focus on Problem Solving

Demonstrations 8.1 and 8.2

Problems

C H A P T E R

8 Tools You Will Need The following items are considered essential background material for this chapter. If you doubt your knowledge of any of these items, you should review the appropriate chapter or section before proceeding.

• z-Scores (Chapter 5) • Distribution of sample means

(Chapter 7) • Expected value • Standard error • Probability and sample means

2 0 4 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

THE LOGIC OF HYPOTHESIS TESTING

It usually is impossible or impractical for a researcher to observe every individual in

a population. Therefore, researchers usually collect data from a sample and then use

the sample data to help answer questions about the population. Hypothesis testing is a

statistical procedure that allows researchers to use sample data to draw inferences about

the population of interest.

Hypothesis testing is one of the most commonly used inferential procedures. In fact,

most of the remainder of this book examines hypothesis testing in a variety of different

situations and applications. Although the details of a hypothesis test change from one

situation to another, the general process remains constant. In this chapter, we introduce

the general procedure for a hypothesis test. You should notice that we use the statisti-

cal techniques that have been developed in the preceding three chapters—that is, we

combine the concepts of z-scores, probability, and the distribution of sample means to

create a new statistical procedure known as a hypothesis test.

A hypothesis test is a statistical method that uses sample data to evaluate a

hypothesis about a population.

In very simple terms, the logic underlying the hypothesis-testing procedure is as

follows:

1. First, we state a hypothesis about a population. Usually the hypothesis concerns

the value of a population parameter. For example, we might hypothesize that

American adults gain an average of � � 7 pounds between Thanksgiving and

New Year’s Day each year.

2. Before we select a sample, we use the hypothesis to predict the characteristics

that the sample should have. For example, if we predict that the average weight

gain for the population is � � 7 pounds, then we would predict that our sample

should have a mean around 7 pounds. Remember: The sample should be similar

to the population, but you always expect a certain amount of error.

3. Next, we obtain a random sample from the population. For example, we might

select a sample of n � 200 American adults and measure the average weight

change for the sample between Thanksgiving and New Year’s Day.

4. Finally, we compare the obtained sample data with the prediction that was made

from the hypothesis. If the sample mean is consistent with the prediction, then

we conclude that the hypothesis is reasonable. But if there is a big discrepancy

between the data and the prediction, then we decide that the hypothesis is wrong.

A hypothesis test is typically used in the context of a research study. That is, a

researcher completes a research study and then uses a hypothesis test to evaluate the

results. Depending on the type of research and the type of data, the details of the hy-

pothesis test change from one research situation to another. In later chapters, we exam-

ine different versions of hypothesis testing that are used for different kinds of research.

For now, however, we focus on the basic elements that are common to all hypothesis

tests. To accomplish this general goal, we examine a hypothesis test as it applies to the

simplest possible situation—using a sample mean to test a hypothesis about a popula-

tion mean.

In the five chapters that follow, we consider hypothesis testing in more complex

research situations involving sample means and mean differences. In Chapter 14,

8.1

D E F I N I T I O N

SECTION 8.1 / THE LOGIC OF HYPOTHESIS TESTING 2 0 5

we look at correlational research and examine how the relationships obtained for

sample data are used to evaluate hypotheses about relationships in the population. In

Chapter 15, we examine how the proportions that exist in a sample are used to test

hypotheses about the corresponding proportions in the population.

Once again, we introduce hypothesis testing with a situation in which a researcher is

using one sample mean to evaluate a hypothesis about one unknown population mean.

The unknown population Figure 8.1 shows the general research situation that we

use to introduce the process of hypothesis testing. Notice that the researcher begins with

a known population. This is the set of individuals as they exist before treatment. For this

example, we are assuming that the original set of scores forms a normal distribution

with � � 80 and � � 20. The purpose of the research is to determine the effect of a

treatment on the individuals in the population.

To simplify the hypothesis-testing situation, one basic assumption is made about the

effect of the treatment: If the treatment has any effect, it is simply to add a constant

amount to (or subtract a constant amount from) each individual’s score. You should re-

call from Chapters 3 and 4 that adding (or subtracting) a constant to each score causes

the mean to change but does not change the shape of the distribution, nor does it change

the standard deviation. Thus, we assume that the population after treatment has the

same shape as the original population and the same standard deviation as the original

population. This assumption is incorporated into the situation shown in Figure 8.1.

Note that the unknown population, after treatment, is the focus of the research ques-

tion. Specifically, the purpose of the research is to determine what would happen if the

treatment were administered to every individual in the population.

The sample in the research study The goal of the hypothesis test is to deter-

mine whether the treatment has any effect on the individuals in the population (see

Figure 8.1). Usually, however, we cannot administer the treatment to the entire popula-

tion, so the actual research study is conducted using a sample. Figure 8.2 shows the

structure of the research study from the point of view of the hypothesis test. The origi-

nal population, before treatment, is shown on the left-hand side. The unknown popula-

tion, after treatment, is shown on the right-hand side. Note that the unknown population

is actually hypothetical (the treatment is never administered to the entire population).

Instead, we are asking what would happen if the treatment were administered to the

entire population. The research study involves selecting a sample from the original

population, administering the treatment to the sample, and then recording scores for

the individuals in the treated sample. Notice that the research study produces a treated

µ = 80

Known population before treatment

σ = 20

µ = ?

Unknown population after treatment

σ = 20

T r e a t

m e n t

FIGURE 8.1

The basic experimental

situation for hypothesis

testing. It is assumed that

the parameter µ is known

for the population before

treatment. The purpose

of the experiment is to

determine whether the

treatment has an effect on

the population mean.

2 0 6 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

sample. Although this sample was obtained indirectly, it is equivalent to a sample that

is obtained directly from the unknown treated population. The hypothesis test uses the

treated sample on the right-hand side of Figure 8.2 to evaluate a hypothesis about the

unknown treated population on the right side of the figure.

A hypothesis test is a formalized procedure that follows a standard series of opera-

tions. In this way, researchers have a standardized method for evaluating the results of

their research studies. Other researchers recognize and understand exactly how the data

were evaluated and how conclusions were reached. To emphasize the formal structure

of a hypothesis test, we present hypothesis testing as a four-step process that is used

throughout the rest of the book. The following example provides a concrete foundation

for introducing the hypothesis-testing procedure.

Researchers have demonstrated that it is possible to improve mathematics skills using

mild electrical brain stimulation (Kadosh, Soskic, Iuculano, Kanai, & Walsh, 2010). In

the study, participants were taught artificial number symbols while researchers applied

mild electrical current across their skulls near the parietal lobes, a part of the brain

that is important for mathematical skill. Other participants learned the same number

symbols with current applied in a different brain location. After six study sessions, the

parietal-lobe group performed significantly better on a test evaluating their knowledge

of the number symbols.

Suppose that a researcher is testing the same kind of brain stimulation on students

who are studying for a standardized mathematics exam. For the general population,

scores on this exam form a normal distribution with a mean of � � 80 and a standard

deviation of � � 20. The researcher’s plan is to obtain a sample of n � 25 students

who are scheduled to take the exam, and have each student study for 30 minutes each

day while current is applied near the parietal lobe. After 4 weeks, the participants are

given the standardized exam. If the mean score for the sample is noticeably different

from the mean for the general population of students, then the researcher can conclude

that the electric stimulation does appear to have an effect on mathematical skill. On the

other hand, if the sample mean is around 80 (the same as the general population mean),

the researcher must conclude that the stimulation does not appear to have any effect.

E X A M P L E 8 . 1

T r e a t

m e n t

Known original

population µ = 80

Treated sample

Sample

Unknown treated

population µ = ?

FIGURE 8.2

From the point of view

of the hypothesis test, the

entire population receives

the treatment and then a

sample is selected from

the treated population. In

the actual research study,

a sample is selected from

the original population and

the treatment is adminis-

tered to the sample. From

either perspective, the

result is a treated sample

that represents the treated

population.

SECTION 8.1 / THE LOGIC OF HYPOTHESIS TESTING 2 0 7

Figure 8.2 depicts the research situation that was described in the preceding ex-

ample. Notice that the population after treatment is unknown. Specifically, we do

not know what will happen to the mean score if the entire population of students is

given the brain stimulation while studying. However, we do have a sample of n � 25

participants who have received the stimulation and we can use this sample to help

draw inferences about the unknown population. The following four steps outline the

hypothesis-testing procedure that allows us to use sample data to answer questions

about an unknown population.

As the name implies, the process of hypothesis testing begins by stating a hypothesis

about the unknown population. Actually, we state two opposing hypotheses. Notice that

both hypotheses are stated in terms of population parameters.

The first, and most important, of the two hypotheses is called the null hypothesis.

The null hypothesis states that the treatment has no effect. In general, the null hypoth-

esis states that there is no change, no effect, no difference—nothing happened, hence

the name null. The null hypothesis is identified by the symbol H 0 . (The H stands for

hypothesis, and the zero subscript indicates that this is the zero-effect hypothesis.) For

the study in Example 8.1, the null hypothesis states that the brain stimulation has no ef-

fect on mathematical skill for the population of students. In symbols, this hypothesis is

H 0 : �

with stimulation � 80 (Even with the stimulation,

the mean test score is still 80.)

The null hypothesis (H 0 ) states that in the general population there is no change,

no difference, or no relationship. In the context of an experiment, H 0 predicts

that the independent variable (treatment) has no effect on the dependent variable

(scores) for the population.

The second hypothesis is the opposite of the null hypothesis, and it is called the

scientific, or alternative, hypothesis (H 1 ). This hypothesis states that the treatment has

an effect on the dependent variable.

The alternative hypothesis (H 1 ) states that there is a change, a difference,

or a relationship for the general population. In the context of an experiment,

H 1 predicts that the independent variable (treatment) does have an effect on

the dependent variable.

For this example, the alternative hypothesis states that the stimulation does have an

effect on mathematical skill for the population and will cause a change in the mean

score. In symbols, the alternative hypothesis is represented as

H 1 : �

with stimulation  80 (With the stimulation,

the mean test score is different from 80.)

Notice that the alternative hypothesis simply states that there will be some type of

change. It does not specify whether the effect will be increased or decreased test scores.

In some circumstances, it is appropriate for the alternative hypothesis to specify the

direction of the effect. For example, the researcher might hypothesize that the stimula-

tion will increase test scores (� . 80). This type of hypothesis results in a directional hypothesis test, which is examined in detail later in this chapter. For now we concentrate

T H E F O U R ST E P S O F A H Y P OT H E S I S T E ST

ST E P 1 : STAT E T H E H Y P OT H E S I S

D E F I N I T I O N

The goal of inferential statis-

tics is to make general state-

ments about the population by

using sample data. Therefore,

when testing hypotheses, we

make our predictions about

the population parameters.

The null hypothesis and the

alternative hypothesis are

mutually exclusive and

exhaustive. They cannot

both be true, and one of

them must be true. The data

determine which one should

be rejected.

2 0 8 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

on nondirectional tests, for which the hypotheses simply state that the treatment has no

effect (H 0 ) or has some effect (H

1 ).

Eventually the researcher uses the data from the sample to evaluate the credibility of the

null hypothesis. The data either provide support for the null hypothesis or tend to refute

the null hypothesis. In particular, if there is a big discrepancy between the data and the

null hypothesis, then we conclude that the null hypothesis is wrong.

To formalize the decision process, we use the null hypothesis to predict the kind

of sample mean that ought to be obtained. Specifically, we determine exactly which

sample means are consistent with the null hypothesis and which sample means are at

odds with the null hypothesis.

For our example, the null hypothesis states that the brain stimulation has no effect

and the population mean is still � � 80. If this is true, then the sample mean should

have a value around 80. Therefore, a sample mean near 80 is consistent with the null

hypothesis. On the other hand, a sample mean that is very different from 80 is not con-

sistent with the null hypothesis. To determine exactly which values are “near” 80 and

which values are “very different from” 80, we examine all of the possible sample means

that could be obtained if the null hypothesis is true. For our example, this is the distribu-

tion of sample means for n � 25. According to the null hypothesis, this distribution is

centered at � � 80. The distribution of sample means is then divided into two sections:

1. Sample means that are likely to be obtained if H 0 is true; that is, sample means

that are close to the null hypothesis

2. Sample means that are very unlikely to be obtained if H 0 is true; that is, sample

means that are very different from the null hypothesis

Figure 8.3 shows the distribution of sample means divided into these two sections.

Notice that the high-probability samples are located in the center of the distribution

ST E P 2 : S E T T H E C R I T E R I A

F O R A D E C I S I O N

The distribution of sample means if the null hypothesis is true (all the possible outcomes)

Sample means close to H0:

high-probability values if H0 is true

Extreme, low- probability values

if H0 is true

Extreme, low- probability values

if H0 is true

from H0

FIGURE 8.3

The set of potential

samples is divided into

those that are likely to be

obtained and those that

are very unlikely to be

obtained if the null

hypothesis is true.

SECTION 8.1 / THE LOGIC OF HYPOTHESIS TESTING 2 0 9

and have sample means close to the value specified in the null hypothesis. On the other

hand, the low-probability samples are located in the extreme tails of the distribution.

After the distribution has been divided in this way, we can compare our sample data

with the values in the distribution. Specifically, we can determine whether our sample

mean is consistent with the null hypothesis (like the values in the center of the distribu-

tion) or whether our sample mean is very different from the null hypothesis (like the

values in the extreme tails).

The alpha level To find the boundaries that separate the high-probability samples

from the low-probability samples, we must define exactly what is meant by “low”

probability and “high” probability. This is accomplished by selecting a specific prob-

ability value, which is known as the level of significance, or the alpha level, for the

hypothesis test. The alpha (a) value is a small probability that is used to identify the

low-probability samples. By convention, commonly used alpha levels are a � .05 (5%),

a � .01 (1%), and a � .001 (0.1%). For example, with a � .05, we separate the most

unlikely 5% of the sample means (the extreme values) from the most likely 95% of the

sample means (the central values).

The extremely unlikely values, as defined by the alpha level, make up what is called

the critical region. These extreme values in the tails of the distribution define outcomes

that are not consistent with the null hypothesis; that is, they are very unlikely to occur

if the null hypothesis is true. Whenever the data from a research study produce a sample

mean that is located in the critical region, we conclude that the data are not consistent

with the null hypothesis, and we reject the null hypothesis.

The alpha level, or the level of significance, is a probability value that is used to

define the concept of “very unlikely” in a hypothesis test.

The critical region is composed of the extreme sample values that are very un-

likely (as defined by the alpha level) to be obtained if the null hypothesis is true.

The boundaries for the critical region are determined by the alpha level. If sample

data fall in the critical region, the null hypothesis is rejected.

Technically, the critical region is defined by sample outcomes that are very

unlikely to occur if the treatment has no effect (that is, if the null hypothesis is true).

Reversing the point of view, we can also define the critical region as sample values

that provide convincing evidence that the treatment really does have an effect. For

our example, the regular population of students has a mean test score of � 5 80.

We selected a sample from this population and administered a treatment (the brain

stimulation) to the individuals in the sample. What kind of sample mean would

convince you that the treatment has an effect? It should be obvious that the most

convincing evidence would be a sample mean that is really different from m 5 80. In

a hypothesis test, the critical region is determined by sample values that are “really

different” from the original population.

The boundaries for the critical region To determine the exact location for the

boundaries that define the critical region, we use the alpha-level probability and the unit

normal table. In most cases, the distribution of sample means is normal, and the unit nor-

mal table provides the precise z-score location for the critical region boundaries. With

a 5 .05, for example, the boundaries separate the extreme 5% from the middle 95%.

Because the extreme 5% is split between two tails of the distribution, there is exactly

2.5% (or 0.0250) in each tail. In the unit normal table, you can look up a proportion

of 0.0250 in column C (the tail) and find that the z-score boundary is z 5 1.96. Thus,

D E F I N I T I O N S

With rare exceptions, an

alpha level is never larger

than .05.

2 1 0 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

for any normal distribution, the extreme 5% is in the tails of the distribution beyond

z � 11.96 and z � –1.96. These values define the boundaries of the critical region for

a hypothesis test using a � .05 (Figure 8.4).

Similarly, an alpha level of a � .01 means that 1%, or .0100, is split between the

two tails. In this case, the proportion in each tail is .0050, and the corresponding z-score

boundaries are z � 62.58 (62.57 is equally good). For a � .001, the boundaries are located at z � 63.30. You should verify these values in the unit normal table and be sure that you understand exactly how they are obtained.

Middle 95%: High-probability values

if H0 is true

z 1.96 z 1.96

Critical region: Extreme 5%

from H0

Reject H0 Reject H0

FIGURE 8.4

The critical region (very

unlikely outcomes) for

a � .05.

1. The city school district is considering increasing class size in the elementary

schools. However, some members of the school board are concerned that larger

classes may have a negative effect on student learning. In words, what would the

null hypothesis say about the effect of class size on student learning?

2. If the alpha level is increased from a � .01 to a � .05, then the boundaries

for the critical region move farther away from the center of the distribution.

(True or false?)

3. If a researcher conducted a hypothesis test with an alpha level of a � .02, what

z-score values would form the boundaries for the critical region?

1. The null hypothesis would say that class size has no effect on student learning.

2. False. A larger alpha means that the boundaries for the critical region move closer to the

center of the distribution.

3. The .02 would be split between the two tails, with .01 in each tail. The z-score boundaries

would be z � 12.33 and z � –2.33.

L E A R N I N G C H E C K

ANSWERS

SECTION 8.1 / THE LOGIC OF HYPOTHESIS TESTING 2 1 1

At this time, we select a sample of students and give them 30 minutes of brain stimula-

tion each day while they study for the mathematics exam. After 4 weeks, the students

take the standardized mathematics test. Notice that the data are collected after the

researcher has stated the hypotheses and established the criteria for a decision. This

sequence of events helps to ensure that a researcher makes an honest, objective evalu-

ation of the data and does not tamper with the decision criteria after the experimental

outcome is known.

Next, the raw data from the sample are summarized with the appropriate statistics:

For this example, the researcher would compute the sample mean. Now it is possible

for the researcher to compare the sample mean (the data) with the null hypothesis. This

is the heart of the hypothesis test: comparing the data with the hypothesis.

The comparison is accomplished by computing a z-score that describes exactly

where the sample mean is located relative to the hypothesized population mean from

H 0 . In step 2, we constructed the distribution of sample means that would be expected

if the null hypothesis were true—that is, the entire set of sample means that could be

obtained if the treatment has no effect (see Figure 8.4). Now we calculate a z-score

that identifies where our sample mean is located in this hypothesized distribution. The

z-score formula for a sample mean is

z M

� �

�

In the formula, the value of the sample mean (M) is obtained from the sample data,

and the value of � is obtained from the null hypothesis. Thus, the z-score formula can

be expressed in words as follows:

z � sample mean hypothesized population mean



ttandard error between andM �

Notice that the top of the z-score formula measures how much difference there is

between the data and the hypothesis. The bottom of the formula measures the standard

distance that ought to exist between a sample mean and the population mean.

In the final step, the researcher uses the z-score value obtained in step 3 to make a deci-

sion about the null hypothesis according to the criteria established in step 2. There are

two possible outcomes:

1. The sample data are located in the critical region. By definition, a sample value

in the critical region is very unlikely to occur if the null hypothesis is true.

Therefore, we conclude that the sample is not consistent with H 0 and our deci-

sion is to reject the null hypothesis. Remember, the null hypothesis states that

there is no treatment effect, so rejecting H 0 means that we are concluding that

the treatment did have an effect.

For the example we have been considering, suppose that the sample produced

a mean of M � 89 after receiving brain stimulation while studying. The null

hypothesis states that the population mean is � � 80 and, with n � 25 and

� � 20, the standard error for the sample mean is

� � �

� � � M

5 4

ST E P 3 : CO L L E C T DATA A N D CO M P U T E SA M P L E

STAT I ST I C S

ST E P 4 : M A K E A D E C I S I O N

2 1 2 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

Thus, a sample mean of M � 89 produces a z-score of

z M

� �

� �

 � �

89 80

4 2.25

With an alpha level of a � .05, this z-score is beyond the boundary of 1.96.

Because the sample z-score is in the critical region, we reject the null hypothesis

and conclude that the brain stimulation did have an effect on mathematics skill.

2. The second possibility is that the sample data are not in the critical region. In

this case, the sample mean is reasonably close to the population mean specified

in the null hypothesis (in the center of the distribution). Because the data do not

provide strong evidence that the null hypothesis is wrong, our conclusion is to

fail to reject the null hypothesis. This conclusion means that the treatment does

not appear to have an effect.

For the research study examining the brain stimulation, if our sample produced

a mean test score of M � 84, then we would obtain a z-score of

z M

� �

� �

 � �

84 80

4 1 00.

The z-score of 1.00 is not in the critical region. Therefore, we would fail to re-

ject the null hypothesis and conclude that the brain stimulation does not appear

to have an effect on mathematical skill.

In general, the final decision is made by comparing our treated sample with the dis-

tribution of sample means that would be obtained for untreated samples. If our treated

sample looks much the same as samples that do not receive the brain stimulation, then

we conclude that the treatment does not appear to have any effect. On the other hand, if

the treated sample is noticeably different from the majority of untreated samples, then

we conclude that the treatment does have an effect.

An analogy for hypothesis testing It may seem awkward to phrase both of the two

possible decisions in terms of rejecting the null hypothesis; either we reject H 0 or we

fail to reject H 0 . These two decisions may be easier to understand if you think of a

research study as an attempt to gather evidence to prove that a treatment works. From

this perspective, the process of conducting a hypothesis test is similar to the process that

takes place during a jury trial. For example,

1. The test begins with a null hypothesis stating that there is no treatment effect.

The trial begins with a null hypothesis that the defendant did not commit a

crime (innocent until proven guilty).

2. The research study gathers evidence to show that the treatment actually does

have an effect, and the police gather evidence to show that the defendant really

did commit a crime. Note that both are trying to refute the null hypothesis.

3. If there is enough evidence, the researcher rejects the null hypothesis and con-

cludes that there really is a treatment effect. If there is enough evidence, the

jury rejects the hypothesis and concludes that the defendant is guilty of a crime.

4. If there is not enough evidence, the researcher fails to reject the null hypothesis.

Note that the researcher does not conclude that there is no treatment effect,

simply that there is not enough evidence to conclude that there is an effect.

Similarly, if there is not enough evidence, the jury fails to find the defendant

guilty. Note that the jury does not conclude that the defendant is innocent,

simply that there is not enough evidence for a guilty verdict.

SECTION 8.2 / UNCERTAINTY AND ERRORS IN HYPOTHESIS TESTING 2 1 3

UNCERTAINTY AND ERRORS IN HYPOTHESIS TESTING

Hypothesis testing is an inferential process, which means that it uses limited informa-

tion as the basis for reaching a general conclusion. Specifically, a sample provides only

limited or incomplete information about the whole population, and yet a hypothesis

test uses a sample to draw a conclusion about the population. In this situation, there is

always the possibility that an incorrect conclusion will be made. Although sample data

are usually representative of the population, there is always a chance that the sample is

misleading and will cause a researcher to make the wrong decision about the research

results. In a hypothesis test, there are two different kinds of errors that can be made.

It is possible that the data will lead you to reject the null hypothesis when in fact the

treatment has no effect. Remember: Samples are not expected to be identical to their

populations, and some extreme samples can be very different from the populations that

they are supposed to represent. If a researcher selects one of these extreme samples by

chance, then the data from the sample may give the appearance of a strong treatment

effect, even though there is no real effect. In the previous section, for example, we

discussed a research study examining how brain stimulation near the parietal lobe

affects the learning of new mathematical skills. Suppose that the researcher selects a

sample of n � 25 students who already have mathematical skills that are well above

average. Even if the stimulation (the treatment) has no effect at all, these people will

still score higher than average on the standardized test. In this case, the researcher is

likely to conclude that the treatment does have an effect, when in fact it really does not.

This is an example of what is called a Type I error.

A Type I error occurs when a researcher rejects a null hypothesis that is actually

true. In a typical research situation, a Type I error means that the researcher con-

cludes that a treatment does have an effect when, in fact, it has no effect.

You should realize that a Type I error is not a stupid mistake in the sense that a

researcher is overlooking something that should be perfectly obvious. On the contrary,

the researcher is looking at sample data that appear to show a clear treatment effect.

The researcher then makes a careful decision based on the available information. The

problem is that the information from the sample is misleading.

8.2

T Y P E I E R R O R S

D E F I N I T I O N

1. A researcher selects a sample of n � 16 individuals from a normal population with

a mean of � � 40 and � � 8. A treatment is administered to the sample and, after

treatment, the sample mean is M � 43. If the researcher uses a hypothesis test to

evaluate the treatment effect, what z-score would be obtained for this sample?

2. A small value (near zero) for the z-score statistic is evidence that the sample data

are consistent with the null hypothesis. (True or false?)

3. A z-score value in the critical region means that you should reject the null

hypothesis. (True or false?)

1. The standard error is 2 points and z � 3

2 � 1.50.

2. True. A z-score near zero indicates that the data support the null hypothesis.

3. True. A z-score value in the critical region means that the sample is not consistent with the

null hypothesis.

L E A R N I N G C H E C K

ANSWERS

2 1 4 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

In most research situations, the consequences of a Type I error can be very serious.

Because the researcher has rejected the null hypothesis and believes that the treatment

has a real effect, it is likely that the researcher will report or even publish the research

results. A Type I error, however, means that this is a false report. Thus, Type I errors

lead to false reports in the scientific literature. Other researchers may try to build theo-

ries or develop other experiments based on the false results. A lot of precious time and

resources may be wasted.

The probability of a Type I error A Type I error occurs when a researcher unknow-

ingly obtains an extreme, nonrepresentative sample. Fortunately, the hypothesis test is

structured to minimize the risk that this will occur. Figure 8.4 shows the distribution of

sample means and the critical region for the research study we have been discussing.

This distribution contains all of the possible sample means for samples of n � 25 if the

null hypothesis is true. Notice that most of the sample means are near the hypothesized

population mean, � 5 80, and that means in the critical region are very unlikely to

occur.

With an alpha level of a 5 .05, only 5% of the samples have means in the critical

region. Therefore, there is only a 5% probability (p 5 .05) that one of these samples

will be obtained. Thus, the alpha level determines the probability of obtaining a sample

mean in the critical region when the null hypothesis is true. In other words, the alpha

level determines the probability of a Type I error.

The alpha level for a hypothesis test is the probability that the test will lead to a

Type I error if the null hypothesis is true. That is, the alpha level determines the

probability of obtaining sample data in the critical region even though there is no

treatment effect.

In summary, whenever the sample data are in the critical region, the appropriate

decision for a hypothesis test is to reject the null hypothesis. Normally this is the correct

decision because the treatment has caused the sample to be different from the original

population. In this case, the hypothesis test has correctly identified a real treatment

effect. Occasionally, however, sample data are in the critical region just by chance, without

any treatment effect. When this occurs, the researcher makes a Type I error; that is, the

researcher concludes that a treatment effect exists when in fact it does not. Fortunately,

the risk of a Type I error is small and is under the control of the researcher. Specifically,

the probability of a Type I error is equal to the alpha level.

Whenever a researcher rejects the null hypothesis, there is a risk of a Type I error.

Similarly, whenever a researcher fails to reject the null hypothesis, there is a risk of a

Type II error. By definition, a Type II error is the failure to reject a false null hypothesis.

In more straightforward English, a Type II error means that a treatment effect really

exists, but the hypothesis test fails to detect it.

A Type II error occurs when a researcher fails to reject a null hypothesis that is

really false. In a typical research situation, a Type II error means that the hypoth-

esis test has failed to detect a real treatment effect.

A Type II error occurs when the sample mean is not in the critical region even though

the treatment has had an effect on the sample. Often this happens when the effect of the

treatment is relatively small. In this case, the treatment does influence the sample, but

D E F I N I T I O N

T Y P E I I E R R O R S

D E F I N I T I O N

SECTION 8.2 / UNCERTAINTY AND ERRORS IN HYPOTHESIS TESTING 2 1 5

the magnitude of the effect is not big enough to move the sample mean into the critical

region. Because the sample is not substantially different from the original population (it

is not in the critical region), the statistical decision is to fail to reject the null hypothesis

and to conclude that there is not enough evidence to say that there is a treatment effect.

The consequences of a Type II error are usually not as serious as those of a

Type I error. In general terms, a Type II error means that the research data do not show

the results that the researcher had hoped to obtain. The researcher can accept this

outcome and conclude that the treatment either has no effect or has only a small effect

that is not worth pursuing, or the researcher can repeat the experiment (usually with

some improvement, such as a larger sample) and try to demonstrate that the treatment

really does work.

Unlike a Type I error, it is impossible to determine a single, exact probability for a

Type II error. Instead, the probability of a Type II error depends on a variety of factors

and therefore is a function, rather than a specific number. Nonetheless, the probability

of a Type II error is represented by the symbol b, the Greek letter beta. In summary, a hypothesis test always leads to one of two decisions:

1. The sample data provide sufficient evidence to reject the null hypothesis and

conclude that the treatment has an effect.

2. The sample data do not provide enough evidence to reject the null hypothesis.

In this case, you fail to reject H 0 and conclude that the treatment does not

appear to have an effect.

In either case, there is a chance that the data are misleading and the decision is

wrong. The complete set of decisions and outcomes is shown in Table 8.1. The risk

of an error is especially important in the case of a Type I error, which can lead to a

false report. Fortunately, the alpha level, which is completely under the control of the

researcher, defines the probability of a Type I error if the null hypothesis is true. At the

beginning of a hypothesis test, the researcher states the hypotheses and selects the alpha

level, which immediately determines the risk that a Type I error will be made.

As you have seen, the alpha level for a hypothesis test serves two very important func-

tions. First, the alpha level helps to determine the boundaries for the critical region by

defining the concept of “very unlikely” outcomes. At the same time, the alpha level

determines the probability of a Type I error if the null hypothesis is true. When you

select a value for alpha at the beginning of a hypothesis test, your decision influences

both of these functions.

The primary concern when selecting an alpha level is to minimize the risk of a

Type I error. Thus, alpha levels tend to be very small probability values. By convention,

the largest permissible value is a � .05. When there is no treatment effect, an alpha

level of .05 means that there is a 5% risk, or a 1-in-20 probability, of rejecting the null

hypothesis and committing a Type I error. Because the consequences of a Type I error

can be relatively serious, many individual researchers and many scientific publications

S E L E C T I N G A N A L P H A L E V E L

Actual Situation

No Effect, H 0 True Effect Exists, H

0 False

Experimenter’s Decision Reject H

0 Type I error Decision correct

Retain H 0

Decision correct Type II error

TABLE 8.1

Possible outcomes

of a statistical decision.

2 1 6 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

prefer to use a more conservative alpha level such as .01 or .001 to reduce the risk

that a false report is published and becomes part of the scientific literature. (For more

information on the origins of the .05 level of significance, see the excellent short

article by Cowles and Davis, 1982.)

At this point, it may appear that the best strategy for selecting an alpha level is to

choose the smallest possible value to minimize the risk of a Type I error. However, there

is a different kind of risk that develops as the alpha level is lowered. Specifically, a

lower alpha level means less risk of a Type I error, but it also means that the hypothesis

test demands more evidence from the research results.

The trade-off between the demands of the test and the risk of a Type I error is con-

trolled by the boundaries of the critical region. For the hypothesis test to conclude that

the treatment does have an effect, the sample data must be in the critical region. If the

treatment really has an effect, it should cause the sample to be different from the origi-

nal population; essentially, the treatment should push the sample into the critical region.

However, as the alpha level is lowered, the boundaries for the critical region move far-

ther out and become more difficult to reach. Figure 8.5 shows how the boundaries for

the critical region move farther into the tails as the alpha level decreases. Notice that

z � 0, in the center of the distribution, corresponds to the value of � specified in the

null hypothesis. The boundaries for the critical region determine how much distance

between the sample mean and � is needed to reject the null hypothesis. As the alpha

level gets smaller, this distance gets larger.

Thus, an extremely small alpha level, such as .000001 (one in a million), would

mean almost no risk of a Type I error but would push the critical region so far out that

it would become essentially impossible to ever reject the null hypothesis; that is, it

would require an enormous treatment effect before the sample data would reach the

critical boundaries.

In general, researchers try to maintain a balance between the risk of a Type I error

and the demands of the hypothesis test. Alpha levels of .05, .01, and .001 are considered

reasonably good values because they provide a low risk of error without placing exces-

sive demands on the research results.

1.96 z

α = .05

α = .01

α = .001

−1.96

2.58−2.58

3.30−3.30

µ from H0

FIGURE 8.5

The locations of the criti-

cal region boundaries for

three different levels of

significance: a � .05,

a5 .01, and a 5 .001.

SECTION 8.3 / MORE ABOUT HYPOTHESIS TESTS 2 1 7

MORE ABOUT HYPOTHESIS TESTS

In section 8.1, we presented a complete example of a hypothesis test evaluating the

effect of brain stimulation on students’ mathematical test scores. The 4-step process

for that hypothesis test is summarized as follows:

State the hypotheses and select an alpha level. For this example, the general population,

without treatment, has an average test score of � � 80 with � � 20. Therefore, the

hypotheses are:

H 0 : �

with stimulation � 80 (the brain stimulation has no effect)

H 1 : �

with stimulation  80 (the brain stimulation does have an effect)

We set a � .05.

Locate the critical region. For a normal distribution with a � .05, the critical region

consists of sample means that produce z-scores in the extreme tails of the distribution

beyond z � 61.96.

Compute the test statistic (the z-score). We obtained a sample mean of M � 89 for

n � 25 participants. With a standard error of � M � 4, we obtain

z M

� �

� �

 � �

89 80

4 2.25

8.3

A S U M M A RY O F T H E H Y P OT H E S I S T E ST

S T E P 1

S T E P 2

S T E P 3

1. Define a Type I error.

2. Define a Type II error.

3. Under what circumstances is a Type II error likely to occur?

4. If a sample mean is in the critical region with a � .05, it would still (always) be

in the critical region if alpha were changed to a � .01. (True or false?)

5. If a sample mean is in the critical region with a � .01, it would still (always) be

in the critical region if alpha were changed to a � .05. (True or false?)

1. A Type I error is rejecting a true null hypothesis—that is, saying that the treatment has

an effect when, in fact, it does not.

2. A Type II error is the failure to reject a false null hypothesis. In terms of a research study,

a Type II error occurs when a study fails to detect a treatment effect that really exists.

3. A Type II error is likely to occur when the treatment effect is very small. In this case,

a research study is more likely to fail to detect the effect.

4. False. With a � .01, the boundaries for the critical region move farther out into the tails of

the distribution. It is possible that a sample mean could be beyond the .05 boundary but not

beyond the .01 boundary.

5. True. With a � .01, the boundaries for the critical region are farther out into the tails of the

distribution than for a � .05. If a sample mean is beyond the .01 boundary it is definitely

beyond the .05 boundary.

L E A R N I N G C H E C K

ANSWERS

2 1 8 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

Make a decision. The z-score is in the critical region, which means that this sample mean

is very unlikely if the null hypothesis is true. Therefore, we reject the null hypothesis

and conclude that the brain stimulation did have an effect on the students’ test scores.

IN THE LITERATURE

REPORTING THE RESULTS OF THE STATISTICAL TEST

A special jargon and notational system are used in published reports of hypothesis

tests. When you are reading a scientific journal, for example, you typically are not told

explicitly that the researcher evaluated the data using a z-score as a test statistic with an

alpha level of .05. Nor are you told that “the null hypothesis is rejected.” Instead, you

see a statement such as:

Electrical stimulation of the scalp near the parietal lobe had a significant effect

on the mathematics test scores for the students, z � 2.25, p , .05.

Let us examine this statement, piece by piece. First, what is meant by the word

significant? In statistical tests, a significant result means that the null hypothesis has

been rejected, which means that the result is very unlikely to have occurred merely by

chance. For this example, the null hypothesis stated that the brain stimulation has no

effect; however, the data clearly indicate that it did have an effect. Specifically, it is very

unlikely that the sample mean, M � 89, would have been obtained if the stimulation

did not have an effect.

A result is said to be significant, or statistically significant, if it is very unlikely

to occur when the null hypothesis is true. That is, the result is sufficient to reject

the null hypothesis. Thus, a treatment has a significant effect if the decision from

the hypothesis test is to reject H 0 .

Next, what is the meaning of z � 2.25? The z indicates that a z-score was used as

the test statistic to evaluate the sample data and that its value is 2.25. Finally, what is

meant by p , .05? This part of the statement is a conventional way of specifying the

alpha level that was used for the hypothesis test. It also acknowledges the possibility

(and the probability) of a Type I error. Specifically, the researcher is reporting that the

treatment had an effect but admits that this could be a false report. That is, it is possible

that the sample mean was in the critical region even though the brain stimulation had

no effect. However, the probability (p) of obtaining a sample mean in the critical region

is extremely small (less than .05) if there is no treatment effect.

In circumstances in which the statistical decision is to fail to reject H 0 , the report

might state that.

The sample did not provide sufficient evidence to conclude that the brain stimulation

had an effect on mathematics test scores, z � 1.30, p . .05.

In that case, we would be saying that the obtained result, z 5 1.30, is not unusual

(not in the critical region) and that it has a relatively high probability of occurring

(greater than .05) even if the null hypothesis is true.

Sometimes students become confused trying to differentiate between p , .05 and

p . .05. Remember that you reject the null hypothesis with extreme, low-probability

values, located in the critical region in the tails of the distribution. Thus, a significant

result that rejects the null hypothesis corresponds to p , .05 (Figure 8.6).

When a hypothesis test is conducted using a computer program, the printout

often includes not only a z-score value but also an exact value for p, the probability

S T E P 4

D E F I N I T I O N

The APA style does not use a

leading zero in a probability

value that refers to a level of

significance.

SECTION 8.3 / MORE ABOUT HYPOTHESIS TESTS 2 1 9

that the result occurred without any treatment effect. In this case, researchers are

encouraged to report the exact p value instead of using the less-than or greater-than

notation. For example, a research report might state that the treatment effect was

significant, with z � 2.45, p � .0142. When using exact values for p, however, you

must still satisfy the traditional criterion for significance; specifically, the p value

must be smaller than .05 to be considered statistically significant. Remember: The

p value is the probability that the result would occur if H 0 were true (there is no

treatment effect), which is also the probability of a Type I error. It is essential that

this probability be very small.

The mathematics used for a hypothesis test are based on a set of assumptions. When

these assumptions are satisfied, you can be confident that the test produces a justi-

fied conclusion. However, if the assumptions are not satisfied, then the hypothesis

test may be compromised. In practice, researchers are not overly concerned with the

assumptions underlying a hypothesis test because the tests usually work well even

when the assumptions are violated. However, you should be aware of the fundamental

conditions that are associated with each type of statistical test to ensure that the test

is being used appropriately. The assumptions for hypothesis tests with z-scores are

summarized as follows.

Random sampling It is assumed that the participants used in the study were selected

randomly. Remember, we wish to generalize our findings from the sample to the popu-

lation. Therefore, the sample must be representative of the population from which it has

been drawn. Random sampling helps to ensure that it is representative.

Independent observations The values in the sample must consist of independent

observations. In everyday terms, two observations are independent if there is no consis-

tent, predictable relationship between the first observation and the second. More pre-

cisely, two events (or observations) are independent if the occurrence of the first event

has no effect on the probability of the second event. Specific examples of independence

and non-independence are examined in Box 8.1. Usually, this assumption is satisfied by

using a random sample, which also helps to ensure that the sample is representative of

the population and that the results can be generalized to the population.

The value of � is unchanged by the treatment A critical part of the z-score

formula in a hypothesis test is the standard error, � M . To compute the value for the

AS S U M P T I O N S F O R H Y P OT H E S I S

T E ST S W I T H z - S CO R E S

p p Fail to reject H0

Reject H0 Reject H0

FIGURE 8.6

Sample means that fall in

the critical region (shaded

areas) have a probability

less than alpha (p , a).

In this case, H 0 should be

rejected. Sample means

that do not fall in the

critical region have a

probability greater than

alpha (p . a).

2 2 0 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

standard error, we must know the sample size (n) and the population standard devia-

tion (�). In a hypothesis test, however, the sample comes from an unknown popula-

tion (see Figure 8.2). If the population is really unknown, it would suggest that we

do not know the standard deviation and, therefore, we cannot calculate the standard

error. To solve this dilemma, we have made an assumption. Specifically, we assume

that the standard deviation for the unknown population (after treatment) is the same

as it was for the population before treatment.

Actually, this assumption is the consequence of a more general assumption that

is part of many statistical procedures. This general assumption states that the effect

of the treatment is to add a constant amount to (or subtract a constant amount from)

every score in the population. You should recall that adding (or subtracting) a constant

changes the mean but has no effect on the standard deviation. You also should note that

this assumption is a theoretical ideal. In actual experiments, a treatment generally does

not show a perfect and consistent additive effect.

Normal sampling distribution To evaluate hypotheses with z-scores, we have used

the unit normal table to identify the critical region. This table can be used only if the

distribution of sample means is normal.

BOX

8.1 INDEPENDENT OBSERVATIONS

Independent observations are a basic requirement for

nearly all hypothesis tests. The critical concern is that

each observation or measurement is not influenced by

any other observation or measurement. An example

of independent observations is the set of outcomes

obtained in a series of coin tosses. Assuming that the

coin is balanced, each toss has a 50–50 chance of

coming up either heads or tails. More important, each

toss is independent of the tosses that came before. On

the fifth toss, for example, there is a 50% chance of

heads no matter what happened on the previous four

tosses; the coin does not remember what happened

earlier and is not influenced by the past. (Note: Many

people fail to believe in the independence of events.

For example, after a series of four tails in a row, it is

tempting to think that the probability of heads must

increase because the coin is overdue to come up

heads. This is a mistake, called the “gambler’s

fallacy.” Remember that the coin does not know

what happened on the preceding tosses and cannot

be influenced by previous outcomes.)

In most research situations, the requirement for

independent observations is satisfied by using a ran-

dom sample of separate, unrelated individuals. Thus,

the measurement obtained for each individual is not

influenced by other participants in the study. The

following two situations demonstrate circumstances in

which the observations are not independent.

1. A researcher is interested in examining television

preferences for children. To obtain a sample of

n � 20 children, the researcher selects 4 children

from family A, 3 children from family B, 5 children

from family C, 2 children from family D, and

6 children from family E.

It should be obvious that the researcher does

not have 20 independent observations. Within each

family, the children probably share television

preference (at least, they watch the same shows).

Thus, the response for each child is likely to be

related to the responses of his or her siblings.

2. The principle of independent observations is

violated if the sample is obtained using sampling

without replacement. For example, if you are

selecting from a group of 20 potential participants,

each individual has a 1 in 20 chance of being

selected first. After the first person is selected,

however, there are only 19 people remaining

and the probability of being selected changes to

1 in 19. Because the probability of the second

selection depends on the first, the two selections

are not independent.

SECTION 8.3 / MORE ABOUT HYPOTHESIS TESTS 2 2 1

The z-score statistic that is used in the hypothesis test is the first specific example of

what is called a test statistic. The term test statistic simply indicates that the sample data

are converted into a single, specific statistic that is used to test the hypotheses. In the

chapters that follow, we introduce several other test statistics that are used in a variety

of different research situations. However, most of the new test statistics have the same

basic structure and serve the same purpose as the z-score. We have already described the

z-score equation as a formal method for comparing the sample data and the population

hypothesis. In this section, we discuss the z-score from two other perspectives that may

give you a better understanding of hypothesis testing and the role that z-scores play in

this inferential technique. In each case, keep in mind that the z-score serves as a general

model for other test statistics that come in future chapters.

The z-score formula as a recipe The z-score formula, like any formula, can be

viewed as a recipe. If you follow instructions and use all of the right ingredients, the

formula produces a z-score. In the hypothesis-testing situation, however, you do not

have all of the necessary ingredients. Specifically, you do not know the value for the

population mean (�), which is one component, or ingredient, in the formula.

This situation is similar to trying to follow a cake recipe in which one of the ingredi-

ents is not clearly listed. For example, the recipe may call for flour but there is a grease

stain that makes it impossible to read how much flour. Faced with this situation, you

might try the following steps:

1. Make a hypothesis about the amount of flour. For example, hypothesize that the

correct amount is 2 cups.

2. To test your hypothesis, add the rest of the ingredients along with the

hypothesized amount of flour and bake the cake.

3. If the cake turns out to be good, you can reasonably conclude that your

hypothesis was correct. But if the cake is terrible, you conclude that your

hypothesis was wrong.

In a hypothesis test with z-scores, we do essentially the same thing. We have a

formula (recipe) for z-scores, but one ingredient (the population mean) is missing.

Therefore, we try the following steps:

1. Make a hypothesis about the value of �. This is the null hypothesis.

2. Plug the hypothesized value in the formula along with the other values

(ingredients).

3. If the formula produces a z-score near zero (which is where z-scores are

supposed to be), we conclude that the hypothesis was correct. On the other

hand, if the formula produces an extreme value (a very unlikely result), we

conclude that the hypothesis was wrong.

The z-score formula as a ratio In the context of a hypothesis test, the z-score

formula has the following structure:

z M

� �

� �

sample mean hypothesized population mean

standard error between andM �

Notice that the numerator of the formula measures the obtained difference

between the sample mean and the hypothesized population mean. The standard

error in the denominator measures the standard amount of distance that exists

A C LO S E R LO O K AT T H E z - S CO R E

I N A H Y P OT H E S I S T E ST

2 2 2 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

naturally between a sample mean and the population mean without any treatment

effect causing the sample to be different. Thus, the z-score formula (and most other

test statistics) forms a ratio

z M

� actual difference between the sample ( ) aand the hypothesis ( )

standard difference

�

bbetween and with no treatment effectM �

Thus, for example, a z-score of z � 3.00 means that the obtained difference between

the sample and the hypothesis is 3 times bigger than would be expected if the treat-

ment had no effect. A discrepancy this large is a strong indication that the hypothesis

is probably wrong.

The final decision in a hypothesis test is determined by the value obtained for the

z-score statistic. If the z-score is large enough to be in the critical region, then we reject

the null hypothesis and conclude that there is a significant treatment effect. Otherwise,

we fail to reject H 0 and conclude that the treatment does not have a significant effect.

The most obvious factor influencing the size of the z-score is the difference between

the sample mean and the hypothesized population mean from H 0 . A big mean difference

indicates that the treated sample is noticeably different from the untreated population

and usually supports a conclusion that the treatment effect is significant. In addition to

the mean difference, however, there are other factors that help determine whether the

z-score is large enough to reject H 0 . In this section, we examine two factors that can

influence the outcome of a hypothesis test.

1. The variability of the scores, which is measured by either the standard deviation

or the variance. The variability influences the size of the standard error in the

denominator of the z-score.

2. The number of scores in the sample. This value also influences the size of the

standard error in the denominator.

We use the research study from Example 8.1 to examine each of these factors.

The study used a sample of n � 25 students and produced a mean test score of

M � 89. Based on these results, the hypothesis test concluded that brain stimu-

lation near the parietal lobe has a significant effect on the ability to learn new

mathematical skills.

The variability of the scores In Chapter 4 (p. 110), we noted that high variabil-

ity can make it very difficult to see any clear patterns in the results from a research

study. In a hypothesis test, higher variability can reduce the chances of finding a

significant treatment effect. For the study in Example 8.1, the standard deviation is

� � 20. With a sample of n � 25, this produced a standard error of � M � 4 points and

a significant z-score of z � 2.25. Now consider what happens if the standard devia-

tion is increased to � � 30. With the increased variability, the standard error becomes

� M 5

30 ��25

5 6 points. Using the same sample mean from the original example, the

new z-score becomes

z M

5 m

s 5

 5 5

89 80

6 1 50.

The z-score is no longer beyond the critical boundary of 1.96, so the statistical de-

cision is to fail to reject the null hypothesis. The increased variability means that the

sample data are no longer sufficient to conclude that the treatment has a significant

FAC TO R S T H AT I N F L U E N C E A

H Y P OT H E S I S T E ST

SECTION 8.3 / MORE ABOUT HYPOTHESIS TESTS 2 2 3

effect. In general, increasing the variability of the scores produces a larger standard

error and a smaller value (closer to zero) for the z-score. If other factors are held con-

stant, then the larger the variability, the lower the likelihood of finding a significant

treatment effect.

The number of scores in the sample The second factor that influences the outcome

of a hypothesis test is the number of scores in the sample. The study in Example 8.1

used a sample of n � 25 students obtained a standard error of � M �

20 ��25

� 4 points

and a significant z-score of z � 2.25. Now consider what happens if we increase

the sample size to n � 100 students. With n � 100, the standard error becomes

� M �

20 ��100

� 2 points, and the z-score becomes

z M

� �

� �

 � �

89 80

2 4.50

Increasing the sample size from n � 25 to n � 100 has doubled the size of the

z-score. In general, increasing the number of scores in the sample produces a smaller

standard error and a larger value for the z-score. If all other factors are held constant,

the larger the sample size, the greater the likelihood of finding a significant treatment

effect. In simple terms, finding a 9-point treatment effect with a large sample is more

convincing than finding a 9-point effect with a small sample.

1. A researcher conducts a hypothesis test with a � .05 to evaluate the effectiveness

of a treatment. Assume that the sample mean produces a z-score of z � 2.17.

a. Do the data indicate that the treatment has a significant effect?

b. Write a sentence describing the outcome of the hypothesis test as it would

appear in a research report.

2. In a research report, the term significant is used when the null hypothesis is

rejected. (True or false?)

3. In a research report, the results of a hypothesis test include the phrase “z � 1.63,

p . .05.” This means that the test failed to reject the null hypothesis. (True or false?)

4. If other factors are held constant, increasing the size of the sample increases the

likelihood of rejecting the null hypothesis. (True or false?)

5. If other factors are held constant, are you more likely to reject the null hypothesis

with a standard deviation of s 5 2 or with s 5 10?

1. a. With a 5 .05, the critical region consists of z-scores in the tails beyond z 5 61.96.

Reject the null hypothesis.

b. The data indicate that the treatment had a significant effect, z 5 2.17, p , .05.

2. True.

3. True. The probability is greater than .05, which means there is a reasonable likelihood that

the result occurred without any treatment effect.

4. True. A larger sample produces a smaller standard error, which leads to a larger z-score.

5. s 5 2. A smaller standard deviation produces a smaller standard error, which leads to a

larger z-score.

L E A R N I N G C H E C K

ANSWERS

2 2 4 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

DIRECTIONAL (ONE-TAILED) HYPOTHESIS TESTS

The hypothesis-testing procedure presented in Section 8.1 is the standard, or two-tailed,

test format. The term two-tailed comes from the fact that the critical region is divided

between the two tails of the distribution. This format is by far the most widely accepted

procedure for hypothesis testing. Nonetheless, there is an alternative that is discussed

in this section.

Usually a researcher begins an experiment with a specific prediction about the

direction of the treatment effect. For example, a special training program is expected

to increase student performance, or alcohol consumption is expected to slow reaction

times. In these situations, it is possible to state the statistical hypotheses in a manner

that incorporates the directional prediction into the statement of H 0 and H

1 . The result

is a directional test, or what commonly is called a one-tailed test.

In a directional hypothesis test, or a one-tailed test, the statistical hypotheses

(H 0 and H

1 ) specify either an increase or a decrease in the population mean. That

is, they make a statement about the direction of the effect.

The following example demonstrates the elements of a one-tailed hypothesis test.

Earlier, in Example 8.1, we discussed a research study that examined the effect of

electrical stimulation near the parietal lobe on the mathematical skills of students. In

the study, each participant in a sample of n � 25 received electrical current near the

parietal lobe for 30 minutes each day while studying for a standardized mathematics

exam. For the general population of students (without any brain stimulation), the test

scores form a normal distribution with a mean of � � 80 and a standard deviation of

� � 20. For this example, the expected effect is that the parietal lobe stimulation will

improve test performance. If the researcher obtains a sample mean of M � 87 for the

n � 25 participants, is the result sufficient to conclude that the stimulation really works?

Because a specific direction is expected for the treatment effect, it is possible for the

researcher to perform a directional test. The first step (and the most critical step) is to

incorporate the directional prediction into the statement of the statistical hypotheses.

Remember that the null hypothesis states that there is no treatment effect and that the alter-

native hypothesis says that there is an effect. For this example, the predicted effect is that

the electrical stimulation will increase test scores. Thus, the two hypotheses would state:

H 0 : Test scores are not increased. (The treatment does not work.)

H 1 : Test scores are increased. (The treatment works as predicted.)

To express directional hypotheses in symbols, it usually is easier to begin with

the alternative hypothesis (H 1 ). Again, we know that the general population has an

average test score of � � 80, and H 1 states that test scores will be increased by the

brain stimulation. Therefore, expressed in symbols, H 1 states,

H 1 : � . 80 (With the stimulation, the average score is greater than 80.)

The null hypothesis states that the stimulation does not increase scores. In symbols,

H 0 : m  80 (With the stimulation, the average score is not greater than 80.)

8.4

D E F I N I T I O N

E X A M P L E 8 . 2

T H E H Y P OT H E S I S F O R A D I R E C T I O N A L

T E ST

SECTION 8.4 / DIRECTIONAL (ONE-TAILED) HYPOTHESIS TESTS 2 2 5

Note again that the two hypotheses are mutually exclusive and cover all of the

possibilities.

The critical region is defined by sample outcomes that are very unlikely to occur if the

null hypothesis is true (that is, if the treatment has no effect). Earlier (p. 209), we noted

that the critical region can also be defined in terms of sample values that provide con-

vincing evidence that the treatment really does have an effect. For a directional test, the

concept of “convincing evidence” is the simplest way to determine the location of the

critical region. We begin with all of the possible sample means that could be obtained

if the null hypothesis is true. This is the distribution of sample means, and it is normal

(because the population of test scores is normal), has an expected value of � � 80 (from

H 0 ), and, for a sample of n � 25, has a standard error of m 5 5

25 4. The distribution

is shown in Figure 8.7.

For this example, the treatment is expected to increase test scores. If untreated stu-

dents average m 5 80 on the test, then a sample mean that is substantially more than

80 would provide convincing evidence that the treatment worked. Thus, the critical

region is located entirely in the right-hand tail of the distribution corresponding to

sample means much greater than m 5 80 (see Figure 8.7). Because the critical region is

contained in one tail of the distribution, a directional test is commonly called a

one-tailed test. Also note that the proportion specified by the alpha level is not

divided between two tails, but rather is contained entirely in one tail. Using a 5 .05, for

example, the whole 5% is located in one tail. In this case, the z-score boundary for

the critical region is z 5 1.65, which is obtained by looking up a proportion of .05 in

column C (the tail) of the unit normal table.

Notice that a directional (one-tailed) test requires changes in the first two steps of

the step-by-step hypothesis-testing procedure.

1. In the first step, the directional prediction is included in the statement of the

hypotheses.

2. In the second step of the process, the critical region is located entirely in one tail

of the distribution.

After these two changes, the remainder of a one-tailed test proceeds exactly

the same as a regular two-tailed test. Specifically, you calculate the z-score statistic

and then make a decision about H 0 depending on whether the z-score is in the

critical region.

T H E C R I T I CA L R E G I O N F O R D I R E C T I O N A L

T E ST S

80 M

0 1.65

M � 4

Reject H0 Data indicate that H0 is wrong

FIGURE 8.7

Critical region for

Example 8.3.

If the prediction is that the

treatment will produce a

decrease in scores, then the

critical region is located

entirely in the left-hand tail

of the distribution.

2 2 6 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

For this example, the researcher obtained a mean of M � 87 for the 25 participants

who received the brain stimulation. This sample mean corresponds to a z-score of

z M

� �

� �

 � �

87 80

4 1 75.

A z-score of z � 1.75 is in the critical region for a one-tailed test (see Figure 8.7).

This is a very unlikely outcome if H 0 is true. Therefore, we reject the null hypothesis

and conclude that the electrical stimulation produces a significant increase in math-

ematics test scores. In the literature, this result would be reported as follows:

The stimulation produced a significant increase in scores, z � 1.75, p , .05,

one tailed.

Note that the report clearly acknowledges that a one-tailed test was used.

The general goal of hypothesis testing is to determine whether a particular treatment has

any effect on a population. The test is performed by selecting a sample, administering

the treatment to the sample, and then comparing the result with the original popula-

tion. If the treated sample is noticeably different from the original population, then we

conclude that the treatment has an effect, and we reject H 0 . On the other hand, if the

treated sample is still similar to the original population, then we conclude that there

is no convincing evidence for a treatment effect, and we fail to reject H 0 . The critical

factor in this decision is the size of the difference between the treated sample and the

original population. A large difference is evidence that the treatment worked; a small

difference is not sufficient to say that the treatment had any effect.

The major distinction between one-tailed and two-tailed tests is the criteria that they

use for rejecting H 0 . A one-tailed test allows you to reject the null hypothesis when the

difference between the sample and the population is relatively small, provided that the

difference is in the specified direction. A two-tailed test, on the other hand, requires

a relatively large difference independent of direction. This point is illustrated in the

following example.

Consider again the one-tailed test evaluating the effect of parietal lobe stimulation. If

we had used a standard two-tailed test, the hypotheses would be

H 0 : � � 80 (The stimulation has no effect on test scores.)

H 1 : �  80 (The stimulation does have an effect on test scores.)

For a two-tailed test with a � .05, the critical region consists of z-scores beyond

61.96. The data from Example 8.3 produced a sample mean of M � 87 and z � 1.75. For the two-tailed test, this z-score is not in the critical region, and we conclude that the

supplement does not have a significant effect.

With the two-tailed test in Example 8.3, the 7-point difference between the sample

mean and the hypothesized population mean (M � 87 and � � 80) is not big enough to

reject the null hypothesis. However, with the one-tailed test, the same 7-point difference

is large enough to reject H 0 and conclude that the treatment had a significant effect.

All researchers agree that one-tailed tests are different from two-tailed tests.

However, there are several ways to interpret the difference. One group of researchers

CO M PA R I S O N O F O N E - TA I L E D V E R S U S

T WO - TA I L E D T E ST S

E X A M P L E 8 . 3

SECTION 8.5 / CONCERNS ABOUT HYPOTHESIS TESTING: MEASURING EFFECT SIZE 2 2 7

contends that a two-tailed test is more rigorous and, therefore, more convincing than a

one-tailed test. Remember that the two-tailed test demands more evidence to reject H 0

and thus provides a stronger demonstration that a treatment effect has occurred.

Other researchers feel that one-tailed tests are preferable because they are more sen-

sitive. That is, a relatively small treatment effect may be significant with a one-tailed

test but fail to reach significance with a two-tailed test. Also, there is the argument that

one-tailed tests are more precise because they test hypotheses about a specific direc-

tional effect instead of an indefinite hypothesis about a general effect.

In general, two-tailed tests should be used in research situations when there is no

strong directional expectation or when there are two competing predictions. For ex-

ample, a two-tailed test would be appropriate for a study in which one theory predicts

an increase in scores but another theory predicts a decrease. One-tailed tests should be

used only in situations in which the directional prediction is made before the research

is conducted and there is a strong justification for making the directional prediction. In

particular, if a two-tailed test fails to reach significance, you should never follow up with

a one-tailed test as a second attempt to salvage a significant result for the same data.

1. A researcher selects a sample from a population with a mean of � � 60 and

administers a treatment to the individuals in the sample. If the researcher predicts

that the treatment will increase scores, then

a. Using symbols, state the hypotheses for a one-tailed test.

b. For the one-tailed test, would the critical region be located in the right-hand tail

or the left-hand tail of the distribution?

2. If a sample is sufficient to reject the null hypothesis for a one-tailed test, then the

same sample would also reject H 0 for a two-tailed test. (True or false?)

3. A researcher obtains z � 2.43 for a hypothesis test. Using a � .01, the researcher

should reject the null hypothesis for a one-tailed test but fail to reject for a two-tailed

test. (True or false?)

1. a. H 0 : �  60 and H

1 : � . 60

b. A large sample mean, in the right-hand tail, would indicate that the treatment worked as

predicted.

2. False. Because a two-tailed test requires a larger mean difference, it is possible for a sample

to be significant for a one-tailed test but not for a two-tailed test.

3. True. The one-tailed critical value is z � 2.33 and the two-tailed value is z � 2.58.

L E A R N I N G C H E C K

ANSWERS

CONCERNS ABOUT HYPOTHESIS TESTING: MEASURING EFFECT SIZE

Although hypothesis testing is the most commonly used technique for evaluating and

interpreting research data, a number of scientists have expressed a variety of concerns

about the hypothesis testing procedure (for example, see Loftus, 1996; Hunter, 1997;

and Killeen, 2005).

There are two serious limitations with using a hypothesis test to establish the sig-

nificance of a treatment effect. The first concern is that the focus of a hypothesis test

8.5

2 2 8 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

is on the data rather than the hypothesis. Specifically, when the null hypothesis is

rejected, we are actually making a strong probability statement about the sample data, not

about the null hypothesis. A significant result permits the following conclusion: “This

specific sample mean is very unlikely (p , .05) if the null hypothesis is true.” Note that the conclusion does not make any definite statement about the probability of the null

hypothesis being true or false. The fact that the sample is very unlikely suggests that the

null hypothesis is also very unlikely, but we do not have any solid grounds for making a

probability statement about the null hypothesis. Specifically, you cannot conclude that

the probability of the null hypothesis being true is less than 5% simply because you

rejected the null hypothesis with a � .05 (see Box 8.2).

A second concern is that demonstrating a significant treatment effect does not neces-

sarily indicate a substantial treatment effect. In particular, statistical significance does

BOX

8.2 A FLAW IN THE LOGIC OF HYPOTHESIS TESTING

Suppose that you do a hypothesis test and reject the null

hypothesis with a � .05. Can you conclude that there

is a 5% probability that you are making a Type I error?

Can you also conclude that there is a 95% probability

that your decision is correct and the treatment does have

an effect? For both questions, the answer is no.

The problem is that the probabilities for a

hypothesis test are well defined only when the null

hypothesis is true. Specifically, a hypothesis test using

a � .05 is structured so that the error rate is p , .05

and the accuracy rate is p  .95 if the null hypothesis

is true. If H 0 is false, however, these probabilities start

to fall apart. When there is a treatment effect (H 0 is

false), the probability that a hypothesis test will detect

it and reject H 0 depends on a variety of factors. For

example, if the treatment effect is very small, then a

hypothesis test is unlikely to detect it. With a large

treatment effect, the hypothesis test is more likely to

detect it and the probability of rejecting H 0 increases.

Thus, whenever there is a treatment effect (H 0 is

false), it becomes impossible to define precisely the

probability of rejecting the null hypothesis.

Most researchers begin research studies believing

that there is a good likelihood that the null hypothesis is

false and there really is a treatment effect. They are hop-

ing that the study will provide evidence of the effect so

they can convince their colleagues. Thus, most research

begins with some probability that the null hypothesis is

false. For the sake of argument, let’s assume that there is

an 80% probability that the null hypothesis is true.

p(there is no treatment effect— H 0 is true) � 0.80 and

p(there is a treatment effect— H 0 is false) � 0.20

In this situation, suppose that 125 researchers are

all doing hypothesis tests with a � .05. Of these

researchers, 80% (n � 100) are testing a true H 0 . For

these researchers, the probability of rejecting the null

hypothesis (and making a Type I error) is a � .05.

Therefore, the 100 hypothesis tests for this group

should produce, on average, 5 tests that reject H 0 .

Meanwhile, the other 20% of the researchers

(n � 25) are testing a false null hypothesis. For this

group, the probability of rejecting the null hypothesis

is unknown. For the sake of argument, however, let’s

assume that the probability of detecting the treatment

effect and correctly rejecting H 0 is 60%. This means

that the 25 hypothesis tests should result in 15 tests

(60%) that reject H 0 and 10 that fail to reject H

0 .

Notice that there should be 20 hypothesis tests that

reject the null hypothesis (5 from the first group

and 15 from the second group). Thus, a total of

20 researchers will find a statistically significant effect.

Of these 20 “significant” results, however, the 5 from

the first group are making a Type I error. Thus, when

the null hypothesis is rejected, the actually likelihood

of a Type I error is 5 out of 20, or p � 5

20 � .25, which

is five times greater than the alpha level of .05.

Based on this kind of argument, many scientists

suspect that a large number of the results and

conclusions published in research journals are simply

wrong. Specifically, the Type I error rate in published

research is almost certainly higher than the alpha

levels used in the hypothesis tests that support the

results (Siegfried, 2010).

SECTION 8.5 / CONCERNS ABOUT HYPOTHESIS TESTING: MEASURING EFFECT SIZE 2 2 9

not provide any real information about the absolute size of a treatment effect. Instead,

the hypothesis test has simply established that the results obtained in the research study

are very unlikely to have occurred if there is no treatment effect. The hypothesis test

reaches this conclusion by (1) calculating the standard error, which measures how much

difference is reasonable to expect between M and �, and (2) demonstrating that the

obtained mean difference is substantially bigger than the standard error.

Notice that the test is making a relative comparison: the size of the treatment effect

is being evaluated relative to the standard error. If the standard error is very small, then

the treatment effect can also be very small and still be large enough to be significant.

Thus, a significant effect does not necessarily mean a big effect. The idea that a hypoth-

esis test evaluates the relative size of a treatment effect, rather than the absolute size, is

illustrated in the following example.

We begin with a population of scores that forms a normal distribution with � � 50 and

� � 10. A sample is selected from the population and a treatment is administered to

the sample. After treatment, the sample mean is found to be M � 51. Does this sample

provide evidence of a statistically significant treatment effect?

Although there is only a 1-point difference between the sample mean and the origi-

nal population mean, the difference may be enough to be significant. In particular, the

outcome of the hypothesis test depends on the sample size.

For example, with a sample of n � 25 the standard error is

� � �

� � � M

5 2 00.

and the z-score for M � 51 is

z M

� �

� �

 � �

51 50

2 0 50.

This z-score fails to reach the critical boundary of z � 1.96, so we fail to reject the

null hypothesis. In this case, the 1-point difference between M and � is not significant

because it is being evaluated relative to a standard error of 2 points.

Now consider the outcome with a sample of n � 400. With a larger sample, the

standard error is

s � s

� � � M

400

20 0 50.

and the z-score for M � 51 is

z M

� �

� �

 � �

51 50

0 5

0 5 2 00

. . .

Now the z-score is beyond the 1.96 boundary, so we reject the null hypothesis and

conclude that there is a significant effect. In this case, the 1-point difference between

M and � is considered statistically significant because it is being evaluated relative to a

standard error of only 0.5 points.

The purpose of Example 8.4 is to demonstrate that a small treatment effect can still

be statistically significant. If the sample size is large enough, any treatment effect, no

matter how small, can be large enough to reject the null hypothesis.

E X A M P L E 8 . 4

2 3 0 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

As noted in the previous section, one concern with hypothesis testing is that a hypothesis

test does not really evaluate the absolute size of a treatment effect. To correct this problem,

it is recommended that whenever researchers report a statistically significant effect, they

also provide a report of the effect size (see the guidelines presented by L. Wilkinson and

the APA Task Force on Statistical Inference, 1999). Therefore, as we present different

hypothesis tests, we also present different options for measuring and reporting effect size.

A measure of effect size is intended to provide a measurement of the absolute

magnitude of a treatment effect, independent of the size of the sample(s)

being used.

One of the simplest and most direct methods for measuring effect size is Cohen’s

d. Cohen (1988) recommended that effect size can be standardized by measuring the

mean difference in terms of the standard deviation. The resulting measure of effect size

is computed as

Cohen's mean difference

standard deviation d � ��

� �

�

treatment no treatment

(8.1)

For the z-score hypothesis test, the mean difference is determined by the difference

between the population mean before treatment and the population mean after treatment.

However, the population mean after treatment is unknown. Therefore, we must use the

mean for the treated sample in its place. Remember, the sample mean is expected to be

representative of the population mean and provides the best measure of the treatment ef-

fect. Thus, the actual calculations are really estimating the value of Cohen’s d as follows

estimated Cohen's mean difference

standard d �

deviation treatment no treatment�

�

�

(8.2)

The standard deviation is included in the calculation to standardize the size of the

mean difference in much the same way that z-scores standardize locations in a distribu-

tion. For example, a 15-point mean difference can be a relatively large treatment effect

or a relatively small effect depending on the size of the standard deviation. This phe-

nomenon is demonstrated in Figure 8.8. The top portion of the figure [part (a)] shows

the results of a treatment that produces a 15-point mean difference in SAT scores;

before treatment, the average SAT score is � � 500, and after treatment the average

is 515. Notice that the standard deviation for SAT scores is � � 100, so the 15-point

difference appears to be small. For this example, Cohen’s d is

Cohen's mean difference

standard deviation d � ��

100 0 15.

Now consider the treatment effect shown in Figure 8.8(b). This time, the treatment pro-

duces a 15-point mean difference in IQ scores; before treatment the average IQ is 100, and

after treatment the average is 115. Because IQ scores have a standard deviation of � � 15,

the 15-point mean difference now appears to be large. For this example, Cohen’s d is

Cohen's mean difference

standard deviation d � ��

15 1 00.

Notice that Cohen’s d measures the size of the treatment effect in terms of the stan-

dard deviation. For example, a value of d � 0.50 indicates that the treatment changed

the mean by half of a standard deviation; similarly, a value of d � 1.00 indicates that the

M E AS U R I N G E F F E C T S I Z E

D E F I N I T I O N

Cohen’s d measures the

distance between two means

and is typically reported as

a positive number even when

the formula produces a

negative value.

SECTION 8.5 / CONCERNS ABOUT HYPOTHESIS TESTING: MEASURING EFFECT SIZE 2 3 1

size of the treatment effect is equal to one whole standard deviation. Cohen (1988) also

suggested criteria for evaluating the size of a treatment effect as shown in Table 8.2.

As one final demonstration of Cohen’s d, consider the two hypothesis tests in

Example 8.4. For each test, the original population had a mean of � � 50 with a stan-

dard deviation of � � 10. For each test, the mean for the treated sample was M � 51.

Although one test used a sample of n � 25 and the other test used a sample of n � 400,

the sample size is not considered when computing Cohen’s d. Therefore, both of the

hypothesis tests would produce the same value:

Cohen's mean difference

standard deviation d � ��

10 0 10.

Notice that Cohen’s d simply describes the size of the treatment effect and is not

influenced by the number of scores in the sample. For both hypothesis tests, the original

population mean was � � 50 and, after treatment, the sample mean was M � 51. Thus,

treatment appears to have increased the scores by 1 point, which is equal to one-tenth

of a standard deviation (Cohen’s d � 0.1).

� � 500

� � 100

Distribution of SAT scores before treatment � � 500 and � � 100

d � 0.15

� � 100

� � 15

Distribution of IQ scores before treatment � � 100 and � � 15

Distribution of SAT scores after treatment � � 515 and � � 100

Distribution of IQ scores after treatment � � 115 and � � 15

d � 1.00

(a)

(b)

FIGURE 8.8

The appearance of a 15-point treatment effect in two different situations. In part (a), the standard

deviation is � � 100 and the 15-point effect is relatively small. In part (b), the standard deviation

is � � 15 and the 15-point effect is relatively large. Cohen’s d uses the standard deviation to help

measure effect size.

2 3 2 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

STATISTICAL POWER

Instead of measuring effect size directly, an alternative approach to determining the

size or strength of a treatment effect is to measure the power of the statistical test. The

power of a test is defined as the probability that the test will reject the null hypothesis

if the treatment really has an effect.

The power of a statistical test is the probability that the test will correctly reject a

false null hypothesis. That is, power is the probability that the test will identify a

treatment effect if one really exists.

Whenever a treatment has an effect, there are only two possible outcomes for a hy-

pothesis test: either fail to reject H 0 or reject H

0 . Because there are only two possible

outcomes, the probability for the first and the probability for the second must add up

to 1.00. The first outcome, failing to reject H 0 when there is a real effect, was defined

earlier (p. 214) as a Type II error with a probability identified as p � b. Therefore,

the second outcome must have a probability of 1 – b. However, the second outcome,

rejecting H 0 when there is a real effect, is the power of the test. Thus, the power of a

hypothesis test is equal to 1 – b. In the examples that follow, we demonstrate the calcu-

lation of power for a hypothesis test; that is, the probability that the test will correctly

reject the null hypothesis. At the same time, however, we are computing the probability

that the test will result in a Type II error. For example, if the power of the test is 70%

(1 – b), then the probability of a Type II error must be 30% (b).

Researchers typically calculate power as a means of determining whether a research

study is likely to be successful. Thus, researchers usually calculate the power of a

8.6

D E F I N I T I O N

Magnitude of d Evaluation of Effect Size

d � 0.2 Small effect (mean difference around 0.2 standard deviation)

d � 0.5 Medium effect (mean difference around 0.5 standard deviation)

d � 0.8 Large effect (mean difference around 0.8 standard deviation)

TABLE 8.2

Evaluating effect size with

Cohen’s d.

1. Explain how increasing the sample size influences the outcome of a hypothesis test

and how it influences the value of Cohen’s d.

2. A researcher selects a sample from a population with � � 45 and � � 8. A treat-

ment is administered to the sample and, after treatment, the sample mean is found

to be M � 47. Compute Cohen’s d to measure the size of the treatment effect.

3. A researcher selects a sample from a population with � � 70 and � � 12. After

administering a treatment to the individuals in the sample, the researcher computes

Cohen’s d � 0.25. What is the mean for the sample?

1. Increasing sample size increases the likelihood of rejecting the null hypothesis but has no

effect on Cohen’s d.

2. d � 2 8 � 0.25

3. There is a 3-point difference between the sample mean and � � 70, so the sample mean is

either 73 or 67.

L E A R N I N G C H E C K

ANSWERS

SECTION 8.6 / STATISTICAL POWER 2 3 3

hypothesis test before they actually conduct the research study. In this way, they can

determine the probability that the results will be significant (reject H 0 ) before investing

time and effort in the actual research. To calculate power, however, it is first necessary

to make assumptions about a variety of factors that influence the outcome of a hypoth-

esis test. Factors such as the sample size, the size of the treatment effect, and the value

chosen for the alpha level can all influence a hypothesis test. The following example

demonstrates the calculation of power for a specific research situation.

We start with a normal-shaped population with a mean of � � 80 and a standard

deviation of � � 10. A researcher plans to select a sample of n � 25 individuals from

this population and administer a treatment to each individual. It is expected that the

treatment will have an 8-point effect; that is, the treatment will add 8 points to each

individual’s score.

Figure 8.9 shows the original population distribution and two possible outcomes:

1. If the null hypothesis is true and there is no treatment effect.

2. If the researcher’s expectation is correct and there is an 8-point effect.

E X A M P L E 8 . 5

With an 8-point treatment effect

� � 88 and � � 10

�M � 2

If H0 is true (no treatment effect)

� � 80 and � � 10

Reject H0

�1.96 0

�1.96 z

Reject H0

Distribution of sample means for n � 25 if H0 is true

Distribution of sample means for n � 25 with 8-point effect

Original Population

Normal with � � 80 and

� � 10

7876 868482 929088

�M � 2

FIGURE 8.9

A demonstration of mea-

suring power for a hypoth-

esis test. The left-hand

side shows the distribution

of sample means that

would occur if the null

hypothesis is true. The

critical region is defined

for this distribution. The

right-hand side shows the

distribution of sample

means that would be

obtained if there were an

8-point treatment effect.

Notice that, if there is an

8-point effect, essentially

all of the sample means

would be in the critical

region. Thus, the

probability of rejecting

H 0 (the power of the test)

would be nearly 100%

for an 8-point treatment

effect.

2 3 4 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

The left-hand side of the figure shows what should happen according to the null

hypothesis. In this case, the treatment has no effect and the population mean is still

� � 80. On the right-hand side of the figure we show what would happen if the treat-

ment has an 8-point effect. If the treatment adds 8 points to each person’s score, then

the population mean after treatment increases to � � 88.

Beneath each of the two populations, Figure 8.9 shows the distribution of sample

means for n � 25. According to the null hypothesis, the sample means are centered

around � � 80. With an 8-point treatment effect, the sample means are centered around

� � 88. Both distributions have a standard error of

� � �

� � � M

5 2

Notice that the distribution on the left shows all of the possible sample means if the

null hypothesis is true. This is the distribution we use to locate the critical region for

the hypothesis test. Using a � .05, the critical region consists of extreme values in this

distribution, specifically sample means beyond z � 1.96 or z � –1.96. These values

are shown in Figure 8.9, and we have shaded all of the sample means located in the

critical region.

Now turn your attention to the distribution on the right, which shows all of the

possible sample means if there is an 8-point treatment effect. Notice that most of

these sample means are located beyond the z � 1.96 boundary. This means that, if

there is an 8-point treatment effect, you are almost guaranteed to obtain a sample

mean in the critical region and reject the null hypothesis. Thus, the power of the

test (the probability of rejecting H 0 ) is close to 100% if there is an 8-point treat-

ment effect.

To calculate the exact value for the power of the test we must determine what por-

tion of the distribution on the right-hand side is shaded. Thus, we must locate the exact

boundary for the critical region, then find the probability value in the unit normal table.

For the distribution on the left-hand side, the critical boundary of z � 11.96 corre-

sponds to a location that is above � � 80 by a distance equal to

1.96� M � 1.96(2) � 3.92 points

Thus, the critical boundary of z � 11.96 corresponds to a sample mean of M � 80

1 3.92 � 83.92. Any sample mean greater than M � 83.92 is in the critical region and

would lead to rejecting the null hypothesis. Next, we determine what proportion of the

treated samples are greater than M � 83.92. For the treated distribution (right-hand

side), the population mean is � � 88 and a sample mean of M � 83.92 corresponds to

a z-score of

z M

� �

� �

 �

 �

83 92 88

4 08

2 2 04

. . .

Finally, look up z � –2.04 in the unit normal table and determine that the shaded

area (z . –2.04) corresponds to p � 0.9793 (or 97.93%). Thus, if the treatment has an

8-point effect, 97.93% of all the possible sample means will be in the critical region and

we will reject the null hypothesis. In other words, the power of the test is 97.93%. In

practical terms, this means that the research study is almost guaranteed to be success-

ful. If the researcher selects a sample of n � 25 individuals, and if the treatment really

does have an 8-point effect, then 97.93% of the time the hypothesis test will conclude

that there is a significant effect.

SECTION 8.6 / STATISTICAL POWER 2 3 5

Logically, it should be clear that power and effect size are related. Figure 8.9 shows the

calculation of power for an 8-point treatment effect. Now consider what would happen

if the treatment effect were only 4 points. With a 4-point treatment effect, the distribu-

tion on the right-hand side would shift to the left so that it is centered at � � 84. In

this new position, only about 50% of the treated sample means would be beyond the

z � 1.96 boundary. Thus, with a 4-point treatment effect, there is only a 50% probabil-

ity of selecting a sample that leads to rejecting the null hypothesis. In other words, the

power of the test is only about 50% for a 4-point effect compared to nearly 98% with an

8-point effect (see Example 8.5). Again, it is possible to find the z-score corresponding

to the exact location of the critical boundary and to look up the probability value for

power in the unit normal table. In this case, you should obtain z � –0.04 and the exact

power of the test is p � 0.5160, or 51.60%.

In general, as the effect size increases, the distribution of sample means on the

right-hand side moves even farther to the right so that more and more of the samples

are beyond the z � 1.96 boundary. Thus, as the effect size increases, the probability

of rejecting H 0 also increases, which means that the power of the test increases. Thus,

measures of effect size such as Cohen’s d and measures of power both provide an indi-

cation of the strength or magnitude of a treatment effect.

Although the power of a hypothesis test is directly influenced by the size of the treat-

ment effect, power is not meant to be a pure measure of effect size. Instead, power is

influenced by several factors, other than effect size, that are related to the hypothesis

test. Some of these factors are considered in the following sections.

Sample size One factor that has a huge influence on power is the size of the sample.

In Example 8.5, we demonstrated power for an 8-point treatment effect using a sample

of n � 25. If the researcher decided to conduct the study using a sample of n � 4, then

the power would be dramatically different. With n � 4, the standard error for the sample

means would be

� � �

� � � M

2 5

Figure 8.10 shows the two distributions of sample means with n � 4 and a standard

error of � M � 5 points. Again, the distribution on the left is centered at � � 80 and

shows all of the possible sample means if H 0 is true. As always, this distribution is used

to locate the critical boundaries for the hypothesis test, z � –1.96 and z � 11.96. The

distribution on the right is centered at � � 88 and shows all of the possible sample

means if there is an 8-point treatment effect. Note that less than half of the treated

sample means in the right-hand distribution are now located beyond the 1.96 boundary.

Thus, with a sample of n � 4, there is less than a 50% probability that the hypoth-

esis test would reject H 0 , even though the treatment has an 8-point effect. Earlier, in

Example 8.5, we found power equal to 97.93% for a sample of n � 25. However, when

the sample size is reduced to n � 4, power decreases to less than 50%. In general, a

larger sample produces greater power for a hypothesis test.

Because power is directly related to sample size, one of the primary reasons for

computing power is to determine what sample size is necessary to achieve a reasonable

probability for a successful research study. Before a study is conducted, researchers can

compute power to determine the probability that their research will successfully reject

the null hypothesis. If the probability (power) is too small, they always have the option

of increasing sample size to increase power.

P OW E R A N D E F F E C T S I Z E

OT H E R FAC TO R S T H AT A F F E C T P OW E R

2 3 6 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

Alpha level Reducing the alpha level for a hypothesis test also reduces the

power of the test. For example, lowering a from .05 to .01 lowers the power of the

hypothesis test. The effect of reducing the alpha level can be seen by referring again

to Figure 8.10. In this figure, the boundaries for the critical region are drawn using

a � .05. Specifically, the critical region on the right-hand side begins at z � 1.96.

If a were changed to .01, the boundary would be moved farther to the right, out to

z � 2.58. It should be clear that moving the critical boundary to the right means that

a smaller portion of the treatment distribution (the distribution on the right-hand

side) will be in the critical region. Thus, there would be a lower probability of reject-

ing the null hypothesis and a lower value for the power of the test.

One-tailed versus two-tailed tests Changing from a regular two-tailed test to a one-tailed

test increases the power of the hypothesis test. Again, this effect can be seen by referring to

Figure 8.10. The figure shows the boundaries for the critical region using a two-tailed test

with a � .05 so that the critical region on the right-hand side begins at z � 1.96. Changing

to a one-tailed test would move the critical boundary to the left to a value of z � 1.65.

Moving the boundary to the left would cause a larger proportion of the treatment distribution

to be in the critical region and, therefore, would increase the power of the test.

With an 8-point treatment effect

� � 88 and � � 10

If H0 is true (no treatment effect)

� � 80 and � � 10

Reject H0

�1.96 0

�1.96 z

Reject H0

Distribution of sample means for n � 4 if H0 is true

Distribution of sample means for n � 4 with 8-point effect

Original Population

Normal with � � 80 and

� � 10

82 84 86 88 90 92 94 9670 72 74 76 78 98

�M � 5�M � 5

FIGURE 8.10

A demonstration of how

sample size affects the

power of a hypothesis test.

As in Figure 8.9, the left-

hand side shows the dis-

tribution of sample means

if the null hypothesis were

true. The critical region is

defined for this distribu-

tion. The right-hand side

shows the distribution of

sample means that would

be obtained if there were

an 8-point treatment

effect. Notice that reduc-

ing the size to n � 4 has

reduced the power of the

test to less than 50% com-

pared to a power of nearly

100% with a sample of

n � 25 in Figure 8.9.

SUMMARY 2 3 7

1. As the power of a test increases, what happens to the probability of a Type II error?

2. For a 5-point treatment effect, a researcher computes power of p � 0.50 for a

two-tailed hypothesis test with a � .05.

a. Will the power increase or decrease for a 10-point treatment effect?

b. Will the power increase or decrease if alpha is changed to a � .01?

c. Will the power increase or decrease if the researcher changes to a one-tailed test?

3. How does sample size influence the power of a hypothesis test?

4. A researcher administers a treatment to a sample of n � 16 individuals selected

from a normal population with � � 60 and � � 12. If the treatment increases

scores by 4 points, what is the power of a two-tailed hypothesis test with a � .05?

1. As power increases, the probability of a Type II error decreases.

2. a. The hypothesis test is more likely to detect a 10-point effect, so power will be greater.

b. Decreasing the alpha level also decreases the power of the test.

c. Switching to a one-tailed test will increase the power.

3. Increasing sample size increases the power of a test

4. With standard error of 3 points, the critical boundary of z � 1.96 corresponds to a sample

mean of M � 65.88. With a 4-point effect, the distribution of sample means has � � 64 and

a sample mean of M � 65.88 corresponds to z � 0.63. Power � p(z . 0.63) � 0.2643.

L E A R N I N G C H E C K

ANSWERS

SUMMARY

1. Hypothesis testing is an inferential procedure that uses the data from a sample to draw a general conclusion about a population. The procedure begins with a hy- pothesis about an unknown population. Then a sample is selected, and the sample data provide evidence that either supports or refutes the hypothesis.

2. In this chapter, we introduced hypothesis testing using the simple situation in which a sample mean is used to test a hypothesis about an unknown population mean; usually the mean for a population that has received a treatment. The question is to determine whether the treatment has an effect on the population mean (see Figure 8.1).

3. Hypothesis testing is structured as a four-step process that is used throughout the remainder of the book.

a. State the null hypothesis (H 0 ), and select an alpha

level. The null hypothesis states that there is no effect or no change. In this case, H

0 states that the

mean for the treated population is the same as the mean before treatment. The alpha level, usually a � .05 or a � .01, provides a definition of the term very unlikely and determines the risk of a Type I error when H

0 is true. Also state an

alternative hypothesis (H 1 ), which is the exact

opposite of the null hypothesis. b. Locate the critical region. The critical region is

defined as extreme sample outcomes that would be very unlikely to occur if the null hypothesis is true. The alpha level defines “very unlikely.”

c. Collect the data, and compute the test statistic. The sample mean is transformed into a z-score by the formula

z M

� �

�

The value of � is obtained from the null hypothesis. The z-score test statistic identifies the location of the sample mean in the distribution of sample means.

d. Make a decision. If the obtained z-score is in the critical region, reject H

0 because it is very un-

likely that these data would be obtained if H 0 were

true. In this case, conclude that the treatment has changed the population mean. If the z-score is not in the critical region, fail to reject H

0 because the

data are not significantly different from the null hypothesis. In this case, the data do not provide

2 3 8 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

sufficient evidence to indicate that the treatment has had an effect.

4. Whatever decision is reached in a hypothesis test, there is always a risk of making the incorrect decision. There are two types of errors that can be committed.

A Type I error is defined as rejecting a true H 0 .

This is a serious error because it results in falsely

reporting a treatment effect. The risk of a Type I

error is determined by the alpha level and, therefore,

is under the experimenter’s control.

A Type II error is defined as the failure to reject

a false H 0 . In this case, the experiment fails to detect

an effect that actually occurred. The probability of

a Type II error cannot be specified as a single value

and depends in part on the size of the treatment

effect. It is identified by the symbol b (beta).

5. When a researcher expects that a treatment will change scores in a particular direction (increase or decrease), it is possible to do a directional, or one-tailed, test. The first step in this procedure is to incorporate the directional prediction into the hypotheses. For example, if the prediction is that a treatment will in- crease scores, the null hypothesis says that there is no increase and the alternative hypothesis states that there is an increase. To locate the critical region, you must determine what kind of data would refute the null hypothesis by demonstrating that the treatment worked as predicted. These outcomes are located entirely in one tail of the distribution, so the entire critical region (5%, 1%, or 0.1% depending on a) is in one tail.

6. In addition to using a hypothesis test to evaluate the significance of a treatment effect, it is recommended that you also measure and report the effect size. One measure of effect size is Cohen’s d, which is a standardized mea- sure of the mean difference. Cohen’s d is computed as

Cohen's mean difference

standard deviation d �

7. The power of a hypothesis test is defined as the probabil- ity that the test will correctly reject the null hypothesis.

8. To determine the power for a hypothesis test, you must first identify the treatment and null distributions. Also, you must specify the magnitude of the treatment ef- fect. Next, you sketch the distribution of sample means predicted by the null hypothesis and the distribution predicted by the specified treatment effect. Locate the critical region in the null hypothesis distribution and then determine the proportion of the treated distribu- tion that is beyond the critical boundaries. This propor- tion is the power of the hypothesis test.

9. As the size of the treatment effect increases, statistical power increases. Also, power is influenced by several factors that can be controlled by the experimenter:

a. A large sample results in more power than a small sample.

b. Increasing the alpha level increases power. c. A one-tailed test has greater power than a

two-tailed test.

KEY TERMS

hypothesis test (204)

null hypothesis (207)

alternative hypothesis (207)

level of significance (209)

alpha level (209)

critical region (209)

Type I error (213)

Type II error (214)

beta (215)

significant (218)

test statistic (221)

directional test (224)

one-tailed test (224)

effect size (230)

Cohen’s d (230)

power (232)

RESOURCES

Go to CengageBrain.com to access Psychology CourseMate, where you will find an

interactive eBook, glossaries, flashcards, quizzes, statistics workshops, and more.

FOCUS ON PROBLEM SOLVING 2 3 9

If your professor has assigned Aplia:

1. Sign in to your account.

2. Complete the corresponding exercises as required by your professor.

3. When finished, click “Grade It Now” to see which areas you have mastered, which

areas need more work, and detailed explanations of every answer.

The statistical computer package SPSS is not structured to conduct hypothesis

tests using z-scores. In truth, the z-score test presented in this chapter is rarely

used in actual research situations. The problem with the z-score test is that it

requires that you know the value of the population standard deviation, and this

information is usually not available. Researchers rarely have detailed information

about the populations that they wish to study. Instead, they must obtain

information entirely from samples. In the following chapters, we introduce

new hypothesis-testing techniques that are based entirely on sample data. These

new techniques are included in SPSS.

FOCUS ON PROBLEM SOLVING

1. Hypothesis testing involves a set of logical procedures and rules that enable us to

make general conclusions about a population based on data from a sample. This

process is reflected in the four steps that have been used throughout this chapter.

S T E P 1 State the hypotheses and set the alpha level.

S T E P 2 Locate the critical region.

S T E P 3 Compute the test statistic (in this case, the z-score) for the sample.

S T E P 4 Make a decision about H 0 based on the result of step 3.

2. Take time to consider the implications of your decision about the null hypothesis.

The null hypothesis states that the treatment has no effect. If your decision is to

reject H 0 , then you are concluding that the sample data provide evidence that a treat-

ment effect exists. However, when you fail to reject the null hypothesis, the results

are inconclusive. In this case, all you can state is that there is not sufficient evidence

to support the existence of a treatment effect.

3. When you are doing a directional hypothesis test, read the problem carefully,

and watch for key words (such as increase or decrease, raise or lower, and more

or less) that tell you which direction the researcher is predicting. The predicted

direction determines the alternative hypothesis (H 1 ) and the critical region.

For example, if a treatment is expected to increase scores, H 1 would contain

a greater than symbol, and the critical region would be in the tail associated

with high scores.

2 4 0 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

DEMONSTRATION 8.1

HYPOTHESIS TEST WITH z

A researcher begins with a known population—in this case, scores on a standardized

test that are normally distributed with � � 65 and � � 15. The researcher suspects that

special training in reading skills will produce a change in the scores for the individu-

als in the population. Because it is not feasible to administer the treatment (the special

training) to everyone in the population, a sample of n � 25 individuals is selected, and

the treatment is given to this sample. Following treatment, the average score for this

sample is M � 70. Is there evidence that the training has an effect on test scores?

State the hypothesis and select an alpha level. The null hypothesis states that the spe- cial training has no effect. In symbols,

H 0 : � � 65 (After special training, the mean is still 65.)

The alternative hypothesis states that the treatment does have an effect.

H 1 : �  65 (After training, the mean is different from 65.)

At this time you also select the alpha level. For this demonstration, we will use a � .05.

Thus, there is a 5% risk of committing a Type I error if we reject H 0 .

Locate the critical region. With a � .05, the critical region consists of sample means that correspond to z-scores beyond the critical boundaries of z � 61.96.

Obtain the sample data, and compute the test statistic. For this example, the distribu- tion of sample means, according to the null hypothesis, is normal with an expected value of � � 65 and a standard error of

� � �

� � � M

5 3

In this distribution, our sample mean of M � 70 corresponds to a z-score of

z M

� �

� �

 � �1

70 65

3 1 67.

Make a decision about H 0 , and state the conclusion. The z-score we obtained is not

in the critical region. This indicates that our sample mean of M � 70 is not an extreme or unusual value to be obtained from a population with � � 65. Therefore, our statisti- cal decision is to fail to reject H

0 . Our conclusion for the study is that the data do not

provide sufficient evidence that the special training changes test scores.

DEMONSTRATION 8.2

EFFECT SIZE USING COHEN’S d

We will compute Cohen’s d using the research situation and the data from

Demonstration 8.1. Again, the original population mean was � � 65 and, after

S T E P 1

S T E P 2

S T E P 3

S T E P 4

PROBLEMS 2 4 1

treatment (special training), the sample mean was M � 70. Thus, there is a 5-point

mean difference. Using the population standard deviation, � � 15, we obtain an effect

size of

Cohen's mean difference

standard deviation d � ��

15 0 33� .

According to Cohen’s evaluation standards (see Table 8.2), this is a medium treat-

ment effect.

PROBLEMS

1. The value of the z-score in a hypothesis test is influ- enced by a variety of factors. Assuming that all other variables are held constant, explain how the value of z is influenced by each of the following:

a. An increase in the difference between the sample mean and the original population mean.

b. An increase in the population standard deviation. c. An increase in the number of scores in the sample.

2. Define the alpha level and the critical region, and explain how they are related.

3. Although there is a popular belief that herbal rem- edies such as Ginkgo biloba and Ginseng may im- prove learning and memory in healthy adults, these effects are usually not supported by well-controlled research (Persson, Bringlov, Nilsson, & Nyberg, 2004). In a typical study, a researcher obtains a sam- ple of n � 16 participants and has each person take the herbal supplements every day for 90 days. At the end of the 90 days, each person takes a standard- ized memory test. For the general population, scores from the test form a normal distribution with a mean of � � 50 and a standard deviation of � � 12. The sample of research participants had an average of M � 54.

a. Assuming a two-tailed test, state the null hypoth- esis in a sentence that includes the two variables being examined.

b. Using the standard 4-step procedure, conduct a two-tailed hypothesis test with a � .05 to evalu- ate the effect of the supplements.

4. Childhood participation in sports, cultural groups, and youth groups appears to be related to improved self-esteem for adolescents (McGee, Williams, Howden-Chapman, Martin, & Kawachi, 2006). In a representative study, a sample of n � 100 adoles- cents with a history of group participation is given a standardized self-esteem questionnaire. For the general population of adolescents, scores on this questionnaire form a normal distribution with a mean of � � 50 and a standard deviation of � � 15. The

sample of group-participation adolescents had an average of M � 53.8.

a. Does this sample provide enough evidence to con- clude that self-esteem scores for these adolescents are significantly different from those of the general population? Use a two-tailed test with a � .05.

b. Compute Cohen’s d to measure the size of the difference.

c. Write a sentence describing the outcome of the hypothesis test and the measure of effect size as it would appear in a research report.

5. A local college requires an English composition course for all freshmen. This year they are evaluating a new online version of the course. A random sample of n � 16 freshmen is selected and the students are placed in the online course. At the end of the semes- ter, all freshmen take the same English composition exam. The average score for the sample is M � 76. For the general population of freshmen who took the traditional lecture class, the exam scores form a normal distribution with a mean of � � 80.

a. If the final exam scores for the population have a standard deviation of � � 12, does the sample provide enough evidence to conclude that the new online course is significantly different from the traditional class? Assume a two-tailed test with a � .05.

b. If the population standard deviation is � � 6, is the sample sufficient to demonstrate a significant difference? Again, assume a two-tailed test with a � .05.

c. Comparing your answers for parts a and b, ex- plain how the magnitude of the standard deviation influences the outcome of a hypothesis test.

6. A random sample is selected from a normal popula- tion with a mean of � � 30 and a standard deviation of � � 8. After a treatment is administered to the individuals in the sample, the sample mean is found to be M � 33.

a. If the sample consists of n � 16 scores, is the sample mean sufficient to conclude that the treat- ment has a significant effect? Use a two-tailed test with a � .05.

2 4 2 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

b. If the sample consists of n � 64 scores, is the sample mean sufficient to conclude that the treat- ment has a significant effect? Use a two-tailed test with a � .05.

c. Comparing your answers for parts a and b, ex- plain how the size of the sample influences the outcome of a hypothesis test.

7. A random sample of n � 25 scores is selected from a normal population with a mean of � � 40. After a treatment is administered to the individuals in the sample, the sample mean is found to be M � 44.

a. If the population standard deviation is � � 5, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with a � .05.

b. If the population standard deviation is � � 15, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with a � .05.

c. Comparing your answers for parts a and b, ex- plain how the magnitude of the standard deviation influences the outcome of a hypothesis test.

8. Brunt, Rhee, and Zhong (2008) surveyed 557 under- graduate college students to examine their weight status, health behaviors, and diet. Using body mass index (BMI), they classified the students into four categories: underweight, healthy weight, overweight, and obese. They also measured dietary variety by counting the number of different foods each student ate from several food groups. Note that the research- ers are not measuring the amount of food eaten, but rather the number of different foods eaten (variety, not quantity). Nonetheless, it was somewhat surpris- ing that the results showed no differences among the four weight categories that were related to eating fatty and/or sugary snacks.

Suppose a researcher conducting a follow up study obtains a sample of n � 25 students classified as healthy weight and a sample of n � 36 students classified as overweight. Each student completes the food variety questionnaire, and the healthy-weight group produces a mean of M � 4.01 for the fatty, sugary snack category compared to a mean of M � 4.48 for the overweight group. The results from the Brunt, Rhee, and Zhong study showed an overall mean variety score of � � 4.22 for the discretionary sweets or fats food group. Assume that the distribu- tion of scores is approximately normal with a stan- dard deviation of � � 0.60.

a. Does the sample of n � 36 indicate that the num- ber of fatty, sugary snacks eaten by overweight students is significantly different from the overall population mean? Use a two-tailed test with a � .05.

b. Based on the sample of n � 25 healthy-weight students, can you conclude that healthy-weight students eat significantly fewer fatty, sugary snacks than the overall population? Use a one- tailed test with a � .05.

9. A random sample is selected from a normal popula- tion with a mean of � � 100 and a standard devia- tion of � � 20. After a treatment is administered to the individuals in the sample, the sample mean is found to be M � 96.

a. How large a sample is necessary for this sample mean to be statistically significant? Assume a two-tailed test with a � .05.

b. If the sample mean were M � 98, what sample size would be needed to be significant for a two- tailed test with a � .05?

10. In a study examining the effect of alcohol on reaction time, Liguori and Robinson (2001) found that even moderate alcohol consumption significantly slowed response time to an emergency situation in a driving simulation. In a similar study, researchers measured reaction time 30 minutes after participants consumed one 6-ounce glass of wine. Again, they used a stan- dardized driving simulation task for which the regular population averages � � 400 msec. The distribution of reaction times is approximately normal with � � 40. Assume that the researcher obtained a sample mean of M � 422 for the n � 25 participants in the study.

a. Are the data sufficient to conclude that the alcohol has a significant effect on reaction time? Use a two-tailed test with a � .01.

b. Do the data provide evidence that the alcohol significantly increased (slowed) reaction time? Use a one-tailed test with a � .05.

c. Compute Cohen’s d to estimate the size of the effect.

11. The researchers cited in the previous problem (Liguori & Robinson, 2001) also examined the effect of caffeine on response time in the driving simulator. In a similar study, researchers measured reaction time 30 minutes after participants consumed one 6-ounce cup of coffee. Using the same driving simulation task, for which the distribution of reaction times is normal with � � 400 msec. and � � 40, they obtained a mean of M � 392 for a sample of n � 36 participants.

a. Are the data sufficient to conclude that caffeine has a significant effect on reaction time? Use a two-tailed test with a � .05.

b. Compute Cohen’s d to estimate the size of the effect. c. Write a sentence describing the outcome of the

hypothesis test and the measure of effect size as it would appear in a research report.

PROBLEMS 2 4 3

12. There is some evidence indicating that people with visible tattoos are viewed more negatively than people without visible tattoos (Resenhoeft, Villa, & Wiseman, 2008). In a similar study, a researcher first obtained overall ratings of attractiveness for a woman with no tattoos shown in a color photograph. On a 7-point scale, the woman received an average rating of � � 4.9, and the distribution of ratings was normal with a standard deviation of � � 0.84. The researcher then modified the photo by adding a tattoo of a butterfly on the woman’s left arm. The modified photo was then shown to a sample of n � 16 stu- dents at a local community college and the students used the same 7-point scale to rate the attractiveness of the woman. The average score for the photo with the tattoo was M � 4.2.

a. Do the data indicate a significant difference in rated attractiveness when the woman appeared to have a tattoo? Use a two-tailed test with a � .05.

b. Compute Cohen’s d to measure the size of the effect.

c. Write a sentence describing the outcome of the hypothesis test and the measure of effect size as it would appear in a research report.

13. Researchers at a National Weather Center in the north- eastern United States recorded the number of 90° days each year since records first started in 1875. The num- bers form a normal shaped distribution with a mean of � � 9.6 and a standard deviation of � � 1.9. To see if the data showed any evidence of global warming, they also computed the mean number of 90° days for the most recent n � 4 years and obtained M � 11.85. Do the data indicate that the past four years have had significantly more 90° days than would be expected for a random sample from this population? Use a one-tailed test with a � .05.

14. Montarello and Martens (2005) found that fifth-grade students completed more mathematics problems cor- rectly when simple problems were mixed in with their regular math assignments. To further explore this phenomenon, suppose that a researcher selects a stan- dardized mathematics achievement test that produces a normal distribution of scores with a mean of � � 100 and a standard deviation of � � 18. The researcher modifies the test by inserting a set of very easy problems among the standardized questions, and gives the modified test to a sample of n � 36 students. If the average test score for the sample is M � 104, is this result sufficient to conclude that inserting the easy questions improves student performance? Use a one-tailed test with a � .01.

15. Researchers have noted a decline in cognitive func- tioning as people age (Bartus, 1990). However, the results from other research suggest that the antioxi- dants in foods such as blueberries can reduce and

even reverse these age-related declines, at least in laboratory rats (Joseph et al., 1999). Based on these results, one might theorize that the same antioxidants might also benefit elderly humans. Suppose a researcher is interested in testing this the- ory. The researcher obtains a sample of n � 16 adults who are older than 65, and gives each participant a daily dose of a blueberry supplement that is very high in antioxidants. After taking the supplement for 6 months, the participants are given a standardized cognitive skills test and produce a mean score of M � 50.2. For the general population of elderly adults, scores on the test average � � 45 and form a normal distribution with � � 9.

a. Can the researcher conclude that the supplement has a significant effect on cognitive skill? Use a two-tailed test with a � .05.

b. Compute Cohen’s d for this study. c. Write a sentence demonstrating how the outcome

of the hypothesis test and the measure of effect size would appear in a research report.

16. A researcher plans to conduct an experiment evalu- ating the effect of a treatment. A sample of n � 9 participants is selected and each person receives the treatment before being tested on a standardized dex- terity task. The treatment is expected to lower scores on the test by an average of 30 points. For the regular population, scores on the dexterity task form a nor- mal distribution with � � 240 and � � 30.

a. If the researcher uses a two-tailed test with a � .05, what is the power of the hypothesis test?

b. Again assuming a two-tailed test with a � .05, what is the power of the hypothesis test if the sample size is increased to n � 25?

17. A sample of n � 40 is selected from a normal popu- lation with � � 75 msec. and � � 12, and a treat- ment is administered to the sample. The treatment is expected to increase scores by an average of 4 msec.

a. If the treatment effect is evaluated with a two- tailed hypothesis test using a � .05, what is the power of the test?

b. What is the power of the test if the researcher uses a one-tailed test with a � .05?

18. Briefly explain how increasing sample size influ- ences each of the following. Assume that all other factors are held constant.

a. The size of the z-score in a hypothesis test. b. The size of Cohen’s d. c. The power of a hypothesis test.

19. Explain how the power of a hypothesis test is influ- enced by each of the following. Assume that all other factors are held constant.

a. Increasing the alpha level from .01 to .05. b. Changing from a one-tailed test to a two-tailed test.

2 4 4 CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING

20. A researcher is investigating the effectiveness of a new medication for lowering blood pressure for individuals with systolic pressure greater than 140. For this population, systolic scores average � � 160 with a standard deviation of � � 20, and the scores form a normal-shaped distribution. The researcher plans to select a sample of n � 25 individuals, and measure their systolic blood pressure after they take the medication for 60 days. If the researcher uses a two-tailed test with a � .05,

a. What is the power of the test if the medication has a 5-point effect?

b. What is the power of the test if the medication has a 10-point effect?

21. A researcher is evaluating the influence of a treatment using a sample selected from a normally distributed population with a mean of � � 80 and a standard deviation of � � 20. The researcher expects a 12-point treatment effect and plans to use a two-tailed hypothesis test with a � .05.

a. Compute the power of the test if the researcher uses a sample of n � 16 individuals.

b. Compute the power of the test if the researcher uses a sample of n � 25 individuals.

1. The ability to transform scores into z-scores to describe locations within a distribution and to standardize entire distributions (Chapter 5).

2. The ability to determine probabilities associated with individual scores selected from a distribution, especially for scores from normal distributions (Chapter 6).

3. The ability to transform sample means into z-scores and to determine the probabilities associated with sample means (Chapter 7).

4. The ability to use a sample mean to evaluate a hypoth- esis about an unknown population mean (Chapter 8).

The general goal of inferential statistics is to use the limited

information from a sample to answer general questions about

an unknown population. In Chapter 8, we introduced hy-

pothesis testing, one of the most commonly used inferential

procedures. The hypothesis test presented in Chapter 8 inte-

grates z-scores from Chapter 5, probability from Chapter 6,

and the distribution of sample means from Chapter 7 into

a single procedure that allows researchers to use a sample

from an unknown population to evaluate a hypothesis about

the population mean. The researcher first obtains a sample

from the unknown population and computes the sample

mean. The sample mean and a hypothesized value for the

population mean are then used to compute a z-score. If the

resulting z-score is a high-probability value, near the center

of the distribution of sample means, then the researcher

concludes that the sample data fit the hypothesis and the

decision is to fail to reject the hypothesis. On the other hand,

if the resulting z-score is a low-probability value, out in the

tails of the distribution of sample means, then the researcher

concludes that the sample data do not fit the hypothesis and

the decision is to reject the hypothesis.

REVIEW EXERCISES

1. Find each of the requested values for a population with a mean of � � 50 and a standard deviation of � � 20.

a. What is the z-score corresponding to X � 52? b. What is the X value corresponding to z � –0.50?

c. If all of the scores in the population are trans- formed into z-scores, what will be the values for the mean and standard deviation for the complete set of z-scores?

d. What is the z-score corresponding to a sample mean of M � 42 for a sample of n � 4 scores?

e. What is the z-score corresponding to a sample mean of M � 42 for a sample of n � 16 scores?

2. A survey of high school seniors shows that the aver- age wake-up time the previous Saturday morning was � � 9:45. Assume that the distribution of times is approximately normal with a standard deviation of � � 65 minutes, and find each of the requested values.

a. What proportion of high school seniors wake up later than 11:00?

b. What is the probability of randomly selecting a high school senior who woke up before 9:00?

c. What is the probability of obtaining a mean wake- up time earlier than M � 9:30 for a sample of n � 25 high school students?

3. Miller (2008) examined the energy drink consumption of college undergraduates and found that males use energy drinks significantly more often than females. To further investigate this phenomenon, suppose that a researcher selects a random sample of n � 36 male undergraduates and a sample of n � 25 females. On average, the males reported consuming M � 2.45 drinks per month and females had an average of M � 1.28. Assume that the overall level of consumption for college undergraduates averages � � 1.85 energy drinks per month, and that the distribution of monthly consumption scores is approxi- mately normal with a standard deviation of � � 1.2.

a. Did this sample of males consume significantly more energy drinks than the overall population average? Use a one-tailed test with a � .01.

b. Did this sample of females consume significantly fewer energy drinks than the overall population average? Use a one-tailed test with a � .01.

REVIEW

After completing this part, you should understand the basic procedures that form the foundation

of inferential statistics. These include:

2 4 5

P A R T II

247

Chapter 9 Introduction to the t Statistic 249

Chapter 10 The t Test for Two Independent Samples 279

Chapter 11 The t Test for Two Related Samples 313

I n Part II, we presented the foundation for inferential statistics. In

this part, we begin to introduce some of the inferential procedures

that are actually used in behavioral science research. Specifically,

we look at a family of t statistics that use sample means and mean

differences to draw inferences about the corresponding population

means and mean differences. The t statistics are all modeled after the

z-score for sample means that was introduced in Chapter 7 and used

for hypothesis testing in Chapter 8. However, the t statistics do not

require any prior knowledge about the population being evaluated.

The three t statistics introduced in this part apply to three distinct

research situations:

1. Using a single sample to draw an inference about the un-

known mean for a single population.

2. Using two separate samples to draw an inference about the

mean difference between two unknown populations.

3. Using one sample, with each individual tested in two different

treatment conditions, to draw an inference about the popula-

tion mean difference between the two conditions.

In addition to the hypothesis testing procedure introduced in

Chapter 8, this part introduces a new inferential technique known

as confidence intervals. Confidence intervals allow researchers to

use sample data to estimate population means or mean differences

by computing a range of values that is highly likely to contain the

unknown parameter.

Using t Statistics for Inferences About Population Means and Mean Differences

P A R T

III

Aplia for Essentials of Statistics for the Behavioral Sciences

After reading, go to “Resources” at the end of this chapter for

an introduction on how to use Aplia’s homework and learning

resources.

Introduction to the t Statistic

9.1 The t Statistic: An Alternative to z

9.2 Hypothesis Tests with the t Statistic

9.3 Measuring Effect Size for the t Statistic

9.4 Directional Hypotheses and One-Tailed Tests

Summary

Focus on Problem Solving

Demonstrations 9.1 and 9.2

Problems

C H A P T E R

9 Tools You Will Need The following items are considered essential background material for this chapter. If you doubt your knowledge of any of these items, you should review the appropriate chapter or section before proceeding.

• Sample variance and standard devia- tion (Chapter 4)

• Standard error (Chapter 7) • Hypothesis testing (Chapter 8)

2 5 0 CHAPTER 9 INTRODUCTION TO THE t STATISTIC

THE t STATISTIC: AN ALTERNATIVE TO z

In the previous chapter, we presented the statistical procedures that permit researchers

to use a sample mean to test hypotheses about an unknown population mean. These

statistical procedures were based on a few basic concepts, which we summarize as

follows:

1. A sample mean (M) is expected to approximate its population mean (�). This

permits us to use the sample mean to test a hypothesis about the population mean.

2. The standard error provides a measure of how well a sample mean approxi-

mates the population mean. Specifically, the standard error determines how

much difference is reasonable to expect between a sample mean (M) and the

population mean (�).

s 5 s

M M n n

or 2

3. To quantify our inferences about the population, we compare the obtained

sample mean (M) with the hypothesized population mean (m) by computing a

z-score test statistic.

z M

5 2m

s 5

obtained difference between data andd hypothesis

standard distance between anM dd m

The goal of the hypothesis test is to determine whether the obtained difference

between the data and the hypothesis is significantly greater than would be expected

by chance. When the z-scores form a normal distribution, we are able to use the unit

normal table (in Appendix B) to find the critical region for the hypothesis test.

The shortcoming of using a z-score for hypothesis testing is that the z-score formula

requires more information than is usually available. Specifically, a z-score requires that

we know the value of the population standard deviation (or variance), which is needed

to compute the standard error. In most situations, however, the standard deviation for

the population is not known. In fact, the whole reason for conducting a hypothesis test

is to gain knowledge about an unknown population. This situation appears to create a

paradox: You want to use a z-score to find out about an unknown population, but you

must know about the population before you can compute a z-score. Fortunately, there

is a relatively simple solution to this problem. When the variance for the population is

not known, we use the corresponding sample value in its place.

In Chapter 4, the sample variance was developed specifically to provide an unbiased

estimate of the corresponding population variance. Recall that the formulas for sample

variance and sample standard deviation are as follows:

sample variance 5 5 2

5s SS

sample standard deviation 5 5 2

5s SS

df1

9.1

T H E P R O B L E M W I T H z - S CO R E S

I N T R O D UC I N G T H E t STAT I ST I C

Remember that the expected

value of the distribution

of sample means is m, the

population mean.

SECTION 9.1 / THE t STATISTIC: AN ALTERNATIVE TO z 2 5 1

Using the sample values, we can now estimate the standard error. Recall from

Chapters 7 and 8 that the value of the standard error can be computed using either

standard deviation or variance:

standard error or5 s 5 s

s 5 s

M M n n

Now we estimate the standard error by simply substituting the sample variance or

standard deviation in place of the unknown population value:

estimated standard error or5 5s s

n M

s s

n M

(9.1)

Notice that the symbol for the estimated standard error of M is s M instead of s

M ,

indicating that the estimated value is computed from sample data rather than from the

actual population parameter.

The estimated standard error (s M ) is used as an estimate of the real standard

error, s M , when the value of s is unknown. It is computed using the sample

variance or sample standard deviation and provides an estimate of the standard

distance between a sample mean, M, and the population mean, m.

Finally, you should recognize that we have shown formulas for standard error

(actual or estimated) using both the standard deviation and the variance. In the past

(Chapters 7 and 8), we concentrated on the formula using the standard deviation.

At this point, however, we shift our focus to the formula based on variance. Thus,

throughout the remainder of this chapter, and in following chapters, the estimated

standard error of M typically is presented and computed using

s s

n M

There are two reasons for making this shift from standard deviation to variance:

1. In Chapter 4 (p. 105), we saw that the sample variance is an unbiased statistic;

on average, the sample variance (s2) provides an accurate and unbiased

estimate of the population variance (s2). Therefore, the most accurate way

to estimate the standard error is to use the sample variance to estimate the

population variance.

2. In future chapters, we encounter other versions of the t statistic that require

variance (instead of standard deviation) in the formulas for estimated standard

error. To maximize the similarity from one version to another, we use variance

in the formula for all of the different t statistics. Thus, whenever we present a

t statistic, the estimated standard error is computed as

estimated standard error sample variance

sam 5

pple size

Now we can substitute the estimated standard error in the denominator of the z-score

formula. The result is a new test statistic called a t statistic:

t M

s M

5 2m

(9.2)

D E F I N I T I O N

The concept of degrees of

freedom, df 5 n – 1, was

introduced in Chapter 4

(p. 103) and is discussed

later in this chapter (p. 252).

2 5 2 CHAPTER 9 INTRODUCTION TO THE t STATISTIC

The t statistic is used to test hypotheses about an unknown population mean,

�, when the value of � is unknown. The formula for the t statistic has the same

structure as the z-score formula, except that the t statistic uses the estimated stan-

dard error in the denominator.

The only difference between the t formula and the z-score formula is that the z-score

uses the actual population variance, �2 (or the standard deviation), and the t formula

uses the corresponding sample variance (or standard deviation) when the population

value is not known.

z M M

n t

sM M �

2�

� �

2�

� �

2� �

2�

2 2 / // n

In this chapter, we have introduced the t statistic as a substitute for a z-score. The basic

difference between these two is that the t statistic uses sample variance (s2) and the

z-score uses the population variance (�2). To determine how well a t statistic approxi-

mates a z-score, we must determine how well the sample variance approximates the

population variance.

In Chapter 4, we introduced the concept of degrees of freedom (p. 103). Reviewing

briefly, you must know the sample mean before you can compute sample variance. This

places a restriction on sample variability such that only n – 1 scores in a sample are

independent and free to vary. The value n – 1 is called the degrees of freedom (or df)

for the sample variance.

degrees of freedom � df � n – 1 (9.3)

Degrees of freedom describe the number of scores in a sample that are indepen-

dent and free to vary. Because the sample mean places a restriction on the value

of one score in the sample, there are n – 1 degrees of freedom for a sample with

n scores (see Chapter 4).

As the value of df for a sample increases, the better the sample variance, s2, represents

the population variance, �2, and the better the t statistic approximates the z-score. This

should make sense because the larger the sample (n) is, the better the sample represents

its population. Thus, the degrees of freedom associated with s2 also describe how well

t represents z.

Every sample from a population can be used to compute a z-score or a t statistic. If

you select all of the possible samples of a particular size (n), and compute the z-score

for each sample mean, then the entire set of z-scores form a z-score distribution. In

the same way, you can compute the t statistic for every sample and the entire set of

t values form a t distribution. As we saw in Chapter 7, the distribution of z-scores for

sample means tends to be a normal distribution. Specifically, if the sample size is large

(around n � 30 or more) or if the sample is selected from a normal population, then

the distribution of sample means is a nearly perfect normal distribution. In these same

situations, the t distribution approximates a normal distribution, just as a t statistic

approximates a z-score. How well a t distribution approximates a normal distribution

is determined by degrees of freedom. In general, as the sample size (n) increases, the

degrees of freedom (n – 1) also increase, and the better the t distribution approximates

the normal distribution (Figure 9.1).

D E F I N I T I O N

D E G R E E S O F F R E E D O M A N D T H E t STAT I ST I C

D E F I N I T I O N

T H E t D I ST R I B U T I O N

SECTION 9.1 / THE t STATISTIC: AN ALTERNATIVE TO z 2 5 3

A t distribution is the complete set of t values computed for every possible

random sample for a specific sample size (n) or a specific degrees of freedom

(df). The t distribution approximates the shape of a normal distribution,

especially for large samples or samples from a normal population.

The exact shape of a t distribution changes with degrees of freedom. In fact, statisti-

cians speak of a “family” of t distributions. That is, there is a different sampling distri-

bution of t for each possible number of degrees of freedom. As df gets very large, the

t distribution gets closer in shape to a normal z-score distribution. Figure 9.1 shows that

distributions of t are bell-shaped and symmetrical and have a mean of zero. However,

the t distribution has more variability than a normal z distribution, especially when

df values are small (see Figure 9.1). The t distribution tends to be flatter and more

spread out, whereas the normal z distribution has more of a central peak.

The reason that the t distribution is flatter and more variable than the normal

z-score distribution becomes clear if you look at the structure of the formulas for

z and t. For both formulas, z and t, the top of the formula, M – �, can take on dif-

ferent values because the sample mean (M) varies from one sample to another. For

z-scores, however, the bottom of the formula does not vary, provided that all of the

samples are the same size and are selected from the same population. Specifically,

all of the z-scores have the same standard error in the denominator, s 5 s M

2 / n ,

because the population variance and the sample size are the same for every sample.

For t statistics, on the other hand, the bottom of the formula varies from one sample

to another. Specifically, the sample variance (s2) changes from one sample to the

next, so the estimated standard error also varies, s s M

5 2

/ n .Thus, only the numer-

ator of the z-score formula varies, but both the numerator and the denominator of the

t statistic vary. As a result, t statistics are more variable than are z-scores, and the

t distribution is flatter and more spread out. As sample size and df increase, how-

ever, the variability in the t distribution decreases, and it more closely resembles

a normal distribution.

D E F I N I T I O N

T H E S H A P E O F T H E t D I ST R I B U T I O N

Normal distribution

t distribution, df � 20

t distribution, df � 5

FIGURE 9.1

Distributions of the

t statistic for different val-

ues of degrees of freedom

are compared to a normal

z-score distribution. Like

the normal distribution,

t distributions are bell-

shaped and symmetrical

and have a mean of zero.

However, t distributions

have more variability,

indicated by the flatter and

more spread-out shape.

The larger the value of

df is, the more closely the

t distribution approximates

a normal distribution.

2 5 4 CHAPTER 9 INTRODUCTION TO THE t STATISTIC

Just as we used the unit normal table to locate proportions associated with z-scores, we use

a t distribution table to find proportions for t statistics. The complete t distribution table is

presented in Appendix B, page 575, and a portion of this table is reproduced in Table 9.1.

The two rows at the top of the table show proportions of the t distribution contained in

either one or two tails, depending on which row is used. The first column of the table lists

degrees of freedom for the t statistic. Finally, the numbers in the body of the table are the

t values that mark the boundary between the tails and the rest of the t distribution.

For example, to find proportions for the t distribution with df � 3, you begin by

locating df � 3 in the first column. Then, use the top two rows of the table to identify a

proportion in either one tail or two tails. For example, to find the t value that separates

5% in the tail from the rest of the distribution, locate a proportion of 0.05 in the one-tail

row at the top of the table. When you line up the 0.05 proportion in the top row with

df � 3 in the first column, you find a value of t � 2.352 in the body of the table (see

highlight in Table 9.1). Because the t distribution is symmetrical, 5% of the distribution

is also located in the tail beyond t � –2.353 (see Figure 9.2). Finally, notice that a total

of 10% (or 0.10) is contained in the two tails beyond t � 62.353 (check the proportion

value in the “two-tails combined” row at the top of the table).

A close inspection of the t distribution table in Appendix B demonstrates a point

we made earlier: As the value for df increases, the t distribution becomes more similar

to a normal distribution. For example, examine the column containing t values for a

0.05 proportion in two tails. You will find that when df 5 1, the t values that separate

the extreme 5% (0.05) from the rest of the distribution are t 5 612.706. As you read

down the column, however, you will find that the critical t values become smaller and

smaller, ultimately reaching 61.96. You should recognize 61.96 as the z-score values

that separate the extreme 5% in a normal distribution. Thus, as df increases, the propor-

tions in a t distribution become more like the proportions in a normal distribution. When

the sample size (and degrees of freedom) is sufficiently large, the difference between a

t distribution and the normal distribution becomes negligible.

Caution: The t distribution table printed in this book has been abridged and does

not include entries for every possible df value. For example, the table lists t values for

df 5 40 and for df 5 60, but does not list any entries for df values between 40 and

60. Occasionally, you will encounter a situation in which your t statistic has a df value

that is not listed in the table. In these situations, you should look up the critical t for

both of the surrounding df values listed and then use the larger value for t. If, for ex-

ample, you have df 5 53 (not listed), look up the critical t value for both df 5 40 and

df 5 60 and then use the larger t value. If your sample t statistic is greater than the

larger value listed, then you can be certain that the data are in the critical region, and

you can confidently reject the null hypothesis.

D E T E R M I N I N G P R O P O R T I O N S

A N D P R O BA B I L I T I E S F O R t D I ST R I B U T I O N S

Proportion in One Tail

0.25 0.10 0.05 0.025 0.01 0.005

Proportion in Two Tails Combined

df 0.50 0.20 0.10 0.05 0.02 0.01

1 1.000 3.078 6.314 12.706 31.821 63.657

2 0.816 1.886 2.920 4.303 6.965 9.925

3 0.765 1.638 2.353 3.182 4.541 5.841

4 0.741 1.533 2.132 2.776 3.747 4.604

5 0.727 1.476 2.015 2.571 3.365 4.032

6 0.718 1.440 1.943 2.447 3.143 3.707

TABLE 9.1

A portion of the t-distribution

table. The numbers in the table

are the values of t that separate

the tail from the main body of the

distribution. Proportions for one

or two tails are listed at the top of

the table, and df values for t are

listed in the first column.

SECTION 9.2 / HYPOTHESIS TESTS WITH THE t STATISTIC 2 5 5

–2.353 0 2.3530

5% 5%

FIGURE 9.2

The t distribution with

df � 3. Note that 5%

of the distribution is

located in the tail

beyond t � 2.353. Also,

5% is in the tail beyond

t � –2.353. Thus, a total

proportion of 10% (0.10)

is in the two tails beyond

t � 62.353.

1. Under what circumstances is a t statistic used instead of a z-score for a hypothesis

test?

2. A sample of n 5 9 scores has SS 5 288.

a. Compute the variance for the sample.

b. Compute the estimated standard error for the sample mean.

3. In general, a distribution of t statistics is flatter and more spread out than the stan-

dard normal distribution. (True or false?)

4. A researcher reports a t statistic with df 5 20. How many individuals participated

in the study?

5. For df 5 15, find the value(s) of t associated with each of the following:

a. The top 5% of the distribution.

b. The middle 95% of the distribution.

c. The middle 99% of the distribution.

1. A t statistic is used instead of a z-score when the population standard deviation and variance

are not known.

2. a. s2 5 36 b. s M 5 2

3. True.

4. n 5 21

5. a. t 5 11.753 b. t 5 62.131 c. t 5 62.947

L E A R N I N G C H E C K

ANSWERS

HYPOTHESIS TESTS WITH THE t STATISTIC

In the hypothesis-testing situation, we begin with a population with an unknown

mean and an unknown variance, often a population that has received some treat-

ment (Figure 9.3). The goal is to use a sample from the treated population (a treated

sample) as the basis for determining whether the treatment has any effect.

9.2

2 5 6 CHAPTER 9 INTRODUCTION TO THE t STATISTIC

As always, the null hypothesis states that the treatment has no effect; specifically,

H 0 states that the population mean is unchanged. Thus, the null hypothesis provides a

specific value for the unknown population mean. The sample data provide a value for

the sample mean. Finally, the variance and estimated standard error are computed from

the sample data. When these values are used in the t formula, the result becomes

t �

2sample mean population mean

(from the daata) (hypothesized from

estimated

H 0 )

standard error

(computed from the sample daata)

As with the z-score formula, the t statistic forms a ratio. The numerator measures the

actual difference between the sample data (M) and the population hypothesis (�). The

estimated standard error in the denominator measures how much difference is reason-

able to expect between a sample mean and the population mean. When the obtained

difference between the data and the hypothesis (numerator) is much greater than

expected (denominator), we obtain a large value for t (either large positive or large

negative). In this case, we conclude that the data are not consistent with the hypothesis,

and our decision is to “reject H 0 .” On the other hand, when the difference between the

data and the hypothesis is small relative to the standard error, we obtain a t statistic near

zero, and our decision is “fail to reject H 0 .”

The unknown population As mentioned earlier, the hypothesis test often concerns

a population that has received a treatment. This situation is shown in Figure 9.3. Note

that the value of the mean is known for the population before treatment. The question

is whether the treatment influences the scores and causes the mean to change. In this

case, the unknown population is the one that exists after the treatment is administered,

and the null hypothesis simply states that the value of the mean is not changed by the

treatment.

Although the t statistic can be used in the “before and after” type of research shown

in Figure 9.3, it also permits hypothesis testing in situations for which you do not have

µ = 30 µ = ?

Known population before treatment

Unknown population after treatment

T r e a t

m e n t

FIGURE 9.3

The basic experimental

situation for using the

t statistic or the z-score is

presented. It is assumed

that the parameter � is

known for the population

before treatment. The

purpose of the experiment

is to determine whether

the treatment has an

effect. Note that the

population after treatment

has unknown values for

the mean and the variance.

We will use a sample to

test a hypothesis about

the population mean.

SECTION 9.2 / HYPOTHESIS TESTS WITH THE t STATISTIC 2 5 7

a known population mean to serve as a standard. Specifically, the t test does not require

any prior knowledge about the population mean or the population variance. All you

need to compute a t statistic is a null hypothesis and a sample from the unknown popu-

lation. Thus, a t test can be used in situations for which the null hypothesis is obtained

from a theory, a logical prediction, or just wishful thinking. For example, many surveys

contain rating-scale questions to determine how people feel about controversial issues.

Participants are presented with a statement and asked to express their opinion on a scale

from 1 to 7, with 1 indicating “strongly agree” and 7 indicating “strongly disagree.” A

score of 4 indicates a neutral position, with no strong opinion one way or the other. In

this situation, the null hypothesis would state that there is no preference in the popula-

tion, H 0 : � � 4. The data from a sample are then used to evaluate the hypothesis. Note

that the researcher has no prior knowledge about the population mean and states a

hypothesis that is based on logic.

The following research situation demonstrates the procedures of hypothesis testing with

the t statistic. Note that this is another example of a null hypothesis that is founded in

logic rather than prior knowledge of a population mean.

Infants, even newborns, prefer to look at attractive faces compared to less attractive

faces (Slater, et al., 1998). In the study, infants from 1 to 6 days old were shown two

photographs of women’s faces. Previously, a group of adults had rated one of the faces

as significantly more attractive than the other. The babies were positioned in front of

a screen on which the photographs were presented. The pair of faces remained on the

screen until the baby accumulated a total of 20 seconds of looking at one or the other.

The number of seconds looking at the attractive face was recorded for each infant.

Suppose that the study used a sample of n � 9 infants and the data produced an average

of M � 13 seconds for the attractive face with SS � 72. Note that all of the available

information comes from the sample. Specifically, we do not know the population mean

or the population standard deviation.

Step 1 State the hypotheses and select an alpha level. Although we have no informa-

tion about the population of scores, it is possible to form a logical hypothesis about the

value of �. In this case, the null hypothesis states that the infants have no preference

for either face. That is, they should average half of the 20 seconds looking at each of

the two faces. In symbols, the null hypothesis states

H 0 : �

attractive � 10 seconds

The alternative hypothesis states that there is a preference and one of the faces is

preferred over the other. A directional, one-tailed test would specify which of the two

faces is preferred, but the nondirectional alternative hypothesis is expressed as follows:

H 1 : �

attractive  10 seconds

We set the level of significance at a � .05 for two tails.

Step 2 Locate the critical region. The test statistic is a t statistic because the popu-

lation variance is not known. Therefore, the value for degrees of freedom must be

determined before the critical region can be located. For this sample

df � n – 1 � 9 – 1 � 8

H Y P OT H E S I S T E ST I N G E X A M P L E

E X A M P L E 9 . 1

2 5 8 CHAPTER 9 INTRODUCTION TO THE t STATISTIC

Consulting the t distribution table or a two-tailed test with a � .05 and df � 8, you should find that the critical region consists of t values greater than �2.306 or

less than 22.306. Figure 9.4 depicts the critical region in this t distribution.

Step 3 Calculate the test statistic. The t statistic typically requires much more com-

putation than is necessary for a z-score. Therefore, we recommend that you divide the

calculations into a three-stage process as follows:

a. First, calculate the sample variance. Remember that the population variance is

unknown, and you must use the sample value in its place. (This is why we are

using a t statistic instead of a z-score.)

s SS

8 9�

� � �

b. Next, use the sample variance (s2) and the sample size (n) to compute the

estimated standard error. This value is the denominator of the t statistic

and measures how much difference is reasonable to expect by chance

between a sample mean and the corresponding population mean if there is

no treatment effect.

s s

n M

� � � �

2 9

9 1 1

Finally, compute the t statistic for the sample data.

t M

s M

� 2�

� 2

� 13 10

1 3 00.

Step 4 Make a decision regarding H 0 . The obtained t statistic of 3.00 falls into the critical

region on the right-hand side of the t distribution (see Figure 9.4). Our statistical decision is

to reject H 0 and conclude that babies do show a preference when given a choice between an

attractive and an unattractive face. Specifically, the average amount of time that the babies

spent looking at the attractive face was significantly different from the 10 seconds that

would be expected if there were no preference. As indicated by the sample mean, there is

a tendency for the babies to spend more time looking at the attractive face.

Reject H 0 Reject H 0

d f = 8

Fail to reject H 0

+2.306–2.306

FIGURE 9.4

The critical region in the

t distribution for a � .05

and df � 8.

SECTION 9.2 / HYPOTHESIS TESTS WITH THE t STATISTIC 2 5 9

Two basic assumptions are necessary for hypothesis tests with the t statistic.

1. The values in the sample must consist of independent observations.

In everyday terms, two observations are independent if there is no consis-

tent, predictable relationship between the first observation and the second.

More precisely, two events (or observations) are independent if the occur-

rence of the first event has no effect on the probability of the second event.

We examined specific examples of independence and non-independence in

Box 8.1 (p. 220).

2. The population that is sampled must be normal.

This assumption is a necessary part of the mathematics underlying the de-

velopment of the t statistic and the t distribution table. However, violating this

assumption has little practical effect on the results obtained for a t statistic,

especially when the sample size is relatively large. With very small samples, a

normal population distribution is important. With larger samples, this assump-

tion can be violated without affecting the validity of the hypothesis test. If you

have reason to suspect that the population distribution is not normal, use a large

sample to be safe.

As we noted in Chapter 8 (p. 222), a variety of factors can influence the outcome of

a hypothesis test. In particular, the number of scores in the sample and the magnitude

of the sample variance both have a large effect on the t statistic and, thereby, influence

the statistical decision. The structure of the t formula makes these factors easier to

understand.

t M

s s

n M

M �

2� �where

Because the estimated standard error, s M , appears in the denominator of the

formula, a larger value for s M produces a smaller value (closer to zero) for t. Thus,

any factor that influences the standard error also affects the likelihood of reject-

ing the null hypothesis and finding a significant treatment effect. The two factors

that determine the size of the standard error are the sample variance, s2, and the

sample size, n.

The estimated standard error is directly related to the sample variance so that the

larger the variance, the larger the error. Thus, large variance means that you are less

likely to obtain a significant treatment effect. In general, large variance is bad for infer-

ential statistics. Large variance means that the scores are widely scattered, which makes

it difficult to see any consistent patterns or trends in the data. In general, high variance

reduces the likelihood of rejecting the null hypothesis.

On the other hand, the estimated standard error is inversely related to the num-

ber of scores in the sample. The larger the sample is, the smaller the error is. If all

other factors are held constant, large samples tend to produce bigger t statistics and

therefore are more likely to produce significant results. For example, a 2-point mean

difference with a sample of n � 4 may not be convincing evidence of a treatment

effect. However, the same 2-point difference with a sample of n � 100 is much more

compelling.

AS S U M P T I O N S O F T H E t T E ST

T H E I N F L U E N C E O F SA M P L E S I Z E

A N D SA M P L E VA R I A N C E

2 6 0 CHAPTER 9 INTRODUCTION TO THE t STATISTIC

MEASURING EFFECT SIZE FOR THE t STATISTIC

In Chapter 8, we noted that one criticism of a hypothesis test is that it does not really

evaluate the size of the treatment effect. Instead, a hypothesis test simply determines

whether the treatment effect is greater than chance, where “chance” is determined by

the standard error and the alpha level. In particular, it is possible for a very small treat-

ment effect to be “statistically significant,” especially when the sample size is very

large. To correct for this problem, it is recommended that the results from a hypothesis

test be accompanied by a report of effect size, such as Cohen’s d.

When Cohen’s d was originally introduced (p. 230), the formula was presented as

Cohen's mean difference

standard deviation d � ��

� 2�

�

treatment no treatment

Cohen defined this measure of effect size in terms of the population mean differ-

ence and the population standard deviation. However, in most situations the population

values are not known and you must substitute the corresponding sample values in their

place. When this is done, many researchers prefer to identify the calculated value as

an “estimated d” or name the value after one of the statisticians who first substituted

sample statistics into Cohen’s formula (e.g., Glass’s g or Hedges’s g). For hypothesis

tests using the t statistic, the population mean with no treatment is the value specified

by the null hypothesis. However, the population mean with treatment and the standard

deviation are both unknown. Therefore, we use the mean for the treated sample and the

standard deviation for the sample after treatment as estimates of the unknown param-

eters. With these substitutions, the formula for estimating Cohen’s d becomes

estimated mean difference

sample standard d �

ddeviation �

2�M

(9.4)

The numerator measures that magnitude of the treatment effect by finding the differ-

ence between the mean for the treated sample and the mean for the untreated population

(� from H 0 ). The sample standard deviation in the denominator standardizes the mean

9.3

E ST I M AT E D CO H E N ’ S d

1. A sample of n � 4 individuals is selected from a population with a mean of

� � 40. A treatment is administered to the individuals in the sample and, after

treatment, the sample has a mean of M � 44 and a variance of s2 � 16.

a. Is this sample sufficient to conclude that the treatment has a significant effect?

Use a two-tailed test with a � .05.

b. If all other factors are held constant and the sample size is increased to n � 16,

is the sample sufficient to conclude that the treatment has a significant effect?

Again, use a two-tailed test with a � .05.

1. a. H 0 : � � 40 even after the treatment. With n � 4, the estimated standard error is 2, and

t � 4 2 � 2.00. With df � 3, the critical boundaries are set at t � 63.182. Fail to reject

H 0 and conclude that the treatment does not have a significant effect.

b. With n 5 16, the estimated standard error is 1 and t 5 4.00. With df 5 15, the critical

boundary is 62.131. The t value is beyond the critical boundary, so we reject H 0 and

conclude that the treatment does have a significant effect.

L E A R N I N G C H E C K

ANSWERS

SECTION 9.3 / MEASURING EFFECT SIZE FOR THE t STATISTIC 2 6 1

difference into standard deviation units. Thus, an estimated d of 1.00 indicates that the

size of the treatment effect is equivalent to one standard deviation. The following ex-

ample demonstrates how the estimated d is used to measure effect size for a hypothesis

test using a t statistic.

For the infant face-preference study in Example 9.1, the babies averaged M � 13 out

of 20 seconds looking at the attractive face. If there were no preference (as stated by

the null hypothesis), the population mean would be � � 10 seconds. Thus, the results

show a 3-second difference between the mean for the sample (M � 13) and the mean

that would be expected if there were no preference between the two faces (� � 10).

Also, for this study the sample standard deviation is

s SS

df � � � �

8 9 3

Thus, Cohen’s d for this example is estimated to be

Cohen's d M

s �

2� �

2 �

13 10

3 1 00.

According to the standards suggested by Cohen (Table 8.2, p. 232), this is a large

treatment effect.

To help you visualize what is measured by Cohen’s d, we have constructed a set of

n � 9 scores with a mean of M � 13 and a standard deviation of s � 3 (the same values

as in Examples 9.1 and 9.2). The set of scores is shown in Figure 9.5. Notice that the

figure also includes an arrow that locates � � 10. Recall that � � 10 is the value speci-

fied by the null hypothesis and identifies what the mean ought to be if the treatment has

no effect. Clearly, our sample is not centered around � � 10. Instead, the scores have

been shifted to the right so that the sample mean is M � 13. This shift, from 10 to 13,

is the 3-point mean difference that was caused by the treatment effect. Also notice that

E X A M P L E 9 . 2

Time spent looking at the attractive face (in seconds)

F re

q u

e n

c y

11 12 14 16 17 188 10 13 159

M 5 13

s 5 3

� � 10 (from H0)

s � 3

FIGURE 9.5

The sample distribution

for the scores that were

used in Examples 9.1 and

9.2. The population mean,

� � 10 seconds, is

the value that would be

expected if attractive-

ness has no effect on the

infants’ behavior. Note

that the sample mean is

displaced away from

� � 10 by a distance

equal to one standard

deviation.

2 6 2 CHAPTER 9 INTRODUCTION TO THE t STATISTIC

the 3-point mean difference is exactly equal to the standard deviation. Thus, the size of

the treatment effect is equal to 1 standard deviation. In other words, Cohen’s d � 1.00.

An alternative method for measuring effect size is to determine how much of the variability

in the scores is explained by the treatment effect. The concept behind this measure is that

the treatment causes the scores to increase (or decrease), which means that the treatment is

causing the scores to vary. If we can measure how much of the variability is explained by

the treatment, we can obtain a measure of the size of the treatment effect.

To demonstrate this concept, we use the data from the hypothesis test in Example 9.1.

Recall that the null hypothesis stated that the treatment (the attractiveness of the faces)

has no effect on the infants’ behavior. According to the null hypothesis, the infants

should show no preference between the two photographs, and therefore should spend

an average of � � 10 out of 20 seconds looking at the attractive face.

However, if you look at the data in Figure 9.5, the scores are not centered around

� � 10. Instead, the scores are shifted to the right so that they are centered around

the sample mean, M � 13. This shift is the treatment effect. To measure the size of

the treatment effect, we calculate deviations from the mean and the sum of squared

deviations, SS, in two different ways.

Figure 9.6(a) shows the original set of scores. For each score, the deviation from

� � 10 is shown as a colored line. Recall that � � 10 comes from the null hypothesis

and represents the population mean if the treatment has no effect. Note that almost all

of the scores are located on the right-hand side of � � 10. This shift to the right is

the treatment effect. Specifically, the preference for the attractive face has caused the

infants to spend more time looking at the attractive photograph, which means that their

M E AS U R I N G T H E P E R C E N TAG E

O F VA R I A N C E E X P L A I N E D, r 2

8 9 10 11 12 13 14 15 16 17 18

No effect µ � 10

5 6 7 8 9 10 11 12 13 14 15

No effect µ � 10

Original scores, including the treatment effect

Adjusted scores with the treatment effect removed

(a)

(b)

FIGURE 9.6

Deviations from � � 10

(no treatment effect) for

the scores in Example 9.1.

The colored lines in part

(a) show the deviations

for the original scores,

including the treatment

effect. In part (b) the

colored lines show the

deviations for the adjusted

scores after the treatment

effect has been removed.

SECTION 9.3 / MEASURING EFFECT SIZE FOR THE t STATISTIC 2 6 3

scores are generally greater than 10. Thus, the treatment has pushed the scores away

from � � 10 and has increased the size of the deviations.

Next, we see what happens if the treatment effect is removed. In this example, the

treatment has a 3-point effect (the average increases from � � 10 to M � 13). To

remove the treatment effect, we simply subtract 3 points from each score. The adjusted

scores are shown in Figure 9.6(b) and, once again, the deviations from � � 10 are

shown as colored lines. First, notice that the adjusted scores are centered at � � 10,

indicating that there is no treatment effect. Also notice that the deviations, the colored

lines, are noticeably smaller when the treatment effect is removed.

To measure how much the variability is reduced when the treatment effect is

removed, we compute the sum of squared deviations, SS, for each set of scores.

The left-hand columns of Table 9.2 show the calculations for the original scores

[Figure 9.6(a)], and the right-hand columns show the calculations for the adjusted

scores [Figure 9.6(b)]. Note that the total variability, including the treatment effect, is

SS � 153. However, when the treatment effect is removed, the variability is reduced to

SS � 72. The difference between these two values, 153 – 72 � 81 points, is the amount

of variability that is accounted for by the treatment effect. This value is usually reported

as a proportion or percentage of the total variability:

variability accounted for

total variability �

881

153 0 5294 52 94� . . %( )

Thus, removing the treatment effect reduces the variability by 52.94%. This value is

called the percentage of variance accounted for by the treatment and is identified as r2.

Rather than computing r2 directly by comparing two different calculations for SS, the

value can be found from a single equation based on the outcome of the t test.

r t

t df

2 2

2 �

�

(9.5)

The letter r is the traditional symbol used for a correlation, and the concept of r2 is

discussed again when we consider correlations in Chapter 14. Also, in the context of

t statistics, the percentage of variance that we are calling r2 is often identified by the

Greek letter omega, squared (2).

Calculation of SS including the treatment effect

Calculation of SS after the treatment effect is removed

Score Deviation

from � � 10 Squared

Deviation Adjusted

Score Deviation

from � � 10 Squared

Deviation

8 22 4 8 2 3 � 5 25 25

10 0 0 10 2 3 � 7 23 9

12 2 4 12 2 3 � 9 21 1

13 3 9 13 2 3 � 10 0 0

15 5 25 15 2 3 � 12 2 4

17 7 49 17 2 3 � 14 4 16

SS � 153 SS � 72

TABLE 9.2

Calculation of SS, the sum of

squared deviations, for the data

in Figure 9.6. The first three

columns show the calculations

for the original scores, includ-

ing the treatment effect. The last

three columns show the calcu-

lations for the adjusted scores

after the treatment effect has

been removed.

2 6 4 CHAPTER 9 INTRODUCTION TO THE t STATISTIC

For the hypothesis test in Example 9.1, we obtained t � 3.00 with df � 8. These

values produce

r 2

3 8

17 0 5294 52 94�

1 � � . . %( )

Note that this is exactly the same as the value we obtained with the direct calculation

of the percentage of variability accounted for by the treatment.

Interpreting r2 In addition to developing the Cohen’s d measure of effect size,

Cohen (1988) also proposed criteria for evaluating the size of a treatment effect that

is measured by r2. The criteria were actually suggested for evaluating the size of a

correlation, r, but are easily extended to apply to r2. Cohen’s standards for interpreting

r2 are shown in Table 9.3.

According to these standards, the data we constructed for Examples 9.1 and 9.2

show a very large effect size with r2 � 0.5294.

As a final note, we should remind you that, although sample size affects the hy-

pothesis test, this factor has little or no effect on measures of effect size. In particular,

estimates of Cohen’s d are not influenced at all by sample size, and measures of r2 are

only slightly affected by changes in the size of the sample. The sample variance, on the

other hand, influences hypothesis tests and measures of effect size. Specifically, high

variance reduces both the likelihood of rejecting the null hypothesis and measures of

effect size.

An alternative technique for describing the size of a treatment effect is to compute an

estimate of the population mean after treatment. For example, if the mean before treat-

ment is known to be � � 80 and the mean after treatment is estimated to be � � 86,

then we can conclude that the size of the treatment effect is around 6 points.

Estimating an unknown population mean involves constructing a confidence

interval. A confidence interval is based on the observation that a sample mean tends

to provide a reasonably accurate estimate of the population mean. The fact that a

sample mean tends to be near to the population mean implies that the population

mean should be near to the sample mean. For example, if we obtain a sample mean

of M � 86, we can be reasonably confident that the population mean is around 86.

Thus, a confidence interval consists of an interval of values around a sample mean,

and we can be reasonably confident that the unknown population mean is located

somewhere in the interval.

A confidence interval is an interval, or range of values, centered around a

sample statistic. The logic behind a confidence interval is that a sample statistic,

such as a sample mean, should be relatively near to the corresponding population

parameter. Therefore, we can confidently estimate that the value of the parameter

should be located in the interval.

CO N F I D E N C E I N T E R VA L S F O R

E ST I M AT I N G 

D E F I N I T I O N

Percentage of Variance Explained, r2

r2 � 0.01 Small effect

r2 � 0.09 Medium effect

r2 � 0.25 Large effect

TABLE 9.3

Criteria for interpreting the

value of r2 as proposed by

Cohen (1988).

SECTION 9.3 / MEASURING EFFECT SIZE FOR THE t STATISTIC 2 6 5

The construction of a confidence interval begins with the observation that every sample

mean has a corresponding t value defined by the equation

t M

s M

� 2�

Although the values for M and s M

are available from the sample data, we cannot use

the equation to calculate t because the value for � is unknown. Instead of calculating the

t value, for a confidence interval we estimate the t value. For example, if the sample has

n � 9 scores, then the t statistic has df � 8, and the distribution of all possible t values

can be pictured as in Figure 9.7. Notice that the t values pile up around t � 0, so we can

estimate that the t value for our sample should have a value around 0. Furthermore, the

t distribution table lists a variety of different t values that correspond to specific propor-

tions of the t distribution. With df � 8, for example, 80% of the t values are located

between t � 11.397 and t � –1.397. To obtain these values, simply look up a two-tailed

proportion of 0.20 (20%) for df � 8. Because 80% of all of the possible t values are

located between 61.397, we can be 80% confident that our sample mean corresponds to

a t value in this interval. Similarly, we can be 95% confident that the mean for a sample

of n � 9 scores corresponds to a t value between 12.306 and –2.306. Notice that we

are able to estimate the value of t with a specific level of confidence. To construct a

confidence interval for �, we plug the estimated t value into the t equation, and then we

can calculate the value of �.

Before we demonstrate the process of constructing a confidence interval for an

unknown population mean, we simplify the calculations by regrouping the terms in the

t equation. Because the goal is to compute the value of �, we use simple algebra to

solve the equation for �. The result is

� � M 6 ts M

(9.5)

The process of using this equation to construct a confidence interval is demonstrated

in the following example.

Example 9.1 describes a study in which infants displayed a preference for the more

attractive face by looking at it, instead of the less attractive face, for the majority of

a 20-second viewing period. Specifically, a sample of n � 9 infants spent an average

CO N ST R UC T I N G A CO N F I D E N C E

I N T E R VA L

E X A M P L E 9 . 3

t distribution df = 8

t = �1.397t = �1.397 t = 0

Middle 80% of t distribution

FIGURE 9.7

The distribution of t statis-

tics with df � 8. The

t values pile up around

t � 0 and 80% of all of the

possible values are located

between t � –1.397 and

t � 11.397.

2 6 6 CHAPTER 9 INTRODUCTION TO THE t STATISTIC

of M � 13 seconds out of a 20-second period looking at the more attractive face. The

data produced an estimated standard error of s M

� 1. We use this sample to construct a

confidence interval to estimate the mean amount of time that the population of infants

spends looking at the more attractive face. That is, we construct an interval of values

that is likely to contain the unknown population mean.

Again, the estimation formula is

� � M 6 t(s M )

In the equation, the value of M � 13 and s M � 1 are obtained from the sample data.

The next step is to select a level of confidence that determines the value of t in the

equation. The most commonly used confidence level is probably 95%, but values of

80%, 90%, and 99% are also common. For this example, we use a confidence level

of 80%, which means that we construct the confidence interval so that we are 80%

confident that the population mean is actually contained in the interval. Because we

are using a confidence level of 80%, the resulting interval is called the 80% confidence

interval for �.

To obtain the value for t in the equation, we simply estimate that the t statistic

for our sample is located somewhere in the middle 80% of the t distribution. With df

� n – 1 � 8, the middle 80% of the distribution is bounded by t values of �1.397

and –1.397 (see Figure 9.7). Using the sample data and the estimated range of

t values, we obtain

� � M 6 t(s M ) � 13 6 1.397(1.00) � 13 6 1.397

At one end of the interval, we obtain � � 13 � 1.397 � 14.397, and at the other end

we obtain � � 13 – 1.397 � 11.603. Our conclusion is that the average time looking

at the more attractive face for the population of infants is between � �11.603 seconds

and � � 14.397 seconds, and we are 80% confident that the true population mean is

located within this interval. The confidence comes from the fact that the calculation was

based on only one assumption. Specifically, we assumed that the t statistic was located

between �1.397 and –1.397, and we are 80% confident that this assumption is correct

because 80% of all of the possible t values are located in this interval. Finally, note that

the confidence interval is constructed around the sample mean. As a result, the sample

mean, M � 13, is located exactly in the center of the interval.

Two characteristics of the confidence interval should be mentioned. First, notice what

happens to the width of the interval when you change the level of confidence (the

percent confidence). To gain more confidence in your estimate, you must increase the

width of the interval. Conversely, to have a smaller, more precise interval, you must give

up confidence. In the estimation formula, the percentage of confidence influences the

value of t. A larger level of confidence (the percentage) produces a larger t value and a

wider interval. This relationship can be seen in Figure 9.7. In the figure, we identified

the middle 80% of the t distribution to find an 80% confidence interval. It should be

obvious that if we were to increase the confidence level to 95%, it would be necessary

to increase the range of t values, and thereby increase the width of the interval.

Second, note what happens to the width of the interval if you change the sample size.

This time the basic rule is as follows: The bigger the sample (n), the smaller the inter-

val. This relationship is straightforward if you consider the sample size as a measure

of the amount of information. A bigger sample gives you more information about the

population and allows you to make a more precise estimate (a narrower interval). The

sample size controls the magnitude of the standard error in the estimation formula. As

FAC TO R S A F F E C T I N G T H E W I D T H

O F A CO N F I D E N C E I N T E R VA L

To have 80% in the middle,

there must be 20% (or .20) in

the tails. To find the t values,

look under two tails, .20 in

the t table.

SECTION 9.3 / MEASURING EFFECT SIZE FOR THE t STATISTIC 2 6 7

the sample size increases, the standard error decreases, and the interval gets smaller.

Finally, we should also note that the width of the confidence interval is also related to

the sample variance. The variance contributes directly to the magnitude of the standard

error so that an increase in variance produces an increase in the standard error and,

thereby, increases the width of the confidence interval.

Because confidence intervals are influenced by sample size, they do not provide

an unqualified measure of absolute effect size and are not an adequate substitute for

Cohen’s d or r2. Nonetheless, they can be used in a research report to provide a descrip-

tion of the size of the treatment effect.

1. If all other factors are held constant, an 80% confidence interval is wider than a

90% confidence interval. (True or false?)

2. If all other factors are held constant, a confidence interval computed from a sample

of n � 25 is wider than a confidence interval computed from a sample of n � 100.

(True or false?)

1. False. Greater confidence requires a wider interval.

2. True. The smaller sample produces a wider interval.

L E A R N I N G C H E C K

ANSWERS

IN THE LITERATURE

REPORTING THE RESULTS OF A t TEST

In Chapter 8, we noted the conventional style for reporting the results of a hypoth-

esis test, according to APA format. First, recall that a scientific report typically

uses the term significant to indicate that the null hypothesis has been rejected and

the term not significant to indicate failure to reject H 0 . Additionally, there is a

prescribed format for reporting the calculated value of the test statistic, degrees of

freedom, and alpha level for a t test. This format parallels the style introduced in

Chapter 8 (p. 218).

In Example 9.1 we calculated a t statistic of 3.00 with df � 8, and we decided to

reject H 0 with alpha set at .05. Using the same data, we obtained r2 � 0.5294 (52.94%)

for the percentage of variance explained by the treatment effect. In a scientific report,

this information is conveyed in a concise statement, as follows:

The infants spent an average of M � 13 out of 20 seconds looking at the attractive

face, with SD � 3.00. Statistical analysis indicates that the time spent looking at

the attractive face was significantly greater than would be expected if there were no

preference, t(8) � 3.00, p , .05, r2 � 0.5294.

The first statement reports the descriptive statistics, the mean (M � 13) and

the standard deviation (SD � 3), as previously described (Chapter 4, p. 109). The

next statement provides the results of the inferential statistical analysis. Note that

the degrees of freedom are reported in parentheses immediately after the symbol

t. The value for the obtained t statistic follows (3.00), and next is the probability

of committing a Type I error (less than 5%). Finally, the effect size is reported,

r2 � 52.94%. If the 80% confidence interval from Example 9.3 were included in

the report as a description of effect size, it would be added after the results of the

hypothesis test as follows:

t(8) � 3.00, p , .05, 80% CI [11.603, 14.397].

2 6 8 CHAPTER 9 INTRODUCTION TO THE t STATISTIC

Often, researchers use a computer to perform a hypothesis test like the one in

Example 9.1. In addition to calculating the mean, standard deviation, and the t statis-

tic for the data, the computer usually calculates and reports the exact probability (or

a level) associated with the t value. In Example 9.1, we determined that any t value

beyond 62.306 has a probability of less than .05 (see Figure 9.4). Thus, the obtained

t value, t � 3.00, is reported as being very unlikely, p , .05. A computer printout, how-

ever, would have included an exact probability for our specific t value.

Whenever a specific probability value is available, you are encouraged to use

it in a research report. For example, the computer analysis of these data reports

an exact p value of p � .017, and the research report would state “t(8) � 3.00,

p � .017” instead of using the less specific “p , .05.” As one final caution, we

note that occasionally a t value is so extreme that the computer reports p � 0.000.

The zero value does not mean that the probability is literally zero; instead, it means

that the computer has rounded off the probability value to three decimal places and

obtained a result of 0.000. In this situation, you do not know the exact probability

value, but you can report p , .001.

1. A sample of n � 16 individuals is selected from a population with a mean of

� � 80. A treatment is administered to the sample and, after treatment, the sample

mean is found to be M � 86 with a standard deviation of s � 8.

a. Does the sample provide sufficient evidence to conclude that the treatment has

a significant effect? Test with a � .05.

b. Compute Cohen’s d and r2 to measure the effect size.

c. Find the 95% confidence interval for the population mean after treatment.

2. How does sample size influence the outcome of a hypothesis test and measures of

effect size? How does the standard deviation influence the outcome of a hypothesis

test and measures of effect size?

1. a. The estimated standard error is 2 points and the data produce t � 6 2 � 3.00. With

df � 15, the critical values are t � 62.131, so the decision is to reject H 0 and conclude

that there is a significant treatment effect.

b. For these data, d � 6 8

� 0.75 and r2 � 9 24 � 0.375 or 37.5%.

c. For 95% confidence and df � 15, use t � 62.131. The confidence interval is � � 86

62.131(2) and extends from 81.738 to 90.262.

2. Increasing sample size increases the likelihood of rejecting the null hypothesis but has little

or no effect on measures of effect size. Increasing the sample variance reduces the likeli-

hood of rejecting the null hypothesis and reduces measures of effect size.

L E A R N I N G C H E C K

ANSWERS

The statement p , .05 was

explained in Chapter 8,

page 218.

DIRECTIONAL HYPOTHESES AND ONE-TAILED TESTS

As noted in Chapter 8, the nondirectional (two-tailed) test is more commonly used than

the directional (one-tailed) alternative. On the other hand, a directional test may be used

in some research situations, such as exploratory investigations or pilot studies or when

there is a priori justification (for example, a theory or previous findings). The follow-

ing example demonstrates a directional hypothesis test with a t statistic, using the same

experimental situation presented in Example 9.1.

9.4

SECTION 9.4 / DIRECTIONAL HYPOTHESES AND ONE-TAILED TESTS 2 6 9

The research question is whether attractiveness affects the behavior of infants looking

at photographs of women’s faces. The researcher is expecting the infants to prefer the

more attractive face. Therefore, the researcher predicts that the infants will spend more

than half of the 20-second period looking at the attractive face. For this example, we

use the same sample data that were used in the original hypothesis test in Example 9.1.

Specifically, the researcher tested a sample of n � 9 infants and obtained a mean of

M � 13 seconds looking at the attractive face with SS � 72.

Step 1 State the hypotheses, and select an alpha level. With most directional tests, it is

usually easier to state the hypothesis in words, including the directional prediction, and

then convert the words into symbols. For this example, the researcher is predicting that

attractiveness will cause the infants to increase the amount of time they spend looking

at the attractive face; that is, more than half of the 20 seconds should be spent looking

at the attractive face. In general, the null hypothesis states that the predicted effect will

not happen. For this study, the null hypothesis states that the infants will not spend more

than half of the 20 seconds looking at the attractive face. In symbols,

H 0 : �

attractive  10 seconds (Not more than half of the 20 seconds looking

at the attractive face)

Similarly, the alternative states that the treatment will work. In this case, H 1 states that

the infants will spend more than half of the time looking at the attractive face. In symbols,

H 1 : �

attractive . 10 seconds (More than half of the 20 seconds looking at

the attractive face)

This time, we set the level of significance at a � .01.

Step 2 Locate the critical region. In this example, the researcher is predicting that the

sample mean (M) will be greater than 10 seconds. Thus, if the infants average more than

10 seconds looking at the attractive face, the data will provide support for the researcher’s

prediction and will tend to refute the null hypothesis. Also note that a sample mean greater

than 10 will produce a positive value for the t statistic. Thus, the critical region for the

one-tailed test will consist of positive t values located in the right-hand tail of the distribu-

tion. To find the critical value, you must look in the t distribution table using the one-tail

proportions. With a sample of n � 9, the t statistic has df � 8; using a � .01, you should

find a critical value of t � 2.896. Therefore, if we obtain a t statistic greater than 2.896,

we will reject the null hypothesis and conclude that the infants show a significant prefer-

ence for the attractive face. Figure 9.8 shows the one-tailed critical region for this test.

Step 3 Calculate the test statistic. The computation of the t statistic is the same for

either a one-tailed or a two-tailed test. Earlier (in Example 9.1), we found that the data

for this experiment produce a test statistic of t � 3.00.

Step 4 Make a decision. The test statistic is in the critical region, so we reject H 0 . In

terms of the experimental variables, we have decided that the infants show a preference

and spend significantly more time looking at the attractive face than they do looking

at the unattractive face. In a research report, the results would be presented as follows:

The time spent looking at the attractive face was significantly greater than would be

expected if there were no preference, t(8) � 3.00, p , .01, one tailed.

Note that the report clearly acknowledges that a one-tailed test was used.

E X A M P L E 9 . 4

2 7 0 CHAPTER 9 INTRODUCTION TO THE t STATISTIC

In step 2 of Example 9.4, we determined that the critical region is in the right-hand tail of the

distribution. However, it is possible to divide this step into two stages that eliminate the need

to determine which tail (right or left) should contain the critical region. The first stage in this

process is simply to determine whether the sample mean is in the direction predicted by the

original research question. For this example, the researcher predicted that the infants would

prefer the attractive face and spend more time looking at it. Specifically, the researcher

expects the infants to spend more than 10 out of 20 seconds focused on the attractive face.

The obtained sample mean, M � 13 seconds, is in the correct direction. This first stage

eliminates the need to determine whether the critical region is in the left- or right-hand tail.

Because we already have determined that the effect is in the correct direction, the sign of

the t statistic (1 or –) no longer matters. The second stage of the process is to determine

whether the effect is large enough to be significant. For this example, the requirement is

that the sample produces a t statistic greater than 2.896. If the magnitude of the t statistic,

independent of its sign, is greater than 2.896, then the result is significant and H 0 is rejected.

T H E C R I T I CA L R E G I O N F O R A O N E - TA I L E D T E ST

t distribution df = 8

t = �2.896 t = 0

Reject H 0

FIGURE 9.8

The one-tailed critical

region for the hypothesis

test in Example 9.4 with

df � 8 and a � .01.

1. A new over-the-counter cold medication includes a warning label stating that

it “may cause drowsiness.” A researcher would like to evaluate this effect. It is

known that under regular circumstances the distribution of reaction times is normal

with � � 200. A sample of n � 9 participants is obtained. Each person is given

the new cold medication, and, 1 hour later, reaction time is measured for each

individual. The average reaction time for this sample is M � 206 with SS � 648.

The researcher would like to use a hypothesis test with a � .05 to evaluate the

effect of the medication.

a. Use a two-tailed test with a � .05 to determine whether the medication has

a significant effect on reaction time.

b. Write a sentence that demonstrates how the outcome of the hypothesis test

would appear in a research report.

c. Use a one-tailed test with a � .05 to determine whether the medication

produces a significant increase in reaction time.

d. Write a sentence that demonstrates how the outcome of the one-tailed

hypothesis test would appear in a research report.

L E A R N I N G C H E C K

SUMMARY 2 7 1

1. a. For the two-tailed test, H 0 : � � 200. The sample variance is 81, the estimated standard

error is 3, and t � 6 3 � 2.00. With df � 8, the critical boundaries are 6 2.306. Fail to

reject the null hypothesis.

b. The result indicates that the medication does not have a significant effect on reaction

time, t(8) � 2.00, p . .05.

c. For a one-tailed test, H 0 : �  200 (no increase). The data produce t �

6 3 � 2.00. With

df � 8, the critical boundary is 1.860. Reject the null hypothesis.

d. The results indicate that the medication produces a significant increase in reaction time,

t(8) � 2.00, p , .05, one tailed.

ANSWERS

SUMMARY

1. The t statistic is used instead of a z-score for hypoth- esis testing when the population standard deviation (or variance) is unknown.

2. To compute the t statistic, you must first calculate the sample variance (or standard deviation) as a substitute for the unknown population value.

sample variance � �s SS

Next, the standard error is estimated by using s2, instead of �2, in the formula for standard error. The estimated standard error is calculated in the following manner:

estimated standard error � �s s

n M

Finally, a t statistic is computed using the estimated standard error. The t statistic is used as a substitute for a z-score, which cannot be computed when the popu- lation variance or standard deviation is unknown.

t M

s M

� 2�

3. The structure of the t formula is similar to that of the z-score in that

z tor sample mean population mean

(estimate �

dd) standard error

For a hypothesis test, you hypothesize a value for the unknown population mean and plug the hypoth- esized value into the equation along with the sample mean and the estimated standard error, which are computed from the sample data. If the hypothesized

mean produces an extreme value for t, then you con- clude that the hypothesis was wrong.

4. There is a family of t distributions, with the exact shape of a particular distribution of t values depending on degrees of freedom (n – 1). Therefore, the critical t values depend on the value for df associated with the t test. As df increases, the shape of the t distribution approaches a normal distribution.

5. When a t statistic is used for a hypothesis test, Cohen’s d can be computed to measure effect size. In this situation, the sample standard deviation is used in the formula to obtain an estimated value for d:

estimated mean difference

standard deviati d �

oon �

2�M

6. A second measure of effect size is r2, which measures the percentage of the variability that is accounted for by the treatment effect. This value is computed as follows:

r t

t df

2 2

2 �

7. An alternative method for describing the size of a treatment effect is to use a confidence interval for �. A confidence interval is a range of values that estimates the unknown population mean. The confidence interval uses the t equation, solved for the unknown mean:

� � M 6 t(s M )

First, select a level of confidence and then look up the corresponding t values to use in the equation. For example, for 95% confidence, use the range of t values that determine the middle 95% of the distribution.

2 7 2 CHAPTER 9 INTRODUCTION TO THE t STATISTIC

KEY TERMS

estimated standard error (251)

t statistic (252)

degrees of freedom (252)

t distribution (252)

estimated d (260)

percentage of variance accounted

for by the treatment (r2) (263)

confidence interval (264)

RESOURCES

Go to CengageBrain.com to access Psychology CourseMate, where you will find an

interactive eBook, glossaries, flashcards, quizzes, statistics workshops, and more.

If your professor has assigned Aplia:

1. Sign in to your account.

2. Complete the corresponding exercises as required by your professor.

3. When finished, click “Grade It Now” to see which areas you have mastered, which

areas need more work, and detailed explanations of every answer.

General instructions for using SPSS are presented in Appendix D. Following are de-

tailed instructions for using SPSS to perform the t Test presented in this chapter.

Data Entry

Enter all of the scores from the sample in one column of the data editor, probably

VAR00001.

Data Analysis

1. Click Analyze on the tool bar, select Compare Means, and click on

One-Sample t Test.

2. Highlight the column label for the set of scores (VAR0001) in the left box and

click the arrow to move it into the Test Variable(s) box.

3. In the Test Value box at the bottom of the One-Sample t Test window, enter the

hypothesized value for the population mean from the null hypothesis. Note: The

value is automatically set at zero until you type in a new value.

4. In addition to performing the hypothesis test, the program computes a confidence

interval for the population mean difference. The confidence level is automatically

set at 95%, but you can select Options and change the percentage.

5. Click OK.

DEMONSTRATION 9.1 2 7 3

SPSS Output

We used the SPSS program to analyze the data from the infants-and-attractive-faces

study in Example 9.1, and the program output is shown in Figure 9.9. The output

includes a table of sample statistics with the mean, standard deviation, and standard

error for the sample mean. A second table shows the results of the hypothesis test,

including the values for t, df, and the level of significance (the p value for the test),

as well as the mean difference from the hypothesized value of � � 10 and a 95%

confidence interval for the mean difference. To obtain a 95% confidence interval for the

mean, simply add � � 10 points to the values in the table.

FOCUS ON PROBLEM SOLVING

1. The first problem we confront in analyzing data is determining the appropriate

statistical test. Remember that you can use a z-score for the test statistic only

when the value for � is known. If the value for � is not provided, then you must

use the t statistic.

2. For the t test, the sample variance is used to find the value for the estimated

standard error. Remember to use n – 1 in the denominator when computing the

sample variance (see Chapter 4). When computing estimated standard error,

use n in the denominator.

DEMONSTRATION 9.1

A HYPOTHESIS TEST WITH THE t STATISTIC

A psychologist has prepared an “Optimism Test” that is administered yearly to

graduating college seniors. The test measures how each graduating class feels about

One-Sample Statistics

VAR00001 9 13.0000

3.000 8 .017 3.00000 .6940 5.3060

1.00000

t df Sig. (2-tailed) Mean

Difference Lower Upper

95% Confidence Interval of the Difference

Mean

3.00000

Std. Deviation Std. Error

Mean

One-Sample Test

VAR00001

Test Value = 10

FIGURE 9.9

The SPSS output for the hypothesis test presented in Example 9.1.

2 7 4 CHAPTER 9 INTRODUCTION TO THE t STATISTIC

its future—the higher the score, the more optimistic the class. Last year’s class

had a mean score of � � 15. A sample of n � 9 seniors from this year’s class was

selected and tested. The scores for these seniors are 7, 12, 11, 15, 7, 8, 15, 9, and 6,

which produce a sample mean of M � 10 with SS � 94.

On the basis of this sample, can the psychologist conclude that this year’s class

has a different level of optimism than last year’s class?

Note that this hypothesis test uses a t statistic because the population variance (�2)

is not known.

State the hypotheses, and select an alpha level. The null hypothesis states that the mean optimism score for this year’s class is the same as the mean for last year’s class.

H 0 : � � 15 (There is no change.)

H 1 : �  15 (This year’s mean is different.)

For this demonstration, we use a � .05, two tails.

Locate the critical region. With a sample of n � 9 students, the t statistic has df � n – 1 � 8. For a two-tailed test with a � .05 and df � 8, the critical t values are t � 62.306. These critical t values define the boundaries of the critical region.

Compute the test statistic. As we have noted, it is easier to separate the calculation of the t statistic into three stages.

Sample variance.

s SS

8 11 75�

2 � � .

Estimated standard error. The estimated standard error for these data is

s s

n M

� � �

2 11 75

9 1 14

. .

The t statistic. Now that we have the estimated standard error and the sample mean,

we can compute the t statistic. For this demonstration,

t M

s M

� 2�

� 2

52 10 15

1 14

1 14 4 39

. . .

Make a decision about H 0 , and state a conclusion. The t statistic we obtained

(t 5 –4.39) is in the critical region. Thus, our sample data are unusual enough to reject the null hypothesis at the .05 level of significance. We can conclude that there is a significant difference in level of optimism between this year’s and last year’s graduating classes, t(8) 5 –4.39, p , .05, two-tailed.

DEMONSTRATION 9.2

EFFECT SIZE: ESTIMATING COHEN’S d AND COMPUTING r 2

We estimate Cohen’s d for the same data used for the hypothesis test in

Demonstration 9.1. The mean optimism score for the sample from this year’s

class was 5 points lower than the mean from last year (M 5 10 versus m 5 15).

S T E P 1

S T E P 2

S T E P 3

S T E P 4

PROBLEMS 2 7 5

In Demonstration 9.1, we computed a sample variance of s2 � 11.75, so the standard

deviation is 11 75. � 3.43. With these values,

estimated mean difference

standard deviati d �

oon � �

3 43 1 46

. .

To calculate the percentage of variance explained by the treatment effect, r2, we need

the value of t and the df value from the hypothesis test. In Demonstration 9.1, we

obtained t � –4.39 with df � 8. Using these values in Equation 9.5, we obtain

r t

t df

2 2

4 39

4 39 8

19 27

27 27 0�

1 �

2 1 � �

. .

( ) ( )

771

PROBLEMS

1. Under what circumstances is a t statistic used instead of a z-score for a hypothesis test?

2. A sample of n � 25 scores has a mean of M � 83 and a standard deviation of s � 15.

a. Explain what is measured by the sample standard deviation.

b. Compute the estimated standard error for the sample mean and explain what is measured by the standard error.

3. Find the estimated standard error for the sample mean for each of the following samples.

a. n � 4 with SS � 48 b. n � 6 with SS � 270 c. n � 12 with SS � 132

4. Explain why t distributions tend to be flatter and more spread out than the normal distribution.

5. Find the t values that form the boundaries of the critical region for a two-tailed test with a � .05 for each of the following sample sizes:

a. n � 6 b. n � 12 c. n � 24

6. The following sample of n � 6 scores was obtained from a population with unknown parameters.

Scores: 7, 1, 6, 3, 6, 7

a. Compute the sample mean and standard deviation. (Note that these are descriptive values that sum- marize the sample data.)

b. Compute the estimated standard error for M. (Note that this is an inferential value that describes how accurately the sample mean represents the unknown population mean.)

7. The following sample was obtained from a population with unknown parameters.

Scores: 6, 12, 0, 13, 4, 7

a. Compute the sample mean and standard deviation. (Note that these are descriptive values that summarize the sample data.)

b. Compute the estimated standard error for M. (Note that this is an inferential value that describes how accurately the sample mean represents the unknown population mean.)

8. A random sample of n � 25 individuals is selected from a population with � � 20, and a treatment is administered to each individual in the sample. After treatment, the sample mean is found to be M � 22.2 with SS � 384.

a. How much difference is there between the mean for the treated sample and the mean for the origi- nal population? (Note: In a hypothesis test, this value forms the numerator of the t statistic.)

b. If there is no treatment effect, how much difference is expected between the sample mean and its popu- lation mean? That is, find the standard error for M. (Note: In a hypothesis test, this value is the denomi- nator of the t statistic.)

c. Based on the sample data, does the treatment have a significant effect? Use a two-tailed test with a � .05.

9. To evaluate the effect of a treatment, a sample is obtained from a population with a mean of � � 30, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M � 31.3 with a standard deviation of s � 3.

a. If the sample consists of n � 16 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with a � .05?

b. If the sample consists of n � 36 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with a � .05?

c. Comparing your answer for parts a and b, how does the size of the sample influence the outcome of a hypothesis test?

2 7 6 CHAPTER 9 INTRODUCTION TO THE t STATISTIC

10. To evaluate the effect of a treatment, a sample of n � 8 is obtained from a population with a mean of � � 40, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M � 35.

a. If the sample variance is s2 � 32, are the data suf- ficient to conclude that the treatment has a signifi- cant effect using a two-tailed test with a � .05?

b. If the sample variance is s2 � 72, are the data suffi- cient to conclude that the treatment has a significant effect using a two-tailed test with a � .05?

c. Comparing your answer for parts a and b, how does the variability of the scores in the sample influence the outcome of a hypothesis test?

11. The spotlight effect refers to overestimating the extent to which others notice your appearance or behavior, especially when you commit a social faux pas. Effectively, you feel as if you are suddenly standing in a spotlight with everyone looking. In one demonstration of this phenomenon, Gilovich, Medvec, and Savitsky (2000) asked college students to put on a Barry Manilow T-shirt that fellow students had previously judged to be embarrassing. The participants were then led into a room in which other students were already participating in an experiment. After a few minutes, the participant was led back out of the room and was allowed to remove the shirt. Later, each participant was asked to estimate how many people in the room had noticed the shirt. The individuals who were in the room were also asked whether they noticed the shirt. In the study, the participants significantly overestimated the actual number of people who had noticed.

a. In a similar study using a sample of n � 9 partici- pants, the individuals who wore the shirt produced an average estimate of M � 6.4 with SS � 162. The average number who said they noticed was 3.1. Is the estimate from the participants significantly different from the actual number? Test the null hypothesis that the true mean is � � 3.1 using a two-tailed test with a � .05.

b. Is the estimate from the participants significantly higher than the actual number (� � 3.1)? Use a one-tailed test with a � .05.

12. Many animals, including humans, tend to avoid direct eye contact and even patterns that look like eyes. Some insects, including moths, have evolved eye-spot patterns on their wings to help ward off predators. Scaife (1976) reports a study examining how eye-spot patterns affect the behavior of birds. In the study, the birds were tested in a box with two chambers and were free to move from one chamber to another. In one chamber, two large eye-spots were painted on one wall. The other chamber had plain walls. The researcher recorded the amount of time each bird

spent in the plain chamber during a 60-minute session. Suppose the study produced a mean of M � 37 minutes in the plain chamber with SS � 288 for a sample of n � 9 birds. (Note: If the eye spots have no effect, then the birds should spend an average of � � 30 minutes in each chamber.)

a. Is this sample sufficient to conclude that the eye-spots have a significant influence on the birds’ behavior? Use a two-tailed test with a � .05.

b. Compute the estimated Cohen’s d to measure the size of the treatment effect.

c. Construct the 95% confidence interval to estimate the mean amount of time spent on the plain side for the population of birds.

13. Standardized measures seem to indicate that the aver- age level of anxiety has increased gradually over the past 50 years (Twenge, 2000). In the 1950s, the aver- age score on the Child Manifest Anxiety Scale was � � 15.1. A sample of n � 16 of today’s children produces a mean score of M � 23.3 with SS � 240.

a. Based on the sample, has there been a significant change in the average level of anxiety since the 1950s? Use a two-tailed test with a � .01.

b. Make a 90% confidence interval estimate of today’s population mean level of anxiety.

c. Write a sentence that demonstrates how the outcome of the hypothesis test and the confidence interval would appear in a research report.

14. The librarian at the local elementary school claims that, on average, the books in the library are more than 20 years old. To test this claim, a student takes a sample of n � 30 books and records the publica- tion date for each. The sample produces an average age of M � 23.8 years with a variance of s2 � 67.5. Use this sample to conduct a one-tailed test with a � .01 to determine whether the average age of the library books is significantly greater than 20 years (� . 20).

15. For several years researchers have noticed that there appears to be a regular, year-by-year increase in the average IQ for the general population. This phenom- enon is called the Flynn effect after the researcher who first reported it (Flynn, 1984, 1999), and it means that psychologists must continuously update IQ tests to keep the population mean at � � 100. To evaluate the size of the effect, a researcher obtained a 10-year-old IQ test that was standardized to produce a mean IQ of � � 100 for the population 10 years ago. The test was then given to a sample of n � 64 of today’s 20-year-old adults. The average score for the sample was M � 107 with a standard deviation of s � 12.

a. Based on the sample, is the average IQ for today’s population significantly different from the average 10 years ago, when the test would have produced

PROBLEMS 2 7 7

a mean of � � 100? Use a two-tailed test with a � .01.

b. Make an 80% confidence interval estimate of to- day’s population mean IQ for the 10-year-old test.

16. Weinstein, McDermott, and Roediger (2010) report that students who were given questions to be answered while studying new material had better scores when tested on the material compared to students who were simply given an opportunity to reread the material. In a similar study, an instruc- tor in a large psychology class gave one group of students questions to be answered while study- ing for the final exam. The overall average for the exam was � � 73.4, but the n � 16 students who answered questions had a mean of M � 78.3 with a standard deviation of s � 8.4. For this study, did answering questions while studying produce significantly higher exam scores? Use a one-tailed test with a � .01.

17. Ackerman and Goldsmith (2011) found that students who studied text from printed hardcopy had bet- ter test scores than students who studied from text presented on a screen. In a related study, a profes- sor noticed that several students in a large class had purchased the e-book version of the course textbook. For the final exam, the overall average for the entire class was � � 81.7, but the n � 9 students who used e-books had a mean of M � 77.2 with a standard deviation of s � 5.7.

a. Is the sample sufficient to conclude that scores for students using e-books were significantly different from scores for the regular class? Use a two-tailed test with a � .05.

b. Construct the 90% confidence interval to estimate the mean exam score if the entire population used e-books.

c. Write a sentence demonstrating how the results from the hypothesis test and the confidence inter- val would appear in a research report.

18. A random sample of n � 16 scores is obtained from a population with a mean of � � 45. After a treat- ment is administered to the individuals in the sample, the sample mean is found to be M � 49.2.

a. Assuming that the sample standard deviation is s � 8, compute r2 and the estimated Cohen’s d to measure the size of the treatment effect.

b. Assuming that the sample standard deviation is s � 20, compute r2 and the estimated Cohen’s d to measure the size of the treatment effect.

c. Comparing your answers from parts a and b, how does the variability of the scores in the sample influence the measures of effect size?

19. A random sample is obtained from a population with a mean of � � 45. After a treatment is administered

to the individuals in the sample, the sample mean is M � 49 with a standard deviation of s � 12.

a. Assuming that the sample consists of n � 9 scores, compute r2 and the estimated Cohen’s d to measure the size of treatment effect.

b. Assuming that the sample consists of n � 16 scores, compute r2 and the estimated Cohen’s d to measure the size of treatment effect.

c. Comparing your answers from parts a and b, how does the number of scores in the sample influence the measures of effect size?

20. An example of the vertical-horizontal illusion is shown in the figure. Although the two lines are exactly the same length, the vertical line appears to be much longer. To examine the strength of this illusion, a researcher prepared an example in which both lines were exactly 10 inches long. The example was shown to individual participants who were told that the horizontal line was 10 inches long and then were asked to estimate the length of the vertical line. For a sample of n � 25 participants, the average estimate was M � 12.2 inches with a standard deviation of s � 1.00.

An example of the vertical- horizontal illusion

a. Use a one-tailed hypothesis test with a � .01 to demonstrate that the individuals in the sample significantly overestimate the true length of the line. (Note: Accurate estimation would produce a mean of � � 10 inches.)

b. Calculate the estimated d and r2, the percentage of variance accounted for, to measure the size of this effect.

c. Construct a 95% confidence interval for the population mean estimated length of the vertical line.

21. In studies examining the effect of humor on interper- sonal attractions, McGee and Shevlin (2009) found that an individual’s sense of humor had a significant effect on how the individual was perceived by oth- ers. In one part of the study, female college students were given brief descriptions of a potential romantic

2 7 8 CHAPTER 9 INTRODUCTION TO THE t STATISTIC

partner. The fictitious male was described positively as being single, ambitious, and having good job prospects. For one group of participants, the descrip- tion also said that he had a great sense of humor. For another group, it said that he had no sense of humor. After reading the description, each participant was asked to rate the attractiveness of the man on a seven-point scale from 1 (very attractive) to 7 (very unattractive). A score of 4 indicates a neutral rating.

a. The females who read the “great sense of humor” description gave the potential partner an average attractiveness score of M � 4.53 with a standard deviation of s � 1.04. If the sample consisted of n � 16 participants, is the average rating signifi- cantly higher than neutral (� � 4)? Use a one- tailed test with a � .05.

b. The females who read the description saying “no sense of humor” gave the potential partner an average attractiveness score of M � 3.30 with a standard deviation of s � 1.18. If the sample con- sisted of n � 16 participants, is the average rating significantly lower than neutral (� � 4)? Use a one-tailed test with a � .05.

22. Oishi and Shigehiro (2010) report that people who move from home to home frequently as children tend to have lower than average levels of well-being as adults. To further examine this relationship, a psychologist obtains a sample of n � 12 young adults who each experienced 5 or more different homes before they were 16 years old. These participants were given a standardized well-being questionnaire for which the general population has

an average score of � � 40. The well-being scores for this sample are as follows: 38, 37, 41, 35, 42, 40, 33, 33, 36, 38, 32, 39.

a. On the basis of this sample, is well-being for frequent movers significantly different from well- being in the general population? Use a two-tailed test with a � .05.

b. Compute the estimated Cohen’s d to measure the size of the difference.

c. Write a sentence showing how the outcome of the hypothesis test and the measure of effect size would appear in a research report.

23. Research examining the effects of preschool child- care has found that children who spent time in day care, especially high-quality day care, perform better on math and language tests than children who stay home with their mothers (Broberg, Wessels, Lamb, & Hwang, 1997). In a typical study, a researcher ob- tains a sample of n � 10 children who attended day care before starting school. The children are given a standardized math test for which the population mean is � � 50. The scores for the sample are as follows: 53, 57, 61, 49, 52, 56, 58, 62, 51, 56.

a. Is this sample sufficient to conclude that the children with a history of preschool day care are significantly different from the general popula- tion? Use a two-tailed test with a � .01.

b. Compute Cohen’s d to measure the size of the preschool effect.

c. Write a sentence showing how the outcome of the hypothesis test and the measure of effect size would appear in a research report.

Aplia for Essentials of Statistics for the Behavioral Sciences

After reading, go to “Resources” at the end of this chapter for

an introduction on how to use Aplia’s homework and learning

resources.

The t Test for Two Independent Samples

10.1 Introduction to the Independent-Measures Design

10.2 The t Statistic for an Independent-Measures Research Design

10.3 Hypothesis Tests and Effect Size with the Independent-Measures t Statistic

10.4 Assumptions Underlying the Independent-Measures t Formula

Summary

Focus on Problem Solving

Demonstrations 10.1 and 10.2

Problems

C H A P T E R

10 Tools You Will Need The following items are considered essential background material for this chapter. If you doubt your knowledge of any of these items, you should review the appropriate chapter or section before proceeding.

• Sample variance (Chapter 4) • Standard error formulas (Chapter 7) • The t statistic (Chapter 9)

• Distribution of t values • df for the t statistic • Estimated standard error

2 8 0 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

INTRODUCTION TO THE INDEPENDENT-MEASURES DESIGN

Until this point, all of the inferential statistics we have considered involve using one

sample as the basis for drawing conclusions about one population. Although these

single-sample techniques are used occasionally in real research, most research studies

require the comparison of two (or more) sets of data. For example, a social psychologist

may want to compare men and women in terms of their political attitudes, an educa-

tional psychologist may want to compare two methods for teaching mathematics, or a

clinical psychologist may want to evaluate a therapy technique by comparing depres-

sion scores for patients before therapy with their scores after therapy. In each case, the

research question concerns a mean difference between two sets of data.

There are two general research designs that can be used to obtain the two sets of

data to be compared:

1. The two sets of data could come from two completely separate groups of par-

ticipants. For example, the study could involve a sample of men compared with

a sample of women. Or the study could compare grades for one group of fresh-

men who are given laptop computers with grades for a second group who are

not given computers.

2. The two sets of data could come from the same group of participants. For ex-

ample, the researcher could obtain one set of scores by measuring depression

for a sample of patients before they begin therapy and then obtain a second set

of data by measuring the same individuals after 6 weeks of therapy.

The first research strategy, using completely separate groups, is called an independent-

measures research design or a between-subjects design. These terms emphasize the fact

that the design involves separate and independent samples and makes a comparison

between two groups of individuals. The structure of an independent-measures research

design is shown in Figure 10.1. Notice that the research study uses two separate samples

to represent the two different populations (or two different treatments) being compared.

10.1

Unknown µ = ?

Sample A

Unknown µ = ?

Sample B

Population A Taught by method A

Population B Taught by method B

FIGURE 10.1

Do the achievement

scores for children taught

by method A differ from

the scores for children

taught by method B? In

statistical terms, are the

two population means the

same or different? Because

neither of the two popula-

tion means is known, it

will be necessary to take

two samples, one from

each population. The first

sample provides informa-

tion about the mean for the

first population, and the

second sample provides

information about the

second population.

SECTION 10.2 / THE t STATISTIC FOR AN INDEPENDENT-MEASURES RESEARCH DESIGN 2 8 1

A research design that uses a separate group of participants for each treatment

condition (or for each population) is called an independent-measures research

design or a between-subjects research design.

In this chapter, we examine the statistical techniques used to evaluate the data from

an independent-measures design. More precisely, we introduce the hypothesis test that

allows researchers to use the data from two separate samples to evaluate the mean dif-

ference between two populations or between two treatment conditions.

The second research strategy, in which the two sets of data are obtained from the same

group of participants, is called a repeated-measures research design or a within-subjects

design. The statistics for evaluating the results from a repeated-measures design are intro-

duced in Chapter 11. Also, at the end of Chapter 11, we discuss some of the advantages

and disadvantages of independent-measures and repeated-measures designs.

THE t STATISTIC FOR AN INDEPENDENT-MEASURES RESEARCH DESIGN

Because an independent-measures study involves two separate samples, we need some

special notation to help specify which data go with which sample. This notation in-

volves the use of subscripts, which are small numbers written beside a sample statistic.

For example, the number of scores in the first sample is identified by n 1 ; for the second

sample, the number of scores is n 2 . The sample means are identified by M

1 and M

2 . The

sums of squares are SS 1 and SS

2 .

The goal of an independent-measures research study is to evaluate the mean difference

between two populations (or between two treatment conditions). Using subscripts to

differentiate the two populations, the mean for the first population is µ 1 , and the second

population mean is µ 2 . The difference between means is simply µ

1 � µ

2 . As always, the

null hypothesis states that there is no change, no effect, or, in this case, no difference.

Thus, in symbols, the null hypothesis for the independent-measures test is

H 0 : µ

1 � µ

2 � 0 (No difference between the population means)

You should notice that the null hypothesis could also be stated as µ 1 � µ

2 . However,

the first version of H 0 produces a specific numerical value (zero) that is used in the

calculation of the t statistic. Therefore, we prefer to phrase the null hypothesis in terms

of the difference between the two population means.

The alternative hypothesis states that there is a mean difference between the two

populations,

H 1 : µ

1 � µ

2 ≠ 0 (There is a mean difference.)

Equivalently, the alternative hypothesis can simply state that the two population

means are not equal: µ 1 ≠ µ

2 .

The independent-measures hypothesis test uses another version of the t statistic. The

formula for this new t statistic has the same general structure as the t statistic formula

that was introduced in Chapter 9. To help distinguish between the two t formulas, we

refer to the original formula (Chapter 9) as the single-sample t statistic and we refer to

D E F I N I T I O N

10.2

T H E H Y P OT H E S I S F O R A N I N D E P E N D E N T-

M E AS U R E S T E ST

T H E F O R M U L AS F O R A N I N D E P E N D E N T-

M E AS U R E S H Y P OT H E S I S T E ST

2 8 2 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

the new formula as the independent-measures t statistic. Because the new independent-

measures t includes data from two separate samples and hypotheses about two popula-

tions, the formulas may appear to be a bit overpowering. However, the new formulas are

easier to understand if you view them in relation to the single-sample t formulas from

Chapter 9. In particular, there are two points to remember:

1. The basic structure of the t statistic is the same for both the independent-

measures and the single-sample hypothesis tests. In both cases,

t � �sample statistic hypothesized population pparameter

estimated standard error

2. The independent-measures t is basically a two-sample t that doubles all the

elements of the single-sample t formulas.

To demonstrate the second point, we examine the two t formulas piece by piece.

The overall t formula The single-sample t uses one sample mean to test a hypothesis

about one population mean. The sample mean and the population mean appear in the

numerator of the t formula, which measures how much difference there is between the

sample data and the population hypothesis.

t � �sample mean population mean

estimated standdard error �

s M

�

The independent-measures t uses the difference between two sample means to

evaluate a hypothesis about the difference between two population means. Thus, the

independent-measures t formula is

t �

�sample mean population mean

difference difference

estimated standard errror �

� � m 2m

M M

s M M

1 2 1 2

1 2

( ) ( )

( )

In this formula, the value of M 1 2 M

2 is obtained from the sample data and the value

for µ 1 2 µ

2 comes from the null hypothesis.

The estimated standard error In each of the t-score formulas, the standard error

in the denominator measures how accurately the sample statistic represents the

population parameter. In the single-sample t formula, the standard error measures

the amount of error expected for a sample mean and is represented by the symbol s M .

For the independent-measures t formula, the standard error measures the amount of

error that is expected when you use a sample mean difference (M 1 2 M

2 ) to represent

a population mean difference (µ 1 2 µ

2 ). The standard error for the sample mean dif-

ference is represented by the symbol s M M1 22( ) .

Caution: Do not let the notation for standard error confuse you. In general, standard

error measures how accurately a statistic represents a parameter. The symbol for stan-

dard error takes the form s statistic

. When the statistic is a sample mean, M, the symbol for

standard error is s M . For the independent-measures test, the statistic is a sample mean

difference (M 1 2 M

2 ), and the symbol for standard error is s M M1 22( ) . In each case, the

standard error tells how much discrepancy is reasonable to expect between the sample

statistic and the corresponding population parameter.

SECTION 10.2 / THE t STATISTIC FOR AN INDEPENDENT-MEASURES RESEARCH DESIGN 2 8 3

Interpreting the estimated standard error The estimated standard error of M 1 – M

that appears in the bottom of the independent-measures t statistic can be interpreted in

two ways. First, the standard error is defined as a measure of the standard, or average,

distance between a sample statistic (M 1 � M

2 ) and the corresponding population pa-

rameter (µ 1 � µ

2 ). As always, samples are not expected to be perfectly accurate and the

standard error measures how much difference is reasonable to expect between a sample

statistic and the population parameter.

Sample mean estimated standard error

Population mean

difference ← → difference

(M 1 � M

2 )

(average distance)

(m 1 � m

2 )

When the null hypothesis is true, however, the population mean difference is zero.

Sample mean estimated standard error

difference ← → 0 (If H 0 is true)

(M 1 � M

2 )

(average distance)

The standard error is measuring how close the sample mean difference is to zero,

which is equivalent to measuring how much difference there is between the two sample

means.

estimated standard error

M 1 ← → M

(average distance)

This produces a second interpretation for the estimated standard error. Specifically,

the standard error can be viewed as a measure of how much difference is reasonable to

expect between two sample means if the null hypothesis is true.

The second interpretation of the estimated standard error produces a simplified ver-

sion of the independent-measures t statistic.

t � sample mean difference

estimated standard eerror

actual difference between and

s �

M M 1 2

ttandard difference (If is true) betweenH 0

andM M 1 2

In this version, the numerator of the t statistic measures how much difference ac-

tually exists between the two sample means, including any difference that is caused

by the different treatments. The denominator measures how much difference should

exist between the two sample means if there is no treatment effect that causes them

to be different. A large value for the t statistic is evidence for the existence of a

treatment effect.

To develop the formula for s M M1 2�( )

we consider the following three points:

1. Each of the two sample means represents it own population mean, but in each

case there is some error.

M 1 approximates µ

1 with some error

M 2 approximates µ

2 with some error

CA L C U L AT I N G T H E E ST I M AT E D STA N DA R D

E R R O R

2 8 4 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

Thus, there are two sources of error.

2. The amount of error associated with each sample mean is measured by the esti-

mated standard error of M. Using Equation 9.1 (p. 251), the estimated standard

error for each sample mean is computed as follows:

For ForM s s

n M s

n M M1

2 2

� � 22

3. For the independent-measures t statistic, we want to know the total amount

of error involved in using two sample means to approximate two population

means. To do this, we find the error from each sample separately and then add

the two errors together. The resulting formula for standard error is

s s

n M M1 2

� � 1

( )

(10.1)

Because the independent-measures t statistic uses two sample means, the formula for

the estimated standard error simply combines the error for the first sample mean and the

error for the second sample mean (Box 10.1).

Although Equation 10.1 accurately presents the concept of standard error for

the independent-measures t statistic, this formula is limited to situations in which

the two samples are exactly the same size (that is, n 1 � n

2 ). For situations in

which the two sample sizes are different, the formula is biased and, therefore, inap-

propriate. The bias comes from the fact that Equation 10.1 treats the two sample

P O O L E D VA R I A N C E

BOX

10.1 THE VARIABILITY OF DIFFERENCE SCORES

It may seem odd that the independent-measures

t statistic adds together the two sample errors when it

subtracts to find the difference between the two sam-

ple means. The logic behind this apparently unusual

procedure is demonstrated here.

We begin with two populations, I and II (Figure

10.2). The scores in population I range from a high of

70 to a low of 50. The scores in population II range

from 30 to 20. We use the range as a measure of how

spread out (variable) each population is:

For population I, the scores cover a range of

20 points.

For population II, the scores cover a range of

10 points.

If we randomly select one score from population I

and one score from population II and compute the differ-

ence between these two scores (X 1 � X

2 ), what range of

values is possible for these differences? To answer this

question, we need to find the biggest possible difference

and the smallest possible difference. As seen in Figure

10.2, the biggest difference occurs when X 1 � 70 and

X 2 � 20. This is a difference of X

1 � X

2 � 50 points.

The smallest difference occurs when X 1 � 50 and

X 2 � 30. This is a difference of X

1 � X

2 � 20 points.

Notice that the differences go from a high of 50 to a low

of 20. This is a range of 30 points:

range for population I (X 1 scores) � 20 points

range for population II (X 2 scores) � 10 points

range for the differences (X 1 � X

2 ) � 30 points

The variability for the difference scores is found

by adding together the variability for each of the two

populations.

In the independent-measures t statistics, we com-

pute the variability (standard error) for a sample mean

difference. To compute this value, we add together the

variability for each of the two sample means.

SECTION 10.2 / THE t STATISTIC FOR AN INDEPENDENT-MEASURES RESEARCH DESIGN 2 8 5

variances equally. However, when the sample sizes are different, the two sample

variances are not equally good and should not be treated equally. In Chapter 7, we

introduced the law of large numbers, which states that statistics obtained from large

samples tend to be better (more accurate) estimates of population parameters than

statistics obtained from small samples. This same fact holds for sample variances:

The variance obtained from a large sample is a more accurate estimate of s2 than

the variance obtained from a small sample.

One method for correcting the bias in the standard error is to combine the two

sample variances into a single value called the pooled variance. The pooled vari-

ance is obtained by averaging, or “pooling,” the two sample variances using a

procedure that allows the bigger sample to carry more weight in determining the

final value.

You should recall that when there is only one sample, the sample variance is computed as

s SS

2 �

For the independent-measures t statistic, there are two SS values and two df values

(one from each sample). The values from the two samples are combined to compute

what is called the pooled variance. The pooled variance is identified by the symbol and

is computed as

pooled variance � � 1

1 s

SS SS

df df p

2 1 2

1 2 (10.2)

With one sample, the variance is computed as SS divided by df. With two samples,

the pooled variance is computed by combining the two SS values and then dividing by

the combination of the two df values.

As we mentioned earlier, the pooled variance is actually an average of the two

sample variances, but the average is computed so that the larger sample carries

more weight in determining the final value. The following examples demonstrate

this point.

Population II Population I

10 30 40 50 60 70 80

Smallest difference 20 points

Biggest difference 50 points

FIGURE 10.2

Two population distribu-

tions. The scores in popu-

lation I vary from 50 to 70

(a 20-point spread), and

the scores in population

II range from 20 to 30 (a

10-point spread). If you

select one score from each

of these two populations,

the closest two values are

X 1 � 50 and X

2 � 30.

The two values that are

farthest apart are X 1 � 70

and X 2 � 20.

An alternative to computing

pooled variance is presented

in Box 10.2, p. 302.

2 8 6 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

Equal sample sizes We begin with two samples that are exactly the same size. The

first sample has n 5 6 scores with SS 5 50, and the second sample has n 5 6 scores

with SS 5 30. Individually, the two sample variances are

Variance for sample 1: s SS

2 50

5 105 5 5

Variance for sample 2: s SS

2 30

5 65 5 5

The pooled variance for these two samples is

s SS SS

df df p

2 1 2

1 2

50 30

5 5

10 8 005

1 5

1 5 5 .

Note that the pooled variance is exactly halfway between the two sample variances.

Because the two samples are exactly the same size, the pooled variance is simply the

average of the two sample variances.

Unequal sample sizes Now consider what happens when the samples are not the

same size. This time the first sample has n 5 3 scores with SS 5 20, and the second

sample has n 5 9 scores with SS 5 48. Individually, the two sample variances are

Variance for sample 1: s SS

2 20

2 105 5 5

Variance for sample 2: s SS

2 48

8 65 5 5

The pooled variance for these two samples is

s SS SS

df df p

2 1 2

1 2

20 48

2 8

10 6 805

1 5

1 5 5 .

This time the pooled variance is not located halfway between the two sample vari-

ances. Instead, the pooled value is closer to the variance for the larger sample (n 5 9

and s2 5 6) than to the variance for the smaller sample (n 5 3 and s2 5 10). The larger

sample carries more weight when the pooled variance is computed.

When computing the pooled variance, the weight for each of the individual sample

variances is determined by its degrees of freedom. Because the larger sample has a

larger df value, it carries more weight when averaging the two variances. This produces

an alternative formula for computing pooled variance.

pooled variance 5 5 1

1 s

df s df s

df df p

2 1 1

2 2

1 2 (10.3)

For example, if the first sample has df 1 5 3 and the second sample has df

2 5 7, then

the formula instructs you to take 3 of the first sample variance and 7 of the second

sample variance for a total of 10 variances. You then divide by 10 to obtain the average.

The alternative formula is especially useful if the sample data are summarized as means

and variances. Finally, you should note that because the pooled variance is an average of

the two sample variances, the value obtained for the pooled variance is always located

between the two sample variances.

SECTION 10.2 / THE t STATISTIC FOR AN INDEPENDENT-MEASURES RESEARCH DESIGN 2 8 7

Using the pooled variance in place of the individual sample variances, we can now

obtain an unbiased measure of the standard error for a sample mean difference. The

resulting formula for the independent-measures estimated standard error is

estimated standard error of M M s M M1 2 1 2

2 5 5 2( )

p p

(10.4)

Conceptually, this standard error measures how accurately the difference between

two sample means represents the difference between the two population means. In a

hypothesis test, H 0 specifies that m

1 2 m

2 5 0, and the standard error also measures

how much difference is expected, on average, between the two sample means. In either

case, the formula combines the error for the first sample mean with the error for the

second sample mean. Also note that the pooled variance from the two samples is used

to compute the standard error for the sample mean difference.

The complete formula for the independent-measures t statistic is as follows:

t M M

s M M

5 2 2 m 2 m

1 2 1 2

1 2

( ) ( )

( )

5 2sample mean difference population mean diffference

estimated standard error (10.5)

In the formula, the estimated standard error in the denominator is calculated using Equation 10.4, and requires calculation of the pooled variance using either

Equation 10.2 or 10.3.

The degrees of freedom for the independent-measures t statistic are determined by

the df values for the two separate samples:

df for the t statistic 5 df for the first sample 1 df for the second sample

5 df 1 1 df

5 (n 1 2 1) 1 (n

2 2 1) (10.6)

Equivalently, the df value for the independent-measures t statistic can be expressed as

df 5 n 1 1 n

2 2 2 (10.7)

Note that the df formula subtracts 2 points from the total number of scores; 1 point

for the first sample and 1 for the second.

The independent-measures t statistic is used for hypothesis testing. Specifically, we use

the difference between two sample means (M 1 2 M

2 ) as the basis for testing hypotheses

about the difference between two population means (µ 1 2 µ

2 ). In this context, the overall

structure of the t statistic can be reduced to the following:

t 5 2data hypothesis

error

This same structure is used for both the single-sample t from Chapter 9 and the new

independent-measures t that was introduced in the preceding pages. Table 10.1 identi-

fies each component of these two t statistics and should help reinforce the point that we

made earlier in the chapter; that is, the independent-measures t statistic simply doubles

each aspect of the single-sample t statistic.

E ST I M AT E D STA N DA R D E R R O R

T H E F I N A L F O R M U L A A N D D E G R E E S O F

F R E E D O M

2 8 8 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

HYPOTHESIS TESTS AND EFFECT SIZE WITH THE INDEPENDENT-MEASURES t STATISTIC

The independent-measures t statistic uses the data from two separate samples to help

decide whether there is a significant mean difference between two populations or

between two treatment conditions. A complete example of a hypothesis test with two

independent samples follows.

10.3

Sample Data

Hypothesized Population Parameter

Estimated Standard

Error Sample

Variance

Single-sample

t statistic

M µ s

2 s

2 �

Independent-

measures

t statistic

(M 1 � M

2 ) (µ

1 � µ

2 )

p p

1 s SS SS

df df p

2 1 2

1 2

� 1

TABLE 10.1

The basic elements of a t statis-

tic for the single-sample t and

the independent-measures t.

1. What is the defining characteristic of an independent-measures research study?

2. Explain what is measured by the estimated standard error in the denominator of

the independent-measures t statistic.

3. One sample from an independent-measures study has n � 4 with SS � 100. The

other sample has n � 8 and SS � 140.

a. Compute the pooled variance. (Note: Equation 10.2 works well with these data.)

b. Compute the estimated standard error for the mean difference.

4. One sample from an independent-measures study has n � 9 with a variance of

s2 � 35. The other sample has n � 3 and s2 � 40.

a. Compute the pooled variance. (Note: Equation 10.3 works well with these data.)

b. Compute the estimated standard error for the mean difference.

5. An independent-measures t statistic is used to evaluate the mean difference between

two treatments with n � 8 in one treatment and n � 12 in the other. What is the df

value for the t statistic?

1. An independent-measures study uses a separate group of participants to represent each of

the populations or treatment conditions being compared.

2. The estimated standard error measures how much difference is expected, on average, between

a sample mean difference and the population mean difference. In a hypothesis test, m 1 – m

2 is

set to zero and the standard error measures how much difference is expected between the two

sample means.

3. a. The pooled variance is 240 10

� 24.

b. The estimated standard error is 3.

4. a. The pooled variance is 36.

b. The estimated standard error is 4.

5. df � df 1 1 df

2 � 7 1 11 � 18.

L E A R N I N G C H E C K

ANSWERS

SECTION 10.3 / HYPOTHESIS TESTS AND EFFECT SIZE WITH THE INDEPENDENT-MEASURES t STATISTIC 2 8 9

Research results suggest a relationship between the TV viewing habits of 5-year-old

children and their future performance in high school. For example, Anderson, Huston,

Wright, and Collins (1998) report that high school students who had regularly watched

Sesame Street as children had better grades in high school than their peers who had not

watched Sesame Street. Suppose that a researcher intends to examine this phenomenon

using a sample of 20 high school students.

The researcher first surveys the students’ parents to obtain information on the family’s

TV-viewing habits during the time that the students were 5 years old. Based on the sur-

vey results, the researcher selects a sample of n � 10 students with a history of watching

Sesame Street and a sample of n � 10 students who did not watch the program. The aver-

age high school grade is recorded for each student and the data are as follows:

Average High School Grade

Watched Sesame Street

Did Not Watch Sesame Street

86 99 90 79

87 97 89 83

91 94 82 86

97 89 83 81

98 92 85 92

n � 10 n � 10

M � 93 M � 85

SS � 200 SS � 160

Note that this is an independent-measures study using two separate samples represent-

ing two distinct populations of high school students. The researcher would like to know

whether there is a significant difference between the two types of high school student.

State the hypotheses and select the alpha level.

H 0 : µ

1 � µ

2 � 0 (No difference.)

H 1 : µ

1 � µ

2 ≠ 0 (There is a difference.)

We set a � .01.

Directional hypotheses could be used and would specify whether the students who

watched Sesame Street should have higher or lower grades.

This is an independent-measures design. The t statistic for these data has degrees of

freedom determined by

df � df 1 1 df

� (n 1 � 1) 1 (n

2 � 1)

� 9 1 9

� 18

The t distribution for df � 18 is presented in Figure 10.3. For a � .01, the critical

region consists of the extreme 1% of the distribution and has boundaries of t � 12.878

and t � –2.878.

E X A M P L E 1 0 . 1

S T E P 1

S T E P 2

2 9 0 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

Obtain the data and compute the test statistic. The data are given, so all that remains is to compute the t statistic. As with the single-sample t test in Chapter 9, we recommend that the calculations be divided into three parts.

First, find the pooled variance for the two samples:

s SS SS

df df p

2 1 2

1 2

� 1

200 160

9 9

� 360

5 20

Second, use the pooled variance to compute the estimated standard error:

s s

n M M

p p

1 2

10 2

5 1 5 1 ( )

5 12 2

5 4

5 2

Third, compute the t statistic:

t M M

s M M

5 2 2 m 2m

5 2 2

1 2 1 2

1 2

93 85 0

( ) ( ) ( )

( )

5 8

5 4

S T E P 3

t = �2.878 t = 0

t = �2.878

Reject H0 Reject H0

t distribution df = 18

FIGURE 10.3

The critical region for the

independent-measures

hypothesis test in

Example 10.1 with

df 5 18 and a 5 .01.

Caution: The pooled variance

combines the two samples to

obtain a single estimate of

variance. In the formula, the

two samples are combined in

a single fraction.

Caution: The standard error

adds the errors from two sep-

arate samples. In the formula,

these two errors are added as

two separate fractions. In this

case, the two errors are equal

because the sample sizes are

the same.

SECTION 10.3 / HYPOTHESIS TESTS AND EFFECT SIZE WITH THE INDEPENDENT-MEASURES t STATISTIC 2 9 1

Make a decision. The obtained value (t � 4.00) is in the critical region. In this example,

the obtained sample mean difference is four times greater than would be expected if

there were no difference between the two populations. In other words, this result is very

unlikely if H 0 is true. Therefore, we reject H

0 and conclude that there is a significant

difference between the high school grades for students who watched Sesame Street and

those who did not. Specifically, the students who watched Sesame Street had signifi-

cantly higher grades than those who did not watch the program.

Note that the Sesame Street study in Example 10.1 is an example of nonexperimental

research (see Chapter 1, p. 16). Specifically, the researcher did not manipulate the TV

programs watched by the children and did not control a variety of variables that could

influence high school grades. As a result, we cannot conclude that watching Sesame Street

causes higher high school grades. In particular, many other, uncontrolled factors, such as

the parents’ level of education or family economic status, might explain the difference be-

tween the two groups. Thus, we do not know exactly why there is a relationship between

watching Sesame Street and high school grades, but we do know that a relationship exists.

As noted in Chapters 8 and 9, a hypothesis test is usually accompanied by a report of

effect size to provide an indication of the absolute magnitude of the treatment effect.

One technique for measuring effect size is Cohen’s d, which produces a standardized

measure of mean difference. In its general form, Cohen’s d is defined as

d � � m 2mmean difference

standard deviation 1 2

In the context of an independent-measures research study, the difference between the

two sample means (M 1 – M

2 ) is used as the best estimate of the mean difference between

the two populations, and the pooled standard deviation (the square root of the pooled

variance) is used to estimate the population standard deviation. Thus, the formula for

estimating Cohen’s d becomes

estimated estimated mean difference

estima d 5

tted standard deviation 5

2M M

s p

1 2

(10.8)

For the data from Example 10.1, the two sample means are 93 and 85, and the pooled

variance is 20. The estimated d for these data is

d M M

s p

5 2

5 5 1 2

93 85

4 7 1 79

. .

Using the criteria established to evaluate Cohen’s d (see Table 8.2 on p. 232), this

value indicates a very large treatment effect.

The independent-measures t test also allows for measuring effect size by comput-

ing the percentage of variance accounted for, r2. As we saw in Chapter 9, r2 measures

how much of the variability in the scores can be explained by the treatment effects.

For example, some of the variability in the high school grades from the Sesame Street

study can be explained by knowing whether a particular student watched the program;

students who watched Sesame Street tend to have higher grades and students who did

not watch the show tend to have lower grades. By measuring exactly how much of the

variability can be explained, we can obtain a measure of how big the treatment effect

S T E P 4

M E AS U R I N G E F F E C T S I Z E F O R

T H E I N D E P E N D E N T- M E AS U R E S t T E ST

2 9 2 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

actually is. The calculation of r2 for the independent-measures t test is exactly the same

as it was for the single-sample t test in Chapter 9.

r t

t df

2 2

2 �

(10.9)

For the data in Example 10.1, we obtained t � 4.00 with df � 18. These values

produce an r2 of

r 2

4 18

16 18

34 0 47�

1 �

1 � � .

According to the standards used to evaluate r2 (see Table 9.3 on p. 264), this value

also indicates a very large treatment effect.

Although the value of r2 is usually obtained by using Equation 10.9, it is possible

to determine the percentage of variability directly by computing SS values for the set

of scores. The following example demonstrates this process using the data from the

Sesame Street study in Example 10.1.

Figure 10.4(a) shows the two samples from Example 10.1 combined into a single fre-

quency distribution of n � 20 scores. The overall mean for the distribution is M � 89

and the scores have SS � 680. However, it is clear that the scores for the students who

watched Sesame Street tend to be clustered at the high end of the distribution. The mean

for these 10 students is M � 93, which is 4 points higher than the overall mean for the

entire group. Also, the scores for the students who did not watch the program are clustered

at the low end of the distribution. For these 10 students the mean is M � 85, which is 4

points below the overall mean. The difference between the two types of student (M � 85

versus M � 93) is the treatment effect produced by the relationship between TV viewing

and high school grades.

Now consider the distribution shown in Figure 10.4(b). To create this distribution, we

started with the original scores and then eliminated the treatment effect. Specifically, we

subtracted 4 points from each score in the Sesame Street group and we added 4 points to

each score for students who did not watch Sesame Street. As a result, both groups now

have the same mean, M � 89, and the treatment effect is eliminated. One consequence of

removing the treatment effect is that the resulting scores are much less variable that the

original scores. Recall that the original scores in Figure 10.4(a) have SS � 680. When

the treatment effect is removed, in Figure 10.4(b), the variability is reduced to SS � 360.

The difference between these two values is 320 points. Thus, the treatment effect accounts

for 320 point of the total variability in the original scores. When expressed as a proportion

of the total variability, we obtain

variability explained by the treatment

total variability � � �

320

680 0 47 47. %

You should recognize that this is exactly the same value we obtained for r2 using

Equation 10.9.

As noted in Chapter 9, it is possible to compute a confidence interval as an alternative

method for measuring and describing the size of the treatment effect. For the single-

sample t, we used a single sample mean, M, to estimate a single population mean. For the

E X A M P L E 1 0 . 2

CO N F I D E N C E I N T E R VA L S F O R

E ST I M AT I N G  1 2 

SECTION 10.3 / HYPOTHESIS TESTS AND EFFECT SIZE WITH THE INDEPENDENT-MEASURES t STATISTIC 2 9 3

independent-measures t, we use a sample mean difference, M 1 � M

2 , to estimate the popu-

lation mean difference, m 1 � m

2 . In this case, the confidence interval literally estimates the

size of the population mean difference between the two populations or treatment conditions.

As with the single-sample t, the first step is to solve the t equation for the unknown

parameter. For the independent-measures t statistic, we obtain

m 2m 5 2  21 2 1 2 1 2

M M ts M M( )

(10.10)

In the equation, the values for M 1 2 M

2 and for s M M1 22( ) are obtained from the

sample data. Although the value for the t statistic is unknown, we can use the degrees

of freedom for the t statistic and the t distribution table to estimate the t value. Using the

estimated t and the known values from the sample, we can then compute the estimated

value for m 1 2 m

2 . The following example demonstrates the process of constructing a

confidence interval for a population mean difference.

Earlier we presented a research study comparing high school grades for students

who had watched Sesame Street as children with the grades for students who had not

watched the program (p. 289). The results of the hypothesis test indicated a significant

mean difference between the two populations of students. Now, we construct a 95%

confidence interval to estimate the size of the population mean difference.

E X A M P L E 1 0 . 3

79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

M 5 89

Average High School Grade

Sesame Street

No Sesame Street

Original scores including the treatment effect

79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

M 5 89

Average High School Grade

Adjusted scores after the treatment effect is removed

(a)

(b)

FIGURE 10.4

The two groups of scores from Example 10.1 combined into a single distribution. The original scores, including the treat-

ment effect, are shown in part (a). Part (b) shows the adjusted scores, after the treatment effect has been removed.

2 9 4 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

The data from the study produced a mean grade of M � 93 for the Sesame Street

group and a mean of M � 85 for the no–Sesame Street group, and the estimated standard

error for the mean difference was s M M1 2�( ) � 2. With n � 10 scores in each sample, the

independent-measures t statistic has df � 18. To have 95% confidence, we simply esti-

mate that the t statistic for the sample mean difference is located somewhere in the middle

95% of all the possible t values. According to the t distribution table, with df � 18, 95%

of the t values are located between t � 12.101 and t � �2.101. Using these values in the

estimation equation, we obtain

m 2 m 5 2 

5 2 

21 2 1 2 1 2

93 85 2 101 2

M M ts M M( )

(. ))

5 8 4 202.

This produces an interval of values ranging from 8 2 4.202 5 3.798 to 8 1 4.202

5 12.202. Thus, our conclusion is that students who watched Sesame Street have

higher grades that those who did not, and the mean difference between the two

populations is somewhere between 3.798 points and 12.202 points. Furthermore,

we are 95% confident that the true mean difference is in this interval because the

only value estimated during the calculations was the t statistic, and we are 95%

confident that the t value is located in the middle 95% of the distribution. Finally,

note that the confidence interval is constructed around the sample mean difference.

As a result, the sample mean difference, M 1 2 M

2 5 93 2 83 5 8 points, is located

exactly in the center of the interval.

As with the confidence interval for the single-sample t (p. 264), the confidence in-

terval for an independent-measures t is influenced by a variety of factors other than the

actual size of the treatment effect. In particular, the width of the interval depends on

the percentage of confidence used, so that a larger percentage produces a wider inter-

val. Also, the width of the interval depends on the sample size, so that a larger sample

produces a narrower interval. Because the interval width is related to sample size, the

confidence interval is not a pure measure of effect size like Cohen’s d or r2.

In addition to describing the size of a treatment effect, estimation can be used

to get an indication of the significance of the effect. Example 10.3 presented an

independent-measures research study examining the effect on high school grades

of having watched Sesame Street as a child. Based on the results of the study, the

95% confidence interval estimated that the population mean difference for the two

groups of students was between 3.798 and 12.202 points. The confidence interval

estimate is shown in Figure 10.5. In addition to the confidence interval for m 1 2 m

2 ,

we have marked the spot where the mean difference is equal to zero. You should

recognize that a mean difference of zero is exactly what would be predicted by the

null hypothesis if we were doing a hypothesis test. You also should realize that a

zero difference (m 1 2 m

2 5 0) is outside of the 95% confidence interval. In other

words, m 1 2 m

2 5 0 is not an acceptable value if we want 95% confidence in our

estimate. To conclude that a value of zero is not acceptable with 95% confidence

is equivalent to concluding that a value of zero is rejected with 95% confidence.

This conclusion is equivalent to rejecting H 0 with a 5 .05. On the other hand, if a

mean difference of zero were included within the 95% confidence interval, then we

would have to conclude that m 1 2 m

2 5 0 is an acceptable value, which is the same

as failing to reject H 0 .

CO N F I D E N C E I N T E R VA L S A N D

H Y P OT H E S I S T E ST S

The hypothesis test for

these data was conducted in

Example 10.1 (p. 289) and

the decision was to reject H 0 .

SECTION 10.3 / HYPOTHESIS TESTS AND EFFECT SIZE WITH THE INDEPENDENT-MEASURES t STATISTIC 2 9 5

IN THE LITERATURE:

REPORTING THE RESULTS OF AN INDEPENDENT-MEASURES t TEST

A research report typically presents the descriptive statistics followed by the results of the

hypothesis test and measures of effect size (inferential statistics). In Chapter 4 (p. 109),

we demonstrated how the mean and the standard deviation are reported in APA format. In

Chapter 9 (p. 267), we illustrated the APA style for reporting the results of a t test. Now

we use the APA format to report the results of Example 10.1, an independent-measures t

test. A concise statement might read as follows:

The students who watched Sesame Street as children had higher high school

grades (M � 93, SD � 4.71) than the students who did not watch the program

(M � 85, SD � 4.22). The mean difference was significant, t(18) � 4.00,

p , .01, d � 1.79.

You should note that standard deviation is not a step in the computations for

the independent-measures t test, yet it is useful when providing descriptive statis-

tics for each treatment group. It is easily computed when doing the t test because

you need SS and df for both groups to determine the pooled variance. Note that the

format for reporting t is exactly the same as that described in Chapter 9 (p. 267)

and that the measure of effect size is reported immediately after the results of the

hypothesis test.

Also, as we noted in Chapter 9, if an exact probability is available from a computer

analysis, it should be reported. For the data in Example 10.1, the computer analysis re-

ports a probability value of p � .001 for t � 4.00 with df � 18. In the research report,

this value would be included as follows:

The difference was significant, t(18) � 4.00, p � .001, d � 1.79.

Finally, if a confidence interval is reported to describe effect size, it appears im-

mediately after the results from the hypothesis test. For the Sesame Street examples

(Example 10.1 and Example 10.3), the report would be as follows:

The difference was significant, t(18) � 4.00, p � .001, 95% CI [3.798,

12.202].

3.798 12.202

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

95% confidence interval estimate for �

1 2 �

� 1 2 �

according to H0

( )

FIGURE 10.5

The 95% confidence interval for the population mean difference (µ 1 – µ

2 ) from Example 10.3. Note that µ

1 – µ

2 � 0 is

excluded from the confidence interval, indicating that a zero difference is not an acceptable value (H 0 would be rejected

in a hypothesis test with a � .05).

2 9 6 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

When planning an independent-measures study, a researcher usually has some

expectation or specific prediction for the outcome. For the Sesame Street study

in Example 10.1, the researcher clearly expects the students who watched Sesame

Street to have higher grades than the students who did not watch it. This kind of

directional prediction can be incorporated into the statement of the hypotheses,

resulting in a directional, or one-tailed, test. Recall from Chapter 8 that one-tailed

tests can lead to rejecting H 0 when the mean difference is relatively small compared

to the magnitude required by a two-tailed test. As a result, one-tailed tests should be

D I R E C T I O N A L H Y P OT H E S E S A N D O N E -

TA I L E D T E ST S

1. An educational psychologist would like to determine whether access to computers

has an effect on grades for high school students. One group of n � 16 students has

home room each day in a computer classroom in which each student has a com-

puter. A comparison group of n � 16 students has home room in a traditional

classroom. At the end of the school year, the average grade is recorded for each

student. The data are as follows:

Computer Traditional

M � 86 M � 82.5

SS � 1005 SS � 1155

a. Is there a significant difference between the two groups? Use a two-tailed test

with a � .05.

b. Compute Cohen’s d to measure the size of the difference.

c. Write a sentence that demonstrates how the outcome of the hypothesis test and

the measure of effect size would appear in a research report.

d. Compute the 90% confidence interval for the population mean difference between

a computer classroom and a regular classroom.

2. A research report states that there is a significant difference between treatments for

an independent-measures design with t(28) � 2.27.

a. How many individuals participated in the research study? (Hint: Start with the

df value.)

b. Should the report state that p . .05 or p , .05?

1. a. The pooled variance is 72, the standard error is 3, and t � 1.17. With a critical value of

t � 2.042, fail to reject the null hypothesis.

b. Cohen’s d � 3 5

. � 0.412

c. The results show no significant difference in grades for students with computers compared

to students without computers, t(30) � 1.17, p . .05, d � 0.412.

d. With df � 30 and 90% confidence, the t values for the confidence interval are ±1.697.

The interval is m 1 � m

2 � 3.5 ± 1.697(3). Thus, the population mean difference is esti-

mated to be between �1.591 and 8.591. The fact that zero is an acceptable value (inside

the interval) is consistent with the decision that there is no significant difference between

the two population means.

2. a. The df � 28, so the total number of participants is 30.

b. A significant result is indicated by p , .05.

L E A R N I N G C H E C K

ANSWERS

SECTION 10.3 / HYPOTHESIS TESTS AND EFFECT SIZE WITH THE INDEPENDENT-MEASURES t STATISTIC 2 9 7

used when clearly justified by theory or previous findings. The following example

demonstrates the procedure for stating hypotheses and locating the critical region for

a one-tailed test using the independent-measures t statistic.

We use the same research situation that was described in Example 10.1. The researcher

is using an independent-measures design to examine the relationship between watching

educational TV as a child and academic performance as a high school student. The pre-

diction is that high school students who watched Sesame Street regularly as 5-year-old

children have higher grades.

State the hypotheses and select the alpha level. As always, the null hypothesis says that

there is no effect, and the alternative hypothesis says that there is an effect. For this ex-

ample, the predicted effect is that the students who watched Sesame Street have higher

grades. Thus, the two hypotheses are as follows.

H 0 : µ

Sesame Street  µ

No Sesame Street (Grades are not higher with Sesame Street.)

H 1 : µ

Sesame Street . µ

No Sesame Street (Grades are higher with Sesame Street.)

Note that it is usually easier to state the hypotheses in words before you try to

write them in symbols. Also, it usually is easier to begin with the alternative hypoth-

esis (H 1 ), which states that the treatment works as predicted. Also note that the equal

sign goes in the null hypothesis, indicating no difference between the two treatment

conditions. The idea of zero difference is the essence of the null hypothesis, and the

numerical value of zero is used for (µ 1 � µ

2 ) during the calculation of the t statistic.

For this test we use a � .01.

Locate the critical region. For a directional test, the critical region is located entirely

in one tail of the distribution. Rather than trying to determine which tail, positive

or negative, is the correct location, we suggest that you identify the criteria for the

critical region in a two-step process as follows. First, look at the data and determine

whether the sample mean difference is in the direction that was predicted. If the

answer is no, then the data obviously do not support the predicted treatment effect,

and you can stop the analysis. On the other hand, if the difference is in the predicted

direction, then the second step is to determine whether the difference is large enough

to be significant. To test for significance, simply find the one-tailed critical value in

the t distribution table. If the calculated t statistic is more extreme (either positive or

negative) than the critical value, then the difference is significant.

For this example, the students who watched Sesame Street had higher grades, as

predicted. With df � 18, the one-tailed critical value for a � .01 is t � 2.552.

Collect the data and calculate the test statistic. The details of the calculations were

shown in Example 10.1. The data produce a t statistic of t � 4.00.

Make a decision. The t statistic of t � 4.00 is well beyond the critical boundary of t �

2.552. Therefore, we reject the null hypothesis and conclude that grades for students

who watched Sesame Street are significantly higher than grades for students who did

not watch the program. In a research report, the one-tailed test would be clearly noted:

Grades were significantly higher for students who watched Sesame Street, t(18) � 4.00,

p , .01, one tailed.

E X A M P L E 1 0 . 4

S T E P 1

S T E P 2

S T E P 3

S T E P 4

2 9 8 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

In Chapter 9 (p. 259), we identified several factors that can influence the outcome

of a hypothesis test. For the independent-measures t, the most obvious factor is

the size of the difference between the two sample means. A larger mean difference

increases the likelihood of rejecting the null hypothesis and increases measures

of effect size. Two other factors that play important roles are the variability of the

scores and the size of the samples. Both influence the magnitude of the estimated

standard error in the denominator of the t statistic. Specifically, the standard error is

directly related to sample variance (larger variance leads to larger error) and is in-

versely related to sample size (larger size leads to smaller error). As a result, larger

variance produces a smaller value for the t statistic (closer to zero) and reduces the

likelihood of finding a significant result. By contrast, a larger sample produces a

larger value for the t statistic (farther from zero) and increases the likelihood of

rejecting H 0 .

Although variance and sample size both influence the hypothesis test, only

variance has a large influence on measures of effect size such as Cohen’s d and

r2; larger variance produces smaller measures of effect size. Sample size, on

the other hand, has no effect on the value of Cohen’s d and only a small influence

on r2.

The following example provides a visual demonstration of how large sample vari-

ance can obscure a mean difference between samples and lower the likelihood of reject-

ing H 0 for an independent-measures study.

We use the data in Figure 10.6 to demonstrate the influence of sample variance. The

figure shows the results from a research study comparing two treatments. Notice

that the study uses two separate samples, each with n � 9, and there is a 5-point

mean difference between the two samples: M � 8 for treatment 1 and M � 13 for

treatment 2. The frequency distribution in Figure 10.6 also shows a clear difference

between the two distributions; the scores for treatment 2 are clearly higher than the

scores for treatment 1.

T H E R O L E O F SA M P L E VA R I A N C E A N D

SA M P L E S I Z E I N T H E I N D E P E N D E N T-

M E AS U R E S t T E ST

E X A M P L E 1 0 . 5

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Treatment 1

n � 9 M � 8 s � 1.22

Treatment 2

n � 9 M � 13 s � 1.22

FIGURE 10.6

Two sample distributions representing two different treatments. These data show a significant difference between treatments,

t(16) � 8.62, p , .01, and both measures of effect size indicate a very large treatment effect, d � 4.10 and r2 � 0.82.

SECTION 10.3 / HYPOTHESIS TESTS AND EFFECT SIZE WITH THE INDEPENDENT-MEASURES t STATISTIC 2 9 9

For the hypothesis test, the data produce a pooled variance of 1.50 and an estimated

standard error of 0.58. The t statistic is

t � � mean difference

estimated standard error

00 58 8 62

. .�

With df � 16, this value is far into the critical region (for a � .05 or a � .01), so we

reject the null hypothesis and conclude that there is a significant difference between the

two treatments. These data also produce Cohen’s d � 4.10 and r2 � 0.82, both indicat-

ing a very large treatment effect.

Now consider the effect of increasing sample variance. Figure 10.7 shows the results

from a second research study comparing two treatments. Notice that there are still n � 9

scores in each sample, and the two sample means are still M � 8 and M � 13. However,

the sample variances have been greatly increased: Each sample now has s2 � 44.25 as

compared with s2 � 1.5 for the data in Figure 10.7. Notice that the increased variance

means that the scores are now spread out over a wider range, with the result that the two

samples are mixed together without any clear distinction between them.

The absence of a clear difference between the two samples is supported by the hy-

pothesis test. The pooled variance is 44.25, the estimated standard error is 3.14, and the

independent-measures t statistic is

t � � mean difference

estimated standard error

33 14 1 59

. .�

With df � 16 and a � .05, this value is not in the critical region. Therefore, we fail

to reject the null hypothesis and conclude that there is no significant difference between

the two treatments. Although there is still a 5-point difference between sample means

(as in Figure 10.7), the 5-point difference is not significant with the increased variance.

The measures of effect size are also substantially smaller. With the increased variance,

Cohen’s d is now 0.75 and r2 � 0.136.

In general, large sample variance can obscure any mean differences that exist in the

data, which reduces the likelihood of obtaining a significant difference in a hypothesis

test and lowers measures of effect size.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Treatment 1

n � 9 M � 8 s � 6.65

Treatment 2

n � 9 M � 13 s � 6.65

FIGURE 10.7

Two sample distributions representing two different treatments. These data show exactly the same mean difference as the

scores in Figure 10.6; however, the variance has been greatly increased. With the increased variance, there is no longer a

significant difference between treatments, t(16) � 1.59, p . .05, and both measures of effect size are substantially reduced,

d � 0.75 and r2 � 0.14.

3 0 0 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

ASSUMPTIONS UNDERLYING THE INDEPENDENT- MEASURES t FORMULA

There are three assumptions that should be satisfied before you use the independent-

measures t formula for hypothesis testing:

1. The observations within each sample must be independent (see p. 220).

2. The two populations from which the samples are selected must be normal.

3. To justify using the pooled variance, the two populations from which the sam-

ples are selected must have equal variances.

The first two assumptions should be familiar from the single-sample t hypothesis test

presented in Chapter 9. As before, the normality assumption is the less important of the

two, especially with large samples. When there is reason to suspect that the populations are

far from normal, you should compensate by ensuring that the samples are relatively large.

The third assumption is referred to as homogeneity of variance and states

that the two populations being compared must have the same variance. You may recall

a similar assumption for the z-score hypothesis test in Chapter 8. For that test, we as-

sumed that the effect of the treatment was to add a constant amount to (or subtract a

constant amount from) each individual score. As a result, the population standard devia-

tion after treatment was the same as it had been before treatment. We now are making

essentially the same assumption, but phrasing it in terms of variances.

Recall that the pooled variance in the t-statistic formula is obtained by averaging

together the two sample variances. It makes sense to average these two values only if

they both are estimating the same population variance—that is, if the homogeneity of

variance assumption is satisfied. If the two sample variances are estimating different

population variances, then the average is meaningless. (Note: If two people are asked

to estimate the same thing—for example, what you weigh—it is reasonable to average

the two estimates. However, it is not meaningful to average estimates of two different

things. If one person estimates your weight and another estimates the number of grapes

in a pound, it is meaningless to average the two numbers.)

Homogeneity of variance is most important when there is a large discrepancy be-

tween the sample sizes. With equal (or nearly equal) sample sizes, this assumption is

less critical, but still important. Violating the homogeneity of variance assumption can

prevent any meaningful interpretation of the data from an independent-measures ex-

periment. Specifically, when you compute the t statistic in a hypothesis test, all of the

numbers in the formula come from the data except for the population mean difference,

which you get from H 0 . Thus, you are sure of all of the numbers in the formula except

one. If you obtain an extreme result for the t statistic (a value in the critical region), then

you conclude that the hypothesized value was wrong. But consider what happens when

the homogeneity assumption is violated. In this case, you have two questionable values

in the formula (the hypothesized population value and the meaningless average of the

two variances). Now if you obtain an extreme t statistic, you do not know which of these

two values is responsible. Specifically, you cannot reject the hypothesis because it may

have been the pooled variance that produced the extreme t statistic. Without satisfying

the homogeneity of variance requirement, you cannot accurately interpret a t statistic,

and the hypothesis test becomes meaningless.

How do you know whether the homogeneity of variance assumption is satisfied? One

simple test involves just looking at the two sample variances. Logically, if the two popula-

tion variances are equal, then the two sample variances should be very similar. When the

10.4

H A R T L E Y ’ S F - M A X T E ST

Remember: Adding a constant

to (or subtracting a constant

from) each score does not

change the standard deviation.

SECTION 10.4 / ASSUMPTIONS UNDERLYING THE INDEPENDENT-MEASURES t FORMULA 3 0 1

two sample variances are reasonably close, you can be reasonably confident that the homo-

geneity assumption has been satisfied and proceed with the test. However, if one sample

variance is more than three or four times larger than the other, then there is reason for

concern. A more objective procedure involves a statistical test to evaluate the homogeneity

assumption. Although there are many different statistical methods for determining whether

the homogeneity of variance assumption has been satisfied, Hartley’s F-max test is one of

the simplest to compute and to understand. An additional advantage is that this test can also

be used to check homogeneity of variance with more than two independent samples. Later,

in Chapter 12, we examine statistical methods for comparing several different samples, and

Hartley’s test is useful again. The following example demonstrates the F-max test for two

independent samples.

The F-max test is based on the principle that a sample variance provides an unbiased

estimate of the population variance. The null hypothesis for this test states that the

population variances are equal, therefore, the sample variances should be very similar.

The procedure for using the F-max test is as follows:

1. Compute the sample variance, s SS

2 � , for each of the separate samples.

2. Select the largest and the smallest of these sample variances and compute

F s

s -max

largest

smallest �

( ) ( )

A relatively large value for F-max indicates a large difference between the

sample variances. In this case, the data suggest that the population variances are

different and that the homogeneity assumption has been violated. On the other

hand, a small value of F-max (near 1.00) indicates that the sample variances are

similar and that the homogeneity assumption is reasonable.

3. The F-max value computed for the sample data is compared with the critical

value found in Table B.3 (Appendix B). If the sample value is larger than the

table value, then you conclude that the variances are different and that the ho-

mogeneity assumption is not valid.

To locate the critical value in the table, you need to know:

a. k � number of separate samples. (For the independent-measures t test, k � 2.)

b. df � n � 1 for each sample variance. The Hartley test assumes that all

samples are the same size.

c. The alpha level. The table provides critical values for a � .05 and a � .01.

Generally a test for homogeneity would use the larger alpha level.

Suppose, for example, that two independent samples each have n � 10 with sample

variances of 12.34 and 9.15. For these data,

F s

s -max

largest

smallest � � �

12 34

9 15 1 3

( ) ( )

. . 55

With a � .05, k � 2, and df � n – 1 � 9, the critical value from the table is 4.03.

Because the obtained F-max is smaller than this critical value, you conclude that the data

do not provide evidence that the homogeneity of variance assumption has been violated.

E X A M P L E 1 0 . 6

3 0 2 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

The goal for most hypothesis tests is to reject the null hypothesis to demonstrate a

significant difference or a significant treatment effect. However, when testing for homo-

geneity of variance, the preferred outcome is to fail to reject H 0 . Failing to reject H

0 with

the F-max test means that there is no significant difference between the two population

variances and the homogeneity assumption is satisfied. In this case, you may proceed with

the independent-measures t test using pooled variance.

If the F-max test rejects the hypothesis of equal variances, or if you simply suspect

that the homogeneity of variance assumption is not justified, you should not compute an

independent-measures t statistic using pooled variance. However, there is an alternative

procedure that does not pool the two sample variances and does not require the homogeneity

assumption. The alternative is presented in Box 10.2.

BOX

10.2 AN ALTERNATIVE TO POOLED VARIANCE

Computing the independent-measures t statis-

tic using pooled variance requires that the data

satisfy the homogeneity of variance assumption.

Specifically, the two distributions from which

the samples are obtained must have equal vari-

ances. To avoid this assumption, many statisticians

recommend an alternative formula for computing

the independent-measures t statistic that does not

require pooled variance or the homogeneity as-

sumption. The alternative procedure consists of

two steps.

1. The standard error is computed using the

two separate sample variances as in

Equation 10.1.

2. The value of degrees of freedom for the t statis-

tic is adjusted using the following equation:

df V V

V s

n �

� 1

�

� 1 2

1 1

( ) where annd V

n 2

�

Decimal values for df should be rounded down to

the next lower integer.

The adjustment to degrees of freedom lowers the

value of df, which pushes the boundaries for the critical

region farther out. Thus, the adjustment makes the test

more demanding and therefore corrects for the same

bias problem that the pooled variance attempts to avoid.

Note: Many computer programs that perform sta-

tistical analysis (such as SPSS) report two versions

of the independent-measures t statistic; one using

pooled variance (with equal variances assumed) and

one using the adjustment shown here (without the as-

sumption of equal variances).

1. A researcher is using an independent-measures design to evaluate the difference

between two treatment conditions with n � 8 in each treatment. The first treatment

produces M � 63 with a variance of s2 � 18, and the second treatment has M � 58

with s2 � 14.

a. Use a one-tailed test with a � .05 to determine whether the scores in the first

treatment are significantly greater than the scores in the second. (Note: Because

the two samples are the same size, the pooled variance is simply the average of

the two sample variances.)

b. Predict how the value for the t statistic would be affected if the two sample

variances were increased to s2 � 68 and s2 � 60. Compute the new t to confirm

your answer.

c. Predict how the value for the t statistic for the original samples would be

affected if each sample had n � 32 scores (instead of n � 8). Compute the

new t to confirm your answer.

L E A R N I N G C H E C K

SUMMARY 3 0 3

1. The independent-measures t statistic uses the data from two separate samples to draw inferences about the mean difference between two populations or be- tween two different treatment conditions.

2. The formula for the independent-measures t statistic has the same structure as the original z-score or the single-sample t:

t � �sample statistic population parameter

estimmated standard error

For the independent-measures t, the sample statistic is the sample mean difference (M

1 � M

2 ). The population

parameter is the population mean difference (µ 1 � µ

2 ).

The estimated standard error for the sample mean dif- ference is computed by combining the errors for the two sample means. The resulting formula is

t M M

s M M

� � � m 2 m

1 2 1 2

1 2

( ) ( )

( )

where the estimated standard error is

s s

n M M

p p

1 2

2 5 1

( )

The pooled variance in the formula, s p

2 , is the weighted mean of the two sample variances:

s SS SS

df df p

2 1 2

1 2

5 1

2. The homogeneity of variance assumption requires that the two sample variances be

equal. (True or false?)

3. When you are using an F-max test to evaluate the homogeneity of variance as-

sumption, you usually do not want to find a significant difference between the

variances. (True or false?)

1. a. The pooled variance is 16, the estimated standard error is 2, and t(14) 5 2.50. With a

one-tailed critical value of 1.761, reject the null hypothesis. Scores in the first treatment

are significantly higher than scores in the second.

b. Increasing the variance should lower the value of t. The new pooled variance is 64, the

estimated standard error is 4, and t(14) 5 1.25.

c. Increasing the sample sizes should increase the value of t. The pooled variance is still

16, but the new standard error is 1, and t(62) 5 5.00.

2. False. The assumption is that the two population variances are equal.

3. True. If there is a significant difference between the two variances, you cannot do the t test

with pooled variance.

ANSWERS

SUMMARY

This t statistic has degrees of freedom determined by the sum of the df values for the two samples:

df 5 df 1 1 df

5 (n 1 2 1) 1 (n

2 2 1)

3. For hypothesis testing, the null hypothesis states that there is no difference between the two population means:

H 0 : m

1 5 m

2 or m

1 2 m

2 5 0

4. When a hypothesis test with an independent-measures t statistic indicates a significant difference, you should also compute a measure of the effect size. One measure of effect size is Cohen’s d, which is a standardized measure of the mean difference. For the independent-measures t statistic, Cohen’s d is esti- mated as follows:

estimated d M M

s p

5 2

1 2

A second common measure of effect size is the percent- age of variance accounted for by the treatment effect. This measure is identified by r2 and is computed as

r t

t df

2 2

2 5

3 0 4 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

5. An alternative method for describing the size of the treatment effect is to construct a confidence interval for the population mean difference, m

1 – m

2 . The con-

fidence interval uses the independent-measures t equa- tion, solved for the unknown mean difference:

m 2m 5 2  21 2 1 2 1 2

M M ts M M( )

First, select a level of confidence and then look up the corresponding t values. For example, for 95% confidence, use the range of t values that determine the middle 95% of the distribution. The t values are then used in the equation along with the values for the

sample mean difference and the standard error, which are computed from the sample data.

6. Appropriate use and interpretation of the t statistic using pooled variance require that the data satisfy the homogeneity of variance assumption. This assumption stipulates that the two populations have equal variances. An informal test of the assumption can be made by verifying that the two sample variances are approxi- mately equal. Hartley’s F-max test provides a statistical technique for determining whether the data satisfy the homogeneity assumption. An alternative technique that avoids pooling variances and eliminates the need for the homogeneity assumption is presented in Box 10.2.

KEY TERMS

independent-measures research design (281)

between-subjects research design (281)

repeated-measures research

design (281)

within-subjects research design (281)

independent-measures t statistic (282)

estimated standard error

of M 1 2 M

2 (283)

pooled variance (285)

homogeneity of variance (300)

RESOURCES

Go to CengageBrain.com to access Psychology CourseMate, where you will find an

interactive eBook, glossaries, flashcards, quizzes, statistics workshops, and more.

If your professor has assigned Aplia:

1. Sign in to your account.

2. Complete the corresponding exercises as required by your professor.

3. When finished, click “Grade It Now” to see which areas you have mastered, which

areas need more work, and detailed explanations of every answer.

General instructions for using SPSS are presented in Appendix D. Following are

detailed instructions for using SPSS to perform The Independent-Measures t Test

presented in this chapter.

Data Entry

1. The scores are entered in what is called stacked format, which means that all of the

scores from both samples are entered in one column of the data editor (probably

FOCUS ON PROBLEM SOLVING 3 0 5

VAR00001). Enter the scores for sample #2 directly beneath the scores from sample #1

with no gaps or extra spaces.

2. Values are then entered into a second column (VAR00002) to identify the sample or

treatment condition corresponding to each of the scores. For example, enter a 1 be-

side each score from sample #1 and enter a 2 beside each score from sample #2.

Data Analysis

1. Click Analyze on the tool bar, select Compare Means, and click on Independent- Samples t Test.

2. Highlight the column label for the set of scores (VAR0001) in the left box and click the arrow to move it into the Test Variable(s) box.

3. Highlight the label from the column containing the sample numbers (VAR0002) in the left box and click the arrow to move it into the Group Variable box.

4. Click on Define Groups.

5. Assuming that you used the numbers 1 and 2 to identify the two sets of scores, enter the values 1 and 2 into the appropriate group boxes.

6. Click Continue.

7. In addition to performing the hypothesis test, the program computes a confidence interval for the population mean difference. The confidence level is automatically set at 95%, but you can select Options to change the percentage.

8. Click OK.

SPSS Output

We used the SPSS program to analyze the data from the Sesame Street study in Example

10.1 and the program output is shown in Figure 10.8. The output includes a table of sam-

ple statistics with the mean, standard deviation, and standard error of the mean for each

group. A second table, which is split into two sections in Figure 10.8, begins with the

results of Levene’s test for homogeneity of variance. This test should not be significant

(you do not want the two variances to be different), so you want the reported Sig. value

to be greater than .05. Next, the results of the independent-measures t test are presented

using two different assumptions. The top row shows the outcome assuming equal vari-

ances, using the pooled variance to compute t. The second row does not assume equal

variances and computes the t statistic using the alternative method presented in Box 10.2.

Each row reports the calculated t value, the degrees of freedom, the level of significance

(the p value for the test), the size of the mean difference, and the standard error for the

mean difference (the denominator of the t statistic). Finally, the output includes a 95%

confidence interval for the mean difference.

FOCUS ON PROBLEM SOLVING

1. As you learn more about different statistical methods, one basic problem is

deciding which method is appropriate for a particular set of data. Fortunately,

it is easy to identify situations in which the independent-measures t statistic is

used. First, the data always consist of two separate samples (two ns, two Ms,

3 0 6 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

two SSs, and so on). Second, this t statistic is always used to answer questions

about a mean difference: On the average, is one group different (better, faster,

smarter) than the other group? If you examine the data and identify the type of

question that a researcher is asking, you should be able to decide whether an

independent-measures t is appropriate.

2. When computing an independent-measures t statistic from sample data, we sug-

gest that you routinely divide the formula into separate stages rather than trying

to do all of the calculations at once. First, find the pooled variance. Second,

compute the standard error. Third, compute the t statistic.

3. One of the most common errors for students involves confusing the formulas

for pooled variance and standard error. When computing pooled variance,

you are “pooling” the two samples together into a single variance. This vari-

ance is computed as a single fraction, with two SS values in the numerator

and two df values in the denominator. When computing the standard error,

you are adding the error from the first sample and the error from the second

sample. These two separate errors are added as two separate fractions under

the square root symbol.

Group Statistics

Independent Samples Test

VAR00001

VAR00002

1.00

2.00

Equal variances assumed

Equal variances not assumed

VAR00001 Equal variances assumed

Equal variances not assumed

.001

8.00000

2.00000

3.79816

3.79443

12.20184

12.20557

4.000

17.780

.543.384

Sig. (2-tailed) Mean

Difference Std. Error Difference

93.0000

85.0000

4.71405

4.21637

1.49071

1.33333

Sig.F

Levene’s Test for Equality of Variances t-test for Equality of Means

t-test for Equality of Means

t df

Lower Upper

95% Confidence

Interval of the Difference

MeanN Std. Deviation Std. Error

Mean

FIGURE 10.8

The SPSS output for the independent-measures hypothesis test in Example 10.1.

DEMONSTRATION 10.1 3 0 7

DEMONSTRATION 10.1

THE INDEPENDENT-MEASURES t TEST

In a study of jury behavior, two samples of participants were provided details about

a trial in which the defendant was obviously guilty. Although group 2 received the

same details as group 1, the second group was also told that some evidence had been

withheld from the jury by the judge. Later, the participants were asked to recommend

a jail sentence. The length of term suggested by each participant is presented here. Is

there a significant difference between the two groups in their responses?

Group 1 Group 2

4 3

4 7

3 8 For Group 1: M � 3 and SS � 16

2 5

5 4 For Group 2: M � 6 and SS � 24

1 7

1 6

4 8

There are two separate samples in this study. Therefore, the analysis uses the inde-

pendent-measures t test.

State the hypothesis, and select an alpha level.

H 0 : m

1 � m

2 � 0 (For the population, knowing that evidence has been

withheld has no effect on the suggested sentence.)

H 1 : m

1 � m

2 ≠ 0 (For the population, knowing that evidence has been

withheld has an effect on the jury’s response.)

We set the level of significance to a � .05, two tails.

Identify the critical region. For the independent-measures t statistic, degrees of free-

dom are determined by

df � df 1 1 df

� 7 1 7

� 14

The t distribution table is consulted, for a two-tailed test with a � .05 and df � 14.

The critical t values are 12.145 and –2.145.

Compute the test statistic. As usual, we recommend that the calculation of the t sta-

tistic be separated into three stages.

Pooled variance: For these data, the pooled variance equals

s SS SS

df df p

2 1 2

1 2

16 24

7 7

14 2 86�

1 �

1 � � .

S T E P 1

S T E P 2

S T E P 3

3 0 8 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

Estimated standard error: Now we can calculate the estimated standard error for

mean differences.

s s

n M M

p p

1 2

2 86

8 0 358 0 35

� � 1 � 1 � 1

( )

. . . . 88 0 716 0 85� �. .

The t statistic: Finally, the t statistic can be computed.

t M M

s M M

� � � m 2m

5 2 2

5 2

1 2 1 2

1 2

3 6 0

0 85

0 8

( ) ( ) ( )

( ) . . 55

3 535 2 .

Make a decision about H 0 , and state a conclusion. The obtained t value of 23.53 falls

in the critical region of the left tail (critical t 5 ±2.145). Therefore, the null hypothesis is rejected. The participants who were informed about the withheld evidence gave sig- nificantly longer sentences, t(14)5 23.53, p , .05, two tails.

DEMONSTRATION 10.2

EFFECT SIZE FOR THE INDEPENDENT-MEASURES t

We compute Cohen’s d and r2 for the jury decision data in Demonstration 10.1. For

these data, the two sample means are M 1 5 3 and M

2 5 6, and the pooled variance is

2.86. Therefore, our estimate of Cohen’s d is

estimated d M M

s p

5 2

5 5 1 2

3 6

2 86

1 69 1 78

. . .

With a t value of t 5 3.53 and df 5 14, the percentage of variance accounted for is

r t

t df

2 2

3 53

3 53 14

12 46

26 46 0 45

1 5

1 5 5

. .

( )

( ) 77 47or %( )

S T E P 4

PROBLEMS

1. Describe the basic characteristics of an independent- measures, or a between-subjects, research study.

2. Describe what is measured by the estimated standard error in the bottom of the independent-measures t statistic.

3. If other factors are held constant, explain how each of the following influences the value of the independent- measures t statistic and the likelihood of rejecting the null hypothesis:

a. An increase in the mean difference between the samples.

b. An increase in the number of scores in each sample. c. An increase in the variance for each sample.

4. Describe the homogeneity of variance assumption and explain why it is important for the independent- measures t test.

5. One sample has SS 5 36 and a second sample has SS 5 18.

a. If n 5 4 for both samples, find each of the sample variances and compute the pooled variance. Because the samples are the same size, you should find that the pooled variance is exactly halfway between the two sample variances.

b. Now assume that n 5 4 for the first sample and n 5 7 for the second. Again, calculate the two sample variances and the pooled variance. You should find that the pooled variance is closer to the variance for the larger sample.

PROBLEMS 3 0 9

6. One sample has SS � 70 and a second sample has SS � 42.

a. If n � 8 for both samples, find each of the sample variances, and calculate the pooled variance. Because the samples are the same size, you should find that the pooled variance is exactly halfway between the two sample variances.

b. Now assume that n � 8 for the first sample and n � 4 for the second. Again, calculate the two sam- ple variances and the pooled variance. You should find that the pooled variance is closer to the vari- ance for the larger sample.

7. As noted on page 283, when the two population means are equal, the estimated standard error for the independent-measures t test provides a measure of how much difference to expect between two sample means. For each of the following situations, assume that µ

1 � µ

2 and calculate how much difference should

be expected between the two sample means. a. One sample has n � 6 scores with SS � 75, and the

second sample has n � 10 scores with SS � 135. b. One sample has n � 6 scores with SS � 310, and the

second sample has n � 10 scores with SS � 530. c. In part b, the samples have larger variability (big-

ger SS values) than in part a, but the sample sizes are unchanged. How does larger variability affect the magnitude of the standard error for the sample mean difference?

8. Two samples are selected from the same population. For each of the following, calculate how much dif- ference is expected, on average, between the two sample means.

a. One sample has n � 4, the second has n � 6, and the pooled variance is 60.

b. One sample has n � 12, the second has n � 15, and the pooled variance is 60.

c. In part b, the sample sizes are larger, but the pooled variance is unchanged. How does larger sample size affect the magnitude of the standard error for the sample mean difference?

9. Two separate samples, each with n � 15 individuals, receive different treatments. After treatment, the first sample has SS � 1740 and the second has SS � 1620.

a. Find the pooled variance for the two samples. b. Compute the estimated standard error for the sample

mean difference. c. If the sample mean difference is 8 points, is this

enough to reject the null hypothesis and conclude that there is a significant difference for a two- tailed test at the .05 level?

10. Two separate samples receive different treatments. After treatment, the first sample has n � 9 with SS � 462, and the second has n � 7 with SS � 420.

a. Compute the pooled variance for the two samples.

b. Calculate the estimated standard error for the sample mean difference.

c. If the sample mean difference is 10 points, is this enough to reject the null hypothesis using a two- tailed test with a � .05?

11. For each of the following, assume that the two samples are obtained from populations with the same mean, and calculate how much difference should be expected, on average, between the two sample means.

a. Each sample has n � 4 scores with s2 � 68 for the first sample and s2 � 76 for the second. (Note: Because the two samples are the same size, the pooled variance is equal to the average of the two sample variances.)

b. Each sample has n � 16 scores with s2 � 68 for

the first sample and s2 � 76 for the second. c. In part b, the two samples are bigger than in part

a, but the variances are unchanged. How does sample size affect the size of the standard error for the sample mean difference?

12. For each of the following, calculate the pooled vari- ance and the estimated standard error for the sample mean difference.

a. The first sample has n � 4 scores and a variance of s2 � 55, and the second sample has n � 6 scores and a variance of s2 � 63.

b. Now the sample variances are increased so that the first sample has n � 4 scores and a variance of s2 � 220, and the second sample has n � 6 scores and a variance of s2 � 252.

c. Comparing your answers for parts a and b, how does increased variance influence the size of the estimated standard error?

13. A researcher conducts an independent-measures study comparing two treatments and reports the t statistic as t(25) � 2.071.

a. How many individuals participated in the entire study?

b. Using a two-tailed test with a � .05, is there a significant difference between the two treatments?

c. Compute r2 to measure the percentage of variance accounted for by the treatment effect.

14. In a recent study, Piff, Kraus, Côté, Cheng, and Keitner (2010) found that people from lower social economic classes tend to display greater prosocial behavior than their higher class counterparts. In one part of the study, participants played a game with an anonymous part- ner. Part of the game involved sharing points with the partner. The lower economic class participants were significantly more generous with their points compared with the upper class individuals. Results similar to those found in the study, show that n � 12 lower class participants shared an average of M � 5.2 points with

3 1 0 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

SS � 11.91, compared to an average of M � 4.3 with SS � 9.21 for the n � 12 upper class participants.

a. Are the data sufficient to conclude that there is a sig- nificant mean difference between the two economic populations? Use a two-tailed test with a � .05.

b. Construct an 80% confidence interval to estimate the size of the population mean difference.

15. Hallam, Price, and Katsarou (2002) investigated the influence of background noise on classroom perfor- mance for children aged 10 to 12. In one part of the study, calming music led to better performance on an arithmetic task compared to a no-music condition. Suppose that a researcher selects one class of n � 18 students who listen to calming music each day while working on arithmetic problems. A second class of n � 18 serves as a control group with no music. Accuracy scores are measured for each child and the average for students in the music condition is M � 86.4 with SS � 1550 compared to an average of M � 78.8 with SS � 1204 for students in the no- music condition.

a. Is there a significant difference between the two music conditions? Use a two-tailed test with a � .05.

b. Compute the 90% confidence interval for the population mean difference.

c. Write a sentence demonstrating how the results from the hypothesis test and the confidence interval would appear in a research report.

16. It appears that there is some truth to the old adage “That which doesn’t kill us makes us stronger.” Seery, Holman, and Silver (2010) found that individuals with some history of adversity report better mental health and higher well-being compared to people with little or no history of adversity. In an attempt to examine this phenomenon, a researcher surveys a group of col- lege students to determine the negative life events that they experienced in the past 5 years and their current feeling of well-being. For n � 18 participants with 2 or fewer negative experiences, the average well-being score is M � 42 with SS � 398, and for n � 16 par- ticipants with 5 to 10 negative experiences the average score is M � 48.6 with SS � 370.

a. Is there a significant difference between the two populations represented by these two samples? Use a two-tailed test with a � .01.

b. Compute Cohen’s d to measure the size of the effect. c. Write a sentence demonstrating how the outcome

of the hypothesis test and the measure of effect size would appear in a research report.

17. Does posting calorie content for menu items affect people’s choices in fast food restaurants? According to results obtained by Elbel, Gyamfi, and Kersh (2011), the answer is no. The researchers monitored the calorie content of food purchases for children and adolescents in four large fast food chains before and

after mandatory labeling began in New York City. Although most of the adolescents reported noticing the calorie labels, apparently the labels had no effect on their choices. Data similar to the results obtained show an average of M � 786 calories per meal with s � 85 for n � 100 children and adolescents before the labeling, compared to an average of M � 772 calories with s � 91 for a similar sample of n � 100 after the mandatory posting.

a. Use a two-tailed test with a � .05 to determine whether the mean number of calories after the posting is significantly different than before calorie content was posted.

b. Calculate r2 to measure effect size for the mean difference.

18. In 1974, Loftus and Palmer conducted a classic study demonstrating how the language used to ask a question can influence eyewitness memory. In the study, college students watched a film of an automobile accident and then were asked questions about what they saw. One group was asked, “About how fast were the cars going when they smashed into each other?” Another group was asked the same question except the verb was changed to “hit” in- stead of “smashed into.” The “smashed into” group reported significantly higher estimates of speed than the “hit” group. Suppose a researcher repeats this study with a sample of today’s college students and obtains the following results.

Estimated Speed

Smashed into Hit

n � 15 n � 15

M � 40.8 M � 34.0

SS � 510 SS � 414

a. Do the results indicate a significantly higher esti- mated speed for the “smashed into” group? Use a one-tailed test with a � .01.

b. Compute the estimated value for Cohen’s d to measure the size of the effect.

c. Write a sentence demonstrating how the results of the hypothesis test and the measure of effect size would appear in a research report.

19. Numerous studies have found that males report higher self-esteem than females, especially for adolescents (Kling, Hyde, Showers, & Buswell, 1999). Typical results show a mean self-esteem score of M � 39.0 with SS � 60.2 for a sample of n � 10 male adoles- cents and a mean of M � 35.4 with SS � 69.4 for a sample of n � 10 female adolescents.

a. Do the results indicate that self-esteem is signifi- cantly higher for males? Use a one-tailed test with a � .01.

PROBLEMS 3 1 1

b. Use the data to make a 95% confidence interval estimate of the mean difference in self-esteem between male and female adolescents.

c. Write a sentence demonstrating how the results from the hypothesis test and the confidence interval would appear in a research report.

20. Recent research has shown that creative people are more likely to cheat than their less creative counterparts (Gino & Ariely, 2010). Participants in the study first completed creativity assessment questionnaires and then returned to the lab several days later for a series of tasks. One task was a multiple-choice general knowledge test for which the participants circled their answers on the test sheet. Afterward, they were asked to transfer their answers to a bubble sheets for computer scoring. However, the experimenter admitted that the wrong bubble sheet had been copied so that the correct answers were still faintly visible. Thus, the partici- pants had an opportunity to cheat and inflate their test scores. Higher scores were valuable because participants were paid based on the number of cor- rect answers. However, the researchers had secretly coded the original tests and the bubble sheets so that they could measure the degree of cheating for each participant. Assuming that the participants were divided into two groups based on their cre- ativity scores, the following data are similar to the cheating scores obtained in the study.

High Creativity Participants

Low Creativity Participants

N � 27 N � 27

M � 7.41 M � 4.78

SS � 749.5 SS � 830

a. Use a one-tailed test with a � .05 to determine whether these data are sufficient to conclude that high creativity people are more likely to cheat than people with lower levels of creativity.

b. Compute Cohen’s d to measure the size of the effect.

c. Write a sentence demonstrating how the results from the hypothesis test and the measure of effect size would appear in a research report.

21. When people learn a new task, their performance usually improves when they are tested the next day, but only if they get at least 6 hours of sleep (Stickgold, Whidbee, Schirmer, Patel, & Hobson, 2000). The following data demonstrate this phe- nomenon. The participants learned a visual discrim- ination task on one day, and then were tested on the task the following day. Half of the participants were allowed to have at least 6 hours of sleep and

the other half were kept awake all night. Is there a significant difference between the two conditions? Use a two-tailed test with a � .05.

Performance Scores

6 Hours Sleep No Sleep

n � 14 n � 14

M � 72 M � 65

SS � 932 SS � 706

22. Recent research has demonstrated that music-based physical training for elderly people can improve balance and walking efficiency and reduce the risk of falls (Trombetti et al., 2011). As part of the training, participants walked in time to music and responded to changes in the music’s rhythm during a 1-hour per week exercise program. After 6 months, participants in the training group increased their walking speed and their stride length compared to individuals in the control group. The following data are similar to the results obtained in the study.

Exercise Group Stride Length

Control Group Stride Length

24 25 22 24 26 23 20 23

26 17 21 22 20 16 21 17

22 19 24 23 18 23 16 20

23 28 25 23 25 19 17 16

Do the results indicate a significant difference in the stride length for the two groups? Use a two-tailed test with a � .05.

23. Downs and Abwender (2002) evaluated soccer players and swimmers to determine whether the routine blows to the head experienced by soccer players produced long-term neurological deficits. In the study, neurological tests were administered to mature soccer players and swimmers and the results indicated significant differences. In a similar study, a researcher obtained the following data.

Swimmers Soccer players

10 7

8 4

7 9

9 3

13 7

3 1 2 CHAPTER 10 THE t TEST FOR TWO INDEPENDENT SAMPLES

Well-Lit Room Dimly-Lit Room

7 9

8 11

10 13

6 10

8 11

5 9

7 15

12 14

5 10

a. Is there a significant difference in reported perfor- mance between the two conditions? Use a two- tailed test with a � .01.

b. Compute Cohen’s d to estimate the size of the treatment effect.

a. Are the neurological test scores significantly lower for the soccer players than for the swimmers in the control group? Use a one-tailed test with a � .05.

b. Compute the value of r2 (percentage of variance accounted for) for these data.

24. Research has shown that people are more likely to show dishonest and self-interested behaviors in dark- ness than in a well-lit environment (Zhong, Bohns, & Gino, 2010). In one experiment, participants were given a set of 20 puzzles and were paid $0.50 for each one solved in a 5-minute period. However, the participants reported their own performance and there was no obvious method for checking their hon- esty. Thus, the task provided a clear opportunity to cheat and receive undeserved money. One group of participants was tested in a room with dimmed light- ing and a second group was tested in a well-lit room. The reported number of solved puzzles was recorded for each individual. The following data represent results similar to those obtained in the study.

Aplia for Essentials of Statistics for the Behavioral Sciences

After reading, go to “Resources” at the end of this chapter for

an introduction on how to use Aplia’s homework and learning

resources.

The t Test for Two Related Samples

11.1 Introduction to Repeated- Measures Designs

11.2 The t Statistic for a Repeated- Measures Research Design

11.3 Hypothesis Tests and Effect Size for the Repeated-Measures Design

11.4 Uses and Assumptions for Repeated-Measures t Tests

Summary

Focus on Problem Solving

Demonstrations 11.1 and 11.2

Problems

C H A P T E R

11 Tools You Will Need The following items are considered essential background material for this chapter. If you doubt your knowledge of any of these items, you should review the appropriate chapter or section before proceeding.

• Introduction to the t statistic (Chapter 9) • Estimated standard error • Degrees of freedom • t Distribution • Hypothesis tests with the t statistic

• Independent-measures design (Chapter 10)

3 1 4 CHAPTER 11 THE t TEST FOR TWO RELATED SAMPLES

INTRODUCTION TO REPEATED-MEASURES DESIGNS

In the previous chapter, we introduced the independent-measures research design as one

strategy for comparing two treatment conditions or two populations. The independent-

measures design is characterized by the fact that two separate samples are used to obtain

the two sets of scores that are to be compared. In this chapter, we examine an alterna-

tive strategy known as a repeated-measures design, or a within-subjects design. With a

repeated-measures design, two separate scores are obtained for each individual in the sam-

ple. For example, a group of patients could be measured before therapy and then measured

again after therapy. Or, response time could be measured in a driving simulation task for

a group of individuals who are first tested when they are sober and then tested again after

two alcoholic drinks. In each case, the same variable is being measured twice for the same

set of individuals; that is, we are literally repeating measurements on the same sample.

A repeated-measures design, or a within-subject design, is one in which the

dependent variable is measured two or more times for each individual in a single

sample. The same group of subjects is used in all of the treatment conditions.

The main advantage of a repeated-measures study is that it uses exactly the same

individuals in all treatment conditions. Thus, there is no risk that the participants in one

treatment are substantially different from the participants in another. With an independent-

measures design, on the other hand, there is always a risk that the results are biased

because the individuals in one sample are systematically different (smarter, faster, more

extroverted, and so on) than the individuals in the other sample. At the end of this chapter,

we present a more detailed comparison of repeated-measures studies and independent-

measures studies, considering the advantages and disadvantages of both types of research.

Occasionally, researchers try to approximate the advantages of a repeated-measures

design by using a technique known as matched subjects. A matched-subjects design

involves two separate samples, but each individual in one sample is matched one-to-

one with an individual in the other sample. Typically, the individuals are matched on

one or more variables that are considered to be especially important for the study. For

example, a researcher studying verbal learning might want to be certain that the two

samples are matched in terms of IQ and gender. In this case, a male participant with

an IQ of 120 in one sample would be matched with another male with an IQ of 120 in

the other sample. Although the participants in one sample are not identical to the par-

ticipants in the other sample, the matched-subjects design at least ensures that the two

samples are equivalent (or matched) with respect to some specific variables.

In a matched-subjects design, each individual in one sample is matched with an

individual in the other sample. The matching is done so that the two individuals

are equivalent (or nearly equivalent) with respect to a specific variable that the

researcher would like to control.

Of course, it is possible to match participants on more than one variable. For example,

a researcher could match pairs of subjects on age, gender, race, and IQ. In this case, for

example, a 22-year-old white female with an IQ of 115 who was in one sample would be

matched with another 22-year-old white female with an IQ of 115 in the second sample.

The more variables that are used, however, the more difficult it is to find matching pairs.

The goal of the matching process is to simulate a repeated-measures design as closely as

11.1

D E F I N I T I O N

T H E M AT C H E D - S U B J E C T S D E S I G N

D E F I N I T I O N

SECTION 11.2 / THE t STATISTIC FOR A REPEATED-MEASURES RESEARCH DESIGN 3 1 5

possible. In a repeated-measures design, the matching is perfect because the same individ-

ual is used in both conditions. In a matched-subjects design, however, the best you can get

is a degree of match that is limited to the variable(s) that are used for the matching process.

In a repeated-measures design or a matched-subjects design comparing two treatment

conditions, the data consist of two sets of scores, which are grouped into sets of two, cor-

responding to the two scores obtained for each individual or each matched pair of subjects

(Table 11.1). Because the scores in one set are directly related, one-to-one, with the scores

in the second set, the two research designs are statistically equivalent and share the com-

mon name related-samples designs (or correlated-samples designs). In this chapter, we

focus our discussion on repeated-measures designs because they are overwhelmingly the

more common example of related-samples designs. However, you should realize that the

statistical techniques used for repeated-measures studies also can be applied directly to

data from matched-subjects studies. We should also note that a matched-subjects study

occasionally is called a matched samples design, but the subjects in the samples must be

matched one-to-one before you can use the statistical techniques in this chapter.

Now we examine the statistical techniques that allow a researcher to use the sample

data from a repeated-measures study to draw inferences about the general population.

Participant or Matched Pair First Score Second Score

#1 12 15 ←The 2 scores for

one participant or

one matched pair

#2 10 14

#3 15 17

#4 17 17

#5 12 18

TABLE 11.1

An example of the data from

a repeated-measures or a

matched-subjects study

using n 5 5 participants (or matched pairs).

THE t STATISTIC FOR A REPEATED-MEASURES RESEARCH DESIGN

The t statistic for a repeated-measures design is structurally similar to the other t statis-

tics we have examined. As we shall see, it is essentially the same as the single-sample t

statistic covered in Chapter 9. The major distinction of the related-samples t is that it is

based on difference scores rather than raw scores (X values). In this section, we examine

difference scores and develop the t statistic for related samples.

Many over-the-counter cold medications include the warning “may cause drowsiness.”

Table 11.2 shows an example of data from a study that examines this phenomenon.

Note that there is one sample of n 5 4 participants, and that each individual is measured twice. The first score for each person (X

1 ) is a measurement of reaction time before the

medication was administered. The second score (X 2 ) measures reaction time 1 hour after

taking the medication. Because we are interested in how the medication affects reaction

time, we have computed the difference between the first score and the second score for

each individual. The difference scores, or D values, are shown in the last column of the

table. Notice that the difference scores measure the amount of change in reaction time

for each person. Typically, the difference scores are obtained by subtracting the first

score (before treatment) from the second score (after treatment) for each person:

difference score 5 D 5 X 2 – X

1 (11.1)

11.2

D I F F E R E N C E S CO R E S : T H E DATA F O R

A R E P E AT E D - M E AS U R E S ST U DY

3 1 6 CHAPTER 11 THE t TEST FOR TWO RELATED SAMPLES

Note that the sign of each D score tells you the direction of the change. Person A,

for example, shows a decrease in reaction time after taking the medication (a negative

change), but person B shows an increase (a positive change).

The sample of difference scores (D values) serves as the sample data for the hypoth-

esis test and all calculations are done using the D scores. To compute the t statistic, for

example, we use the number of D scores (n) as well as the sample mean (M D ) and the

value of SS for the sample of D scores.

The researcher’s goal is to use the sample of difference scores to answer questions

about the general population. In particular, the researcher would like to know whether

there is any difference between the two treatment conditions for the general popu-

lation. Note that we are interested in a population of difference scores. That is, we

would like to know what would happen if every individual in the population were

measured in two treatment conditions (X 1 and X

2 ) and a difference score (D) were

computed for everyone. Specifically, we are interested in the mean for the population

of difference scores. We identify this population mean difference with the symbol � D

(using the subscript letter D to indicate that we are dealing with D values rather than

X scores).

As always, the null hypothesis states that, for the general population, there is no

effect, no change, or no difference. For a repeated-measures study, the null hypothesis

states that the mean difference for the general population is zero. In symbols,

H 0 : �

D � 0

Again, this hypothesis refers to the mean for the entire population of difference

scores. Figure 11.1(a) shows an example of a population of difference scores with a

mean of � D � 0. Although the population mean is zero, the individual scores in the

population are not all equal to zero. Thus, even when the null hypothesis is true, we

still expect some individuals to have positive difference scores and some to have nega-

tive difference scores. However, the positives and negatives are unsystematic and in the

long run balance out to � D � 0. Also note that a sample selected from this population

probably will not have a mean exactly equal to zero. As always, there will be some error

between a sample mean and the population mean, so even if � D � 0 (H

0 is true), we do

not expect M D to be exactly equal to zero.

The alternative hypothesis states that there is a treatment effect that causes the scores

in one treatment condition to be systematically higher (or lower) than the scores in the

other condition. In symbols,

H 1 : �

D ≠ 0

T H E H Y P OT H E S E S F O R A R E L AT E D -

SA M P L E S ST U DY

Person Before Medication (X 1 ) After Medication (X

2 ) Difference D

A 215 210 25

B 221 242 21

C 196 219 23

D 203 228 25

∑ D � 64

M D

� � � o D

4 16

TABLE 11.2

Reaction-time measurements

taken before and after

taking an over-the-counter

cold medication.

Note that M D is the mean for the

sample of D scores.

SECTION 11.2 / THE t STATISTIC FOR A REPEATED-MEASURES RESEARCH DESIGN 3 1 7

According to H 1 , the difference scores for the individuals in the population tend to

be systematically positive (or negative), indicating a consistent, predictable difference

between the two treatments.

Figure 11.1(b) shows an example of a population of difference scores with a posi-

tive mean difference, � D . 0. This time, most of the individuals in the population have

difference scores that are greater than zero. A sample selected from this population will

contain primarily positive difference scores and will probably have a mean difference

that is greater than zero, M D . 0. See Box 11.1 for further discussion of H

0 and H

1 .

0 +5–5

D = 0

0 +5 +10

D > 0 (a) (b)

FIGURE 11.1

(a) A population of difference scores for which the mean is µ D 5 0. Note that the typical difference

score (D value) is not equal to zero. (b) A population of difference scores for which the mean is

greater than zero. Note that most of the difference scores are also greater than zero.

BOX

11.1 ANALOGIES FOR H

0 AND H

1 IN THE REPEATED-MEASURES TEST

An Analogy for H 0 : Intelligence is a fairly stable

characteristic; that is, you do not get notice-

ably smarter or dumber from one day to the next.

However, if we gave you an IQ test every day for a

week, we probably would get seven different num-

bers. The day-to-day changes in your IQ score are

caused by random factors such as your health, your

mood, and your luck at guessing answers you do not

know. Some days your IQ score is slightly higher,

and some days it is slightly lower. On average, the

day-to-day changes in IQ should balance out to zero.

This is the situation that is predicted by the null

hypothesis for a repeated-measures test. According to

H 0 , any changes that occur either for an individual or

for a sample are just due to chance, and in the long

run, they will average out to zero.

An Analogy for H 1 : On the other hand, suppose

that we evaluate your performance on a new video

game by measuring your score every day for a week.

Again, we probably will find small differences in

your scores from one day to the next, just as we did

with the IQ scores. However, the day-to-day changes

in your game score will not be random. Instead,

there should be a general trend toward higher scores

as you gain more experience with the new game.

Thus, most of the day-to-day changes should show

an increase. This is the situation that is predicted by

the alternative hypothesis for the repeated-measures

test. According to H 1 , the changes that occur are

systematic and predictable and will not average

out to zero.

3 1 8 CHAPTER 11 THE t TEST FOR TWO RELATED SAMPLES

Figure 11.2 shows the general situation that exists for a repeated-measures hypoth-

esis test. You may recognize that we are facing essentially the same situation that we

encountered in Chapter 9. In particular, we have a population for which the mean

and the standard deviation are unknown, and we have a sample that will be used to

test a hypothesis about the unknown population. In Chapter 9, we introduced the

single-sample t statistic, which allowed us to use a sample mean as a basis for testing

hypotheses about an unknown population mean. This t-statistic formula is used again

here to develop the repeated-measures t test. To refresh your memory, the single-

sample t statistic (Chapter 9) is defined by the formula

t M

s M

5 2m

In this formula, the sample mean, M, is calculated from the data, and the value for the

population mean, m, is obtained from the null hypothesis. The estimated standard error,

s M , is also calculated from the data and provides a measure of how much difference it

is reasonable to expect between a sample mean and the population mean if there is no

treatment effect.

For the repeated-measures design, the sample data are difference scores and

are identified by the letter D, rather than X. Therefore, we modify the t formula

by adding Ds to emphasize that we are dealing with difference scores instead

of X values. Specifically, we are using the mean for a sample of difference

scores, M D , to test a hypothesis about the mean for the population of difference

T H E t STAT I ST I C F O R R E L AT E D SA M P L E S

As noted earlier, the repeated-

measures t formula is also

used for matched-subjects

designs.

FIGURE 11.2

A sample of n 5 4 people

is selected from the popu-

lation. Each individual is

measured twice, once in

treatment I and once in

treatment II, and a differ-

ence score, D, is computed

for each individual. This

sample of difference scores

is intended to represent the

population. Note that we are

using a sample of difference

scores to represent a popu-

lation of difference scores.

Also note that the mean for

the population of difference

scores is unknown. The

null hypothesis states that,

for the general population,

there is no consistent or sys-

tematic difference between

the two treatments, so the

population mean difference

is µ D 5 0.

µD

Population of difference scores

= ?

Sample of difference scores

Subject

A B C D

I II

10 15 12 11

14 13 15 12

4 −2 3 1

SECTION 11.2 / THE t STATISTIC FOR A REPEATED-MEASURES RESEARCH DESIGN 3 1 9

1. For the following data from a repeated-measures study, find the difference scores

and compute the sample mean difference, the variance for the difference scores,

and the standard error for the sample mean difference.

Participant Treatment 1 Treatment 2

A 5 8

B 6 13

C 7 10

D 9 12

2. If the sample in the previous problem were being used to test for a significant differ-

ence between the two treatments, state the null hypothesis in words and in symbols.

1. The difference scores are 3, 7, 3, and 3. The mean difference is M D 5 4 with SS 5 12,

s2 5 4 and the standard error for the sample mean difference is 1 point.

2. The null hypothesis states that, for the general population, the average difference between

the two treatments is zero. In symbols, m D 5 0.

L E A R N I N G C H E C K

ANSWERS

scores m D . With this simple change, the t formula for the repeated-measures

design becomes

t M

D D

M D

5 2m

(11.2)

In this formula, the estimated standard error for M D , s

M D , is computed in exactly the

same way as it is computed for the single-sample t statistic. The first step is to compute

the variance (or the standard deviation) for the sample of D scores.

s SS

df s

1 5

2 5 5or

The estimated standard error is then computed using the sample variance (or sample

standard deviation) and the sample size, n.

s s

n s

n M MD D

5 5 2

(11.3)

Notice that all of the calculations are done using the difference scores (the D scores)

and that there is only one D score for each subject. With a sample of n subjects, there

are exactly n D scores, and the t statistic has df 5 n – 1. Remember that n refers to the

number of D scores, not the number of X scores in the original data.

You should also note that the repeated-measures t statistic is conceptually similar to

the t statistics that we have previously examined:

t 5 2sample statistic population parameter

estimmated standard error

In this case, the sample data are represented by the sample mean of the difference

scores (M D ), the population parameter is the value predicted by H

0 (m

D 5 0), and the

estimated standard error is computed from the sample data using Equation 11.3.

3 2 0 CHAPTER 11 THE t TEST FOR TWO RELATED SAMPLES

HYPOTHESIS TESTS AND EFFECT SIZE FOR THE REPEATED-MEASURES DESIGN

In a repeated-measures study, each individual is measured in two different treatment

conditions and we are interested in whether there is a systematic difference between the

scores in the first treatment condition and the scores in the second treatment condition.

A difference score (D value) is computed for each person and the hypothesis test uses

the difference scores from the sample to evaluate the overall mean difference, m D , for

the entire population. The hypothesis test with the repeated-measures t statistic follows

the same four-step process that we have used for other tests. The complete hypothesis-

testing procedure is demonstrated in Example 11.1.

Swearing is a common, almost reflexive, response to pain. Whether you knock your shin

into the edge of a coffee table or smash your thumb with a hammer, most of us respond

with a streak of obscenities. One question, however, is whether swearing focuses attention

on the pain and, thereby, increases its intensity, or serves as a distraction that reduces pain.

To address this issue, Stephens, Atkins, and Kingston (2009) conducted an experiment

comparing swearing with other responses to pain. In the study, participants were asked to

place one hand in icy cold water for as long as they could bear the pain. Half of the partici-

pants were told to repeat their favorite swear word over and over for as long as their hands

were in the water. The other half repeated a neutral word. The researchers recorded how

long each participant was able to tolerate the ice water. After a brief rest, the two groups

switched words and repeated the ice water plunge. Thus, all the participants experienced

both conditions (swearing and neutral) with half swearing on their first plunge and half on

their second. The results clearly showed that swearing significantly increased pain tolerance

and decreased the perceived level of pain. The data in Table 11.3 are representative of the

results obtained in the study and represented the reports of pain level of n 5 9 participants.

State the hypotheses, and select the alpha level.

H 0 : m

D 5 0 (There is no difference between the two conditions.)

H 1 : m

D  0 (There is a difference.)

For this test, we use a 5 .05.

11.3

E X A M P L E 1 1 . 1

S T E P 1

Participant Neutral Word Swearing D D2

A 9 7 –2 4

B 8 7 –1 1

C 7 3 –4 16

D 7 8 11 1

E 8 6 –2 4

F 9 4 –5 25

G 7 6 –1 1

H 7 7 0 0

I 8 4 –4 16

D 5 –18 D2 5 68

M SS D

N D

5 2

5 2 5 2 5 2 2

5 2 18

9 68

9 68 36

2 2

o o

D ( ) ( )

55 32

TABLE 11.3

Ratings of pain level on a

scale from 1 to 10 for partici-

pants who were swearing or

repeating a neutral word.

SECTION 11.3 / HYPOTHESIS TESTS AND EFFECT SIZE FOR THE REPEATED-MEASURES DESIGN 3 2 1

Locate the critical region. For this example, n 5 9, so the t statistic has df 5 n – 1 5 8. For a 5 .01, the critical value listed in the t distribution table is 62.306. The critical region is shown in Figure 11.3.

Calculate the t statistic. Table 11.3 shows the sample data and the calculations of

M D 5 –2 and SS 5 32. Note that all calculations are done with the difference scores.

As we have done with the other t statistics, we present the calculation of the t statistic

as a three-step process.

First, compute the sample variance

s SS

8 45

2 5=

Next, use the sample variance to compute the estimated standard error.

s s

n M D

5 5 5 2

9 0 667.

Finally, use the sample mean (M D ) and the hypothesized population mean (m

D ) along

with the estimated standard error to compute the value for the t statistic.

t M

D D

M D

5 2m

5 2 2

52 2 0

0 667 3 00

. .

Make a decision. The t value we obtained falls in the critical region (see Figure 11.3).

The researcher rejects the null hypothesis and concludes that cursing, as opposed to

repeating a neutral work, has a significant effect on pain perception.

As we noted with other hypothesis tests, whenever a treatment effect is found to be sta-

tistically significant, it is recommended that you also report a measure of the absolute

magnitude of the effect. The most commonly used measures of effect size are Cohen’s

d and r2, the percentage of variance accounted for. The size of the treatment effect also

can be described with a confidence interval estimating the population mean difference,

m D . Using the data from Example 11.1, we demonstrate how these values are calculated

to measure and describe effect size.

S T E P 2

S T E P 3

S T E P 4

M E AS U R I N G E F F E C T S I Z E F O R

T H E R E P E AT E D - M E AS U R E S t

2.306 2.3060

Reject H0

FIGURE 11.3

The critical region for the

t distribution with df 5 8

and a 5 .05.

3 2 2 CHAPTER 11 THE t TEST FOR TWO RELATED SAMPLES

Cohen’s d In Chapters 8 and 9, we introduced Cohen’s d as a standardized measure

of the mean difference between treatments. The standardization simply divides the

population mean difference by the standard deviation. For a repeated-measures study,

Cohen’s d is defined as

d 5 population mean difference

standard deviatiion 5

Because the population mean and standard deviation are unknown, we use the sample

values instead. The sample mean, M D , is the best estimate of the actual mean difference, and

the sample standard deviation (square root of sample variance) provides the best estimate

of the actual standard deviation. Thus, we are able to estimate the value of d as follows:

estimated sample mean difference

sample me d 5

aan deviation 5

(11.4)

For the repeated-measures study in Example 11.1, M D 5 22 and the sample variance

is s2 5 4.00, so the data produce

estimated d M

D5 5 2

5 2

5 2 2

4 00

2 1 00

. .

Any value greater than 0.8 is considered to be a large effect, and these data are

clearly in that category (see Table 8.2 on p. 232).

The percentage of variance accounted for, r2 Percentage of variance is computed

using the obtained t value and the df value from the hypothesis test, exactly as was done

for the single-sample t (see p. 263) and for the independent-measures t (see p. 292). For

the data in Example 11.1, we obtain

r t

t df

2 2

3 00

3 00 8

17 0 5295

1 5

1 5 5

. .

( ) ( )

or 552 9. %

For these data, 52.9% of the variance in the scores is explained by the effect of curs-

ing. More specifically, swearing caused the estimated pain ratings to be consistently

negative. Thus, the deviations from zero are largely explained by the treatment.

Confidence intervals for estimating m D As noted in the previous two chapters, it is

possible to compute a confidence interval as an alternative method for measuring and

describing the size of the treatment effect. For the repeated-measures t, we use a sample

mean difference, M D , to estimate the population mean difference, m

D . In this case, the

confidence interval literally estimates the size of the treatment effect by estimating the

population mean difference between the two treatment conditions.

As with the other t statistics, the first step is to solve the t equation for the unknown

parameter. For the repeated-measures t statistic, we obtain

m 5 6 D D M

M ts D

(11.5)

In the equation, the values for M D and for sM D are obtained from the sample data.

Although the value for the t statistic is unknown, we can use the degrees of freedom

for the t statistic and the t distribution table to estimate the t value. Using the estimated

t and the known values from the sample, we can then compute the value of m D . The

following example demonstrates the process of constructing a confidence interval for a

population mean difference.

Because we are measuring

the size of the effect and not

the direction, it is customary

to ignore the minus sign and

report Cohen’s d as a positive

value.

SECTION 11.3 / HYPOTHESIS TESTS AND EFFECT SIZE FOR THE REPEATED-MEASURES DESIGN 3 2 3

In Example 11.1, we presented a research study demonstrating how swearing influenced

the perception of pain. In the study, a sample of n 5 9 participants rated their level of pain significantly lower when they were repeating a swear word than when they repeated

a neutral word. The mean difference between treatments was M D

5 22 points and the estimated standard error for the mean difference was s

M D 5 0.667. Now, we construct a

95% confidence interval to estimate the size of the population mean difference.

With a sample of n 5 9 participants, the repeated-measures t statistic has df 5 8. To

have 95% confidence, we simply estimate that the t statistic for the sample mean differ-

ence is located somewhere in the middle 95% of all the possible t values. According to

the t distribution table, with df 5 8, 95% of the t values are located between t 5 12.306

and t 5 22.306. Using these values in the estimation equation, together with the values

for the sample mean and the standard error, we obtain

m 5 6 D D M

M ts D

5 22 6 2.306(0.667)

5 22 6 1.538

This produces an interval of values ranging from 22 – 1.538 5 23.538 to 22 1

1.538 5 20.462. Our conclusion is that for the general population, swearing instead of

repeating a neutral word decreases the perceived pain between 0.462 and 3.538 points.

We are 95% confident that the true mean difference is in this interval because the only

value estimated during the calculations was the t statistic, and we are 95% confident

that the t value is located in the middle 95% of the distribution. Finally, note that the

confidence interval is constructed around the sample mean difference. As a result, the

sample mean difference, M D 5 22 points, is located exactly in the center of the interval.

As with the other confidence intervals presented in Chapters 9 and 10, the confi-

dence interval for a repeated-measures t is influenced by a variety of factors other than

the actual size of the treatment effect. In particular, the width of the interval depends

on the percentage of confidence used, so that a larger percentage produces a wider

interval. Also, the width of the interval depends on the sample size, so that a larger

sample produces a narrower interval. Because the interval width is related to sample

size, the confidence interval is not a pure measure of effect size like Cohen’s d or r2.

Finally, we should note that the 95% confidence interval computed in Example 11.2

does not include the value m D

5 0. In other words, we are 95% confident that the popula-

tion mean difference is not m D 5 0. This is equivalent to concluding that a null hypoth-

esis specifying that m D 5 0 would be rejected with a test using a 5 .05. If m

D 5 0 were

included in the 95% confidence interval, it would indicate that a hypothesis test would

fail to reject H 0 with a 5 .05.

E X A M P L E 1 1 . 2

1. A researcher is investigating the effect of a treatment by measuring performance

for a sample of n 5 9 participants before and after they receive the treatment.

For this sample, performance scores increased after treatment by an average of

M 5 1.9 points with SS 5 200.

a. Are the data sufficient to conclude that treatment has a significant effect on

performance? Use a two-tailed test with a 5 .05.

b. Compute the effect size for the treatment using both Cohen’s d and r2.

c. Compute the 90% confidence for the population mean difference.

L E A R N I N G C H E C K

3 2 4 CHAPTER 11 THE t TEST FOR TWO RELATED SAMPLES

IN THE LITERATURE

REPORTING THE RESULTS OF A REPEATED-MEASURES t TEST

As we have seen in Chapters 9 and 10, the APA format for reporting the results of t tests

consists of a concise statement that incorporates the t value, degrees of freedom, and

alpha level. One typically includes values for means and standard deviations, either in a

statement or a table (Chapter 4). For Example 11.1, we observed a mean difference of

M D 5 22.00 with s 5 2.00. Also, we obtained a t statistic of t 5 23.00 with df 5 8, and

our decision was to reject the null hypothesis at the .05 level of significance. Finally, we

measured effect size by computing the percentage of variance explained and obtained

r2 5 0.529. A published report of this study might summarize the results as follows:

Changing from a neutral word to a swear word reduced the perceived level of pain by

an average of M 5 2.00 points with SD 5 2.00. The treatment effect was statistically

significant, t(8) 5 23.00, p , .05, r2 5 0.529.

When the hypothesis test is conducted with a computer program, the printout typi-

cally includes an exact probability for the level of significance. The p-value from the

printout is then stated as the level of significance in the research report. For example,

the data from Example 11.1 produced a significance level of p 5 .017, and the results

would be reported as “statistically significant, t(8) 5 23.00, p 5 .017, r2 5 0.529.”

Occasionally, a probability is so small that the computer rounds it off to 3 decimal

points and produces a value of zero. In this situation you do not know the exact prob-

ability value and should report p , .001.

If the confidence interval from Example 11.2 is reported as a description of effect

size together with the results from the hypothesis test, it would appear as follows:

Changing from a neutral word to a swear word reduced the perceived level of pain,

t(8) 5 –3.00, p , .05, 95% CI [–0.462, –3.538].

Often, a close look at the sample data from a research study makes it easier to see the size

of the treatment effect and to understand the outcome of the hypothesis test. In Example

11.1, we obtained a sample of n 5 9 participants who produce a mean difference of

M D 5 –2.00 with a standard deviation of s 5 2.00 points. The sample mean and standard

deviation describe a set of scores centered at M D 5 –2.00 with most of the scores located

within 2.00 points of the mean. Figure 11.4 shows the actual set of difference scores that

were obtained in Example 11.1. In addition to showing the scores in the sample, we have

highlighted the position of m D 5 0; that is, the value specified in the null hypothesis.

Notice that the scores in the sample are displaced away from zero. Specifically, the data

are not consistent with a population mean of m D 5 0, which is why we rejected the null

hypothesis. In addition, note that the sample mean is located 1 standard deviation below

zero. This distance corresponds to the effect size measured by Cohen’s d 5 –1.00. For

these data, the picture of the sample distribution (see Figure 11.4) should help you to

understand the measure of effect size and the outcome of the hypothesis test.

D E S C R I P T I V E STAT I ST I C S A N D

T H E H Y P OT H E S I S T E ST

1. a. The null hypothesis states that the treatment has no effect, m D 5 0. The sample variance is

25, the standard error is 1.67 and t(8) 5 1.14. With df 5 24, the critical value is t 5 2.064.

Reject the null hypothesis and conclude that the treatment has a significant effect.

b. Cohen’s d 5 1 9 5 .

5 0.38 and r2 5 10 05 34 10

. 5 0.295.

c. For 90% confidence, use t 5 61.711. The interval is m D 5 1.9 61.711(0.6) and extends

from 0.87 to 2.93.

ANSWERS

SECTION 11.3 / HYPOTHESIS TESTS AND EFFECT SIZE FOR THE REPEATED-MEASURES DESIGN 3 2 5

In Chapters 9 and 10 (p. 259 and 298), we identified several factors that can influ-

ence the outcome of a hypothesis test with the t statistic. The same factors apply

to the repeated-measures t. The most obvious factor is the size of the sample

mean difference. A larger mean difference increases the likelihood of rejecting the

null hypothesis and increases measures of effect size. The two other factors are the

variability of the scores and the size of the sample, which both influence the magni-

tude of the estimated standard error in the denominator of the t statistic. Specifically,

the standard error is inversely related to sample size (larger size leads to smaller

error) and is directly related to sample variance (larger variance leads to larger

error). As a result, a larger sample produces a larger value for the t statistic (farther

from zero) and increases the likelihood of rejecting H 0 . Larger variance, on the other

hand, produces a smaller value for the t statistic (closer to zero) and reduces the

likelihood of finding a significant result.

Although variance and sample size both influence the hypothesis test, only variance

has a large influence on measures of effect size such as Cohen’s d and r2; larger variance

produces smaller measures of effect size. Sample size, on the other hand, has no effect

on the value of Cohen’s d and only a small influence on r2.

In a repeated-measures study, the variability of the difference scores becomes a rela-

tively concrete and easy-to-understand concept. In particular, the sample variability

describes the consistency of the treatment effect. For example, if a treatment consis-

tently adds a few points to each individual’s score, then the set of difference scores are

clustered together with relatively small variability. This is the situation that we observed

in Example 11.1 (see Figure 11.4) in which nearly all of the participants had lower rat-

ings of pain in the swearing condition. In this situation, with small variability, it is easy

to see the treatment effect and it is likely to be significant.

Now consider what happens when the variability is large. Suppose that the swear-

ing study in Example 11.1 produced a sample of n 5 9 difference scores consisting of –11, –10, –7, –2, 0, 0, 13, 14, and 15. These difference scores also have a mean of

M D 5 –2.00, but now the variability is substantially increased so that SS 5 288 and

the standard deviation is s 5 6.00. Figure 11.5 shows the new set of difference scores.

Again, we have highlighted the position of m D 5 0, which is the value specified in

the null hypothesis. Notice that the high variability means that there is no consistent

SA M P L E VA R I A N C E A N D SA M P L E S I Z E I N T H E R E P E AT E D - M E AS U R E S t T E ST

VA R I A B I L I T Y AS A M E AS U R E O F

CO N S I ST E N CY F O R T H E T R E AT M E N T E F F E C T

–4 –3 –2 –1 0 +1–5 D

MD � –2

s � 2

mD � 0

FIGURE 11.4

The sample of difference

scores from Example 11.1.

The mean is M D 5 –2 and

the standard deviation is

s 5 2. The difference

scores are consistently

negative, indicating a

decrease in perceived pain,

suggest that m D 5 0 (no

effect) is not a reasonable

hypothesis.

3 2 6 CHAPTER 11 THE t TEST FOR TWO RELATED SAMPLES

treatment effect. Some participants rate the pain higher while swearing (the positive

differences) and some rate it lower (the negative differences). In the hypothesis test,

the high variability increases the size of the estimated standard error and results in a

hypothesis test that produces t 5 1.50, which is not in the critical region. With these data, we would fail to reject the null hypothesis and conclude that swearing has no

effect on the perceived level of pain.

With small variability (see Figure 11.4), the 2-point treatment effect is easy to see

and is statistically significant. With large variability (see Figure 11.5), the 2-point

effect is not easy to see and is not significant. As we have noted several times in the

past, large variability can obscure patterns in the data and reduces the likelihood of

finding a significant treatment effect.

In many repeated-measures and matched-subjects studies, the researcher has a specific

prediction concerning the direction of the treatment effect. For example, in the study

described in Example 11.1, the researcher expects the level of perceived pain to be rated

lower when the participant is cursing. This kind of directional prediction can be incor-

porated into the statement of the hypotheses, resulting in a directional, or one-tailed,

hypothesis test. The following example demonstrates how the hypotheses and critical

region are determined for a directional test.

We reexamine the experiment presented in Example 11.1. The researcher is using

a repeated-measures design to investigate the effect of swearing on perceived pain.

The researcher predicts that the pain ratings for the ice water will decrease when the

participants are swearing compared to repeating a neutral word.

State the hypotheses and select the alpha level. For this example, the researcher

predicts that pain ratings will decrease when the participants are swearing. The null

D I R E C T I O N A L H Y P OT H E S I S A N D O N E -

TA I L E D T E ST S

E X A M P L E 1 1 . 3

S T E P 1

0–10 –9 –8 –7 –6 –5 –4 –3 –2 –1–11 +5+4+3+2+1 D

MD � –2

s � 6

mD � 0

FIGURE 11.5

A sample of difference scores with a mean of M D 5 –2 and a standard deviation of s 5 6. The data

do not show a consistent increase or decrease in scores. Because there is no consistent treatment

effect, m D 5 0 is a reasonable hypothesis.

SECTION 11.3 / HYPOTHESIS TESTS AND EFFECT SIZE FOR THE REPEATED-MEASURES DESIGN 3 2 7

hypothesis, on the other hand, says that the pain ratings will not decrease but rather will

be unchanged or even increased with the swearing. In symbols,

H 0 : m

D $ 0 (There is no decrease with swearing.)

The alternative hypothesis says that the treatment does work. For this example,

H 1 says that swearing will decrease the pain ratings.

H 1 : m

D , 0 (The ratings are decreased.)

We use a 5 .01.

Locate the critical region. As we demonstrated with the independent-measures t sta-

tistic (p. 297), the critical region for a one-tailed test can be located using a two-stage

process. Rather than trying to determine which tail of the distribution contains the

critical region, you first look at the sample mean difference to verify that it is in the

predicted direction. If not, then the treatment clearly did not work as expected and you

can stop the test. If the change is in the correct direction, then the question is whether

it is large enough to be significant. For this example, change is in the predicted direc-

tion (the researcher predicted lower ratings and the sample mean shows a decrease).

With n 5 9, we obtain df 5 8 and a critical value of t 5 2.896 for a one-tailed test

with a 5 .01. Thus, any t statistic beyond 2.896 (positive or negative) is sufficient to

reject the null hypothesis.

Compute the t statistic. We calculated the t statistic in Example 11.1 and obtained

t 5 23.00.

Make a decision. The obtained t statistic is beyond the critical boundary. Therefore, we

reject the null hypothesis and conclude that swearing significantly reduced the pain rat-

ings. In a research report, the use of a one-tailed test would be clearly noted as follows:

Swearing, compared to repeating a neutral word, significantly decreased the pain ratings,

t(8) 5 23.00, p , .01, one tailed.

S T E P 2

S T E P 3

S T E P 4

1. A researcher is investigating the effectiveness of acupuncture treatment for chronic

back pain. A sample of n 5 4 participants is obtained from a pain clinic. Each

individual ranks the current level of pain and then begins a 6-week program of

acupuncture treatment. At the end of the program, the pain level is rated again and

the researcher records the amount of difference between the two ratings. For this

sample, pain level decreased by an average of M 5 4.5 points with SS 5 27.

a. Are the data sufficient to conclude that acupuncture has a significant effect on

back pain? Use a two-tailed test with a 5 .05.

b. Can you conclude that acupuncture significantly reduces back pain? Use a

one-tailed test with a 5 .05.

2. Compute the effect size using both Cohen’s d and r2 acupuncture study in the

previous question.

3. A computer printout for a repeated-measures t test reports a p value of p 5 .021.

a. Can the researcher claim a significant effect with a 5 .01?

b. Is the effect significant with a 5 .05?

L E A R N I N G C H E C K

3 2 8 CHAPTER 11 THE t TEST FOR TWO RELATED SAMPLES

USES AND ASSUMPTIONS FOR REPEATED-MEASURES t TESTS

In many research situations, it is possible to use either a repeated-measures design or an

independent-measures design to compare two treatment conditions. The independent-

measures design would use two separate samples (one in each treatment condition) and

the repeated-measures design would use only one sample with the same individuals

participating in both treatments. The decision about which design to use is often made

by considering the advantages and disadvantages of the two designs. In general, the

repeated-measures design has most of the advantages.

Number of subjects A repeated-measures design typically requires fewer subjects

than an independent-measures design. The repeated-measures design uses the subjects

more efficiently because each individual is measured in both of the treatment condi-

tions. This can be especially important when there are relatively few subjects available

(for example, when you are studying a rare species or individuals in a rare profession).

Study changes over time The repeated-measures design is especially well suited for

studying learning, development, or other changes that take place over time. Remember

that this design involves measuring individuals at one time and then returning to mea-

sure the same individuals at a later time. In this way, a researcher can observe behaviors

that change or develop over time.

Individual differences The primary advantage of a repeated-measures design is that it

reduces or eliminates problems caused by individual differences. Individual differences

are characteristics such as age, IQ, gender, and personality that vary from one individual

to another. These individual differences can influence the scores obtained in a research

study, and they can affect the outcome of a hypothesis test. Consider the data in Table 11.4.

The first set of data represents the results from a typical independent-measures study, and

the second set represents a repeated-measures study. Note that we have identified each

participant by name to help demonstrate the effects of individual differences.

For the independent-measures data, note that every score represents a different

person. For the repeated-measures study, on the other hand, the same participants are

measured in both of the treatment conditions. This difference between the two designs

has some important consequences.

1. We have constructed the data so that both research studies have exactly the same

scores and they both show the same 5-point mean difference between treatments.

In each case, the researcher would like to conclude that the 5-point difference

11.4

R E P E AT E D - M E AS U R E S V E R S U S I N D E P E N D E N T-

M E AS U R E S D E S I G N S

1. a. For these data, the sample variance is 9, the standard error is 1.50, and t 5 3.00. With df 5 3, the critical values are t 5 63.182. Fail to reject the null hypothesis.

b. For a one-tailed test, the critical value is t 5 2.353. Reject the null hypothesis and con- clude that acupuncture treatment significantly reduces pain.

2. d 5 4.5/3 5 1.50 and r2 5 9/12 5 0.75.

3. a. The exact p value, p 5 .021, is not less than a 5 .01. Therefore, the effect is not signifi- cant for a 5 .01 (p . .01).

b. The p value is less than .05, so the effect is significant with a 5 .05.

ANSWERS

SECTION 11.4 / USES AND ASSUMPTIONS FOR REPEATED-MEASURES t TESTS 3 2 9

was caused by the treatments. However, with the independent-measures design,

there is always the possibility that the participants in treatment 1 have different

characteristics than those in treatment 2. For example, the three participants in

treatment 1 may be more intelligent than those in treatment 2 and their higher in-

telligence caused them to have higher scores. Note that this problem disappears

with the repeated-measures design. Specifically, with repeated measures there

is no possibility that the participants in one treatment are different from those in

another treatment because the same participants are used in all of the treatments.

2. Although the two sets of data contain exactly the same scores and have exactly

the same 5-point mean difference, you should realize that they are very different

in terms of the variance used to compute standard error. For the independent-

measures study, you calculate the SS or variance for the scores in each of the

two separate samples. Note that in each sample there are big differences between

participants. In treatment 1, for example, Bill has a score of 33 and John’s score

is only 18. These individual differences produce a relatively large sample variance

and a large standard error. For the independent-measures study, the standard error

is 5.77, which produces a t statistic of t 5 0.87. For these data, the hypothesis test concludes that there is no significant difference between treatments.

In the repeated-measures study, the SS and variance are computed for the

difference scores. If you examine the repeated-measures data in Table 11.4,

you will see that the big differences between John and Bill that exist in treat-

ment 1 and in treatment 2 are eliminated when you get to the difference scores.

Because the individual differences are eliminated, the variance and standard

error are dramatically reduced. For the repeated-measures study, the standard

error is 1.15 and the t statistic is t 5 –4.35. With the repeated-measures t, the data show a significant difference between treatments. Thus, one big advantage

of a repeated-measures study is that it reduces variance by removing individual

differences, which increases the chances of finding a significant result.

The primary disadvantage of a repeated-measures design is that the structure of the design

allows for factors other than the treatment effect to cause a participant’s score to change

from one treatment to the next. Specifically, in a repeated-measures design, each indi-

vidual is measured in two different treatment conditions, usually at two different times. In

this situation, outside factors that change over time may be responsible for changes in the

participants’ scores. For example, a participant’s health or mood may change over time and

cause a difference in the participant’s scores. Outside factors such as the weather can also

change and may have an influence on participants’ scores. Because a repeated-measures

study typically takes place over time, it is possible that time-related factors (other than the

two treatments) are responsible for causing changes in the participants’ scores.

T I M E - R E L AT E D FAC TO R S A N D O R D E R

E F F E C T S

Independent-Measures Study (2 Separate Samples)

Repeated-Measures Study (Same Sample in Both Treatments)

Treatment 1 Treatment 2 Treatment 1 Treatment 2 D

(John) X 5 18 (Sue) X 5 15 (John) X 5 18 (John) X 5 15 –3

(Mary) X 5 27 (Tom) X 5 20 (Mary) X 5 27 (Mary) X 5 20 –7

(Bill) X 5 33 (Dave) X 5 28 (Bill) X 5 33 (Bill) X 5 28 –5

M 5 26 M 5 21 M D 5 –5

SS 5 114 SS 5 86 SS 5 8

TABLE 11.4

Hypothetical data showing the

results from an independent-

measures study and a repeated-

measures study. The two sets

of data use exactly the same

numerical scores and they both

show the same 5-point mean

difference between treatments.

3 3 0 CHAPTER 11 THE t TEST FOR TWO RELATED SAMPLES

Also, it is possible that participation in the first treatment influences the individual’s

score in the second treatment. If the researcher is measuring individual performance, for

example, the participants may gain experience during the first treatment condition, and this

extra practice may help their performance in the second condition. In this situation, the

researcher would find a mean difference between the two conditions; however, the differ-

ence would not be caused by the treatments, instead it would be caused by practice effects.

Changes in scores that are caused by participation in an earlier treatment are called order

effects and can distort the mean differences found in repeated-measures research studies.

Counterbalancing One way to deal with time-related factors and order effects is

to counterbalance the order of presentation of treatments. That is, the participants are

randomly divided into two groups, with one group receiving treatment 1 followed by

treatment 2, and the other group receiving treatment 2 followed by treatment 1. The

goal of counterbalancing is to distribute any outside effects evenly over the two treat-

ments. For example, if practice effects are a problem, then half of the participants

gain experience in treatment 1, which then helps their performance in treatment 2.

However, the other half gain experience in treatment 2, which helps their performance

in treatment 1. Thus, prior experience helps the two treatments equally.

Finally, if there is reason to expect strong time-related effects or strong order effects,

your best strategy is not to use a repeated-measures design. Instead, use independent-

measures (or a matched-subjects design) so that each individual participates in only one

treatment and is measured only one time.

The related-samples t statistic requires two basic assumptions:

1. The observations within each treatment condition must be independent

(see p. 220). Notice that the assumption of independence refers to the scores

within each treatment. Inside each treatment, the scores are obtained from

different individuals and should be independent of one another.

2. The population distribution of difference scores (D values) must be normal. As

before, the normality assumption is not a cause for concern unless the sample

size is relatively small. In the case of severe departures from normality, the

validity of the t test may be compromised with small samples. However, with

relatively large samples (n . 30), this assumption can be ignored.

AS S U M P T I O N S O F T H E R E L AT E D - SA M P L E S t T E ST

1. What are the basic assumptions underlying a hypothesis test with the repeated-

measures t?

2. A repeated-measures study and an independent-measures study are used to

compare two treatments and both use a total of 20 participants. What are the

df values for the two t tests?

3. Compared to an independent-measures study, a repeated-measures study tends

to have less variance and a greater likelihood of detecting a treatment effect.

(True or false?)

1. The observations within a treatment are independent. The population distribution of

D scores is assumed to be normal.

2. The repeated-measures t has df 5 19 and the independent-measures t has df 5 18.

3. True

L E A R N I N G C H E C K

ANSWERS

SUMMARY 3 3 1

SUMMARY

1. In a related-samples research study, the individuals in one treatment condition are directly related, one-to-one, with the individuals in the other treatment condition(s). The most common related-samples study is a repeated-measures design, in which the same sample of individuals is tested in all of the treatment conditions. This design literally repeats measurements on the same subjects. An alternative is a matched- subjects design, in which the individuals in one sample are matched one-to-one with individuals in another sample. The matching is based on a variable relevant to the study.

2. The repeated-measures t test begins by computing a dif- ference between the first and second measurements for each subject (or the difference for each matched pair). The difference scores, or D scores, are obtained by

D 5 X 2 – X

The sample mean, M D , and sample variance, s2, are

used to summarize and describe the set of difference scores.

3. The formula for the repeated-measures t statistic is

t M

D D

M D

5 2m

In the formula, the null hypothesis specifies m D 5 0,

and the estimated standard error is computed by

s s

n M D

5 2

4. A repeated-measures design may be preferred to an independent-measures study when one wants to observe changes in behavior in the same subjects, as in learning or developmental studies. An important

advantage of the repeated-measures design is that it removes or reduces individual differences, which lowers sample variability and tends to increase the chances for obtaining a significant result.

5. For a repeated-measures design, effect size can be measured using either r2 (the percentage of variance accounted for) or Cohen’s d (the standardized mean difference). The value of r2 is computed the same way for both independent- and repeated-measures designs.

r t

t df

2 2

2 5

Cohen’s d is defined as the sample mean difference divided by standard deviation for both repeated- and independent-measures designs. For repeated-measures studies, Cohen’s d is estimated as

estimated d M

6. An alternative method for describing the size of the treatment effect is to construct a confidence interval for the population mean difference, m

D . The confi-

dence interval uses the repeated-measures t equation, solved for the unknown mean difference:

m 5 6 D D M

M ts D

KEY TERMS

repeated-measures design (314)

within-subjects design (314)

matched-subjects design (314)

related-samples design (315)

difference scores (315)

estimated standard error for M D (319)

repeated-measures t statistic (319)

individual differences (328)

order effects (330)

3 3 2 CHAPTER 11 THE t TEST FOR TWO RELATED SAMPLES

RESOURCES

Go to CengageBrain.com to access Psychology CourseMate, where you will find an

interactive eBook, glossaries, flashcards, quizzes, statistics workshops, and more.

If your professor has assigned Aplia:

1. Sign in to your account.

2. Complete the corresponding exercises as required by your professor.

3. When finished, click “Grade It Now” to see which areas you have mastered, which

areas need more work, and detailed explanations of every answer.

General instructions for using SPSS are presented in Appendix D. Following are

detailed instructions for using SPSS to perform The Repeated-Measures t Test

presented in this chapter.

Data Entry

Enter the data into two columns (VAR0001 and VAR0002) in the data editor with the first

score for each participant in the first column and the second score in the second column.

The two scores for each participant must be in the same row.

Data Analysis

1. Click Analyze on the tool bar, select Compare Means, and click on Paired-

Samples T Test.

2. One at a time, highlight the column labels for the two data columns and click

the arrow to move them into the Paired Variables box.

3. In addition to performing the hypothesis test, the program computes a confi-

dence interval for the population mean difference. The confidence level is auto-

matically set at 95%, but you can select Options and change the percentage.

4. Click OK.

SPSS Output

We used the SPSS program to analyze the data from the swearing experiment in

Example 11.1 and the program output is shown in Figure 11.6. The output includes a table

of sample statistics with the mean and standard deviation for each treatment. A second

table shows the correlation between the two sets of scores (correlations are presented in

Chapter 14). The final table, which is split into two sections in Figure 11.6, shows the

results of the hypothesis test, including the mean and standard deviation for the difference

scores, the standard error for the mean, a 95% confidence interval for the mean difference,

and the values for t, df, and the level of significance (the p value for the test).

FOCUS ON PROBLEM SOLVING

1. Once data have been collected, we must then select the appropriate statistical

analysis. How can you tell whether the data call for a repeated-measures

DEMONSTRATION 11.1 3 3 3

t test? Look at the experiment carefully. Is there only one sample of subjects?

Are the same subjects tested a second time? If your answers are yes to both of

these questions, then a repeated-measures t test should be done. There is only

one situation in which the repeated-measures t can be used for data from two

samples, and that is for matched-subjects studies (p. 314).

2. The repeated-measures t test is based on difference scores. In finding difference

scores, be sure that you are consistent with your method. That is, you may use

either X 2 – X

1 or X

1 – X

2 to find D scores, but you must use the same method for

all subjects.

DEMONSTRATION 11.1

A REPEATED-MEASURES t TEST

A major oil company would like to improve its tarnished image following a large oil

spill. Its marketing department develops a short television commercial and tests it on

Paired Samples Statistics

Paired Samples Correlations

Paired Samples Test

VAR00001

VAR00002

VAR00001 & VAR00002

Pair 1

Pair 1 VAR00001 - VAR00002 .46266 3.53734 3.000 8 .017

.746–.1269

.83333

1.71594

.27778

.57198

CorrelationN

Paired Differences

Sig.

Lower Upper t df Sig. (2-tailed)

95% Confidence Interval of the Difference

7.7778

5.7778

Mean Std. Deviation Std. Error

Mean

Paired Samples Test

VAR00001 - VAR00002Pair 1 .666672.000002.00000

Std. Deviation

Paired Differences

Mean Std. Error

Mean

FIGURE 11.6

The SPSS output for the repeated-measures hypothesis test in Example 11.1.

3 3 4 CHAPTER 11 THE t TEST FOR TWO RELATED SAMPLES

a sample of n 5 7 participants. People’s attitudes about the company are measured with a short questionnaire, both before and after viewing the commercial. The data

are as follows:

Person X 1 (Before) X

2 (After) D (Difference)

A 15 15 0

B 11 13 12 ∑D 5 21

C 10 18 18

D 11 12 11 M D 5

21 7 5 3.00

E 14 16 12

F 10 10 0 SS 5 74

G 11 19 18

Was there a significant change? Note that participants are being tested twice—once

before and once after viewing the commercial. Therefore, we have a repeated-

measures design.

State the hypotheses, and select an alpha level. The null hypothesis states that the commercial has no effect on people’s attitude, or, in symbols,

H 0 : m

D 5 0 (The mean difference is zero.)

The alternative hypothesis states that the commercial does alter attitudes about the

company, or

H 1 : m

D ≠ 0 (There is a mean change in attitudes.)

For this demonstration, we use an alpha level of .05 for a two-tailed test.

Locate the critical region. Degrees of freedom for the repeated-measures t test are

obtained by the formula

df 5 n – 1

For these data, degrees of freedom equal

df 5 7 – 1 5 6

The t distribution table is consulted for a two-tailed test with a 5 .05 for df 5 6. The

critical t values for the critical region are t 5 62.447.

Compute the test statistic. Once again, we suggest that the calculation of the t statistic

be divided into a three-part process.

Variance for the D scores: The variance for the sample of D scores is

s SS

6 12 335

2 5 5 .

Estimated standard error for M D : The estimated standard error for the sample mean

difference is computed as follows:

s s

n M D

5 5 5 5 2

12 33

7 1 76 1 33

. . .

S T E P 1

S T E P 2

S T E P 3

PROBLEMS 3 3 5

The repeated-measures t statistic: Now we have the information required to calculate

the t statistic.

t M

D D

M D

5 2m

5 2

5 3 0

1 33 2 26

. .

Make a decision about H 0 , and state the conclusion. The obtained t value is not ex-

treme enough to fall in the critical region. Therefore, we fail to reject the null hypoth- esis. We conclude that there is not enough evidence to conclude that the commercial changes people’s attitudes, t(6) 5 2.26, p . .05, two-tailed. (Note that we state that p is greater than .05 because we failed to reject H

0 .)

DEMONSTRATION 11.2

EFFECT SIZE FOR THE REPEATED-MEASURES t

We estimate Cohen’s d and calculate r2 for the data in Demonstration 11.1. The

data produced a sample mean difference of M D 5 3.00 with a sample variance of

s2 5 12.33. Based on these values, Cohen’s d is

estimated mean difference

standard deviati d 5

oon 5 5 5 5

D 3 00

12 33

3 00

3 51 0 86

. .

The hypothesis test produced t 5 2.26 with df 5 6. Based on these values,

r t

t df

2 2

2 26

2 26 6

5 11

11 11 0 465

1 5

1 5 5

. .

( ) ( )

or 46%( )

S T E P 4

PROBLEMS

1. What is the defining characteristic of a repeated- measures or within-subjects research design?

2. Participants enter a research study with unique charac- teristics that produce different scores from one person to another. For an independent-measures study, these individual differences can cause problems. Identify the problems and briefly explain how they are eliminated or reduced with a repeated-measures study.

3. Explain the difference between a matched-subjects design and a repeated-measures design.

4. A researcher conducts an experiment comparing two treatment conditions with 20 scores in each treatment condition.

a. If an independent-measures design is used, how many subjects are needed for the experiment?

b. If a repeated-measures design is used, how many subjects are needed for the experiment?

c. If a matched-subjects design is used, how many subjects are needed for the experiment?

5. A sample of n 5 9 individuals participates in a repeated-measures study that produces a sample mean difference of M

D 5 4.25 with SS 5 128 for

the difference scores. a. Calculate the standard deviation for the sample

of difference scores. Briefly explain what is measured by the standard deviation.

b. Calculate the estimated standard error for the sample mean difference. Briefly explain what is measured by the estimated standard error.

6. a. A repeated-measures study with a sample of n 5 16 participants produces a mean difference of M

D 5 3 with a standard deviation of s 5 4.

Use a two-tailed hypothesis test with a 5 .05 to determine whether this sample provides evidence of a significant treatment effect.

b. Now assume that the sample standard deviation is s 5 12 and repeat the hypothesis test.

c. Explain how the size of the sample standard deviation influences the likelihood of finding a significant mean difference.

3 3 6 CHAPTER 11 THE t TEST FOR TWO RELATED SAMPLES

7. a. A repeated-measures study with a sample of n 5 9 participants produces a mean difference of M

D 5 3 with a standard deviation of s 5 6.

Use a two-tailed hypothesis test with a 5 .05 to determine whether it is likely that this sample came from a population with m

D 5 0.

b. Now assume that the sample mean difference is M

D 5 12, and once again visualize the sample

distribution. Use a two-tailed hypothesis test with a 5 .05 to determine whether it is likely that this sample came from a population with m

D 5 0.

c. Explain how the size of the sample mean difference influences the likelihood of finding a significant mean difference.

8. A sample of difference scores from a repeated- measures experiment has a mean of M

D 5 4 with

a standard deviation of s 5 6. a. If n 5 4, is this sample sufficient to reject the null

hypothesis using a two-tailed test with a 5 .05? b. Would you reject H

0 if n 5 16? Again, assume a

two-tailed test with a 5 .05. c. Explain how the size of the sample influences the

likelihood of finding a significant mean difference.

9. When you get a surprisingly low price on a product do you assume that you got a really good deal or that you bought a low-quality product? Research indi- cates that you are more likely to associate low price and low quality if someone else makes the purchase rather than yourself (Yan & Sengupta, 2011). In a similar study, n 5 16 participants were asked to rate the quality of low-priced items under two scenarios: purchased by a friend or purchased yourself. The results produced a mean difference of M

D 5 2.6 and

SS 5 135, with self-purchases rated higher. a. Is the judged quality of objects significantly different

for self-purchases than for purchases made by others? Use a two-tailed test with a 5 .05.

b. Compute Cohen’s d to measure the size of the treatment effect.

10. Research has shown that losing even one night’s sleep can have a significant effect on performance of complex tasks, such as problem solving (Linde & Bergstroem, 1992). To demonstrate this phenom- enon, a sample of n 5 25 college students was given a problem-solving task at noon on one day and again at noon on the following day. The students were not permitted any sleep between the two tests. For each student, the difference between the first and second score was recorded. For this sample, the students averaged M

D 5 4.7 points better on the first test with

a variance of s2 5 64 for the difference scores. a. Do the data indicate a significant change in

problem-solving ability? Use a two-tailed test with a 5 .05.

b. Compute an estimated Cohen’s d to measure the size of the effect.

11. Strack, Martin, and Stepper (1988) reported that people rate cartoons as funnier when holding a pen in their teeth (which forced them to smile) than when holding a pen in their lips (which forced them to frown). A researcher attempted to replicate this result using a sample of n 5 25 adults between the ages of 40 and 45. For each person, the researcher recorded the difference between the rating obtained while smiling and the rating obtained while frowning. On average the cartoons were rated as funnier when the participants were smiling, with an average difference of M

D 5 1.6 with SS 5 150.

a. Do the data indicate that the cartoons are rated significantly funnier when the participants are smiling? Use a one-tailed test with a 5 .01.

b. Compute r2 to measure the size of the treatment effect.

c. Write a sentence describing the outcome of the hypothesis test and the measure of effect size as it would appear in a research report.

12. Masculine-themed words (such as competitive, independent, analyze, strong) are commonly used in job recruitment materials, especially for job adver- tisements in male-dominated areas (Gaucher, Friesen, & Kay, 2011). The same study found that these words also make the jobs less appealing to women. In a similar study, female participants were asked to read a series of job advertisements and then rate how interesting or appealing the job appeared to be. Half of the advertisements were constructed to include several masculine-themed words and the others were worded neutrally. The average rating for each type of advertisement was obtained for each participant. For n 5 25 participants, the mean difference between the two types of advertisements is M

D 5 1.32 points

(neutral ads rated higher) with SS 5 150 for the difference scores.

a. Is this result sufficient to conclude that there is a significant difference in the ratings for two types of advertisements? Use a two-tailed test with a 5 .05.

b. Compute r2 to measure the size of the treatment effect.

c. Write a sentence describing the outcome of the hypothesis test and the measure of effect size as it would appear in a research report.

13. Research results indicate that physically attractive people are also perceived as being more intelligent (Eagly, Ashmore, Makhijani, & Longo, 1991). As a demonstration of this phenomenon, a researcher obtained a set of 10 photographs, 5 showing men who were judged to be attractive and 5 showing men who were judged to be unattractive. The photographs were shown to a sample of n 5 25 college students and the students were asked to rate the intelligence of the person in the photo on a scale from 1 to 10.

PROBLEMS 3 3 7

For each student, the researcher determined the average rating for the 5 attractive photos and the average for the 5 unattractive photos, and then com- puted the difference between the two scores. For the entire sample, the average difference was M

D 5 2.7

(attractive photos rated higher) with s 5 2.00. Are the data sufficient to conclude that there was a significant difference in perceived intelligence for the two sets of photos? Use a two-tailed test at the .05 level of significance.

14. Researchers have noted a decline in cognitive func- tioning as people age (Bartus, 1990). However, the results from other research suggest that the antioxi- dants in foods such as blueberries may reduce and even reverse these age-related declines (Joseph et al., 1999). To examine this phenomenon, suppose that a researcher obtains a sample of n 5 16 adults who are between the ages of 65 and 75. The researcher uses a standardized test to measure cognitive performance for each individual. The participants then begin a 2-month program in which they receive daily doses of a blueberry supplement. At the end of the 2-month period, the researcher again measures cognitive per- formance for each participant. The results show an average increase in performance of M

D 5 7.4 with

SS 5 1215. a. Does this result support the conclusion that the

antioxidant supplement has a significant effect on cognitive performance? Use a two-tailed test with a 5 .05.

b. Construct a 95% confidence interval to estimate the average cognitive performance improvement for the population of older adults.

15. The following data are from a repeated-measures study examining the effect of a treatment by measur- ing a group of n 5 6 participants before and after they receive the treatment.

a. Calculate the difference scores and M D .

b. Compute SS, sample variance, and estimated standard error.

c. Is there a significant treatment effect? Use a 5 .05, two tails.

Participant Before

Treatment After

Treatment

A 7 8

B 2 9

C 4 6

D 5 7

E 5 6

F 3 8

16. A researcher for a cereal company wanted to demon- strate the health benefits of eating oatmeal. A sample of 9 volunteers was obtained and each participant

ate a fixed diet without any oatmeal for 30 days. At the end of the 30-day period, cholesterol was measured for each individual. Then the participants began a second 30-day period in which they repeated exactly the same diet except that they added 2 cups of oatmeal each day. After the second 30-day pe- riod, cholesterol levels were measured again and the researcher recorded the difference between the two scores for each participant. For this sample, choles- terol scores averaged M

D 5 16 points lower with the

oatmeal diet with SS 5 538 for the difference scores. a. Are the data sufficient to indicate a significant

change in cholesterol level? Use a two-tailed test with a 5 .01.

b. Compute r2, the percentage of variance accounted for by the treatment, to measure the size of the treatment effect.

c. Write a sentence describing the outcome of the hypothesis test and the measure of effect size as it would appear in a research report.

17. Research indicates that the color red increases men’s attraction to women (Elliot & Niesta, 2008). In the original study, men were shown women’s photographs presented on either a white or a red background. Photographs presented on red were rated significantly more attractive than the same photographs mounted on white. In a similar study, a researcher prepares a set of 30 women’s photographs, with 15 mounted on a white background and 15 mounted on red. One picture is identified as the test photograph and appears twice in the set, once on white and once on red. Each male participant looks through the entire set of photographs and rates the attractiveness of each woman on a 10-point scale. The following table summarizes the ratings of the test photograph for a sample of n 5 9 men. Are the ratings for the test photograph significantly different when it is presented on a red background compared to a white background? Use a two-tailed test with a 5 .01.

Participant White

Background Red

Background

A 4 7

B 6 7

C 5 8

D 5 9

E 6 9

F 4 7

G 3 9

H 8 9

I 6 9

18. One of the primary advantages of a repeated- measures design, compared to independent-measures, is that it reduces the overall variability by removing

3 3 8 CHAPTER 11 THE t TEST FOR TWO RELATED SAMPLES

variance caused by individual differences. The following data are from a research study comparing two treatment conditions.

a. Assume that the data are from an independent- measures study using two separate samples, each with n 5 6 participants. Compute the pooled variance and the estimated standard error for the mean difference.

b. Now assume that the data are from a repeated- measures study using the same sample of n 5 6 participants in both treatment conditions. Compute the variance for the sample of difference scores and the estimated standard error for the mean difference. (You should find that the repeated-measures design substantially reduces the variance and the standard error.)

Treatment 1 Treatment 2 Difference

10 13 3

12 12 0

8 10 2

6 10 4

5 6 1

7 9 2

M 5 8 M 5 10 M D 5 2

SS 5 34 SS 5 30 SS 5 10

19. Problem 18 shows that removing individual differ- ences can substantially reduce variance and lower the standard error. However, this benefit only occurs if the individual differences are consistent across treatment conditions. In problem 18, for example, the first two participants (top two rows) consistently had the highest scores in both treatment conditions. Similarly, the last two participants consistently had the lowest scores in both treatments. To construct the following data, we started with the scores in problem 18 and scrambled the scores in treatment 1 to eliminate the consistency of the individual differences.

b. Now assume that the data are from a repeated- measures study using the same sample of n 5 6 participants in both treatment conditions. Compute the variance for the sample of differ- ence scores and the estimated standard error for the mean difference. (This time you should find that removing the individual differences does not reduce the variance or the standard error.)

Treatment 1 Treatment 2 Difference

6 13 7

7 12 5

8 10 2

10 10 0

5 6 1

12 9 23

M 5 8 M 5 10 M D 5 2

SS 5 34 SS 5 30 SS 5 64

20. A researcher uses a matched-subjects design to in- vestigate whether single people with pets are happier than singles without pets. A mood inventory survey is given to a group of 20- to 29-year-old non–pet owners and a similar age group of pet owners. The pet owners are matched one to one with the non–pet owners for income, number of close friendships, and general health. The data follow:

Matched Pair Non–Pet Owner Pet Owner

A 12 14

B 8 7

C 10 13

D 9 9

E 7 13

F 10 12

a. Is there a significant difference in the mood scores for non–pet owners versus pet owners? Test with a 5 .05 for two tails.

b. Construct the 95% confidence interval to estimate the size of the mean difference in mood between the population of pet owners and the population of non–pet owners. (You should find that a mean difference of m

D 5 0 is an acceptable value, which

is consistent with the conclusion from the hypoth- esis test.)

21. Some evidence suggests that you are likely to improve your test score if you rethink and change answers on a multiple-choice exam (Johnston, 1975). To examine this phenomenon, a teacher gave the same final exam to two sections of a course. Students in one section were told to turn in their exams immediately after finishing, without changing any answers. In the other section, students were encouraged to reconsider each question and to change answers when they felt it was appropriate. Before the final, the teacher matched 9 students in the first section with 9 students in the sec- ond section based on their midterm grades. For exam- ple, a student in the no-change section with an 89 on the midterm was matched with a student in the change section who also had an 89 on the midterm. The final

PROBLEMS 3 3 9

exam grades for the 9 matched pairs of students are presented in the table below.

a. Do the data indicate a significant difference be- tween the two conditions? Use a two-tailed test with a 5 .05.

b. Construct a 95% confidence interval to estimate the size of the population mean difference.

c. Write a sentence demonstrating how the results of the hypothesis test and the confidence interval would appear in a research report.

Matched Pair

No-Change Section

Change Section

#1 71 86

#2 68 80

#3 91 88

#4 65 74

#5 73 82

#6 81 89

#7 85 85

#8 86 88

#9 65 76

22. The teacher from Problem 21 also tried a different ap- proach to determining whether changing answers helps or hurts exam grades. In another class, students were told to review their final exams and change any answers they wanted to before turning them in. However, the students had to indicate both the original answer and the changed answer for each question. The teacher graded each exam twice, once using the set of original answers and once with the changes. In the class of n 5 22 stu- dents, the exam scores improved by an average of M

5 2.5 points with the changed answers. The standard deviation for the difference scores was s 5 3.1. Are the data sufficient to conclude that rethinking and changing answers can significantly improve scores? Use a one- tailed test at the .01 level of significance.

23. At the Olympic level of competition, even the small- est factors can make the difference between winning and losing. For example, Pelton (1983) has shown that Olympic marksmen shoot much better if they fire between heartbeats, rather than squeezing the trigger during a heartbeat. The small vibration caused by a heartbeat seems to be sufficient to affect the marksman’s aim. The following hypothetical data demonstrate this phenomenon. A sample of n 5 8 Olympic marksmen fires a series of rounds while a researcher records heartbeats. For each marksman, a score is recorded for shots fired during heartbeats

and for shots fired between heartbeats. Do these data indicate a significant difference? Test with a 5 .05.

Participant During

Heartbeats Between

Heartbeats

A 93 98

B 90 94

C 95 96

D 92 91

E 95 97

F 91 97

G 92 95

H 93 97

24. Example 11.1 in this chapter presented a repeated- measures research study demonstrating that swearing can help reduce ratings of pain (Stephens, Atkins, & Kingston, 2009). In the study, each participant was asked to plunge a hand into icy water and keep it there as long as the pain would allow. In one condition, the participants repeated their favorite curse words while their hands were in the water. In the other condition, the participants repeated a neutral word. In addition to lowering the participants’ perception of pain, swearing also increased the amount of time that they were able to tolerate the pain. Data similar to the results obtained in the study are shown in the following table.

a. Do these data indicate a significant difference in pain tolerance between the two conditions? Use a two-tailed test with a 5 .05.

b. Compute r2, the percentage of variance accounted for, to measure the size of the treatment effect.

c. Write a sentence demonstrating how the results of the hypothesis test and the measure of effect size would appear in a research report.

Amount of Time (in Seconds)

Participant Swear words Neutral words

1 94 59

2 70 61

3 52 47

4 83 60

5 46 35

6 117 92

7 69 53

8 39 30

9 51 56

10 73 61

SECTION 11.1 / RHR 3 4 1

1. The single-sample t introduced in Chapter 9. 2. The independent-measures t introduced in Chapter 10. 3. The repeated-measures t introduced in Chapter 11.

In this part, we considered a set of three t statistics that

are used to draw inferences about the means and mean

differences for unknown populations. Because the popula-

tions are completely unknown, we rely on sample data to

provide all of the necessary information. In particular, each

inferential procedure begins by computing sample means

and sample variances (or the corresponding SS values or

standard deviations). Therefore, a good understanding of

the definitions and formulas from Chapters 3 and 4 is a

critical foundation for this section.

With three different t statistics available, the first prob-

lem is often deciding which one is appropriate for a specific

research situation. Perhaps the best approach is to begin

with a close look at the sample data.

1. For the single-sample t (Chapter 9), there is only one group of participants and only one score for each individual. With a single sample mean and a single sample variance, the t statistic can be used to test a hypothesis about a single unknown population mean or construct a confidence interval to estimate the population mean.

2. For the independent-measures t, there are two separate groups of participants who produce two groups of scores. The mean and variance are computed for each group, producing two sample means and two sample variances. After pooling the two variances, the t statis- tic uses the difference between the two sample means to test a hypothesis about the corresponding differ- ence between the two unknown population means or estimate the population mean difference with a confidence interval. The null hypothesis always states that there is no difference between the two population means; m

1 – m

2 5 0.

3. For the repeated-measures t, there is only one group of participants but each individual is measured twice, at two different times and/or under two different treat- ment conditions. The two scores are then used to find a difference score for each person, and the mean and variance are computed for the sample of difference scores. The t statistic uses the sample mean difference to test a hypothesis about the corresponding population mean difference or estimate the population mean dif- ference with a confidence interval. The null hypothesis always states that the mean for the population of differ- ence scores is zero; m

D 5 0.

REVIEW EXERCISES

1. Belsky, Weinraub, Owen, and Kelly (2001) reported on the effects of preschool childcare on the development of young children. One result suggests that children who spend more time away from their mothers are more likely to show behavioral problems in kinder- garten. Using a standardized scale, the average rating of behavioral problems for kindergarten children is m 5 35. A sample of n 5 16 kindergarten children who had spent at least 20 hours per week in child care during the previous year produced a mean score of M 5 42.7 with a standard deviation of s 5 6.

a. Are the data sufficient to conclude that children with a history of childcare show significantly more behavioral problems than the average kindergarten child? Use a one-tailed test with a 5 .01.

b. Compute the 90% confidence interval for the mean rating of behavioral problems for the popu- lation of kindergarten children who have a history of daycare.

c. Write a sentence showing how the outcome of the hypothesis test and the confidence interval would appear in a research report.

2. Do you view a chocolate bar as delicious or as fattening? Your attitude may depend on your gender. In a study of American college students, Rozin, Bauer, and Catanese (2003) examined the importance of food as a source of pleasure versus concerns about food associated with weight gain and health. The following results are similar to those obtained in the study. The scores are a measure of concern about the negative aspects of eating.

Males Females

n 5 9 n 5 15

M 5 33 M 5 42

SS 5 740 SS 5 1240

a. Based on these results, is there a significant dif- ference between the attitudes for males and for females? Use a two-tailed test with a 5 .05.

b. Compute r2, the percentage of variance accounted for by the gender difference, to measure effect size for this study.

c. Write a sentence demonstrating how the result of the hypothesis test and the measure of effect size would appear in a research report.

REVIEW

After completing this part, you should be able to perform hypothesis tests and compute confi-

dence intervals using t statistics. These include:

3 4 1

P A R T III

3 4 2 CHAPTER 11 THE t TEST FOR TWO RELATED SAMPLES

3. The stimulant Ritalin has been shown to increase at- tention span and improve academic performance in children with ADHD (Evans et al., 2001). To demon- strate the effectiveness of the drug, a researcher selects a sample of n 5 20 children diagnosed with the disorder and measures each child’s attention span before and after taking the drug. The data show an average increase

of attention span of M D 5 4.8 minutes with a variance

of s2 5 125 for the sample of difference scores. a. Is this result sufficient to conclude that Ritalin

significantly improves attention span? Use a one- tailed test with a 5 .05.

b. Compute the 80% confidence interval for the mean change in attention span for the population.

343

Chapter 12 Introduction to Analysis of Variance 345

Chapter 13 Repeated-Measures and Two-Factor Analysis of Variance 393

I n Part III, we presented a set of t statistics that use sample means

and mean differences to draw inferences about the corresponding

population means and mean differences. However, the t statistics

are limited to situations that compare no more than two population

means. Often, a research question involves the differences among

more than two means and, in these situations, t tests are not appro-

priate. In this part, we introduce a new hypothesis testing technique

known as analysis of variance (ANOVA). ANOVA permits research-

ers to evaluate the mean differences among two or more populations

using sample data. We present three different applications of ANOVA

that apply to three distinct research situations:

1. Independent-measures designs: Using two or more separate

samples to draw an inference about the mean differences

between two or more unknown populations.

2. Repeated-measures designs: Using one sample, with each in-

dividual tested in two or more different treatment conditions,

to draw an inference about the population mean differences

among the conditions.

3. Two-factor designs: Allowing two independent variables to

change simultaneously within one study to create combinations

of treatment conditions involving both variables. The ANOVA

then evaluates the mean differences attributed to each variable

acting independently and to combinations of the two variables

interacting together.

In the next two chapters, we continue to examine statistical meth-

ods that use sample means as the foundation for drawing inferences

about population means. The primary application of these inferential

Analysis of Variance: Tests for Differences Among Two or More Population Means

P A R T

methods is to help researchers interpret the outcome of their

research studies. In a typical study, the goal is to demonstrate

a difference between two or more treatment conditions. For

example, a researcher hopes to demonstrate that a group of

children who are exposed to violent TV programs behave

more aggressively than children who are shown nonviolent

TV programs. In this situation, the data consist of one sample

mean representing the scores in one treatment condition and

another sample mean representing the scores from a different

treatment. The researcher hopes to find a difference between

the sample means and would like to generalize the mean dif-

ference to the entire population.

The problem is that sample means can be different even

when there are no differences whatsoever among the popu-

lation means. As you saw in Chapter 1 (see Figure 1.2),

two samples can have different means even when they are

selected from the same population. Thus, even though a re-

searcher may obtain a sample mean difference in a research

study, it does not necessarily indicate that there is a mean

difference in the population. As with the t tests presented

in Part III, a hypothesis test is needed to determine whether

the mean differences found in sample data are statistically

significant. With more than two sample means, the appro-

priate hypothesis test is ANOVA.

344

Aplia for Essentials of Statistics for the Behavioral Sciences

After reading, go to “Resources” at the end of this chapter for

an introduction on how to use Aplia’s homework and learning

resources.

Introduction to Analysis of Variance

12.1 Introduction

12.2 The Logic of ANOVA

12.3 ANOVA Notation and Formulas

12.4 The Distribution of F-Ratios

12.5 Examples of Hypothesis Testing and Effect Size with ANOVA

12.6 Post Hoc Tests

12.7 The Relationship Between ANOVA and t Tests

Summary

Focus on Problem Solving

Demonstrations 12.1 and 12.2

Problems

C H A P T E R

12 Tools You Will Need The following items are considered essential background material for this chapter. If you doubt your knowledge of any of these items, you should review the appropriate chapter or section before proceeding.

• Variability (Chapter 4) • Sum of squares • Sample variance • Degrees of freedom

• Introduction to hypothesis testing (Chapter 8) • The logic of hypothesis testing

• Independent-measures t statistic (Chapter 10)

3 4 6 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

INTRODUCTION

Analysis of variance (ANOVA) is a hypothesis-testing procedure that is used to evaluate

mean differences between two or more treatments (or populations). As with all inferential

procedures, ANOVA uses sample data as the basis for drawing general conclusions about

populations. It may appear that ANOVA and t tests are simply two different ways of doing

exactly the same job: testing for mean differences. In some respects, this is true—both

tests use sample data to test hypotheses about population means. However, ANOVA has

a tremendous advantage over t tests. Specifically, t tests are limited to situations in which

there are only two treatments to compare. The major advantage of ANOVA is that it can be

used to compare two or more treatments. Thus, ANOVA provides researchers with much

greater flexibility in designing experiments and interpreting results.

Figure 12.1 shows a typical research situation for which ANOVA would be used.

Note that the study involves three samples representing three populations. The goal of

the analysis is to determine whether the mean differences observed among the samples

provide enough evidence to conclude that there are mean differences among the three

populations. Specifically, we must decide between two interpretations:

1. There really are no differences between the populations (or treatments). The

observed differences between the sample means are caused by random, unsys-

tematic factors (sampling error) that differentiate one sample from another.

2. The populations (or treatments) really do have different means, and these popula-

tion mean differences are responsible for causing systematic differences between

the sample means.

You should recognize that these two interpretations correspond to the two hypotheses

(null and alternative) that are part of the general hypothesis-testing procedure.

Before we continue, it is necessary to introduce some of the terminology that is used to

describe the research situation shown in Figure 12.1. Recall (from Chapter 1) that when a

researcher manipulates a variable to create the treatment conditions in an experiment, the

variable is called an independent variable. For example, Figure 12.1 could represent a study

12.1

T E R M I N O LO G Y I N A N OVA

Population 2 (Treatment 2)

Population 1 (Treatment 1)

Population 3 (Treatment 3)

µ 3

= ?µ 2

= ?µ 1

= ?

Sample 3Sample 2Sample 1

n 5 15 M 5 23.1 SS 5 114

n 5 15 M 5 28.5 SS 5 130

n 5 15 M 5 20.8 SS 5 101

FIGURE 12.1

A typical situation in

which ANOVA would

be used. Three separate

samples are obtained to

evaluate the mean differ-

ences among three popu-

lations (or treatments)

with unknown means.

SECTION 12.1 / INTRODUCTION 3 4 7

examining driving performance under three different telephone conditions: driving with

no phone, talking on a hands-free phone, and talking on a hand-held phone. Note that the

three conditions are created by the researcher. On the other hand, when a researcher uses

a nonmanipulated variable to designate groups, the variable is called a quasi-independent

variable. For example, the three groups in Figure 12.1 could represent 6-year-old, 8-year-

old, and 10-year-old children. In the context of ANOVA, an independent variable or a quasi-

independent variable is called a factor. Thus, Figure 12.1 could represent an experimental

study in which the telephone condition is the factor being evaluated or it could represent a

nonexperimental study in which age is the factor being examined.

In ANOVA, the variable (independent or quasi-independent) that designates the

groups being compared is called a factor.

The individual groups or treatment conditions that are used to make up a factor are

called the levels of the factor. For example, a study that examined performance under

three different telephone conditions would have three levels of the factor.

The individual conditions or values that make up a factor are called the levels of

the factor.

Like the t tests presented in Chapters 10 and 11, ANOVA can be used with either an in-

dependent-measures or a repeated-measures design. Recall that an independent-measures

design means that there is a separate group of participants for each of the treatments (or

populations) being compared. In a repeated-measures design, on the other hand, the same

group is tested in all of the different treatment conditions. In addition, ANOVA can be

used to evaluate the mean differences from a research study that involves more than one

factor. For example, a researcher may want to compare two different therapy techniques,

examining their immediate effectiveness as well as the persistence of their effectiveness

over time. In this situation, the research study could involve two different groups of par-

ticipants, one for each therapy, and measure each group at several different points in time.

The structure of this design is shown in Figure 12.2. Notice that the study uses two factors,

one independent-measures factor and one repeated-measures factor:

1. Factor 1: Therapy technique. A separate group is used for each technique (inde-

pendent measures).

2. Factor 2: Time. Each group is tested at three different times (repeated measures).

In this case, the ANOVA would evaluate mean differences between the two therapies

as well as mean differences between the scores obtained at different times. A study that

combines two factors, like the one in Figure 12.2, is called a two-factor design or a

factorial design.

The ability to combine different factors and to mix different designs within one study

provides researchers with the flexibility to develop studies that address scientific ques-

tions that could not be answered by a single design using a single factor.

Although ANOVA can be used in a wide variety of research situations, this chapter

introduces ANOVA in its simplest form. Specifically, we consider only single-factor de-

signs. That is, we examine studies that have only one independent variable (or only one

quasi-independent variable). Second, we consider only independent-measures designs;

that is, studies that use a separate group of participants for each treatment condition.

The basic logic and procedures that are presented in this chapter form the foundation

for more complex applications of ANOVA. For example, in Chapter 13, we extend the

D E F I N I T I O N

3 4 8 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

analysis to single-factor, repeated-measures designs and we introduce two-factor de-

signs. But for now, in this chapter, we limit our discussion of ANOVA to single-factor,

independent-measures research studies.

The following example introduces the statistical hypotheses for ANOVA. Suppose

that a researcher examined driving performance under three different telephone

conditions: no phone, a hands-free phone, and a hand-held phone. Three samples

of participants are selected, one sample for each treatment condition. The purpose

of the study is to determine whether using a telephone affects driving performance.

In statistical terms, we want to decide between two hypotheses: the null hypothesis

(H 0 ), which states that the telephone condition has no effect, and the alternative

hypothesis (H 1 ), which states that the telephone condition does affect driving. In

symbols, the null hypothesis states

H 0 : �

1 � �

2 � �

In words, the null hypothesis states that the telephone condition has no effect on

driving performance. That is, the population means for the three telephone conditions

are all the same. In general, H 0 states that there is no treatment effect.

The alternative hypothesis states that the population means are not all the same:

H 1 : There is at least one mean difference among the populations.

In general, H 1 states that there is a real treatment effect. As always, the hypotheses are

stated in terms of population parameters, even though we use sample data to test them.

STAT I ST I CA L H Y P OT H E S E S F O R

A N OVA

Scores for group 1

measured before

Therapy I

Before Therapy

Therapy I (Group 1)

THERAPY TECHNIQUE

Therapy II (Group 2)

TIME

After Therapy

6 Months After Therapy

Scores for group 1

measured after

Therapy I

Scores for group 1

measured 6 months after

Therapy I

Scores for group 2

measured before

Therapy II

Scores for group 2

measured after

Therapy II

Scores for group 2

measured 6 months after

Therapy II

FIGURE 12.2

A research design with two factors. The research study uses two factors: One factor uses two levels of therapy technique

(I versus II), and the second factor uses three levels of time (before, after, and 6 months after). Also notice that the ther-

apy factor uses two separate groups (independent measures) and the time factor uses the same group for all three levels

(repeated measures).

SECTION 12.1 / INTRODUCTION 3 4 9

Notice that we are not stating a specific alternative hypothesis. This is because many

different alternatives are possible, and it would be tedious to list them all. One alterna-

tive, for example, is that the first two populations are identical, but that the third is dif-

ferent. Another alternative states that the last two means are the same, but that the first

is different. Other alternatives might be

H 1 : �

1 ≠ �

2 ≠ �

3 (All three means are different.)

H 1 : �

1 � �

3 , but �

2 is different.

We should point out that a researcher typically entertains only one (or at most a few) of

these alternative hypotheses. Usually a theory or the outcomes of previous studies dictate

a specific prediction concerning the treatment effect. For the sake of simplicity, we state a

general alternative hypothesis rather than try to list all of the possible specific alternatives.

The test statistic for ANOVA is very similar to the independent-measures t statistic used in

Chapter 10. For the t statistic, we first computed the standard error, which measures how

much difference is expected between two sample means if there is no treatment effect (that

is, if H 0 is true). Then we computed the t statistic with the following structure:

t � obtained difference between two sample meaans

standard error (the difference expected with no treatment effect)

For ANOVA, however, we want to compare differences among two or more sample

means. With more than two samples, the concept of “difference between sample means”

becomes difficult to define or measure. For example, if there are only two samples and they

have means of M � 20 and M � 30, then there is a 10-point difference between the sample

means. Suppose, however, that we add a third sample with a mean of M � 35. Now how

much difference is there between the sample means? It should be clear that we have a prob-

lem. The solution to this problem is to use variance to define and measure the size of the

differences among the sample means. Consider the following two sets of sample means:

Set 1 Set 2

M 1 � 20 M

1 � 28

M 2 � 30 M

2 � 30

M 3 � 35 M

3 � 31

If you compute the variance for the three numbers in each set, then the variance for

set 1 is s2 � 58.33 and the variance for set 2 is s2 � 2.33. Notice that the two variances

provide an accurate representation of the size of the differences. In set 1, there are rela-

tively large differences between sample means and the variance is relatively large. In

set 2, the mean differences are small and the variance is small.

Thus, we can use variance to measure sample mean differences when there are two

or more samples. The test statistic for ANOVA uses this fact to compute an F-ratio with

the following structure:

F � variance (differences) between sample meanns

variance (differences) expected with no ttreatment effect

Note that the F-ratio has the same basic structure as the t statistic but is based on variance

instead of sample mean difference. The variance in the numerator of the F-ratio provides a

T H E T E ST STAT I ST I C F O R A N OVA

3 5 0 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

single number that measures the differences among all of the sample means. The variance

in the denominator of the F-ratio, like the standard error in the denominator of the t statistic,

measures the mean differences that would be expected if there were no treatment effect.

Thus, the t statistic and the F-ratio provide the same basic information. In each case, a large

value for the test statistic provides evidence that the sample mean differences (numerator)

are larger than would be expected if there were no treatment effects (denominator).

If we already have t tests for comparing mean differences, you might wonder why

ANOVA is necessary. Why create a whole new hypothesis-testing procedure that sim-

ply duplicates what the t tests can already do? The answer to this question is based in

a concern about Type I errors.

Remember that each time you do a hypothesis test, you select an alpha level that

determines the risk of a Type I error. With a � .05, for example, there is a 5%, or a

1-in-20, risk of a Type I error. Often a single experiment requires several hypothesis

tests to evaluate all of the mean differences. However, each test has a risk of a Type I

error, and the more tests you do, the more risk there is.

For this reason, researchers often make a distinction between the testwise alpha level

and the experimentwise alpha level. The testwise alpha level is simply the alpha level that

you select for each individual hypothesis test. The experimentwise alpha level is the total

probability of a Type I error accumulated from all of the separate tests in the experiment.

As the number of separate tests increases, so does the experimentwise alpha level.

The testwise alpha level is the risk of a Type I error, or alpha level, for an individual

hypothesis test.

When an experiment involves several different hypothesis tests, the experimentwise

alpha level is the total probability of a Type I error that is accumulated from all of

the individual tests in the experiment. Typically, the experimentwise alpha level is

substantially greater than the value of alpha used for any one of the individual tests.

For example, an experiment involving three treatments would require three separate

t tests to compare all of the mean differences:

Test 1 compares treatment I with treatment II.

Test 2 compares treatment I with treatment III.

Test 3 compares treatment II with treatment III.

If all tests use a � .05, then there is a 5% risk of a Type I error for the first test, a 5%

risk for the second test, and another 5% risk for the third test. The three separate tests

accumulate to produce a relatively large experimentwise alpha level. The advantage of

ANOVA is that it performs all three comparisons simultaneously in one hypothesis test.

Thus, no matter how many different means are being compared, ANOVA uses one test

with one alpha level to evaluate the mean differences, and thereby avoids the problem of

an inflated experimentwise alpha level.

THE LOGIC OF ANOVA

The formulas and calculations required in ANOVA are somewhat complicated, but

the logic that underlies the whole procedure is fairly straightforward. Therefore, this

section gives a general picture of ANOVA before we start looking at the details. We

introduce the logic of ANOVA with the help of the hypothetical data in Table 12.1.

T Y P E I E R R O R S A N D M U LT I P L E - H Y P OT H E S I S

T E ST S

D E F I N I T I O N S

12.2

SECTION 12.2 / THE LOGIC OF ANOVA 3 5 1

These data represent the results of an independent-measures experiment comparing

performance in a driving simulator under three telephone conditions.

One obvious characteristic of the data in Table 12.1 is that the scores are not all the

same. In everyday language, the scores are different; in statistical terms, the scores are

variable. Our goal is to measure the amount of variability (the size of the differences)

and to explain why the scores are different.

The first step is to determine the total variability for the entire set of data. To com-

pute the total variability, we combine all of the scores from all of the separate samples

to obtain one general measure of variability for the complete experiment. Once we have

measured the total variability, we can begin to break it apart into separate components.

The word analysis means dividing into smaller parts. Because we are going to analyze

variability, the process is called analysis of variance. This analysis process divides the

total variability into two basic components.

1. Between-Treatments Variance. Looking at the data in Table 12.1, we

clearly see that much of the variability in the scores results from general

differences between treatment conditions. For example, the scores in the

no-phone condition tend to be much higher (M � 4) than the scores in the

hand-held condition (M � 1). We calculate the variance between treatments

to provide a measure of the overall differences between treatment conditions.

Notice that the variance between treatments is really measuring the differ-

ences between sample means.

2. Within-Treatment Variance. In addition to the general differences between

treatment conditions, there is variability within each sample. Looking again

at Table 12.1, we see that the scores in the no-phone condition are not all the

same; they are variable. The within-treatments variance provides a measure of

the variability inside each treatment condition.

Analyzing the total variability into these two components is the heart of ANOVA.

We now examine each of the components in more detail.

Remember that calculating variance is simply a method for measuring how big

the differences are for a set of numbers. When you see the term variance, you

can automatically translate it into the term differences. Thus, the between-treat-

ments variance simply measures how much difference exists between the treatment

conditions. There are two possible explanations for these between-treatment

differences:

1. The differences between treatments are not caused by any treatment effect but

are simply the naturally occurring, random, and unsystematic differences that

B E T W E E N - T R E AT M E N T S VA R I A N C E

TABLE 12.1

Hypothetical data from an

experiment examining driv-

ing performance under three

telephone conditions.*

Treatment 1: No Phone (Sample 1)

Treatment 2: Hand-Held Phone

(Sample 2)

Treatment 3: Hands-Free Phone

(Sample 3)

4 0 1

3 1 2

6 3 2

3 1 0

4 0 0

M � 4 M � 1 M � 1

*Note that there are three separate samples, with n � 5 in each sample. The depen-

dent variable is a measure of performance in a driving simulator.

3 5 2 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

exist between one sample and another. That is, the differences are the result of

sampling error.

2. The differences between treatments have been caused by the treatment effects.

For example, if using a telephone really does interfere with driving performance,

then scores in the telephone conditions should be systematically lower than

scores in the no-phone condition.

Thus, when we compute the between-treatments variance, we are measuring differences

that could be caused by a systematic treatment effect or could simply be random and un-

systematic mean differences caused by sampling error. To demonstrate that there really is a

treatment effect, we must establish that the differences between treatments are bigger than

would be expected by sampling error alone. To accomplish this goal, we determine how

big the differences are when there is no systematic treatment effect; that is, we measure

how much difference (or variance) can be explained by random and unsystematic factors.

To measure these differences, we compute the variance within treatments.

Inside each treatment condition, we have a set of individuals who all receive exactly the same

treatment; that is, the researcher does not do anything that would cause these individuals to

have different scores. In Table 12.1, for example, the data show that five individuals were

tested while talking on a hand-held phone (sample 2). Although these five individuals all

received exactly the same treatment, their scores are different. Why are the scores different?

The answer is that there is no specific cause for the differences. Instead, the differences that

exist within a treatment represent random and unsystematic differences that occur when there

are no treatment effects causing the scores to be different. Thus, the within-treatments vari-

ance provides a measure of how big the differences are when H 0 is true.

Figure 12.3 shows the overall ANOVA and identifies the sources of variability that

are measured by each of the two basic components.

Once we have analyzed the total variability into two basic components (between treat-

ments and within treatments), we simply compare them. The comparison is made by

W I T H I N - T R E AT M E N T S VA R I A N C E

T H E F - R AT I O : T H E T E ST STAT I ST I C F O R A N OVA

Total variability

Measures differences caused by 1. Systematic treatment effects 2. Random, unsystematic factors

Between- treatments variance

Measures differences caused by 1. Random, unsystematic factors

Within- treatments variance

FIGURE 12.3

The independent-

measures ANOVA parti-

tions, or analyzes, the

total variability into two

components: variance

between treatments

and variance within

treatments.

SECTION 12.2 / THE LOGIC OF ANOVA 3 5 3

computing an F-ratio. For the independent-measures ANOVA, the F-ratio has the

following structure:

F � variance between treatments

variance withinn treatments

differences including any tre �

aatment effects

differences with no treatmentt effects (12.1)

When we express each component of variability in terms of its sources (see Figure 12.3),

the structure of the F-ratio is

F � 1systematic treatment effects random, unsysstematic differences

random, unsystematic diifferences

(12.2)

The value obtained for the F-ratio helps to determine whether any treatment effects

exist. Consider the following two possibilities:

1. When there are no systematic treatment effects, the differences between treat-

ments (numerator) are entirely caused by random, unsystematic factors. In this

case, the numerator and the denominator of the F-ratio are both measuring

random differences and should be roughly the same size. With the numerator

and denominator roughly equal, the F-ratio should have a value around 1.00. In

terms of the formula, when the treatment effect is zero, we obtain

F = +0 random, unsystematic differences

random,, unsystematic differences

Thus, an F-ratio near 1.00 indicates that the differences between treatments

(numerator) are random and unsystematic, just like the differences in the de-

nominator. With an F-ratio near 1.00, we conclude that there is no evidence to

suggest that the treatment has any effect.

2. When the treatment does have an effect, causing systematic differences

between samples, then the combination of systematic and random differ-

ences in the numerator should be larger than the random differences alone

in the denominator. In this case, the numerator of the F-ratio should be

noticeably larger than the denominator, and we should obtain an F-ratio

that is substantially larger than 1.00. Thus, a large F-ratio is evidence for

the existence of systematic treatment effects; that is, there are consistent

differences between treatments.

Because the denominator of the F-ratio measures only random and unsystematic

variability, it is called the error term. The numerator of the F-ratio always includes the

same unsystematic variability as in the error term, but it also includes any systematic

differences caused by the treatment effect. The goal of ANOVA is to find out whether

a treatment effect exists.

For ANOVA, the denominator of the F-ratio is called the error term. The

error term provides a measure of the variance caused by random, unsystem-

atic differences. When the treatment effect is zero (H 0 is true), the error term

measures the same sources of variance as the numerator of the F-ratio, so the

value of the F-ratio is expected to be nearly equal to 1.00.

D E F I N I T I O N

3 5 4 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

ANOVA NOTATION AND FORMULAS

Because ANOVA typically is used to examine data from more than two treatment con-

ditions (and more than two samples), we need a notational system to keep track of all

of the individual scores and totals. To help introduce this notational system, we use the

hypothetical data from Table 12.1 again. The data are reproduced in Table 12.2 along

with some of the notation and statistics that are described in the following list.

1. The letter k is used to identify the number of treatment conditions—that is, the

number of levels of the factor. For an independent-measures study, k also speci-

fies the number of separate samples. For the data in Table 12.2, there are three

treatments, so k � 3.

2. The number of scores in each treatment is identified by a lowercase letter n.

For the example in Table 12.2, n � 5 for all the treatments. If the samples are

of different sizes, you can identify a specific sample by using a subscript. For

example, n 2 is the number of scores in treatment 2.

3. The total number of scores in the entire study is specified by a capital letter N.

When all of the samples are the same size (n is constant), N � kn. For the data

in Table 12.2, there are n � 5 scores in each of the k � 3 treatments, so we

have a total of N � 3(5) � 15 scores in the entire study.

4. The sum of the scores (oX) for each treatment condition is identified by the capital letter T (for treatment total). The total for a specific treatment can be

identified by adding a numerical subscript to the T. For example, the total for

the second treatment in Table 12.2 is T 2 � 5.

5. The sum of all of the scores in the research study (the grand total) is identified by

G. You can compute G by adding up all N scores or by adding up the treatment

totals: G � oT.

12.3

1. Explain the difference between the testwise alpha level and the experimentwise

alpha level.

2. The term “analysis” means separating or breaking a whole into parts. What is the

basic analysis that takes place in analysis of variance?

3. If there is no systematic treatment effect, then what value is expected, on average,

for the F-ratio in an ANOVA?

4. What is the implication when an ANOVA produces a very large value for the F-ratio?

1. When a single research study involves several hypothesis tests, the testwise alpha level is the

value selected for each individual test and the experimentwise alpha level is the total risk of

a Type I error that is accumulated for all of the separate tests.

2. In ANOVA, the total variability for a set of scores is separated into two components: between-

treatments variability and within-treatments variability.

3. When H 0 is true, the expected value for the F-ratio is 1.00 because the top and bottom of the

ratio are both measuring the same variance.

4. A large F-ratio indicates the existence of a treatment effect because the differences between

treatments (numerator) are much bigger than the differences that would be expected if there

were no effect (denominator).

L E A R N I N G C H E C K

ANSWERS

SECTION 12.3 / ANOVA NOTATION AND FORMULAS 3 5 5

6. Although there is no new notation involved, we also have computed SS and

M for each sample, and we have calculated oX2 for the entire set of N � 15 scores. These values are given in Table 12.2 and are important in the formulas

and calculations for ANOVA.

Finally, we should note that there is no universally accepted notation for ANOVA.

Although we are using Gs and Ts, for example, you may find that other sources use

other symbols.

Because ANOVA requires extensive calculations and many formulas, one common

problem for students is simply keeping track of the different formulas and numbers.

Therefore, we examine the general structure of the procedure and look at the organiza-

tion of the calculations before we introduce the individual formulas.

1. The final calculation for ANOVA is the F-ratio, which is composed of

two variances:

F � variance between treatments

variance withinn treatments

2. Each of the two variances in the F-ratio is calculated using the basic formula

for sample variance.

sample variance � �s SS

Therefore, we need to compute an SS and a df for the variance between treat-

ments (numerator of F), and we need another SS and df for the variance within

treatments (denominator of F). To obtain these SS and df values, we must go

through two separate analyses: First, compute SS for the total study, and analyze

it in two components (between and within). Then compute df for the total study,

and analyze it in two components (between and within).

Thus, the entire process of ANOVA requires nine calculations: three values for SS,

three values for df, two variances (between and within), and a final F-ratio. However,

these nine calculations are all logically related and are all directed toward finding the

final F-ratio. Figure 12.4 shows the logical structure of ANOVA calculations.

A N OVA F O R M U L AS

Telephone Conditions

Treatment 1 No Phone (Sample 1)

Treatment 2 Hand-Held Phone

(Sample 2)

Treatment 3 Hands-Free Phone

(Sample 3)

4 0 1 oX2 � 106

3 1 2 G � 30

6 3 2 N � 15

3 1 0 k � 3

4 0 0

T 1 � 20 T

2 � 5 T

3 � 5

SS 1 � 6 SS

2 � 6 SS

3 � 4

n 1 � 5 n

2 � 5 n

3 � 5

M 1 � 4 M

2 � 1 M

3 � 1

TABLE 12.2

The same data that appeared

in Table 12.1 with summary

values and notation appropri-

ate for an ANOVA.

Because ANOVA formulas

require oX for each treatment

and oX for the entire set of

scores, we have introduced

new notation (T and G) to

help identify which oX is

being used. Remember: T

stands for treatment total,

and G stands for grand total.

3 5 6 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

The ANOVA requires that we first compute a total sum of squares and then partition this

value into two components: between treatments and within treatments. This analysis is

outlined in Figure 12.5. We examine each of the three components separately.

1. Total Sum of Squares, SS total

. As the name implies, SS total

is the sum of squares

for the entire set of N scores. As described in Chapter 4 (pp. 96–98), this SS value

can be computed using either a definitional or a computational formula. However,

ANOVA typically involves a large number of scores and the mean is often not a

whole number. Therefore, it is usually much easier to calculate SS total

using the

computational formula:

SS N

5 2o o

X X2

2 ( )

ANALYSIS OF THE SUM OF SQUARES ( S S )

To obtain each of the SS and df values, the total variability is analyzed into the two components

Each variance in the F-ratio is computed as SS/df

The final goal for the ANOVA is an F-ratio F 5

Variance between treatments

Variance within treatments

5 Variance between

treatments

SS between

SS between SS within

SS total

df between 5

Variance within

treatments

SS within

df within

df between df within

df total

FIGURE 12.4

The structure and sequence

of calculations for the

ANOVA.

SS within treatments

ΣSS inside each treatment

SS between treatments

n (SS for the treatment means)

Σ G2

T 2

SS Total

G2 Σ X

2 �

�

FIGURE 12.5

Partitioning the sum

of squares (SS) for the

independent-measures

ANOVA.

SECTION 12.3 / ANOVA NOTATION AND FORMULAS 3 5 7

To make this formula consistent with the ANOVA notation, we substitute the letter G

in place of oX and obtain

SS G

N total

5 2o X 2

(12.3)

Applying this formula to the set of data in Table 12.2, we obtain

SS total

5 2106 30

5 106 2 60

5 46

2. Within-Treatments Sum of Squares, SS within treatments

. Now we are looking at the

variability inside each of the treatment conditions. We already have computed the

SS within each of the three treatment conditions (Table 12.2): SS 1 5 6, SS

2 5 6,

and SS 3 5 4. To find the overall within-treatment sum of squares, we simply add

these values together:

SS within treatments

5 oSS inside each treatment

(12.4)

For the data in Table 12.2, this formula gives

SS within treatments

5 6 1 6 1 4

5 16

3. Between-Treatments Sum of Squares, SS within treatments

. Before we introduce any

equations for SS between treatments

, consider what we have found so far. The total vari-

ability for the data in Table 12.2 is SS total

5 46. We intend to partition this total into two parts (see Figure 12.5). One part, SS

within treatments , has been found to be

equal to 16. This means that SS between treatments

must be equal to 30 so that the two

parts (16 and 30) add up to the total (46). Thus, the value for SS between treatments

can

be found simply by subtraction:

SS between

5 SS total

2 SS within

(12.5)

Computing SS between

Instead of finding SS between

by subtraction, you can compute

SS between

independently, and then check your calculations by ensuring that the two com-

ponents, between and within, add up to the total. We present two different formulas for

calculating SS between

directly from the data.

Recall that the variability between treatments is measuring the differences between

treatment means. Conceptually, the most direct way of measuring the amount of variability

among the treatment means is to compute the sum of squares for the set of sample means,

SS means

. For the data in Table 12.2, the sample means are 4, 1, and 1. These three values

produce SS means

5 6. However, each of the three means represents a group of n 5 5 scores. Therefore, the final value for SS

between is obtained by multiplying SS

means by n.

SS between

5 n(SS means

) (12.6)

For the data in Table 12.2, we obtain

SS between

5 n(SS means

) 5 5(6) 5 30

To simplify the notation we

use the subscripts between

and within in place of be-

tween treatments and within

treatments.

3 5 8 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

Unfortunately, Equation 12.6 can only be used when all of the samples are exactly

the same size (equal ns), and the equation can be very awkward, especially when the

treatment means are not whole numbers. Therefore, we also present a computational

formula for SS between

that uses the treatment totals (T) instead of the treatment means.

SS G

N between

5 2 o T

(12.7)

For the data in Table 12.2, this formula produces:

SS between

5 1 1 2 20

2 2 2 2

5 80 1 5 1 5 2 60

5 90 2 60

5 30

Note that all three techniques (Equations 12.5, 12.6, and 12.7) produce the same

result, SS between

5 30. We have presented three different equations for computing SS

between . Rather than mem-

orizing all three, however, we suggest that you pick one formula and use it consistently.

There are two reasonable alternatives to use. The simplest is Equation 12.5, which finds

SS between

simply by subtraction: First you compute SS total

and SS within

, then subtract:

SS between

5 SS total

2 SS within

The second alternative is to use Equation 12.7, which computes SS between

using the

treatment totals (the T values). The advantage of this alternative is that it provides a way

to check your arithmetic: Calculate SS total

, SS between

, and SS within

separately, and then check

to be sure that the two components add up to equal SS total

Using Equation 12.6, which computes SS for the set of sample means, is usually

not a good choice. Unless the sample means are all whole numbers, this equation can

produce very tedious calculations. In most situations, one of the other two equations is

a better alternative.

The analysis of degrees of freedom (df) follows the same pattern as the analysis of SS.

First, we find df for the total set of N scores, and then we partition this value into two

components: degrees of freedom between treatments and degrees of freedom within

treatments. In computing degrees of freedom, there are two important considerations

to keep in mind:

1. Each df value is associated with a specific SS value.

2. Normally, the value of df is obtained by counting the number of items that were

used to calculate SS and then subtracting 1. For example, if you compute SS for

a set of n scores, then df 5 n 2 1.

With this in mind, we examine the degrees of freedom for each part of the analysis.

1. Total Degrees of Freedom, df total

. To find the df associated with SS total

, you must

first recall that this SS value measures variability for the entire set of N scores.

Therefore, the df value is

df total

5 N 2 1 (12.8)

T H E A N A LYS I S O F D E G R E E S O F F R E E D O M

( d f )

SECTION 12.3 / ANOVA NOTATION AND FORMULAS 3 5 9

For the data in Table 12.2, the total number of scores is N 5 15, so the total degrees of freedom are

df total

5 15 2 1

5 14

2. Within-Treatments Degrees of Freedom, df within

. To find the df associated with

SS within

, we must look at how this SS value is computed. Remember, we first find

SS inside of each of the treatments and then add these values together. Each of

the treatment SS values measures variability for the n scores in the treatment,

so each SS has df 5 n 2 1. When all of these individual treatment values are added together, we obtain

df within

5 o(n 2 1) 5 odf in each treatment

(12.9)

For the experiment we have been considering, each treatment has n 5 5 scores. This means there are n 2 1 5 4 degrees of freedom inside each treatment. Because there are three different treatment conditions, this gives a total of 12

for the within-treatments degrees of freedom.

Notice that this formula for df simply adds up the number of scores in each

treatment (the n values) and subtracts 1 for each treatment. If these two stages

are done separately, you obtain

df within

5 N 2 k (12.10)

(Adding up all the n values gives N. If you subtract 1 for each treatment, then

altogether you have subtracted k because there are k treatments.) For the data in

Table 12.2, N 5 15 and k 5 3, so

df within

5 15 2 3

5 12

3. Between-Treatments Degrees of Freedom, df between

. The df associated with

SS between

can be found by considering how the SS value is obtained. This SS

formula measure the variability for the set of treatments (totals or means). To

find df between

, simply count the number of treatments and subtract 1. Because the

number of treatments is specified by the letter k, the formula for df is

df between

5 k 2 1 (12.11)

For the data in Table 12.2, there are three different treatment conditions and

df between

5 3 2 1

5 2

Notice that the two parts we obtained from this analysis of degrees of freedom

add up to equal the total degrees of freedom:

df total

5 df within

1 df between

14 5 12 1 2

The complete analysis of degrees of freedom is shown in Figure 12.6.

3 6 0 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

As you are computing the SS and df values for ANOVA, keep in mind that the labels

that are used for each value can help you understand the formulas. Specifically,

1. The term total refers to the entire set of scores. We compute SS for the whole

set of N scores, and the df value is simply N 2 1.

2. The term within treatments refers to differences that exist inside of the

individual treatment conditions. Thus, we compute SS and df inside each of

the separate treatments.

3. The term between treatments refers to differences from one treatment to another.

With three treatments, for example, we are comparing three different means (or

totals) and have df 5 3 2 1 5 2.

The next step in the ANOVA procedure is to compute the variance between treatments and

the variance within treatments, which are used to calculate the F-ratio (see Figure 12.4).

In ANOVA, it is customary to use the term mean square, or simply MS, in place of

the term variance. Recall (from Chapter 4) that variance is defined as the mean of the

squared deviations. In the same way that we use SS to stand for the sum of the squared

deviations, we now use MS to stand for the mean of the squared deviations. For the

final F-ratio, we need an MS (variance) between treatments for the numerator and an

MS (variance) within treatments for the denominator. In each case

MS s SS

df variance( ) 5 52

(12.12)

For the data we have been considering,

MS s SS

df between between

between

5 5 5 5 2 30

2 115

and

MS s SS

df within within

within

5 5 5 5 2 16

12 1 3. 33

CA L C U L AT I O N O F VA R I A N C E S ( M S ) A N D

T H E F - R AT I O

N �� 1

df within treatmentsdf between treatments

k Σ(n �1) = k

df total

N � 1

FIGURE 12.6

Partitioning degrees

of freedom (df) for the

independent-measures

ANOVA.

SECTION 12.3 / ANOVA NOTATION AND FORMULAS 3 6 1

We now have a measure of the variance (or differences) between the treatments and

a measure of the variance within the treatments. The F-ratio simply compares these

two variances:

F s

MS 5 5between

within

between

within

(12.13)

For the experiment we have been examining, the data give an F-ratio of

F 5 5 15

1 33 11 28

. .

For this example, the obtained value of F 5 11.28 indicates that the numerator of the F-ratio is substantially bigger than the denominator. If you recall the conceptual structure

of the F-ratio as presented in Equations 12.1 and 12.2, the F value we obtained indicates

that the differences between treatments are more than 11 times bigger than what would be

expected if there were no treatment effect. Stated in terms of the experimental variables:

using a telephone while driving does appear to have an effect on driving performance.

However, to properly evaluate the F-ratio, we must select an a level and consult the F-distribution table that is discussed in the next section.

ANOVA summary tables It is useful to organize the results of the analysis in one

table called an ANOVA summary table. The table shows the source of variability (between

treatments, within treatments, and total variability), SS, df, MS, and F. For the previous

computations, the ANOVA summary table is constructed as follows:

Source SS df MS

Between treatments 30 2 15 F 5 11.28

Within treatments 16 12 1.33

Total 46 14

Although these tables are no longer used in published reports, they are a common

part of computer printouts, and they do provide a concise method for presenting the

results of an analysis. (Note that you can conveniently check your work: Adding the

first two entries in the SS column, 30 1 16, produces SS total

. The same applies to the df

column.) When using ANOVA, you might start with a blank ANOVA summary table

and then fill in the values as they are calculated. With this method, you are less likely

to “get lost” in the analysis, wondering what to do next.

1. Calculate SS total

, SS between

, and SS within

for the following set of data:

Treatment 1 Treatment 2 Treatment 3

n 5 5 n 5 5 n 5 5 N 5 15

T 5 10 T 5 15 T 5 35 G 5 60

SS 5 21 SS 5 16 SS 5 23 oX2 5 370

2. A researcher reports an F-ratio with df between

5 3 and df within

5 28 for an independent- measures ANOVA. How many treatment conditions were compared in the experi-

ment? If all the treatments have the same number of participants, then how many

are in each treatment?

L E A R N I N G C H E C K

3 6 2 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

THE DISTRIBUTION OF F-RATIOS

In ANOVA, the F-ratio is constructed so that the numerator and denominator of the

ratio are measuring exactly the same variance when the null hypothesis is true (see

Equation 12.2). In this situation, we expect the value of F to be around 1.00. If we

obtain an F-ratio that is much greater than 1.00, then it is evidence that a treatment

effect exists and the null hypothesis is false. The problem now is to define precisely

which values are “around 1.00” and which are “much greater than 1.00.” To answer this

question, we need to look at all of the possible F values when H 0 is true—that is, the

distribution of F-ratios.

Before we examine this distribution in detail, you should note two obvious

characteristics:

1. Because F-ratios are computed from two variances (the numerator and denomi-

nator of the ratio), F values always are positive numbers. Remember that

variance is always positive.

2. When H 0 is true, the numerator and denominator of the F-ratio are measur-

ing the same variance. In this case, the two sample variances should be about

the same size, so the ratio should be near 1. In other words, the distribution of

F-ratios should pile up around 1.00.

With these two factors in mind, we can sketch the distribution of F-ratios. The

distribution is cut off at zero (all positive values), piles up around 1.00, and then

tapers off to the right (Figure 12.7). The exact shape of the F distribution depends

on the degrees of freedom for the two variances in the F-ratio. You should recall that

the precision of a sample variance depends on the number of scores or the degrees

of freedom. In general, the variance for a large sample (large df) provides a more

accurate estimate of the population variance. Because the precision of the MS values

12.4

3. A researcher conducts an experiment comparing three treatment conditions with

a separate sample of n 5 8 in each treatment. An ANOVA is used to evaluate the data, and the results of the ANOVA are presented in the following table. Complete

all missing values in the table. Hint: Begin with the values in the df column.

Source SS df MS

Between treatments __ __ ___ F 5 ____

Within treatments __ __ 2

Total 62 __

1. SS total

5 130; SS between

5 70; SS within

5 60

2. There were 4 treatment conditions (df between

5 k 2 1 5 3). A total of N 5 32 individuals participated (df

within 5 30 5 N 2 k) with n 5 8 in each treatment.

ANSWERS

Source SS df MS

Between treatments 20 2 10 F 5 5.00

Within treatments 42 21 2

Total 62 23

SECTION 12.4 / THE DISTRIBUTION OF F-RATIOS 3 6 3

depends on df, the shape of the F distribution also depends on the df values for the

numerator and denominator of the F-ratio. With very large df values, nearly all of the

F-ratios are clustered very near to 1.00. With the smaller df values, the F distribution

is more spread out.

For ANOVA, we expect F near 1.00 if H 0 is true, and we expect a large value for

F if H 0 is not true. In the F distribution, we need to separate those values that are

reasonably near 1.00 from the values that are significantly greater than 1.00. These

critical values are presented in an F distribution table in Appendix B. A portion of

the F distribution table is shown in Table 12.3. To use the table, you must know the

df values for the F-ratio (numerator and denominator), and you must know the alpha

level for the hypothesis test. It is customary for an F table to have the df values for

the numerator of the F-ratio printed across the top of the table. The df values for

the denominator of F are printed in a column on the left-hand side. For the experi-

ment we have been considering, the numerator of the F-ratio (between treatments)

has df 5 2, and the denominator of the F-ratio (within treatments) has df 5 12. This F-ratio is said to have “degrees of freedom equal to 2 and 12.” The degrees

of freedom would be written as df 5 2, 12. To use the table, you would first find df 5 2 across the top of the table and df 5 12 in the first column. When you line up these two values, they point to a pair of numbers in the middle of the table. These

numbers give the critical cutoffs for a 5 .05 and a 5 .01. With df 5 2, 12, for example, the numbers in the table are 3.88 and 6.93. Thus, only 5% of the distribu-

tion (a 5 .05) corresponds to values greater than 3.88 and only1% of the distribution (a 5 .01) corresponds to values greater than 6.93 (see Figure 12.7).

In the experiment comparing driving performance under different telephone condi-

tions, we obtained an F-ratio of 11.28. According to the critical cutoffs in Figure 12.7,

this value is extremely unlikely (it is in the most extreme 1%). Therefore, we would

reject H 0 with an a level of either .05 or .01 and conclude that the different telephone

conditions significantly affect driving performance.

T H E F D I ST R I B U T I O N TA B L E

F 0 1 2 3 4 5 6 7

3.88 6.93

FIGURE 12.7

The distribution of F-ratios

with df 5 2, 12. Of all the values in the distribution,

only 5% are larger than F

5 3.88, and only 1% are larger than F 5 6.93.

3 6 4 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

EXAMPLES OF HYPOTHESIS TESTING AND EFFECT SIZE WITH ANOVA

Although we have now seen all the individual components of ANOVA, the following

example demonstrates the complete ANOVA process using the standard four-step pro-

cedure for hypothesis testing.

What is the best strategy for studying before a quiz or exam? A partial answer to this

question comes from a research study comparing three different strategies (Weinstein,

McDermott, & Roediger, 2010). In the study, students read a passage knowing that they

would be tested on the material. In one condition, participants simply reread the passage.

In a second condition, the students answered prepared comprehension questions about the

material, and in a third condition, the students generated and answered their own questions.

The results showed that answering comprehension questions significantly improved exam

performance but it did not matter whether the students had to generate their own questions.

The data in Table 12.4 are similar to those obtained in the study but use four separate

groups of students and add a fourth condition in which the passage is read only once.

We use an ANOVA to determine whether there are any significant differences among

the four study strategies.

Before we begin the hypothesis test, note that we have already computed several

summary statistics for the data in Table 12.4. Specifically, the treatment totals (T) and

SS values are shown for each sample, and the grand total (G) as well as N and oX2 are shown for the entire set of data. Having these summary values simplifies the computa-

tions in the hypothesis test, and we suggest that you always compute these summary

statistics before you begin an ANOVA.

12.5

E X A M P L E 1 2 . 1

TABLE 12.3

A portion of the F distribu-

tion table. Entries in roman

type are critical values for

the .05 level of significance,

and entries in bold type are

for the .01 level of signifi-

cance. The critical values

for df 5 2, 12 have been highlighted (see text).

Degrees of Freedom: Numerator

Degrees of Freedom: Denominator 1 2 3 4 5 6

10 4.96 4.10 3.71 3.48 3.33 3.22

10.04 7.56 6.55 5.99 5.64 5.39

11 4.84 3.98 3.59 3.36 3.20 3.09

9.65 7.20 6.22 5.67 5.32 5.07

12 4.75 3.88 3.49 3.26 3.11 3.00

9.33 6.93 5.95 5.41 5.06 4.82

13 4.67 3.80 3.41 3.18 3.02 2.92

9.07 6.70 5.74 5.20 4.86 4.62

14 4.60 3.74 3.34 3.11 2.96 2.85

8.86 6.51 5.56 5.03 4.69 4.46

1. A researcher obtains F 5 4.10 with df 5 2, 14. Is this value sufficient to reject H 0

with a 5 .05? Is it big enough to reject H 0 if a 5 .01?

2. For an ANOVA evaluating the mean differences among three treatment conditions

with n 5 10 in each treatment, what is the critical value for the F-ratio with a 5 .05?

1. For a 5 .05, the critical value is 3.74 and you should reject H 0 . For a 5 .01, the critical

value is 6.51 and you should fail to reject H 0 .

2. The F-ratio has df 5 2, 27 and the critical value is 3.35.

L E A R N I N G C H E C K

ANSWERS

SECTION 12.5 / EXAMPLES OF HYPOTHESIS TESTING AND EFFECT SIZE WITH ANOVA 3 6 5

State the hypotheses and select an alpha level.

H 0 : �

1 5 �

2 5 �

3 5 �

4 (There is no treatment effect.)

H 1 : At least one of the treatment means is different.

We use a 5 .01.

Locate the critical region.

We first must determine degrees of freedom for MS between treatments

and MS within treatments

(the numerator and denominator of the F-ratio), so we begin by analyzing the degrees

of freedom. For these data, the total degrees of freedom are

df total

5 N 2 1

5 24 2 1

5 23

Analyzing this total into two components, we obtain

df between

5 k 2 1 5 4 2 1 5 3

df within

5 odf inside each treatment

5 5 1 5 1 5 1 5 5 20

The F-ratio for these data has df 5 3, 20. The distribution of all the possible F-ratios with df 5 3, 20 is presented in Figure 12.8. Note that F-ratios larger than 4.94 are extremely rare (p , .01) if H

0 is true and, therefore, form the critical region

for the test.

Compute the F-ratio.

The series of calculations for computing F is presented in Figure 12.4 and can be

summarized as follows:

a. Analyze the SS to obtain SS between

and SS within

b. Use the SS values and the df values (from step 2) to calculate the two variances,

MS between

and MS within

c. Finally, use the two MS values (variances) to compute the F-ratio.

S T E P 1 :

S T E P 1 :S T E P 2 :

S T E P 3 :

TABLE 12.4

Quiz scores for students

using four different study

strategies.

Read Once

Read and Reread

Answer Prepared Questions

Create and Answer Questions

3 5 8 8

3 3 5 9 N 5 24

4 5 8 7 G 5 168

6 7 9 10 oX2 5 1298

6 8 8 10

8 8 10 10

T 5 30 T 5 36 T 5 48 T 5 54

M 5 5 M 5 6 M 5 8 M 5 9

SS 5 20 SS 5 20 SS 5 14 SS 5 8

Often it is easier to postpone

finding the critical region

until after step 3, where you

compute the df values as part

of the calculations for the

F-ratio.

3 6 6 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

Analysis of SS. First, we compute the total SS and then the two components, as

indicated in Figure 12.5.

SS total

is simply the SS for the total set of N 5 24 scores.

SS G

N total

5 2 o X 2

5 21298 168

5 1298 2 1176

5 122

SS within

combines the SS values from inside each of the treatment conditions.

SS within

5 oSS inside each treatment

5 20 1 20 1 14 1 85 62

SS between

measures the differences among the four treatment means (or treatment

totals). Because we have already calculated SS total

and SS within

, the simplest way to obtain

SS between

is by subtraction (see Equation 12.5).

SS between

5 SS total

2 SS within

5 122 2 62

5 60

Calculation of mean squares. Because we already found df between

5 3 and df within

5 16 (Step 2), we now can compute the variance or MS value for each of the two components.

MS SS

df M

between between

between

5 5 5 60

3 20 SS

df within

within

5 5 5 62

20 3 1.

Calculation of F. We compute the F-ratio:

F MS

MS 5 5 5between

within

3 1 6 45

. .

4.94

1 2 3 4 5 6

df = 3, 20

FIGURE 12.8

The distribution of F-ratios

with df 5 3, 20. The critical value for a 5 .01 is 4.94.

SECTION 12.5 / EXAMPLES OF HYPOTHESIS TESTING AND EFFECT SIZE WITH ANOVA 3 6 7

Make a decision.

The F value we obtained, F 5 6.45, is in the critical region (see Figure 12.8). It is very unlikely (p , .01) that we would obtain a value this large if H

0 is true. Therefore,

we reject H 0 and conclude that there is a significant treatment effect.

Example 12.1 demonstrated the complete, step-by-step application of the ANOVA

procedure. There are two additional points that can be made using this example.

First, you should look carefully at the statistical decision. We have rejected H 0 and

concluded that not all the treatments are the same. But we have not determined which

ones are different. Is generating your own questions different from using prepared ques-

tions? Is rereading different from reading only once? Unfortunately, these questions

remain unanswered. We do know that at least one difference exists (we rejected H 0 ), but

additional analysis is necessary to find out exactly where this difference is. We address

this problem in Section 12.6.

Second, as noted earlier, all of the components of the analysis (the SS, df, MS, and F)

can be presented together in one summary table. The summary table for the analysis in

Example 12.1 is as follows:

Source SS df MS

Between treatments 60 3 20.00 F 5 6.45

Within treatments 62 20 3.10

Total 122 23

Although these tables are very useful for organizing the components of an ANOVA,

they are not commonly used in published reports. The current method for reporting the

results from an ANOVA is presented on page 368.

As we noted previously, a significant mean difference simply indicates that the difference

observed in the sample data is very unlikely to have occurred just by chance. Thus, the

term significant does not necessarily mean large, it simply means larger than expected by

chance. To provide an indication of how large the effect actually is, researchers should

report a measure of effect size in addition to the measure of significance.

For ANOVA, the simplest and most direct way to measure effect size is to compute

the percentage of variance accounted for by the treatment conditions. Like the r2 value

used to measure effect size for the t tests in Chapters 9, 10, and 11, this percentage

measures how much of the variability in the scores is accounted for by the differences

between treatments. For ANOVA, the calculation and the concept of the percentage of

variance is extremely straightforward. Specifically, we determine how much of the total

SS is accounted for by the SS between treatments

The percentage of variance accounted for 5 SS

bbetween treatments

total SS

(12.14)

For the data in Example 12.1, we obtain

The percentage of variance accounted for 5 60

1122 0 4925 .

In published reports of ANOVA results, the percentage of variance accounted for by

the treatment effect is usually called h2 (the Greek letter eta squared) instead of using r2. Thus, for the study in Example 12.1, h2 5 0.492.

S T E P 4 :

M E AS U R I N G E F F E C T S I Z E F O R A N OVA

3 6 8 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

IN THE LITERATURE

REPORTING THE RESULTS OF ANOVA

The APA format for reporting the results of ANOVA begins with a presentation of the

treatment means and standard deviations in the narrative of the article, a table, or a

graph. These descriptive statistics are not part of the calculations for the ANOVA, but

you can easily determine the treatment means from n and T (M = T/n) and the standard

deviations from the SS and n – 1 values for each treatment. Next, report the results

of the ANOVA. For the study described in Example 12.1, the report might state the

following:

The means and standard deviations are presented in Table 1. The analysis of

variance indicates that there are significant differences among the four study

strategies, F(3, 20) 5 6.45, p , .01, h2 5 0.492.

TABLE 1

Quiz scores for students using four different study strategies

Read Once Read and Reread Answer Prepared

Questions Create and

Answer Questions

M 5.00 6.00 8.00 9.00

SD 2.00 2.00 1.67 1.26

Note how the F-ratio is reported. In this example, degrees of freedom for between

and within treatments are df 5 3, 20, respectively. These values are placed in parenthe- ses immediately following the symbol F. Next, the calculated value for F is reported,

followed by the probability of committing a Type I error (the alpha level) and the

measure of effect size.

When an ANOVA is done using a computer program, the F-ratio is usually accompa-

nied by an exact value for p. The data from Example 12.1 were analyzed using the SPSS

program (see Resources at the end of this chapter) and the computer output included a sig-

nificance level of p 5 .003. Using the exact p value from the computer output, the research report would conclude, “The analysis of variance revealed significant differences among

the four viewing distances, F(3, 20) 5 6.45, p 5 .003, h2 5 0.492.”

Because ANOVA requires relatively complex calculations, students encountering this

statistical technique for the first time often tend to be overwhelmed by the formulas

and arithmetic and lose sight of the general purpose for the analysis. The following two

examples are intended to minimize the role of the formulas and shift attention back to

the conceptual goal of the ANOVA process.

The following data represent the outcome of an experiment using two separate samples

to evaluate the mean difference between two treatment conditions. Take a minute to

look at the data and, without doing any calculations, try to predict the outcome of an

A CO N C E P T UA L V I E W O F A N OVA

E X A M P L E 1 2 . 2

SECTION 12.5 / EXAMPLES OF HYPOTHESIS TESTING AND EFFECT SIZE WITH ANOVA 3 6 9

ANOVA for these values. Specifically, predict what values should be obtained for the

between-treatments variance (MS) and the F-ratio. If you do not “see” the answer after

20 or 30 seconds, try reading the hints that follow the data.

Treatment I Treatment II

4 2 N 5 8

0 1 G 5 16

1 0 oX2 5 56

3 5

T 5 8 T 5 8

SS 5 10 SS 5 14

If you are having trouble predicting the outcome of the ANOVA, read the following

hints, and then go back and look at the data.

Hint 1: Remember: SS between

and MS between

provide a measure of how much differ-

ence there is between treatment conditions.

Hint 2: Find the mean or total (T) for each treatment, and determine how much

difference there is between the two treatments.

You should realize by now that the data have been constructed so that there is zero

difference between treatments. The two sample means (and totals) are identical, so

SS between

5 0, MS between

5 0, and the F-ratio is zero.

Conceptually, the numerator of the F-ratio always measures how much difference

exists between treatments. In Example 12.2, we constructed an extreme set of scores

with zero difference. However, you should be able to look at any set of data and quickly

compare the means (or totals) to determine whether there are big differences or small dif-

ferences between treatments. We should also note that the size of the differences between

treatments also directly influences measures of effect size. The data in example 12.2 have

SS between

5 0, which means that h2 5 0. As the difference between treatments increases, h2 also increases.

Being able to estimate the magnitude of between-treatment differences is a good

first step in understanding ANOVA and should help you to predict the outcome of an

ANOVA. However, the between-treatment differences are only one part of the analysis.

You must also understand the within-treatment differences that form the denominator of

the F-ratio. The following example is intended to demonstrate the concepts underlying

SS within

and MS within

. In addition, the example should give you a better understanding of

how the between-treatment differences and the within-treatment differences act together

within the ANOVA.

The purpose of this example is to present a visual image for the concepts of between-

treatments variability and within-treatments variability. In this example, we compare

two hypothetical outcomes for the same experiment. In each case, the experiment

uses two separate samples to evaluate the mean difference between two treatments.

The following data represent the two outcomes, which we call experiment A and

experiment B.

E X A M P L E 1 2 . 3

3 7 0 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

Experiment A Experiment B

Treatment Treatment

I II I II

8 12 4 12

8 13 11 9

7 12 2 20

9 11 17 6

8 13 0 16

9 12 8 18

7 11 14 3

M 5 8 M 5 12 M 5 8 M 5 12

s 5 0.82 s 5 0.82 s 5 6.35 s 5 6.35

The data from experiment A are displayed in a frequency distribution graph in

Figure 12.9(a). Notice that there is a 4-point difference between the treatment means

(M 1 5 8 and M

2 5 12). This is the between-treatments difference that contributes to

the numerator of the F-ratio. Also notice that the scores in each treatment are clustered

closely around the mean, indicating that the variance inside each treatment is relatively

small. This is the within-treatments variance that contributes to the denominator of the

F-ratio. Finally, you should realize that it is easy to see the mean difference between the

Between treatments

Treatment 1

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Treatment 2

18 19 20

F re

q u

e n

c y

Between treatments

Treatment 1

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Treatment 2

18 19 20

F re

q u

e n

c y

M1 5 8 SS1 5 4

M2 5 12 SS2 5 4

M2 5 12 SS2 5 242

M1 5 8 SS1 5 242

Experiment B

Experiment A

FIGURE 12.9

A visual representation of

the between-treatments

variability and the within-

treatments variability that

form the numerator and

denominator, respectively,

of the F-ratio. In (a), the

difference between treat-

ments is relatively large

and easy to see. In (b),

the same 4-point differ-

ence between treatments

is relatively small and

is overwhelmed by the

within-treatments vari-

ability.

(a)

(b)

SECTION 12.5 / EXAMPLES OF HYPOTHESIS TESTING AND EFFECT SIZE WITH ANOVA 3 7 1

two samples. The fact that there is a clear mean difference between the two treatments

is confirmed by computing the F-ratio for experiment A.

F 5 between-treatments difference

within-treatmments differences between

within

5 5 MS

0 66. 77 83 965 .

An F-ratio of F 5 83.96 is sufficient to reject the null hypothesis, so we conclude that there is a significant difference between the two treatments.

Now consider the data from experiment B, which are shown in Figure 12.9(b) and

present a very different picture. This experiment has the same 4-point difference be-

tween treatment means that we found in experiment A (M 1 5 8 and M

2 5 12). However,

for these data, the scores in each treatment are scattered across the entire scale, indicat-

ing relatively large variance inside each treatment. In this case, the large variance within

treatments overwhelms the relatively small mean difference between treatments. In the

figure, it is almost impossible to see the mean difference between treatments. For these

data, the F-ratio confirms that there is no clear mean difference between treatments.

F 5 between-treatments difference

within-treatmments differences between

within

5 5 MS

40 3. 33 1 395 .

For experiment B, the F-ratio is not large enough to reject the null hypothesis, so

we conclude that there is no significant difference between the two treatments. Once

again, the statistical conclusion is consistent with the appearance of the data in Figure

12.9(b). Looking at the figure, we see that the scores from the two samples appear to

be intermixed randomly with no clear distinction between treatments.

Sample variance also influences measures of effect size. The two samples in Figure

12.9(a), with small variance, produce h2 5 56/64 5 0.875. When the variance is in- creased in Figure 12.9(b), the measure of effect size is reduced to h2 5

540 5 0.104.

As we have noted in previous chapters, high variability makes it difficult to see any

patterns in the data. In Figure 12.9(a), the 4-point mean difference between treatments

is easy to see because the sample variance is small. In Figure 12.9(b), the 4-point differ-

ence gets lost because the sample variance is large. In general, you can think of variance

as measuring the amount of “noise” or “confusion” in the data. With large variance

there is a lot of noise and confusion and it is difficult to see any clear patterns.

Although Examples 12.2 and 12.3 present somewhat simplified demonstrations with

exaggerated data, the general point of the examples is to help you see what happens

when you perform an ANOVA. Specifically:

1. The numerator of the F-ratio (MS between

) measures how much difference exists

between the treatment means. Bigger mean differences produce larger F-ratios

and larger measures of effect size.

2. The denominator of the F-ratio (MS within

) measures the variance of the scores

inside each treatment; that is, the variance for each of the separate samples. In

general, larger sample variance produces smaller F-ratios and smaller measures

of effect size.

We should note that the number of scores in the samples also can influence the

outcome of an ANOVA. As with most other hypothesis tests, if other factors are held

constant, increasing the sample size tends to increase the likelihood of rejecting the

null hypothesis. However, changes in sample size have little or no effect on measures

of effect size such as h2.

3 7 2 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

You may have recognized that the two research outcomes presented in Example 12.3

are similar to those presented earlier in Example 10.5 in Chapter 10. Both examples are

intended to demonstrate the role of variance in a hypothesis test. Both examples show

that large values for sample variance can obscure any patterns in the data and reduce

the potential for finding significant differences between means.

For the independent-measures t statistic in Chapter 10, the sample variance con-

tributed directly to the standard error in the bottom of the t formula. Now, the sample

variance contributes directly to the value of MS within

in the bottom of the F-ratio. In the

t-statistic and in the F-ratio, the variances from the separate samples are pooled together

to create one average value for sample variance. For the independent-measures t statis-

tic, we pooled two samples together to compute

pooled variance 5 5 1

1 s

SS SS

df df p

2 1 2

1 2

Now, in ANOVA, we are combining two or more samples to calculate

MS SS

SS SS SS within

within

5 5 5 1 1o

1 2 33

1 2 3

1 1 1

. . .

. . .df df df

Notice that the concept of pooled variance is the same whether you have exactly

two samples or more than two samples. In either case, you simply add the SS values

and divide by the sum of the df values. The result is an average of all of the different

sample variances.

When the samples are all the same size, MS within

is literally the average of the sample

variances. In Example, 12.1, the four samples all have n 5 6 scores. The four SS values are 20, 20, 14, and 8, and each has df 5 5. Therefore, the four sample variances are 20

5 5 4.00,

5 5 4.00, 14

5 5 2.80 and

5 5 1.60. The average of the four variances is

(4.0 1 4.0 1 2.8 11.6)/4 5 12.4/4 5 3.10, which is exactly the value we obtained for MS

within . If the samples are not all the same size, MS

within is still the average of the sample

variances, however, the average is computed so each sample variance is weighted by

its sample size.

In the previous examples, all of the samples were exactly the same size (equal ns).

However, the formulas for ANOVA can be used when the sample size varies within a re-

search study. You also should note, however, that the general ANOVA procedure is most

accurate when used to examine data with equal sample sizes. Therefore, researchers

generally try to plan research studies with equal ns. However, there are circumstances

in which it is impossible or impractical to have an equal number of subjects in every

treatment condition. In these situations, ANOVA still provides a valid test, especially

when the samples are relatively large and when the discrepancy between sample sizes

is not extreme.

The following example demonstrates an ANOVA with samples of different sizes.

A researcher is interested in the amount of homework required by different academic

majors. Students are recruited from Biology, English, and Psychology to participate in

the study. The researcher randomly selects one course that each student is currently tak-

ing and asks the student to record the amount of out-of-class work required each week

for the course. The researcher used all of the volunteer participants, which resulted in

unequal sample sizes. The data are summarized in Table 12.5.

M S W I T H I N

A N D P O O L E D VA R I A N C E

A N E X A M P L E W I T H U N E Q UA L SA M P L E

S I Z E S

E X A M P L E 1 2 . 4

SECTION 12.5 / EXAMPLES OF HYPOTHESIS TESTING AND EFFECT SIZE WITH ANOVA 3 7 3

State the hypotheses, and select the alpha level.

H 0 : �

1 5 �

2 5 �

H 1 : At least one population is different.

a 5 .05

Locate the critical region.

To find the critical region, we first must determine the df values for the F-ratio:

df total

5 N – 1 5 20 2 1 5 19

df between

5 k 2 1 5 3 2 1 5 2

df within

5 N 2 k 5 20 2 3 5 17

The F-ratio for these data has df 5 2, 17. With a 5 .05, the critical value for the F-ratio is 3.59.

Compute the F-ratio.

First, compute the three SS values. As usual, SS total

is the SS for the total set of N 5 20 scores, and SS

within combines the SS values from inside each of the treatment conditions.

SS G

N total

5 2 o X 2

within 5 oSS

inside each treatment

5 3377 2 3125 5 37 1 90 1 60

5 252 5 187

SS between

can be found by subtraction (Equation 12.5).

SS between

5 SS total

2 SS within

5 252 2 187

5 65

Or, SS between

can be calculated using the computation formula (see Equation 12.7). If

you use the computational formula, be careful to match each treatment total (T) with

the appropriate sample size (n) as follows:

SS T

N between

5 2

5 1 1 2

 2 2

2 2 2 2 36

130

250

220

324 1690 1176 3125

5 1 1 2

S T E P 1 :

S T E P 2 :

S T E P 3 :

TABLE 12.5

Average hours of homework

per week for one course for

students in three academic

majors.

Biology English Psychology

n 5 4 n 5 10 n 5 6 N 5 20

M 5 9 M 5 13 M 5 14 G 5 250

T 5 36 T 5 130 T 5 84 oX2 5 3377

SS 5 37 SS 5 90 SS 5 60

3 7 4 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

Finally, compute the MS values and the F-ratio:

MS SS

between

within

5 5 5

2 32 5

187

111

32 5

11 2 95F

. .5 5 5between

within

Make a decision.

Because the obtained F-ratio is not in the critical region, we fail to reject the

null hypothesis and conclude that there are no significant differences among

the three populations of students in terms of the average amount of homework each

week.

S T E P 4 :

1. A researcher used ANOVA and computed F 5 4.25 for the following data.

Treatments

I II III

n 5 10 n 5 10 n 5 10

M 5 20 M 5 28 M 5 35

SS 5 1005 SS 5 1391 SS 5 1180

a. If the mean for treatment III were changed to M 5 25, what would happen to the size of the F-ratio (increase or decrease)?

b. If the SS for treatment I were changed to SS 5 1400, what would happen to the size of the F-ratio (increase or decrease)?

c. If the sample means and variances were held constant but each sample size were

increased to n 5 20, then what would happen to the size of the F-ratio (increase or decrease)?

2. A research study comparing three treatment conditions produces M 5 5 with n 5 4 for the first treatment, M 5 2 with n 5 5 for the second treatment, and M 5 5 with n 5 6 for the third treatment. Calculate SS

between treatments for these data.

1. a. If the mean for treatment III were changed to M 5 25, it would reduce the size of the mean differences (the three means would be closer together). This would reduce the size

of MS between

and would reduce the size of the F-ratio.

b. If the SS in treatment I were increased to SS 5 1400, it would increase the size of the variability within treatments. This would increase MS

within and would reduce the size of the

F-ratio.

c. If other factors are held constant, increasing the size of the samples increases the F-ratio

and increases the likelihood of rejecting the null hypothesis.

2. The three sample totals are T 1 5 20, T

2 5 10, and T

3 5 30, which produces G 5 60 and

N 5 15. SS between

5 30.

L E A R N I N G C H E C K

ANSWERS

SECTION 12.6 / POST HOC TESTS 3 7 5

POST HOC TESTS

As noted earlier, the primary advantage of ANOVA (compared to t tests) is that it allows

researchers to test for significant mean differences when there are more than two treat-

ment conditions. ANOVA accomplishes this feat by comparing all of the individual mean

differences simultaneously within a single test. Unfortunately, the process of combining

several mean differences into a single test statistic creates some difficulty when it is time

to interpret the outcome of the test. Specifically, when you obtain a significant F-ratio

(reject H 0 ), it simply indicates that somewhere among the entire set of mean differences

there is at least one that is statistically significant. In other words, the overall F-ratio

only tells you that a significant difference exists; it does not tell exactly which means are

significantly different and which are not.

Consider, for example, a research study that uses three samples to compare three treat-

ment conditions. Suppose that the three sample means are M 1 5 3, M

2 5 5, and M

3 5 10.

In this hypothetical study, there are three mean differences:

1. There is a 2-point difference between M 1 and M

2 .

2. There is a 5-point difference between M 2 and M

3 .

3. There is a 7-point difference between M 1 and M

3 .

If an ANOVA were used to evaluate these data, a significant F-ratio would indi-

cate that at least one of the sample mean differences is large enough to satisfy the

criterion of statistical significance. In this example, the 7-point difference is the

biggest of the three and, therefore, it must indicate a significant difference between

the first treatment and the third treatment (� 1 ≠ �

3 ). But what about the 5-point

difference? Is it also large enough to be significant? And what about the 2-point

difference between M 1 and M

2 ? Is it also significant? The purpose of post hoc tests

is to answer these questions.

Post hoc tests (or posttests) are additional hypothesis tests that are done after

an ANOVA to determine exactly which mean differences are significant and

which are not.

As the name implies, post hoc tests are done after an ANOVA. More specifically,

these tests are done after ANOVA when

1. You reject H 0 and

2. There are three or more treatments (k ≥ 3).

Rejecting H 0 indicates that at least one difference exists among the treatments.

If there are only two treatments, then there is no question about which means are

different and, therefore, no need for posttests. However, with three or more treat-

ments (k ≥ 3), the problem is to determine exactly which means are significantly

different.

In general, a post hoc test enables you to go back through the data and compare the

individual treatments two at a time. In statistical terms, this is called making pairwise

comparisons. For example, with k 5 3, we would compare � 1 versus �

2 , then �

2 versus

� 3 , and then �

1 versus �

3 . In each case, we are looking for a significant mean difference.

12.6

D E F I N I T I O N

P O ST T E ST S A N D T Y P E I E R R O R S

3 7 6 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

The process of conducting pairwise comparisons involves performing a series of sepa-

rate hypothesis tests, and each of these tests includes the risk of a Type I error. As you

do more and more separate tests, the risk of a Type I error accumulates and is called the

experimentwise alpha level (see p. 350).

Whenever you are conducting posttests, you must be concerned about the experi-

mentwise alpha level. Statisticians have worked with this problem and have developed

several methods for trying to control Type I errors in the context of post hoc tests. We

consider two alternatives.

The first post hoc test we consider is Tukey’s HSD test. We selected Tukey’s HSD

test because it is a commonly used test in psychological research. Tukey’s test

allows you to compute a single value that determines the minimum difference

between treatment means that is necessary for significance. This value, called

the honestly significant difference, or HSD, is then used to compare any two treat-

ment conditions. If the mean difference exceeds Tukey’s HSD, then you conclude

that there is a significant difference between the treatments. Otherwise, you cannot

conclude that the treatments are significantly different. The formula for Tukey’s

HSD is

HSD q MS

n 5 within

(12.15)

where the value of q is found in Table B.5 (Appendix B, p. 580), MS within

is the

within-treatments variance from the ANOVA, and n is the number of scores in each

treatment. Tukey’s test requires that the sample size, n, be the same for all treat-

ments. To locate the appropriate value of q, you must know the number of treat-

ments in the overall experiment (k), the degrees of freedom for MS within

(the error

term in the F-ratio), and you must select an alpha level (generally the same a used for the ANOVA).

To demonstrate the procedure for conducting post hoc tests with Tukey’s HSD, we use the

hypothetical data shown in Table 12.6. The data represent the results of a study compar-

ing scores in three different treatment conditions. Note that the table displays summary

statistics for each sample and the results from the overall ANOVA. With k 5 3 treatments, df

within 5 24, and a 5 .05, you should find that the value of q for the test is q 5 3.53 (see

Table B.5). Therefore, Tukey’s HSD is

HSD q MS

n .

. .5 5 5within 3 53

4 00

9 2 36

T U K E Y ’ S H O N E ST LY S I G N I F I CA N T

D I F F E R E N C E ( H S D ) T E ST

E X A M P L E 1 2 . 5

The q value used in Tukey’s

HSD test is called a

Studentized range statistic.

TABLE 12.6

Hypothetical results from a

research study comparing

three treatment conditions.

Summary statistics are pre-

sented for each treatment

along with the outcome from

the ANOVA.

Source SS df MS

Between 73.19 2 36.60

Within 96.00 24 4.00

Total 169.19 26

Overall F(2, 24) 5 9.15

Treatment A

Treatment B

Treatment C

n 5 9 n 5 9 n 5 9

T 5 27 T 5 49 T 5 63

M 5 3.00 M 5 5.44 M 5 7.00

SECTION 12.6 / POST HOC TESTS 3 7 7

Thus, the mean difference between any two samples must be at least 2.36 to be sig-

nificant. Using this value, we can make the following conclusions:

1. Treatment A is significantly different from treatment B (M A 2 M

B 5 2.44).

2. Treatment A is also significantly different from treatment C (M A – M

C 5 4.00).

3. Treatment B is not significantly different from treatment C (M B – M

C 5 1.56).

Because it uses an extremely cautious method for reducing the risk of a Type I error,

the Scheffé test has the distinction of being one of the safest of all possible post hoc

tests (smallest risk of a Type I error). The Scheffé test uses an F-ratio to evaluate the

significance of the difference between any two treatment conditions. The numerator of

the F-ratio is an MS between

that is calculated using only the two treatments you want to

compare. The denominator is the same MS within

that was used for the overall ANOVA.

The “safety factor” for the Scheffé test comes from the following two considerations:

1. Although you are comparing only two treatments, the Scheffé test uses the

value of k from the original experiment to compute df between treatments.

Thus, df for the numerator of the F-ratio is k – 1.

2. The critical value for the Scheffé F-ratio is the same as was used to evaluate

the F-ratio from the overall ANOVA. Thus, Scheffé requires that every posttest

satisfy the same criterion that was used for the complete ANOVA. The follow-

ing example uses the data from Table 12.6 to demonstrate the Scheffé posttest

procedure.

Remember that the Scheffé procedure requires a separate SS between

, MS between

, and F-ratio

for each comparison being made. Although Scheffé computes SS between

using the regular

computational formula (Equation 12.7), you must remember that the numbers in the

formula are entirely determined by the two treatment conditions being compared. We

begin by comparing treatment A (with T 5 27 and n 5 9) and treatment B (with T 5 49 and n 5 9). The first step is to compute SS

between for these two groups. In the formula for

SS, notice that the grand total for the two groups is G 5 27 1 49 5 76, and the total number of scores for the two groups is N 5 9 1 9 5 18.

SS T

N between

5 2

5 1 2

5 1

 2 2

2 2 2 27

81 266 .. .

78 320 89

26 89

Although we are comparing only two groups, these two were selected from a study

consisting of k 5 3 samples. The Scheffé test uses the overall study to determine the degrees of freedom between treatments. Therefore, df

between 5 3 – 1 5 2, and the

MS between

MS SS

df between

between

5 5 5 26 89

2 1

. 33 45.

T H E S C H E F F É T E ST

E X A M P L E 1 2 . 6

3 7 8 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

Finally, the Scheffé procedure uses the error term from the overall ANOVA to com-

pute the F-ratio. In this case, MS within

5 4.00 with df within

5 24. Thus, the Scheffé test produces an F-ratio of

F MS

MS A Bversus

between

within

5 5 5 13 45

4 00 3 3

. . 66

With df 5 2, 24 and a 5 .05, the critical value for F is 3.40 (see Table B.4). Therefore, our obtained F-ratio is not in the critical region, and we must conclude that

these data show no significant difference between treatment A and treatment B.

The second comparison involves treatment B (T 5 49) and treatment C (T 5 63). This time the data produce SS

between 5 10.89, MS

between 5 5.45, and F(2, 24) 5 1.36 (check the

calculations for yourself). Once again the critical value for F is 3.40, so we must conclude

that the data show no significant difference between treatment B and treatment C.

The final comparison is treatment A (T 5 27) and treatment C (T 5 63). This time the data produce SS

between 5 72, MS

between 5 36, and F(2, 24) 5 9.00 (check the calcula-

tions for yourself). Once again the critical value for F is 3.40, and this time we conclude

that the data show a significant difference.

Thus, the Scheffé posttest indicates that the only significant difference is between

treatment A and treatment C.

There are two interesting points to be made from the posttest outcomes presented

in the preceding two examples. First, the Scheffé test was introduced as being one

of the safest of the posttest techniques because it provides the greatest protection

from Type I errors. To provide this protection, the Scheffé test simply requires a

larger difference between sample means before you may conclude that the differ-

ence is significant. For example, using Tukey’s test in Example 12.5, we found

that the difference between treatment A and treatment B was large enough to be

significant. However, this same difference failed to reach significance according to

the Scheffé test (Example 12.6). The discrepancy between the results is an example

of the Scheffé test’s extra demands: The Scheffé test simply requires more evidence

and, therefore, it is less likely to lead to a Type I error.

The second point concerns the pattern of results from the three Scheffé tests in

Example 12.6. You may have noticed that the posttests produce what are apparently

contradictory results. Specifically, the tests show no significant difference between

A and B and they show no significant difference between B and C. This combina-

tion of outcomes might lead you to suspect that there is no significant difference

between A and C. However, the test did show a significant difference. The answer

to this apparent contradiction lies in the criterion of statistical significance. The dif-

ferences between A and B and between B and C are too small to satisfy the criterion

of significance. However, when these differences are combined, the total difference

between A and C is large enough to meet the criterion for significance.

1. With k 5 2 treatments, are post hoc tests necessary when the null hypothesis is rejected? Explain why or why not.

2. Three treatments, each with a sample of n 5 8 participants, have treatment totals of T

1 5 16, T

2 5 32, and T

3 5 40, and produce MS

within 5 2.5.

a. Use Tukey’s HSD test with a 5 .05 to evaluate the significance of the mean difference between treatments 1 and 2.

L E A R N I N G C H E C K

SECTION 12.7 / THE RELATIONSHIP BETWEEN ANOVA AND t TESTS 3 7 9

THE RELATIONSHIP BETWEEN ANOVA AND t TESTS

When you are evaluating the mean difference from an independent-measures study

comparing only two treatments (two separate samples), you can use either an

independent-measures t test (Chapter 10) or the ANOVA presented in this chapter.

In practical terms, it makes no difference which you choose. These two statistical

techniques always result in the same statistical decision. In fact, the two methods

use many of the same calculations and are very closely related in several other

respects. The basic relationship between t statistics and F-ratios can be stated in

an equation:

F 5 t2

This relationship can be explained by first looking at the structure of the formulas

for F and t. The t statistic compares distances: the distance between two sample means

(numerator) and the distance computed for the standard error (denominator). The F-ratio,

on the other hand, compares variances. You should recall that variance is a measure of

squared distance. Hence, the relationship: F 5 t2. There are several other points to consider in comparing the t statistic to the

F-ratio.

1. It should be obvious that you are testing the same hypotheses whether you

choose a t test or an ANOVA. With only two treatments, the hypotheses for

either test are

H 0 : �

1 5 �

H 1 : �

1 ≠ �

2. The degrees of freedom for the t statistic and the df for the denominator

of the F-ratio (df within

) are identical. For example, if you have two samples,

each with six scores, the independent-measures t statistic has df 5 10, and the F-ratio has df 5 1, 10. In each case, you are adding the df from the first sample (n 2 1) and the df from the second sample (n 2 1).

3. The distribution of t and the distribution of F-ratios match perfectly if you take

into consideration the relationship F 5 t2. Consider the t distribution with df 5 18

12.7

b. Use the Scheffé test with a 5 .05 to evaluate the significance of the mean difference between treatments 1 and 2.

1. No. Post hoc tests are used to determine which treatments are different. With only two treat-

ment conditions, there is no uncertainty as to which two treatments are different.

2. a. For this test, q 5 3.53 and HSD 5 1.97. There is a 2-point mean difference between treatments 1 and 2, which is large enough to be significant.

b. For these two treatments, Scheffé produces SS between

5 16, MS between

5 8, and F 5 3.2. With df 5 2, 21 the critical value is 3.47. Conclude that the mean difference between treatments 1 and 2 is not significant.

ANSWERS

3 8 0 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

and the corresponding F distribution with df 5 1, 18 that are presented in Figure 12.10. Notice the following relationships:

a. If each of the t values is squared, then all of the negative values become posi-

tive. As a result, the whole left-hand side of the t distribution (below zero)

is flipped over to the positive side. This creates an asymmetrical, positively

skewed distribution—that is, the F distribution.

b. For a 5 .05, the critical region for t is determined by values greater than 12.101 or less than 22.101. When these boundaries are squared, you get ±2.1012 5 4.41.

Notice that 4.41 is the critical value for a 5 .05 in the F distribution. Any value that is in the critical region for t ends up in the critical region for F-ratios after it is squared.

The independent-measures ANOVA requires the same three assumptions that were

necessary for the independent-measures t hypothesis test:

1. The observations within each sample must be independent (see p. 220).

2. The populations from which the samples are selected must be normal.

3. The populations from which the samples are selected must have equal variances

(homogeneity of variance).

Ordinarily, researchers are not overly concerned with the assumption of normality,

especially when large samples are used, unless there are strong reasons to suspect that

the assumption has not been satisfied. The assumption of homogeneity of variance is

an important one. If a researcher suspects that it has been violated, it can be tested by

Hartley’s F-max test for homogeneity of variance (Chapter 10, p. 300).

AS S U M P T I O N S F O R T H E I N D E P E N D E N T- M E AS U R E S A N OVA

FIGURE 12.10

The distribution of t

statistics with df 5 18 and the corresponding

distribution of F-ratios

with df 5 1, 18. Notice that the critical values for

a 5 .05 are t 5 ±2.101 and F 5 2.1012 5 4.41.

0 1 2 3 4 5

4.41

(2.101 2 )

95 %

–2.101 0 2.101

SUMMARY 3 8 1

1. Analysis of variance (ANOVA) is a statistical tech- nique that is used to test the significance of mean dif- ferences among two or more treatment conditions. The null hypothesis for this test states that, in the general population, there are no mean differences among the treatments. The alternative states that at least one mean is different from another.

2. The test statistic for ANOVA is a ratio of two vari- ances called an F-ratio. The variances in the F-ratio are called mean squares, or MS values. Each MS is computed by

MS SS

df 5

3. For the independent-measures ANOVA, the F-ratio is

F MS

MS 5 between

within

The MS between

measures differences between the treat- ments by computing the variance for the treatment means or totals. These differences are assumed to be produced by

a. Treatment effects (if they exist)

b. Random, unsystematic differences (chance)

The MS within

measures variance inside each of the treatment conditions. Because individuals inside a treatment condition are all treated exactly the same, any differences within treatments cannot be caused by treatment effects. Thus, the within-treatments MS is produced only by random, unsystematic dif- ferences. With these factors in mind, the F-ratio has the following structure:

1. A researcher uses an independent-measures t test to evaluate the mean difference

obtained in a research study, and obtains a t statistic of t 5 3.00. If the researcher used an ANOVA to evaluate the results, then what F-ratio would be obtained?

2. An ANOVA produces an F-ratio with df 5 1, 34. Could the data have been analyzed with a t test? What would be the degrees of freedom for the t statistic?

1. F 5 9 because F 5 t2

2. If the F-ratio has df 5 1, 34, then the experiment compared only two treatments, and you could use a t statistic to evaluate the data. The t statistic would have df 5 34.

L E A R N I N G C H E C K

ANSWERS

SUMMARY

F 5 1treatment effect random, unsystematic diffferences

random, unsystematic differences

When there is no treatment effect (H 0 is true), the

numerator and the denominator of the F-ratio are measuring the same variance, and the obtained ratio should be near 1.00. If there is a significant treatment effect, then the numerator of the ratio should be larger than the denominator, and the obtained F value should be much greater than 1.00.

4. The formulas for computing each SS, df, and MS value are presented in Figure 12.11, which also shows the general structure for the ANOVA.

5. The F-ratio has two values for degrees of freedom, one associated with the MS in the numerator and one asso- ciated with the MS in the denominator. These df values are used to find the critical value for the F-ratio in the F distribution table.

6. Effect size for the independent-measures ANOVA is measured by computing eta squared, the percentage of variance accounted for by the treatment effect.

h 2 5 1

5 SS

SS SS

between

between within

between

SS total

7. When the decision from an ANOVA is to reject the null hypothesis and when the experiment contains more than two treatment conditions, it is necessary to continue the analysis with a post hoc test, such as Tukey’s HSD test or the Scheffé test. The purpose of these tests is to determine exactly which treatments are significantly different and which are not.

3 8 2 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

KEY TERMS

analysis of variance (ANOVA) (346)

factor (347)

levels (347)

F-ratio (349)

testwise alpha level (350)

experimentwise alpha level (350)

between-treatments variance (351)

within-treatments variance (352)

treatment effect (352)

error term (353)

mean square (MS) (360)

ANOVA summary table (361)

distribution of F-ratios (362)

eta squared (h2) (367)

post hoc tests (375)

pairwise comparisons (375)

Tukey’s HSD test (376)

Scheffé test (377)

RESOURCES

Go to CengageBrain.com to access Psychology CourseMate, where you will find an

interactive eBook, glossaries, flashcards, quizzes, statistics workshops, and more.

If your professor has assigned Aplia:

1. Sign in to your account.

2. Complete the corresponding exercises as required by your professor.

3. When finished, click “Grade It Now” to see which areas you have mastered, which

areas need more work, and detailed explanations of every answer.

Within treatments

SS = ΣSSeach treatment

Between treatments

SS = n(SS for the means)

or SS =

df = k � 1

MS =

Total

df = N � 1

G2 SS = Σ X

2 �

Σ G2

T 2

n �

df = N � k

MS = SS

MS between treatments

MS within treatments F-ratio =

FIGURE 12.11

Formulas for ANOVA.

FOCUS ON PROBLEM SOLVING 3 8 3

General instructions for using SPSS are presented in Appendix D. Following are

detailed instructions for using SPSS to perform The Single-Factor, Independent-

Measures Analysis of Variance (ANOVA) presented in this chapter.

Data Entry

1. The scores are entered in a stacked format in the data editor, which means that

all of the scores from all of the different treatments are entered in a single col-

umn (VAR00001). Enter the scores for treatment #2 directly beneath the scores

from treatment #1 with no gaps or extra spaces. Continue in the same column

with the scores from treatment #3, and so on.

2. In the second column (VAR00002), enter a number to identify the treatment

condition for each score. For example, enter a 1 beside each score from

the first treatment, enter a 2 beside each score from the second treatment,

and so on.

Data Analysis

1. Click Analyze on the tool bar, select Compare Means, and click on One-Way

ANOVA.

2. Highlight the column label for the set of scores (VAR0001) in the left box and

click the arrow to move it into the Dependent List box.

3. Highlight the label for the column containing the treatment numbers

(VAR0002) in the left box and click the arrow to move it into the

Factor box.

4. If you want descriptive statistics for each treatment, click on the Options box,

select Descriptives, and click Continue.

5. Click OK.

SPSS Output

We used the SPSS program to analyze the data from the studying strategy experiment

in Example 12.1, and the program output is shown in Figure 12.12. The output begins

with a table showing descriptive statistics (number of scores, mean, standard deviation,

standard error for the mean, a 95% confidence interval for the mean, maximum, and

minimum scores) for each sample. The second part of the output presents a summary

table showing the results from the ANOVA.

FOCUS ON PROBLEM SOLVING

1. It can be helpful to compute all three SS values separately, and then check

to verify that the two components (between and within) add up to the total.

However, you can greatly simplify the calculations if you simply find SS total

and

SS within

, then obtain SS between

by subtraction.

3 8 4 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

2. Remember that an F-ratio has two separate values for df: a value for the numer-

ator and one for the denominator. Properly reported, the df between

value is stated

first. You will need both df values when consulting the F distribution table for

the critical F value.

3. When you encounter an F-ratio and its df values reported in the literature, you

should be able to reconstruct much of the original experiment. For example, if you

see “F(2, 36) 5 4.80,” you should realize that the experiment compared k 5 3 treatment groups (because df

between 5 k 2 1 5 2), with a total of N 5 39 subjects

participating in the experiment (because df within

5 N 2 k 5 36).

DEMONSTRATION 12.1

ANALYSIS OF VARIANCE

A human-factors psychologist studied three computer keyboard designs. Three

samples of individuals were given material to type on a particular keyboard, and

VAR00001

Descriptives

1.00

2.00

3.00

4.00

Total

Between Groups

Within Groups

Total

60.000

62.000

122.000

20.000

3.100

6.452 .003

5.0000

6.0000

8.0000

9.0000

7.0000

2.00000

1.67332

1.26491

2.30312

.81650

.68313

.51640

.47012

2.9011

3.9011

6.2440

7.6726

6.0275

7.0989

8.0989

9.7560

10.3274

7.9725

3.00

5.00

7.00

3.00

8.00

10.00

N Mean Std. Deviation Std. Error Lower Bound Upper Bound Minimum Maximum

95% Confidence Interval for Mean

VAR00001

ANOVA

df Sum of Squares Mean Square F Sig.

FIGURE 12.12

SPSS output of the ANOVA for the studying strategy experiment in Example 12.1.

DEMONSTRATION 12.1 3 8 5

the number of errors committed by each participant was recorded. The data are as

follows:

Keyboard A Keyboard B Keyboard C

0 6 6 N 5 15

4 8 5 G 5 60

0 5 9 oX2 5 356

1 4 4

0 2 6

T 5 5 T 5 25 T 5 30

SS 5 12 SS 5 20 SS 5 14

Are these data sufficient to conclude that there are significant differences in typing

performance among the three keyboard designs?

State the hypotheses, and specify the alpha level. The null hypothesis states that there are no differences among the keyboards in terms of number of errors committed. In symbols, we would state

H 0 : �

1 5 �

2 5 �

3 (Type of keyboard used has no effect.)

As noted previously in this chapter, there are a number of possible statements for the

alternative hypothesis. Here we state the general alternative hypothesis:

H 1 : At least one of the treatment means is different.

We set alpha at a 5 .05.

Locate the critical region. To locate the critical region, we must obtain the values for df

between and df

within .

df between

5 k – 1 5 3 – 1 5 2 df

within 5 N – k 5 15 – 3 5 12

The F-ratio for this problem has df 5 2, 12, and the critical F value for a 5 .05 is F 5 3.88.

Perform the analysis. The analysis involves the following steps:

1. Perform the analysis of SS.

2. Perform the analysis of df.

3. Calculate mean squares.

4. Calculate the F-ratio.

Perform the analysis of SS. We compute SS total

followed by its two components.

SS X G

N total

5 2 5 2 5 2 2

2 2

356 60

15 356

000

356 240 116 5 2 5

S T E P 1

S T E P 2

S T E P 3

3 8 6 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

SS SS within inside

each treatment5 

5 1 1

12 20 14

By subtraction, SS between

5 SS total

2 SS within

5 116 2 46 5 70

Analyze degrees of freedom. We compute df total

. Its components, df between

and df within

were previously calculated (see step 2).

df total

5 N 2 1 5 15 2 1 5 14 df

between 5 2

df within

5 12

Calculate the MS values. We determine the values for MS between

and MS within

MS SS

between between

between

within

5 5 5 70

2 35

55 5 5 SS

within

12 3 83.

Compute the F-ratio. Finally, we can compute F.

F MS

MS 5 5 5between

within

3 83 9 14

. .

Make a decision about H 0 , and state a conclusion. The obtained F of 9.14 exceeds the

critical value of 3.88. Therefore, we can reject the null hypothesis. The type of keyboard

used has a significant effect on the number of errors committed, F(2, 12) 5 9.14, p , .05. The following table summarizes the results of the analysis:

Source SS df MS

Between treatments 70 2 35 F 5 9.14

Within treatments 46 12 3.83

Total 116 14

DEMONSTRATION 12.2

COMPUTING EFFECT SIZE FOR ANOVA

We compute eta squared (h2), the percentage of variance explained, for the data that were analyzed in Demonstration 12.1. The data produced a between-treatments SS of

70 and a total SS of 116. Thus,

h 2 70

116 0 60 605 5 5

between

total

(or ). %

S T E P 4

PROBLEMS 3 8 7

PROBLEMS

1. Explain why the F-ratio is expected to be near 1.00 when the null hypothesis is true.

2. Several factors influence the size of the F-ratio. For each of the following, indicate whether it influences the numerator or the denominator of the F-ratio, and indicate whether the size of the F-ratio would increase or decrease. In each case, assume that all other factors are held constant.

a. An increase in the differences between the sample means.

b. An increase in the size of the sample variances.

3. Why should you use ANOVA instead of several t tests to evaluate mean differences when an experi- ment consists of three or more treatment conditions?

4. Posttests are done after an ANOVA. a. What is the purpose of posttests? b. Explain why you do not need posttests if the

analysis is comparing only two treatments. c. Explain why you do not need posttests if the

decision from the ANOVA is to fail to reject the null hypothesis.

5. An independent-measures study comparing four treatment conditions with a sample of n 5 8 in each condition produces sample means of M

1 5 2, M

2 5 3,

M 3 5 1, and M

1 5 6.

a. Compute SS for the set of 4 treatment means. (Treat the means as a set of n 5 4 scores and compute SS.)

b. Using the result from part a, compute n(SS means

). Note that this value is equal to SS

between (see Equation 12.6).

c. Now, find the 4 treatment totals and compute SS

between with the computational formula using the

T values (see Equation 12.7). You should obtain the same result as in part b.

6. The following data summarize the results from an independent-measures study comparing three treatment conditions.

Treatment

I II III

3 5 6 N 5 12

5 5 10 G 5 60

3 1 10 oX2 5 392

1 5 6

M 5 3 M 5 4 M 5 8

T 5 12 T 5 16 T 5 32

SS 5 8 SS 5 12 SS 5 16

a. Use an ANOVA with a 5 .05 to determine whether there are any significant differences among the three treatment means.

b. Calculate h2 to measure the effect size for this study. c. Write a sentence demonstrating how a research

report would present the results of the hypothesis test and the measure of effect size.

7. For the preceding problem you should find that there are significant differences among the three treatments. The primary reason for the significance is that the mean for treatment III is substantially larger than the means for the other two treatments. To create the following data, we started with the values from problem 6 and subtracted 3 points to each score in treatment III. Notice that subtracting a constant causes the mean to change but has no influence on the variability of the sample. In the resulting data, the mean differences are much smaller than those in problem 6.

Treatment

I II III

3 5 3 N 5 12

5 5 7 G 5 48

3 1 7 oX2 5 236

1 5 3

M 5 3 M 5 4 M 5 5

T 5 12 T 5 16 T 5 20

SS 5 8 SS 5 12 SS 5 16

a. Before you begin any calculations, predict how the change in the data should influence the out- come of the analysis. That is, how will the F-ratio and the value of h2 for these data compare with the values obtained in problem 6?

b. Use an ANOVA with a 5 .05 to determine whether there are any significant differences among the three treatment means. (Does your answer agree with your prediction in part a?)

c. Calculate h2 to measure the effect size for this study. (Does your answer agree with your prediction in part a?)

8. The following data summarize the results from an independent-measures study comparing three treatment conditions.

Treatment

I II III

4 3 8 N 5 12

3 1 4 G 5 48

5 3 6 oX2 5 238

4 1 6

M 5 4 M 5 2 M 5 6

T 5 16 T 5 8 T 5 24

SS 5 2 SS 5 4 SS 5 8

3 8 8 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

a. Calculate the sample variance for each of the three samples.

b. Use an ANOVA with a 5 .05 to determine whether there are any significant differences among the three treatment means.

9. For the preceding problem you should find that there are significant differences among the three treat- ments. One reason for the significance is that the sample variances are relatively small. To create the following data, we kept the same sample means that appeared in problem 8 but increased the SS values within each sample.

Treatment

I II III

4 4 9 N 5 12

2 0 3 G 5 48

6 3 6 oX2 5 260

4 1 6

M 5 4 M 5 2 M 5 6

T 5 16 T 5 8 T 5 24

SS 5 8 SS 5 10 SS 5 18

a. Calculate the sample variance for each of the three samples. Describe how these sample vari- ances compare with those from problem 8.

b. Predict how the increase in sample variance should influence the outcome of the analysis. That is, how will the F-ratio for these data compare with the value obtained in problem 8?

c. Use an ANOVA with a 5 .05 to determine whether there are any significant differences among the three treatment means. (Does your answer agree with your prediction in part b?)

10. The following data summarize the results from an independent-measures study comparing three treat- ment conditions.

M 5 2 M 5 3 M 5 4

n 5 10 n 5 10 n 5 10

T 5 20 T 5 30 T 5 40

s2 5 2.67 s2 5 2.00 s2 5 1.33

a. Use an ANOVA with a 5 .05 to determine whether there are any significant differences among the three treatment means. Note: Because the samples are all the same size, MS

within is the

average of the three sample variances. b. Calculate h2 to measure the effect size for this study.

11. To create the following data we started with the same sample means and variances that appeared in problem 10 but increased the sample size to n 5 25.

M 5 2 M 5 3 M 5 4

n 5 25 n 5 25 n 5 25

T 5 50 T 5 75 T 5 100

s2 5 2.67 s2 5 2.00 s2 5 1.33

a. Predict how the increase in sample size should affect the F-ratio for these data compared to the F-ratio in problem 10. Use an ANOVA to check your prediction. Note: Because the samples are all the same size, MS

within is the average of the three

sample variances. b. Predict how the increase in sample size should

affect the value of h2 for these data compared to the h2 in problem 10. Calculate h2 to check your prediction.

12. The following values are from an independent-measures study comparing three treatment conditions.

Treatment

I II III

n 5 10 n 5 10 n 5 10

SS 5 63 SS 5 66 SS 5 87

a. Compute the variance for each sample. b. Compute MS

within , which would be the denomina-

tor of the F-ratio for an ANOVA. Because the samples are all the same size, you should find that MS

within is equal to the average of the three sample

variances.

13. There is some evidence that high school students justify cheating in class on the basis of poor teacher skills or low levels of teacher caring (Murdock, Miller, & Kohlhardt, 2004). Students appear to rationalize their illicit behavior based on perceptions of how their teachers view cheating. Poor teachers are thought not to know or care whether students cheat, so cheating in their classes is okay. Good teachers, on the other hand, do care and are alert to cheating, so students tend not to cheat in their classes. Following are hypothetical data similar to the actual research results. The scores represent judgments of the acceptability of cheating for the students in each sample.

Poor Teacher

Average Teacher

Good Teacher

n 5 6 n 5 8 n 5 10 N 5 24

M 5 6 M 5 2 M 5 2 G 5 72

SS 5 30 SS 5 33 SS 5 42 oX2 5 393

a. Use an ANOVA with a 5 .05 to determine whether there are significant differences in student judg- ments depending on how they see their teachers.

PROBLEMS 3 8 9

b. Calculate h2 to measure the effect size for this study.

c. Write a sentence demonstrating how a research report would present the results of the hypothesis test and the measure of effect size.

14. A researcher reports an F-ratio with df 5 2, 27 from an independent-measures research study.

a. How many treatment conditions were compared in the study?

b. What was the total number of participants in the study?

15. A research report from an independent-measures study states that there are significant differences between treatments, F(3, 48) 5 2.95, p , .05.

a. How many treatment conditions were compared in the study?

b. What was the total number of participants in the study?

16. The following summary table presents the results from an ANOVA comparing three treatment condi- tions with n 5 8 participants in each condition. Complete all missing values. (Hint: Start with the df column.)

Source SS df MS

Between treatments _____ _____ 15 F 5 _____

Within treatments _____ _____ _____

Total 93 ____

17. A pharmaceutical company has developed a drug that is expected to reduce hunger. To test the drug, two samples of rats are selected with n 5 20 in each sample. The rats in the first sample receive the drug every day and those in the second sample are given a placebo. The dependent variable is the amount of food eaten by each rat over a 1-month period. An ANOVA is used to evaluate the differ- ence between the two sample means and the results are reported in the following summary table. Fill in all missing values in the table. (Hint: Start with the df column.)

Source SS df MS

Between treatments ____ ____ 20 F 5 4.00

Within treatments ____ ____ ____

Total ____ ____

18. A developmental psychologist is examining the development of language skills from age 2 to age 4. Three different groups of children are obtained, one for each age, with n 5 16 children in each group.

Each child is given a language-skills assessment test. The resulting data were analyzed with an ANOVA to test for mean differences between age groups. The results of the ANOVA are presented in the following table. Fill in all missing values.

Source SS df MS

Between treatments 20 ____ ____ F 5 ____

Within treatments ____ ____ ____

Total 200 ____

19. The following data were obtained from an independent- measures research study comparing three treatment conditions. Use an ANOVA with a 5 .05 to determine whether there are any significant mean differences among the treatments.

Treatment

I II III

n 5 8 n 5 6 n 5 4 N 5 18

T 5 16 T 5 24 T 5 32 G 5 72

SS 5 40 SS 5 24 SS 5 16 oX2 5 464

20. The following values summarize the results from an independent-measures study comparing two treat- ment conditions.

a. Use an independent-measures t test with a 5 .05 to determine whether there is a significant mean difference between the two treatments. You should find that F 5 t2.

b. Use an ANOVA with a 5 .05 to determine whether there is a significant mean difference between the two treatments.

Treatment

I II

n 5 8 n 5 4

M 5 4 M 5 10 N 5 12

T 5 32 T 5 40 G 5 72

SS 5 45 SS 5 15 oX2 5 588

21. The following data are from an independent-measures study comparing two treatment conditions.

a. Use an independent-measures t test with a 5 .05 to determine whether there is a significant mean difference between the two treatments.

b. Use an ANOVA with a 5 .05 to determine whether there is a significant mean difference between the two treatments. You should find that F 5 t2.

3 9 0 CHAPTER 12 INTRODUCTION TO ANALYSIS OF VARIANCE

Treatment

I II

8 2 N 5 10

7 3 G 5 50

6 3 oX2 5 306

5 5

9 2

M 5 7 M 5 3

T 5 35 T 5 15

SS 5 10 SS 5 6

22. One possible explanation for why some birds mi- grate and others maintain year round residency in a single location is intelligence. Specifically, birds with small brains, relative to their body size, are simply not smart enough to find food during the winter and must migrate to warmer climates where food is easily available (Sol, Lefebvre, & Rodriguez-Teijeiro, 2005). Birds with bigger brains, on the other hand, are more creative and can find food even when the weather turns harsh. Following are hypothetical data similar to the actual research results. The numbers represent relative brain size for the individual birds in each sample.

Non- Migrating

Short- Distance Migrants

Long Distance Migrants

18 6 4 N 5 18

13 11 9 G 5 180

19 7 5 oX2 5 2150

12 9 6

16 8 5

12 13 7

M 5 15 M 5 9 M 5 6

T 5 90 T 5 54 T 5 36

SS 5 48 SS 5 34 SS 5 16

a. Use an ANOVA with a 5 .05 to determine whether there are any significant mean differences among the three groups of birds.

b. Compute h2, the percentage of variance explained by the group differences, for these data.

c. Write a sentence demonstrating how a research report would present the results of the hypothesis test and the measure of effect size.

d. Use the Tukey HSD posttest to determine which groups are significantly different.

23. There is some research indicating that college students who use Facebook while studying tend to have lower grades than non-users (Kirschner & Karpinski, 2010). A representative study surveys students to determine the amount of Facebook use during the time they are studying or doing home- work. Based on the amount of time spent on Facebook, students are classified into three groups and their grade point averages are recorded. The following data show the typical pattern of results.

Facebook Use While Studying

Non-User Rarely Use Regularly Use

3.70 3.51 3.02

3.45 3.42 2.84

2.98 3.81 3.42

3.94 3.15 3.10

3.82 3.64 2.74

3.68 3.20 3.22

3.90 2.95 2.58

4.00 3.55 3.07

3.75 3.92 3.31

3.88 3.45 2.80

a. Use an ANOVA with a 5 .05 to determine whether there are significant mean differences among the three groups.

b. Compute h2 to measure the size of the effect. c. Write a sentence demonstrating how the result

from the hypothesis test and the measure of effect size would appear in a research report.

24 New research suggests that watching television, especially medical shows such as Grey’s Anatomy and House, can result in increased concern about personal health (Ye, 2010). Surveys administered to college students measure television viewing habits and health concerns such as fear of developing the diseases and disorders seen on television. For the following data, students are classified into three categories based on their television viewing patterns

PROBLEMS 3 9 1

and health concerns are measured on a 10-point scale with 0 indicating “none.”

Television Viewing

Little or None Moderate Substantial

4 5 5

2 7 7

5 3 6

1 4 6

3 8 8

7 6 9

4 2 6

4 7 4

8 3 6

2 5 8

a. Use an ANOVA with a 5 .05 to determine whether there are significant mean differences among the three groups.

b. Compute h2 to measure the size of the effect. c. Use Tukey’s HSD test with a 5 .05 to determine

which groups are significantly different.

Aplia for Essentials of Statistics for the Behavioral Sciences

After reading, go to “Resources” at the end of this chapter for

an introduction on how to use Aplia’s homework and learning

resources.

Repeated- Measures and Two-Factor Analysis of Variance 13.1 Overview

13.2 Repeated-Measures ANOVA

13.3 Two-Factor ANOVA (Independent Measures)

Summary

Focus on Problem Solving

Demonstrations 13.1 and 13.2

Problems

C H A P T E R

13 Tools You Will Need The following items are considered essential background material for this chapter. If you doubt your knowledge of any of these items, you should review the appropriate chapter or section before proceeding.

• Independent-measures analysis of variance (Chapter 12)

• Repeated-measures designs (Chapter 11)

• Individual differences

3 9 4 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

OVERVIEW

In the preceding chapter, we introduced ANOVA as a hypothesis-testing procedure

for evaluating differences among two or more sample means. The specific advantage

of ANOVA, especially in contrast to t tests, is that ANOVA can be used to evaluate

the significance of mean differences in situations in which there are more than two

sample means being compared. However, the presentation of ANOVA in Chapter 12

was limited to single-factor, independent-measures research designs. Recall that single

factor indicates that the research study involves only one independent variable (or only

one quasi-independent variable), and the term independent-measures indicates that the

study uses a separate sample for each of the different treatment conditions being com-

pared. In fact, ANOVA is an extremely flexible technique, with applications far beyond

this single research design. In this chapter, we begin to explore some more sophisticated

research situations in which ANOVA is used. Specifically, we introduce the following

ANOVA topics:

1. Repeated-Measures ANOVA. It is possible to compare several different treat-

ment conditions using a repeated-measures research design in which the same

group of individuals participates in every treatment. We demonstrate how the

ANOVA procedure can be adapted to test for mean differences from a repeated-

measures study.

2. Two-Factor ANOVA. Often, research questions are concerned with how

behavior is influenced by several different variables acting simultaneously.

For example, a researcher may want to examine how weight loss is related to

different combinations of diet and exercise. In this situation, two variables are

manipulated (diet and exercise) while a third variable is observed (weight loss).

In statistical terminology, the research study has two independent variables,

or two factors. In the final section of this chapter, we show how the general

ANOVA procedure from Chapter 12 can be used to test for mean differences

in a two-factor research study.

REPEATED-MEASURES ANOVA

Chapter 12 introduced the general logic underlying ANOVA and presented the equa-

tions used for analyzing data from a single-factor, independent-measures research

study. As we noted, the defining characteristic of an independent-measures research

design is that the study uses a separate sample for each of the different treatment

conditions. One concern with an independent-measures design is that the partici-

pants in one treatment condition may have characteristics that are noticeably differ-

ent from participants in another condition. For example, the individuals in treatment

1 may be smarter than the individuals in treatment 2. In this case, it is impossible

to explain any differences that are found between the two treatments. For example:

(1) It could be that treatment 1 causes people to have higher scores, or (2) It could

be that smarter people have higher scores. To avoid this problem, researchers often

chose to use a repeated-measures design. You should recall (Chapter 11) that a

repeated-measures design uses the same group of participants in all of the treatment

conditions. In a repeated-measures study, it is impossible for the participants in one

group to be different from those in another, because exactly the same group is in

every treatment condition.

13.1

13.2

An independent variable

is a manipulated variable

in an experiment. A quasi-

independent variable is not

manipulated but defines the

groups of scores in a nonex-

perimental design.

SECTION 13.2 / REPEATED-MEASURES ANOVA 3 9 5

In this section we extend the ANOVA procedure to single-factor, repeated-measures de-

signs. The analysis is used to evaluate mean differences in two general research situations:

1. An experimental study, in which the researcher manipulates an independent

variable to create two or more treatment conditions, with the same group of

individuals tested in all of the conditions.

2. A nonexperimental study, in which the same group of individuals is simply

observed at two or more different times.

Examples of these two research situations are presented in Table 13.1. Table 13.1(a)

shows data from a study in which the researcher changes the type of distraction to

create three treatment conditions. One group of participants is then tested in all three

conditions. In this study, the factor being examined is the type of distraction.

Table 13.1(b) shows a study in which a researcher observes depression scores for the

same group of individuals at three different times. In this study, the time of measure-

ment is the factor being examined. Another common example of this type of design

is found in developmental psychology when the participants’ age is the factor being

studied. For example, a researcher could study the development of vocabulary skill by

measuring vocabulary for a sample of 3-year-old children, then measuring the same

children again at ages 4 and 5.

The hypotheses for the repeated-measures ANOVA are exactly the same as those for

the independent-measures ANOVA presented in Chapter 12. Specifically, the null

hypothesis states that, for the general population, there are no mean differences among

the treatment conditions being compared. In symbols,

H 0 : �

1 5 �

2 5 �

3 5 …

H Y P OT H E S E S F O R T H E R E P E AT E D -

M E AS U R E S A N OVA

(a) Data from an experimental study evaluating the effects of different types of

distraction on the performance of a visual detection task.

Visual Detection Scores

Participant No Distraction Visual Distraction Auditory

Distraction

A 47 22 41

B 57 31 52

C 38 18 40

D 45 32 43

(b) Data from a nonexperimental design evaluating the effectiveness of a clinical

therapy for treating depression.

Depression Scores

Participant Before Therapy After Therapy 6-Month

Follow-Up

A 71 53 55

B 62 45 44

C 82 56 61

D 77 50 46

E 81 54 55

TABLE 13.1

Two sets of data representing

typical examples of single-

factor, repeated-measures

research designs.

3 9 6 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

The null hypothesis states that, on average, all of the treatments have exactly the

same effect. According to the null hypothesis, any differences that may exist among the

sample means are not caused by systematic treatment effects but rather are the result of

random and unsystematic factors.

The alternative hypothesis states that there are mean differences among the treat-

ment conditions. Rather than specifying exactly which treatments are different, we use

a generic version of H 1 , which simply states that differences exist:

H 1 : At least one treatment mean (�) is different from another.

Notice that the alternative says that, on average, the treatments do have different ef-

fects. Thus, the treatment conditions may be responsible for causing mean differences

among the samples. As always, the goal of the ANOVA is to use the sample data to

determine which of the two hypotheses is more likely to be correct.

The F-ratio for the repeated-measures ANOVA has the same structure that was used

for the independent-measures ANOVA in Chapter 12. In each case, the F-ratio com-

pares the actual mean differences between treatments with the amount of difference

that would be expected if there were no treatment effect. The numerator of the F-ratio

measures the actual mean differences between treatments. The denominator measures

how big the differences should be if there is no treatment effect. As always, the F-ratio

uses variance to measure the size of the differences. Thus, the F-ratio for both ANOVAs

has the general structure

F 5 variance (differences) between treatments

vvariance (differences) expected if there iss no treatment effect

A large value for the F-ratio indicates that the differences between treatments are

greater than would be expected without any treatment effect. If the F-ratio is larger than

the critical value in the F distribution table, then we can conclude that the differences

between treatments are significantly larger than would be caused by chance.

Individual differences in the F-ratio Although the structure of the F-ratio is the

same for independent-measures and repeated-measures designs, there is a fundamental

difference between the two designs that produces a corresponding difference in the two

F-ratios. Specifically, individual differences are a part of one ratio but are eliminated

from the other.

You should recall that the term individual differences refers to participant char-

acteristics such as age, personality, and gender that vary from one person to another

and may influence the measurements that you obtain for each person. Suppose, for

example, that you are measuring reaction time. The first participant in your study is

a 19-year-old female with an IQ of 136 who is on the college varsity volleyball team.

The next participant is a 42-year-old male with an IQ of 111 who returned to college

after losing his job and comes to the research study with a head cold. Would you

expect to obtain the same reaction time score for these two individuals even if they

receive the same treatment?

Individual differences are a part of the variance in the numerator and in the

denominator of the F-ratio for the independent-measures ANOVA. However, individual

difference are eliminated or removed from the variances in the F-ratio for the repeated

measures ANOVA. The idea of removing individual differences was first presented in

Chapter 11 when we introduced the repeated-measures design (p. 329), but we review

it briefly now.

T H E F - R AT I O F O R R E P E AT E D - M E AS U R E S

A N OVA

SECTION 13.2 / REPEATED-MEASURES ANOVA 3 9 7

In a repeated-measures study, exactly the same individuals participate in all of the

treatment conditions. Therefore, if there are any mean differences between treatments,

they cannot be explained by individual differences. Thus, individual differences are

automatically eliminated from the numerator of the repeated-measures F-ratio.

A repeated-measures design also allows you to remove individual differences from

the variance in the denominator of the F-ratio. Because the same individuals are mea-

sured in every treatment condition, it is possible to measure the size of the individual

differences. In Table 13.1(a), for example, participant A has scores that are consistently

around 10 points lower than the scores for participant B. Because the individual dif-

ferences are systematic and predictable, they can be measured and separated from the

random, unsystematic differences in the denominator of the F-ratio.

Thus, individual differences are automatically eliminated from the numerator of the

repeated-measures F-ratio. In addition, they can be measured and removed from the

denominator. As a result, the structure of the final F-ratio is as follows:

F 5

variance between treatments

(without indiviidual differences)

variance with no treatmennt effect

(individual differences removed)

The process of removing the variance caused individual differences is an important

part of the procedure for a repeated-measures ANOVA.

In summary, the F-ratio for a repeated-measures ANOVA has the same basic struc-

ture as the F-ratio for independent measures (Chapter 12) except that it includes no

variance caused by individual differences. The individual differences are automatically

eliminated from the variance between treatments (numerator) because the repeated-

measures design uses the same individuals in all treatments. In the denominator, the

individual differences are subtracted during the analysis. As a result, the repeated-

measures F-ratio has the following structure:

F 5 between-treatments variance

error variance

55 1treatment effects random, unsystematic diffferences

random, unsystematic differences

(13.1)

When there is no treatment effect, the F-ratio is balanced because the numerator and

denominator are both measuring exactly the same variance. In this case, the F-ratio should

have a value near 1.00. When research results produce an F-ratio near 1.00, we conclude

that there is no evidence of a treatment effect and we fail to reject the null hypothesis. On

the other hand, when a treatment effect does exist, it contributes only to the numerator and

should produce a large value for the F-ratio. Thus, a large value for F indicates that there

is a real treatment effect and, therefore, we should reject the null hypothesis.

1. The F-ratio for the repeated-measures ANOVA is structured so that variance

caused by individual differences is eliminated.

a. Explain why individual differences are not part of the between-treatments

variance in the numerator of the F-ratio.

b. Explain why individual differences are not part of the error variance in the

denominator of the F-ratio.

L E A R N I N G C H E C K

3 9 8 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

The overall structure of the repeated-measures ANOVA is shown in Figure 13.1. Note

that the ANOVA can be viewed as a two-stage process. In the first stage, the total

variance is partitioned into two components: between-treatments variance and within-

treatments variance. This stage is identical to the analysis that we conducted for an

independent-measures design in Chapter 12.

The second stage of the analysis is intended to remove the individual differences from

the denominator of the F-ratio. In the second stage, we begin with the variance within

treatments and then measure and subtract out the between-subject variance, which

measures the size of the individual differences. The remaining variance, often called

the residual variance, or error variance, provides a measure of how much variance is

reasonable to expect after the treatment effects and individual differences have been

removed. The second stage of the analysis is what differentiates the repeated-measures

ANOVA from the independent-measures ANOVA. Specifically, the repeated-measures

design requires that the individual differences be removed.

T H E ST R UC T U R E O F T H E R E P E AT E D - M E AS U R E S A N OVA

1. a. Because the individuals in one treatment are exactly the same as the individuals in every

other treatment, there are no individual differences between treatments.

b. Variance caused by individual differences is measured and subtracted from the within-

treatments variance to produce a measure of error variance that does not include any

treatment effects or individual differences.

ANSWERS

Stage 2

Stage 1

Between-treatments variance

Numerator of F - ratio

Denominator of F - ratio

1. Treatment effect 2. Error or chance (excluding individual differences)

Between-subjects variance

1. Individual differences

Error variance

1. Error (excluding individual differences)

Total variance

Within-treatments variance

1. Individual differences 2. Other error

FIGURE 13.1

The partitioning of

variance for a repeated-

measures experiment.

SECTION 13.2 / REPEATED-MEASURES ANOVA 3 9 9

In a repeated-measures ANOVA, the denominator of the F-ratio is called the

residual variance, or the error variance, and measures how much variance is

expected if there are no systematic treatment effects and no individual differences

contributing to the variability of the scores.

We use the data in Table 13.2 to introduce the notation for the repeated-measures

ANOVA. The data represent the results of a study similar to one reported by Weinstein,

McDermott, and Roediger (2010), comparing different strategies for studying text pas-

sages in preparation for a quiz or exam. Four strategies were evaluated—reading once,

rereading, answering prepared comprehension questions, creating and answering your

own comprehension questions. After studying a passage, each participant was given a

10-point quiz on the material. After a short break, the participant moved on to a new

passage and a new studying strategy. This process continued until all participants had

completed all four strategies. You may notice that this research study and the numerical

values in the table are identical to those used to demonstrate the independent-measures

ANOVA in the previous chapter (Example 12.1, page 364). In this case, however, the

data represent a repeated-measures study in which the same group of n 5 6 individuals is tested in all four treatment conditions.

You should recognize that most of the notation in Table 13.2 is identical to the no-

tation used in an independent-measures analysis (Chapter 12) For example, there are

n 5 6 participants who are tested in k 5 4 treatment conditions, producing a total of N 5 24 scores that add up to a grand total of G 5 168. Note, however, that N 5 24 now refers to the total number of scores in the study, not the number of participants.

The repeated-measures ANOVA introduces only one new notational symbol.

The letter P is used to represent the total of all of the scores for each individual in

the study. You can think of the P values as “Person totals” or “Participant totals.”

In Table 13.2, for example, participant A had scores of 3, 5, 8, and 8 for a total of

P 5 24. The P values are used to define and measure the magnitude of the individual differences in the second stage of the analysis.

We use the data in Table 13.2 to demonstrate the repeated-measures ANOVA. Again,

the goal of the test is to determine whether there are any significant differences among

the four strategies being compared. Specifically, are any of the mean differences in the

D E F I N I T I O N

N OTAT I O N F O R T H E R E P E AT E D -

M E AS U R E S A N OVA

E X A M P L E 1 3 . 1

Strategies for Studying Text Passages

Student Read Once

Read and

Reread

Answer Prepared Questions

Create and Answer

Questions Person Totals

A 3 5 8 8 P 5 24 n 5 6

B 3 3 5 9 P 5 20 k 5 4

C 4 5 8 7 P 5 24 N 5 24

D 6 7 9 10 P 5 32 G 5 168

E 6 8 8 10 P 5 32 oX2 5 1298

F 8 8 10 10 P 5 36

T 5 30 T 5 36 T 5 48 T 5 54

M 5 5 M 5 6 M 5 8 M 5 9

SS 5 20 SS 5 20 SS 5 14 SS 5 8

TABLE 13.2

Quiz scores for students

using four different strategies

for studying text passages.

Note: For comparison, the

scores are identical to the

values in Example 12.1.

4 0 0 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

data greater than would be expected if there were no systematic differences among the

four strategies?

The first stage of the repeated-measures analysis is identical to the independent-

measures ANOVA that was presented in Chapter 12. Specially, the SS and df for

the total set of scores are analyzed into within-treatments and between-treatments

components.

Because the numerical values in Table 13.2 are the same as the values used in

Example 12.1 (p. 364), the computations for the first stage of the repeated-measures

analysis are identical to those in Example 12.1. Rather than repeating the same arith-

metic, the results of the first stage of the repeated-measures analysis can be summarized

as follows:

Total:

SS G

N total

2 5 2 5 2 5 2 5o X

2 2

1298 168

24 1298 1176 122

ddf N total

5 2 51 23

Within treatments:

SS within treatments

5 oSS inside each treatment

5 20 1 20 1 14 1 8 5 62

df within treatments

5 odf inside each treatment

5 5 1 5 1 5 1 5 5 20

Between treatments:

For this example we find SS between treatments

by subtraction.

SS between treatments

5 SS total

– SS within treatments

5 122 – 62 5 60

df between treatments

5 k – 1 5 3

For more details on the formulas and calculations, see Example 12.1, pages 364–366.

This completes the first stage of the repeated-measures ANOVA. Note that the two

components, between and within, add up to the total for the SS values and for the df

values. Also note that the between-treatments SS and df values provide a measure of

the mean differences between treatments and are used to compute the variance in the

numerator of the final F-ratio.

The second stage of the analysis involves removing the individual differences from the

denominator of the F-ratio. Because the same individuals are used in every treatment, it

is possible to measure the size of the individual differences. For the data in Table 13.2,

for example, participants A, B, and C tend to have the lowest scores and participants D,

E, and F tend to have the highest scores. These individual differences are reflected in

the P values, or person totals, in the right-hand column. We use these P values to create

a computational formula for SS between subjects

in much the same way that we used the treat-

ment totals, the T values, in the computational formula for SS between treatments

. Specifically,

the formula for the between-subjects SS is

SS G

N between subjects

5 2o P

(13.2)

STAG E 1 O F T H E R E P E AT E D - M E AS U R E S A N OVA

STAG E 2 O F T H E R E P E AT E D - M E AS U R E S A N OVA

SECTION 13.2 / REPEATED-MEASURES ANOVA 4 0 1

Notice that the formula for the between-subjects SS has exactly the same struc-

ture as the computational formula for the between-treatments SS (see the calcula-

tion on p. 358). In this case, we use the person totals (P values) instead of the

treatment totals (T values). Each P value is squared and divided by the number of

scores that were added to obtain the total. In this case, each person has k scores,

one for each treatment. Box 13.1 presents another demonstration of the similarity

of the formulas for SS between subjects and SS between treatments. For the data

in Table 13.2,

SS between subjects

5 1 1 1 1 24

2 2 2 2 2

11 2

5 1 1

168

2 2

144 100 1144 256 256 324 11761 1 1 2

5 448

The value of SS between subjects

provides a measure of the size of the individual

differences—that is, the differences between subjects. In the second stage of the

analysis, we simply subtract the individual differences to obtain the measure of error

that forms the denominator of the F-ratio. Thus, the final step in the analysis of SS is

SS error

5 SS within treatments

– SS between subjects

(13.3)

We have already computed SS within treatments

5 62 and SS between subjects

5 48, therefore

SS error

5 62 – 48 5 14

BOX

13.1 SS

between subjects AND SS

between treatments

The data for a repeated-measures study are normally

presented in a matrix, with the treatment conditions

determining the columns and the participants defining

the rows. The data in Table 13.2 demonstrate

this normal presentation. The calculation of

SS between treatments

provides a measure of the differences

between treatment conditions—that is, a measure of

the mean differences between the columns in the data

matrix. For the data in Table 13.2, the column totals

are 30, 36, 48, and 54. These values are variable,

and SS between treatments

measures the amount of variability.

The following table reproduces the data from

Table 13.2, but now we have turned the data matrix

on its side so that the participants define the columns

and the treatment conditions define the rows.

In this new format, the differences between the

columns represent the between-subjects variance. The

column totals are now P values (instead of T values)

and the number of scores in each column is now

identified by k (instead of n). With these changes in no-

tation, the formula for SS between subjects

has exactly the same

structure as the formula for SS between treatments

. If you exam-

ine the two equations, the similarity should be clear.

Participant

A B C D E F

Read Once

3 3 4 6 6 8 T 5 30

Read and Reread

5 3 5 7 8 8 T 5 36

Answer Prepared Questions

8 5 8 9 8 10 T 5 48

Create and Answer Questions

8 9 7 10 10 10 T 5 54

P 5 24 P 5 20 P 5 24 P 5 32 P 5 32 P 5 36

4 0 2 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

The analysis of degrees of freedom follows exactly the same pattern that was used to

analyze SS. Remember that we are using the P values to measure the magnitude of the

individual differences. The number of P values corresponds to the number of subjects,

n, so the corresponding df is

df between subjects

5 n – 1 (13.4)

For the data in Table 13.2, there are n 5 6 subjects and

df between subjects

5 6 – 1 5 5

Next, we subtract the individual differences from the within-subjects component to

obtain a measure of error. In terms of degrees of freedom,

df error

5 df within treatments

– df between subjects

(13.5)

For the data in Table 13.2,

df error

5 20 – 5 5 15

An algebraically equivalent formula for df error

uses only the number of treatment

conditions (k) and the number of participants (n):

df error

5 (k – 1)(n – 1) (13.6)

The usefulness of equation 13.6 is discussed in Box 13.2.

Remember: The purpose for the second stage of the analysis is to measure the

individual differences and then remove the individual differences from the denomi-

nator of the F-ratio. This goal is accomplished by computing SS and df between

subjects (the individual differences) and then subtracting these values from the

within-treatments values. The result is a measure of variability resulting from error

with the individual differences removed. This error variance (SS and df) is used in the

denominator of the F-ratio.

BOX

13.2 USING THE ALTERNATIVE FORMULA FOR df

error

The statistics presented in a research report not only

describe the significance of the results but typically

provide enough information to reconstruct the research

design. The alternative formula for df error

is particularly

useful for this purpose. Suppose, for example, that a

research report for a repeated-measures study includes

an F-ratio with df 5 2, 10. How many treatment conditions were compared in the study and how many

individuals participated?

To answer these question, begin with the first df

value, which is df between treatments

5 2 5 k – 1. From this value, it is clear that k 5 3 treatments. Next, use the

second df value, which is df error

5 10. Using this value and the fact that k – 1 5 2, use equation 13.6 to find the number of participants.

df error

5 10 5 (k – 1)(n – 1) 5 2(n – 1)

If 2(n – 1) 5 10, then n – 1 must equal 5. Therefore, n 5 6.

Therefore, we conclude that a repeated-measures

study producing an F-ratio with df 5 2, 10 must have compared 3 treatment conditions using a sample of

6 participants.

SECTION 13.2 / REPEATED-MEASURES ANOVA 4 0 3

The final calculation in the analysis is the F-ratio, which is a ratio of two variances.

Each variance is called a mean square, or MS, and is obtained by dividing the appropri-

ate SS by its corresponding df value. The MS in the numerator of the F-ratio measures

the size of the differences between treatments and is calculated as

MS SS

df between treatments

between treatments5 bbetween treatments

(13.7)

For the data in Table 13.2,

MS SS

df between treatments

between treatments5 bbetween treatments

5 5 60

3 20

The denominator of the F-ratio measures how much difference is reasonable to

expect if there are no systematic treatment effects and the individual differences have

been removed. This is the error variance, or the residual variance, obtained in stage 2

of the analysis.

MS SS

df error

error

(13.8)

For the data in Table 13.2,

MS SS

df error

error

15 0.9335 5 5

Finally, the F-ratio is computed as

F MS

MS 5

between treatments

error

(13.9)

For the data in Table 13.2,

F MS

MS 5 5 5

between treatments

error

0 933 21 43

. .

Once again, notice that the repeated-measures ANOVA uses MS error

in the denomina-

tor of the F-ratio. This MS value is obtained in the second stage of the analysis, after the

individual differences have been removed. As a result, individual differences are com-

pletely eliminated from the repeated-measures F-ratio, so that the general structure is

F 5 1treatment effects unsystematic differencess (without individual diffs)

unsystematic diifferences (without individual diffs)

For the data we have been examining, the F-ratio is F 5 21.43, indicating that the

differences between treatments (numerator) are 22 times bigger than you would expect

without any treatment effects (denominator). A ratio this large provides clear evidence that

there is a real treatment effect. To verify this conclusion you must consult the F distribution

CA L C U L AT I O N O F T H E VA R I A N C E S

(M S VA L U E S ) A N D T H E F - R AT I O

4 0 4 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

table to determine the appropriate critical value for the test. The degrees of freedom for the

F-ratio are determined by the two variances that form the numerator and the denominator.

For a repeated-measures ANOVA, the df values for the F-ratio are reported as

df 5 df between treatments

, df error

For the example we are considering, the F-ratio has df 5 3, 15 (“degrees of freedom equal three and fifteen”). Using the F distribution table (p. 577) with a 5 .05, the criti- cal value is F 5 3.29, and with a 5 .01 the critical value is F 5 5.42. Our obtained F-ratio, F 5 21.43, is well beyond either of the critical values, so we can conclude that the differences between treatments are significantly greater than expected by chance

using either a 5 .05 or a 5 .01.

The summary table for the repeated-measures ANOVA from Example 13.1 is

presented in Table 13.3. Although these tables are no longer commonly used in

research reports, they provide a concise format for displaying all of the elements

of the analysis.

The most common method for measuring effect size with ANOVA is to compute the

percentage of variance that is explained by the treatment differences. In the context

of ANOVA, the percentage of variance is commonly identified as h2 (eta squared). In

Chapter 12, for the independent-measures analysis, we computed h2 as

h 5 2 SS

between treatments

between treatments 11

5 SS

S within treatments

between treatments

SS total

The intent is to measure how much of the total variability is explained by the

differences between treatments. With a repeated-measures design, however, there is

another component that can explain some of the variability in the data. Specifically,

part of the variability is caused by differences between individuals. In Table 13.2,

for example, person A consistently scored lower than person F. This consistent dif-

ference explains some of the variability in the data. When computing the size of the

treatment effect, it is customary to remove any variability that can be explained by

other factors, and then compute the percentage of the remaining variability that can

be explained by the treatment effects. Thus, for a repeated-measures ANOVA, the

variability from the individual differences is removed before computing h2. As a

result, h2 is computed as

h 2

5 2

SS SS

between treatments

total between subbjects

(13.10)

M E AS U R I N G E F F E C T S I Z E F O R

T H E R E P E AT E D - M E AS U R E S A N OVA

Source SS df MS F

Between treatments 60 3 20.00 F(3,15) 5 21.43

Within treatments 62 20

Between subjects 48 5

Error 14 15 0.933

Total 122 23

TABLE 13.3

A summary table for

the repeated-measures

ANOVA for the data

from Example 13.1.

SECTION 13.2 / REPEATED-MEASURES ANOVA 4 0 5

Because Equation 13.10 computes a percentage that is not based on the total

variability of the scores (one part, SS between subjects

, is removed), the result is often called

a partial eta squared.

The general goal of Equation 13.10 is to calculate a percentage of the variabil-

ity that has not already been explained by other factors. Thus, the denominator of

Equation 13.10 is limited to variability from the treatment differences and variability

that is exclusively from random, unsystematic factors. With this in mind, an equivalent

version of the h2 formula is

h 5 2 SS

between treatments

between treatments 11 SS

error (13.11)

In this new version of the eta-squared formula, the denominator consists of the

variability that is explained by the treatment differences plus the other unexplained

variability. Using either formula, the data from Example 13.1 produce

h 5 5 2 60

74 0 811 81 1. . %or( )

This result means that 81.1% of the variability in the data (except for the individual

differences) is accounted for by the differences between treatments.

IN THE LITERATURE

REPORTING THE RESULTS OF A REPEATED-MEASURES ANOVA

As described in Chapter 12 (p. 368), the format for reporting ANOVA results in journal

articles consists of

1. A summary of descriptive statistics (at least treatment means and standard

deviations, and tables or graphs as needed)

2. A concise statement of the outcome of the ANOVA

For the study in Example 13.1, the report could state:

The means and variances of the quiz scores for the four strategies are shown in

Table 1. A repeated-measures analysis of variance indicated significant mean

differences in the four methods for studying text passages, F (3, 15) 5 21.43,

p , .01, h2 5 0.811.

TABLE 1

Quiz scores for students using four different study strategies

Read Once Read and Reread Answer Prepared

Questions Create and

Answer Questions

M 5.00 6.00 8.00 9.00

SD 2.00 2.00 1.67 1.26

4 0 6 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

Recall that ANOVA provides an overall test of significance for the mean differences

between treatments. When the null hypothesis is rejected, it indicates only that there is

a difference between at least two of the treatment means. If k 5 2, it is obvious which two treatments are different. However, when k is greater than 2, the situation becomes

more complex. To determine exactly where significant differences exist, the researcher

must follow the ANOVA with post hoc tests. In Chapter 12, we used Tukey’s HSD and

the Scheffé test to make these multiple comparisons among treatment means. These

two procedures attempt to control the overall alpha level by making adjustments for the

number of potential comparisons.

For a repeated-measures ANOVA, Tukey’s HSD and the Scheffé test can be used in

the exact same manner as was done for the independent-measures ANOVA, provided

that you substitute MS error

in place of MS within treatments

in the formulas and use df error

in place

of df within treatments

when locating the critical value in a statistical table. For example, the

study in Example 13.1 compared four treatments and produced MS error

5 0.933 with df

error 5 15 and n 5 6. Using Tukey’s HSD posttest, we find q 5 4.08 and obtain

HSD 4.08 0.933

6 4.08(0.394) 1error5 5 5 5q

n ..61

Thus, any mean difference greater than 1.61 points is significant with a 5 .05. For

this study, reading once is not significantly different from rereading, and answering

prepared questions is not significantly different from creating and answering your

own questions. However, answering either type of comprehension questions produces

significantly better performance than simply reading. Note: This is the same pattern of

results that was obtained in the original study.

The basic assumptions for the repeated-measures ANOVA are identical to those

required for the independent-measures ANOVA.

1. The observations within each treatment condition must be independent (see p. 220).

2. The population distribution within each treatment must be normal. (As before,

the assumption of normality is important only with small samples.)

3. The variances of the population distributions for each treatment should be

equivalent.

For the repeated-measures ANOVA, there is an additional assumption, called homo-

geneity of covariance. Basically, it refers to the requirement that the relative standing of

each subject be maintained in each treatment condition. This assumption is violated if

the effect of the treatment is not consistent for all of the subjects or if order effects exist

for some, but not other, subjects. This issue is very complex and is beyond the scope

of this book. However, methods do exist for dealing with violations of this assumption

(for a discussion, see Keppel, 1973).

As we noted in Chapter 11 (p. 328), a repeated-measures design has some distinct ad-

vantages and some disadvantages compared to an independent-measures design. On the

positive side, a repeated-measures design typically requires fewer subjects than an inde-

pendent-measures design. The repeated-measures design uses only one group of subjects,

which can be an asset if relatively few subjects are available. The primary advantage of

repeated-measures designs, however, is that they remove variance caused by individual dif-

ferences from the analysis. If individual differences are relatively large, they may obscure a

treatment effect in an independent-measures design. In this situation, a repeated-measures

design is more likely to detect the treatment effect and produce a significant result.

P O ST H O C T E ST S W I T H R E P E AT E D -

M E AS U R E S A N OVA

AS S U M P T I O N S O F T H E R E P E AT E D - M E AS U R E S A N OVA

A DVA N TAG E S A N D D I SA DVA N TAG E S

O F T H E R E P E AT E D - M E AS U R E S D E S I G N

SECTION 13.2 / REPEATED-MEASURES ANOVA 4 0 7

The primary disadvantage of a repeated-measures design is that it often creates the

opportunity for factors other than the treatment effect to cause a participant’s score to

change from one treatment to the next. Specifically, when participants are measured in

different treatment conditions at different times, outside factors that change over time,

such as the weather, may cause changes in the participants’ scores. It also is possible that

experience gained in one treatment condition may influence performance in later treat-

ments. For example, practicing a task in one condition may lead to improved performance

in a later treatment. In this case, the participant’s scores are changing from one treatment

to another but the changes are not being caused by the treatments.

1. A repeated-measures study is used to evaluate the mean differences among four

treatment conditions using a sample of n 5 10 participants. What are the df values for the F-ratio?

2. A research report includes a repeated-measures F-ratio with df 5 3, 24. How many treatment conditions were compared and how many individuals participated in the

study? (See Box 13.2.)

3. For the following data, compute SS within treatments

, SS between subjects

and SS error

Treatment

Subject 1 2 3 4

A 2 2 2 2 G 5 32

B 4 0 0 4 oX2 5 96

C 2 0 2 0

D 4 2 2 4

T 5 12 T 5 4 T 5 6 T 5 10

SS 5 4 SS 5 4 SS 5 3 SS 5 11

1. df 5 3, 27

2. There were 4 treatment conditions (k – 1 5 3) and 9 participants (n – 1 5 8).

3. SS within treatments

5 22, SS between subjects

5 8, and SS error

5 22 – 8 5 14.

L E A R N I N G C H E C K

ANSWERS

As we noted in Chapter 12 (pp. 379–380), whenever you are evaluating the difference

between two sample means, you can use either a t test or ANOVA. In Chapter 12 we

demonstrated that the two tests are related in many respects, including:

1. The two tests always reach the same conclusion about the null hypothesis.

2. The basic relationship between the two test statistics is F 5 t2.

3. The df value for the t statistic is identical to the df value for the denominator of

the F-ratio.

4. If you square the critical value for the two-tailed t test, you obtain the critical

value for the F-ratio. Again, the basic relationship is F 5 t2.

In Chapter 12, these relationships were demonstrated for the independent-measures

tests, but they are also true for repeated-measures designs comparing two treatment

conditions. The following example demonstrates the relationships.

R E P E AT E D - M E AS U R E S A N OVA A N D

T H E R E P E AT E D - M E AS U R E S t T E ST

4 0 8 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

The following table shows the data from a repeated-measures study comparing two

treatment conditions. We have structured the data in a format that is compatible with

the repeated-measures t test. Note that the calculations for the t test are based on the

difference scores (D values) in the final column.

Treatment

Participant I II D

A 3 5 2

B 4 14 10

C 5 7 2

D 4 6 2

M D 5 4

SS D 5 48

The repeated-measures t test The null hypothesis for the t test states that, for the

general population, there is no mean difference between the two treatment conditions.

H 0 : �

D 5 0

With n 5 4 participants, the test has df 5 3 and the critical boundaries for a two- tailed test with a 5 .05 are t 5 ±3.182.

For these data, the sample mean difference is M D 5 4, the variance for the differ-

ence scores is s2 5 16, and the standard error is sM D 5 2 points. These values produce a t statistic of

t s

M D

5 2 m

5 2

5 M

D D 4

2 00 0

2 .

The t value is not in the critical region, so we fail to reject H 0 and conclude that there

is no significant difference between the two treatments.

The repeated-measures ANOVA Now we reorganize the data into a format that

is compatible with a repeated-measures ANOVA. Notice that the ANOVA uses the

original scores (not the difference scores) and requires the P totals for each participant.

Treatment

Participant I II P

A 3 5 8 G 5 48

B 4 14 18 oX 2 5 372

C 5 7 12 N 5 8

D 4 6 10

Again, the null hypothesis states that, for the general population, there is no mean

difference between the two treatment conditions.

H 0 : m

1 5 m

For this study, df between treatments

5 1, df within treatments

5 6, df between subjects

5 3, which produce

df error

5 (6 – 3) 5 3. Thus, the F-ratio has df 5 1, 3 and the critical value for a 5 .05

is F 5 10.13. Note that the denominator of the F-ratio has the same df value as the t

statistic (df 5 3) and that the critical value for F is equal to the squared critical value

for t (10.13 5 3.1822).

E X A M P L E 1 3 . 2

SECTION 13.3 / TWO-FACTOR ANOVA (INDEPENDENT MEASURES) 4 0 9

For these data,

SS total

5 84

SS within

5 52

SS between treatments

5 (84 – 52) 5 32

SS between subjects

5 28

SS error

5 (52 – 28) 5 24

The two variances in the F-ratio are

MS between treatments

between treatments 5

dff between treatments

1 325 5

and 24

3 8

error error

error

MS 5 5 5 SS

and the -ratio is between treatmentsF F MS

MS 5

eerror

8 4.005 5

Notice that the F-ratio and the t statistic are related by the equation F 5 t2 (4 5 22).

The F-ratio (like the t statistic) is not in the critical region so, once again, we fail to reject

H 0 and conclude that there is no significant difference between the two treatments.

1. A repeated-measures study is used to evaluate the mean differences between

two treatment conditions using a sample of n 5 20 participants.

a. If a repeated-measures t is used for the hypothesis test, what are the df values

for the t statistic?

b. If a repeated-measures ANOVA is used for the hypothesis test, what are the

df values for the F-ratio?

2. A research report includes a repeated-measures F-ratio of F 5 4.00 with df 5 1, 24.

a. Could the researchers have used a t statistic instead of an F-ratio to evaluate the

mean difference? Explain your answer.

b. If a t statistic were used, then what value would be obtained for t?

1. a. df 5 19

b. df 51, 19

2. a. The study compared only 2 treatments (df between treatments

5 1), so a t statistic could be used.

b. t 5 54 2

L E A R N I N G C H E C K

ANSWERS

TWO-FACTOR ANOVA (INDEPENDENT MEASURES)

In most research situations, the goal is to examine the relationship between two

variables. Typically, the research study attempts to isolate the two variables to eliminate

or reduce the influence of any outside variables that may distort the relationship being

13.3

4 1 0 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

studied. A typical experiment, for example, focuses on one independent variable (which

is expected to influence behavior) and one dependent variable (which is a measure of

the behavior). In real life, however, variables rarely exist in isolation. That is, behavior

usually is influenced by a variety of different variables acting and interacting simulta-

neously. To examine these more complex, real-life situations, researchers often design

research studies that include more than one independent variable. Thus, researchers

systematically change two (or more) variables and then observe how the changes influ-

ence another (dependent) variable.

In Chapter 12 and earlier in this chapter, we examined ANOVA for single-factor

research designs—that is, designs that included only one independent variable or only

one quasi-independent variable. When a research study involves more than one factor,

it is called a factorial design. In this chapter, we consider the simplest version of a

factorial design. Specifically, we examine ANOVA as it applies to research studies with

exactly two factors. In addition, we limit our discussion to studies that use a separate

sample for each treatment condition—that is, independent-measures designs. Finally,

we consider only research designs for which the sample size (n) is the same for all

treatment conditions. In the terminology of ANOVA, this chapter examines two-factor,

independent-measures, equal n designs. The following example demonstrates the

general elements of this kind of research design.

Imagine that you are seated at your desk, ready to take the final exam in statistics. Just

before the exams are handed out, a television crew appears and sets up a camera and

lights aimed directly at you. They explain they are filming students during exams for

a television special. You are told to ignore the camera and go ahead with your exam.

Would the presence of a TV camera affect your performance on an exam? For some

of you, the answer to this question is “definitely yes,” and for others, “probably not.” In

fact, both answers are right; whether or not the TV camera affects performance depends

on your personality. Some of you would become terribly distressed and self-conscious,

while others really could ignore the camera and go on as if everything were normal.

In an experiment that duplicates the situation we have described, Shrauger (1972)

tested participants on a concept formation task. Half the participants worked alone

(no audience), and half worked with an audience of people who claimed to be

interested in observing the experiment. Shrauger also divided the participants into

two groups on the basis of personality: those high in self-esteem and those low in

self-esteem. Table 13.4 shows the structure of Shrauger’s study. Note that the study

E X A M P L E 1 3 . 3

An independent variable

is a manipulated variable

in an experiment. A quasi-

independent variable is not

manipulated but defines

the groups of scores in a

nonexperimental study.

Factor B: Audience Condition

No Audience Audience

Factor A: Self-Esteem

Low

Scores for a group

of participants who

are classified as low

self-esteem and are

tested with no audience.

Scores for a group

of participants who

are classified as low

self-esteem and are

tested with an audience.

High

Scores for a group

of participants who

are classified as high

self-esteem and are

tested with no audience.

Scores for a group

of participants who

are classified as high

self-esteem and are

tested with an audience.

TABLE 13.4

The structure of a two-factor

experiment presented as a

matrix. The two factors are

self-esteem and presence/

absence of an audience, with

two levels for each factor.

SECTION 13.3 / TWO-FACTOR ANOVA (INDEPENDENT MEASURES) 4 1 1

involves two separate factors: One factor is manipulated by the researcher, changing

from no-audience to audience, and the second factor is self-esteem, which varies

from high to low. The two factors are used to create a matrix with the different levels

of self-esteem defining the rows and the different audience conditions defining the

columns. The resulting two-by-two matrix shows four different combinations of the

variables, producing four different conditions. Thus, the research study would require

four separate samples, one for each cell, or box, in the matrix. The dependent vari-

able for the study is the number of errors on the concept formation task for people

observed in each of the four conditions.

The two-factor ANOVA tests for mean differences in research studies that are struc-

tured like the audience-and-self-esteem example in Table 13.4. For this example, the

two-factor ANOVA evaluates three separate sets of mean differences:

1. What happens to the mean number of errors when the audience is added or

taken away?

2. Is there a difference in the mean number of errors for participants with high

self-esteem compared to those with low self-esteem?

3. Is the mean number of errors affected by specific combinations of self-

esteem and audience? (For example, an audience may have a large effect on

participants with low self-esteem but only a small effect for those with high

self-esteem.)

Thus, the two-factor ANOVA allows us to examine three types of mean differences

within one analysis. In particular, we conduct three separate hypotheses tests for the

same data, with a separate F-ratio for each test. The three F-ratios have the same basic

structure:

F 5 variance (differences) between treatments

vvariance (differences) expected if there iss no treatment effect

In each case, the numerator of the F-ratio measures the actual mean differences in

the data, and the denominator measures the differences that would be expected if there

is no treatment effect. As always, a large value for the F-ratio indicates that the sample

mean differences are greater than would be expected by chance alone, and, therefore,

provides evidence of a treatment effect. To determine whether the obtained F-ratios

are significant, we need to compare each F-ratio with the critical values found in the

F-distribution table in Appendix B.

As noted in the previous section, a two-factor ANOVA actually involves three distinct

hypothesis tests. In this section, we examine these three tests in more detail.

Traditionally, the two independent variables in a two-factor experiment are identified

as factor A and factor B. For the study presented in Table 13.4, self-esteem is factor A,

and the presence or absence of an audience is factor B. The goal of the study is to evaluate

the mean differences that may be produced by either of these factors acting independently

or by the two factors acting together.

Main effects One purpose of the study is to determine whether differences in self-

esteem (factor A) result in differences in performance. To answer this question, we

compare the mean score for all of the participants with low self-esteem with the mean

M A I N E F F E C T S A N D I N T E R AC T I O N S

4 1 2 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

for those with high self-esteem. Note that this process evaluates the mean difference

between the top row and the bottom row in Table 13.4.

To make this process more concrete, we present a set of hypothetical data in

Table 13.5. The table shows the mean score for each of the treatment conditions (cells)

as well as the overall mean for each column (each audience condition) and the overall

mean for each row (each self-esteem group). These data indicate that the low self-

esteem participants (the top row) had an overall mean of M 5 8 errors. This overall mean was obtained by computing the average of the two means in the top row. In con-

trast, the high self-esteem participants had an overall mean of M 5 4 errors (the mean for the bottom row). The difference between these means constitutes what is called the

main effect for self-esteem, or the main effect for factor A.

Similarly, the main effect for factor B (audience condition) is defined by the mean

difference between the columns of the matrix. For the data in Table 13.5, the two groups

of participants tested with no audience had an overall mean score of M 5 5 errors. Participants tested with an audience committed an overall average of M 5 7 errors. The difference between these means constitutes the main effect for the audience conditions,

or the main effect for factor B.

The mean differences among the levels of one factor are referred to as the main

effect of that factor. When the design of the research study is represented as a

matrix with one factor determining the rows and the second factor determining

the columns, then the mean differences among the rows describe the main effect

of one factor, and the mean differences among the columns describe the main

effect for the second factor.

The mean differences between columns or rows simply describe the main effects for a

two-factor study. As we have observed in earlier chapters, the existence of sample mean

differences does not necessarily imply that the differences are statistically significant.

In general, two samples are not expected to have exactly the same means. There will al-

ways be small differences from one sample to another, and you should not automatically

assume that these differences are an indication of a systematic treatment effect. In the

case of a two-factor study, any main effects that are observed in the data must be evalu-

ated with a hypothesis test to determine whether they are statistically significant effects.

Unless the hypothesis test demonstrates that the main effects are significant, you must

conclude that the observed mean differences are simply the result of sampling error.

The evaluation of main effects accounts for two of the three hypothesis tests in a two-

factor ANOVA. We state hypotheses concerning the main effect of factor A and the main

effect of factor B and then calculate two separate F-ratios to evaluate the hypotheses.

For the example we are considering, factor A involves the comparison of two dif-

ferent levels of self-esteem. The null hypothesis would state that there is no difference

between the two levels; that is, self-esteem has no effect on performance. In symbols,

H 0 1 2

: m 5 m A A

D E F I N I T I O N

No Audience Audience

Low Self-Esteem M 5 7 M 5 9 M 5 8

High Self-Esteem M 5 3 M 5 5 M 5 4

M 5 5 M 5 7

TABLE 13.5

Hypothetical data for an

experiment examining the

effect of an audience on

participants with different

levels of self-esteem.

SECTION 13.3 / TWO-FACTOR ANOVA (INDEPENDENT MEASURES) 4 1 3

The alternative hypothesis is that the two different levels of self-esteem do produce

different scores:

H 1 1 2

: � � A A



To evaluate these hypotheses, we compute an F-ratio that compares the actual mean

differences between the two self-esteem levels versus the amount of difference that

would be expected without any systematic treatment effects.

F 5 variance for (differences between) the meaans for factor

variance (differences) exp

eected if there is no treatment effect

var F 5

iiance for (differences between) the row meaans

variance (differences) expected if theree is no treatment effect

Similarly, factor B involves the comparison of the two different audience conditions.

The null hypothesis states that there is no difference in the mean number of errors

between the two conditions. In symbols,

H 0 1 2

: � � B B

As always, the alternative hypothesis states that the means are different:

H 1 1 2

: � � B B



Again, the F-ratio compares the obtained mean difference between the two audi-

ence conditions versus the amount of difference that would be expected if there is no

systematic treatment effect.

F 5 variance for (differences between) the meaans for factor

variance (differences) exp

eected if there is no treatment effect

var F 5

iiance for (differences between) the column means

variance (differences) expected if thhere is no treatment effect

Interactions In addition to evaluating the main effect of each factor individually,

the two-factor ANOVA allows you to evaluate other mean differences that may result

from unique combinations of the two factors. For example, specific combinations of

self-esteem and an audience acting together may have effects that are different from the

effects of self-esteem or an audience acting alone. Any “extra” mean differences

that are not explained by the main effects are called an interaction, or an interaction

between factors. The real advantage of combining two factors within the same study is

the ability to examine the unique effects caused by an interaction.

An interaction between two factors occurs whenever the mean differences

between individual treatment conditions, or cells, are different from what would

be predicted from the overall main effects of the factors.

To make the concept of an interaction more concrete, we reexamine the data shown

in Table 13.5. For these data, there is no interaction; that is, there are no extra mean

differences that are not explained by the main effects. For example, within each audi-

ence condition (each column of the matrix), the average number of errors for the low

self-esteem participants is 4 points higher than the average for the high self-esteem

participants. This 4-point mean difference is exactly what is predicted by the overall

main effect for self-esteem.

D E F I N I T I O N

4 1 4 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

Now consider a different set of data shown in Table 13.6. These new data show

exactly the same main effects that existed in Table 13.5 (the column means and the

row means have not been changed). But now there is an interaction between the two

factors. For example, for the low self-esteem participants (top row), there is a 4-point

difference in the number of errors committed with an audience and without an audi-

ence. This 4-point difference cannot be explained by the 2-point main effect for the

audience factor. Also, for the high self-esteem participants (bottom row), the data show

no difference between the two audience conditions. Again, the zero difference is not

what would be expected based on the 2-point main effect for the audience factor. Mean

differences that are not explained by the main effects are an indication of an interaction

between the two factors.

To evaluate the interaction, the two-factor ANOVA first identifies mean differences

that are not explained by the main effects. The extra mean differences are then evaluated

by an F-ratio with the following structure:

F 5 variance (mean differences) not explained by the main effects

variance (mean differennces) expected if there are no treatment efffects

The null hypothesis for this F-ratio simply states that there is no interaction:

H 0 : There is no interaction between factors A and B. All of the mean differences

between treatment conditions are explained by the main effects of the two factors.

The alternative hypothesis states that there is an interaction between the two factors:

H 1 : There is an interaction between factors. The mean differences between treat-

ment conditions are not what would be predicted from the overall main effects of

the two factors.

More about interactions In the previous section, we introduced the concept of an

interaction as the unique effect produced by two factors working together. This section

presents two alternative definitions of an interaction. These alternatives are intended to

help you understand the concept of an interaction and to help you identify an interaction

when you encounter one in a set of data. You should realize that the new definitions

are equivalent to the original and simply present slightly different perspectives on the

same concept.

The first new perspective on the concept of an interaction focuses on the notion

of independence for the two factors. More specifically, if the two factors are inde-

pendent, so that one factor does not influence the effect of the other, then there is no

interaction. On the other hand, when the two factors are not independent, so that the

effect of one factor depends on the other, then there is an interaction. The notion of

dependence between factors is consistent with our earlier discussion of interactions.

If one factor influences the effect of the other, then unique combinations of the factors

produce unique effects.

The data in Table 13.6

show the same pattern of

results that was obtained in

Shrauger’s research study.

No Audience Audience

Low Self-Esteem M 5 6 M 5 10 M 5 8

High Self-Esteem M 5 4 M 5 4 M 5 4

M 5 5 M 5 7

TABLE 13.6

Hypothetical data for an

experiment examining the ef-

fect of an audience on partic-

ipants with different levels of

self-esteem. The data show

the same main effects as the

values in Table 13.5, but the

individual treatment means

have been modified to create

an interaction.

SECTION 13.3 / TWO-FACTOR ANOVA (INDEPENDENT MEASURES) 4 1 5

When the effect of one factor depends on the different levels of a second factor,

then there is an interaction between the factors.

This definition of an interaction should be familiar in the context of a “drug interac-

tion.” Your doctor and pharmacist are always concerned that the effect of one medica-

tion may be altered or distorted by a second medication that is being taken at the same

time. Thus, the effect of one drug (factor A) depends on a second drug (factor B), and

you have an interaction between the two drugs.

Returning to Table 13.5, you will notice that the size of the audience effect (first

column versus second column) does not depend on the self-esteem of the participants.

For these data, adding an audience produces the same 2-point increase in errors for

both groups of participants. Thus, the audience effect does not depend on self-esteem,

and there is no interaction. Now consider the data in Table 13.6. This time, the effect of

adding an audience depends on the self-esteem of the participants. For example, there

is a 4-point increase in errors for the low self-esteem participants but adding an audi-

ence has no effect on the errors for the high self-esteem participants. Thus, the audience

effect depends on the level of self-esteem, which means that there is an interaction

between the two factors.

The second alternative definition of an interaction is obtained when the results of a

two-factor study are presented in a graph. In this case, the concept of an interaction can

be defined in terms of the pattern displayed in the graph. Figure 13.2 shows the two

sets of data we have been considering. The original data from Table 13.5, where there

is no interaction, are presented in Figure 13.2(a). To construct this figure, we selected

one of the factors to be displayed on the horizontal axis; in this case, the different levels

of the audience factor are displayed. The dependent variable, the number of errors, is

shown on the vertical axis. Note that the figure actually contains two separate graphs:

The top line shows the relationship between the audience factor and errors for the low

self-esteem participants, and the bottom line shows the relationship for the high self-

esteem participants. In general, the picture in the graph matches the structure of the data

D E F I N I T I O N

No audience

Audience

Low self-esteem

High self-esteem

M e

a n

e rr

o rs

No audience

Audience

Low self-esteem

High self-esteem

M e

a n

e rr

o rs

FIGURE 13.2

(a) Graph showing the treatment means from Table 13.5, for which there is no interaction.

(b) Graph for Table 13.6, for which there is an interaction.

(a) (b)

4 1 6 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

matrix; the columns of the matrix appear as values along the X-axis, and the rows of the

matrix appear as separate lines in the graph.

For the original set of data, Figure 13.2(a), note that the two lines are parallel; that is,

the distance between lines is constant. In this case, the distance between lines reflects

the 2-point difference in mean errors between low and high self-esteem participants,

and this 2-point difference is the same for both audience conditions.

Now look at a graph that is obtained when there is an interaction in the data.

Figure 13.2(b) shows the data from Table 13.6. This time, note that the lines in the

graph are not parallel. The distance between the lines changes as you scan from

left to right. For these data, the distance between the lines corresponds to the self-

esteem effect—that is, the mean difference in errors for low versus high self-esteem

participants. The fact that this difference depends on the audience condition is an

indication of an interaction between the two factors.

When the results of a two-factor study are presented in a graph, the existence of

nonparallel lines (lines that cross or diverge) indicates an interaction between

the two factors.

For many students, the concept of an interaction is easiest to understand using the

perspective of interdependency; that is, an interaction exists when the effects of one

variable depend on another factor. However, the easiest way to identify an interaction

within a set of data is to draw a graph showing the treatment means. The presence of

nonparallel lines is an easy way to spot an interaction.

The two-factor ANOVA consists of three hypothesis tests, each evaluating specific

mean differences: the A effect, the B effect, and the A 3 B interaction. As we have

noted, these are three separate tests, but you should also realize that the three tests are

independent. That is, the outcome for any one of the three tests is totally unrelated to the

outcome for either of the other two. Thus, it is possible for data from a two-factor study

to display any possible combination of significant and/or not significant main effects

and interactions. The data sets in Table 13.7 show several possibilities.

Table 13.7(a) shows data with mean differences between levels of factor A (an

A effect) but no mean differences for factor B and no interaction. To identify the

A effect, notice that the overall mean for A 1 (the top row) is 10 points higher than the

overall mean for A 2 (the bottom row). This 10-point difference is the main effect for

factor A. To evaluate the B effect, notice that both columns have exactly the same

overall mean, indicating no difference between levels of factor B; hence, there is no

B effect. Finally, the absence of an interaction is indicated by the fact that the overall

A effect (the 10-point difference) is constant within each column; that is, the A effect

does not depend on the levels of factor B. (Alternatively, the data indicate that the

overall B effect is constant within each row.)

Table 13.7(b) shows data with an A effect and a B effect but no interaction. For these

data, the A effect is indicated by the 10-point mean difference between rows, and the

B effect is indicated by the 20-point mean difference between columns. The fact that the

10-point A effect is constant within each column indicates no interaction.

Finally, Table 13.7(c) shows data that display an interaction but no main effect

for factor A or for factor B. For these data, there is no mean difference between rows

(no A effect) and no mean difference between columns (no B effect). However, within

each row (or within each column), there are mean differences. The “extra” mean differ-

ences within the rows and columns cannot be explained by the overall main effects and

therefore indicate an interaction.

D E F I N I T I O N

I N D E P E N D E N C E O F M A I N E F F E C T S

A N D I N T E R AC T I O N S

The A 3 B interaction

typically is called “A by B”

interaction. If there is an

interaction between an

audience and self-esteem, it

may be called the “audience

by self-esteem” interaction.

SECTION 13.3 / TWO-FACTOR ANOVA (INDEPENDENT MEASURES) 4 1 7

1. The following matrix shows the results from a two-factor experiment.

a. What means are compared to evaluate the main effect for factor A?

b. What means are compared to evaluate the main effect for factor B?

c. Does there appear to be an interaction between the two factors? Explain your

answer.

B 1

B 2

A 1 M 5 20 M 5 30 M 5 25

A 2 M 5 10 M 5 10 M 5 10

M 5 15 M 5 25

2. It is impossible to have an interaction unless you also have main effects for at least

one of the two factors. (True or false?)

L E A R N I N G C H E C K

a. Data showing a main effect for factor A but no B effect and no interaction

B 1

B 2

A 1

20 20 A 1 mean 5 20

m n 10-point difference

A 2

10 10 A 2 mean 5 10

B 1 mean

5 15

B 2 mean

5 15 m888n

No difference

b. Data showing main effects for both factor A and factor B but no interaction

B 1

B 2

A 1

10 30 A 1 mean 5 20

m n 10-point difference

A 2

20 40 A 2 mean 5 30

B 1 mean

5 15

B 2 mean

5 35

m888n

20-point difference

c. Data showing no main effect for either factor but an interaction

B 1

B 2

A 1

10 20 A 1 mean 5 15

m n No difference

A 2

20 10 A 2 mean 5 15

B 1 mean

5 15

B 2 mean

5 15 m888n

No difference

TABLE 13.7

Three sets of data showing

different combinations of

main effects and interaction

for a two-factor study. (The

numerical value in each cell

of the matrices represents

the mean value obtained for

the sample in that treatment

condition.)

4 1 8 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

1. a. The main effect for factor A compares M 5 25 and M 5 10.

b. The main effect for factor B compares M 5 15 and M 5 25.

c. Yes, there is an interaction. In A 1 there is a 10-point difference between the two levels

of factor B, but in A 2 there is no mean difference. Thus, the B effect depends on the levels

of factor A.

2. False. The existence of an interaction is completely independent of the main effects.

ANSWERS

The two-factor ANOVA is composed of three distinct hypothesis tests:

1. The main effect of factor A (often called the A-effect). Assuming that factor A is

used to define the rows of the matrix, the main effect of factor A evaluates the

mean differences between rows.

2. The main effect of factor B (called the B-effect). Assuming that factor B is used

to define the columns of the matrix, the main effect of factor B evaluates the

mean differences between columns.

3. The interaction (called the A 3 B interaction). The interaction evaluates mean differences between treatment conditions that are not predicted from the overall

main effects from factor A or factor B.

For each of these three tests, we are looking for mean differences between treatments

that are larger than would be expected if there were no treatment effects. In each case,

the significance of the treatment effect is evaluated by an F-ratio. All three F-ratios have

the same basic structure:

F 5 variance (mean differences) between treatmments

variance (mean differences) expected iif there is no treatment effect

(13.12)

The general structure of the two-factor ANOVA is shown in Figure 13.3. Note that

the overall analysis is divided into two stages. In the first stage, the total variability is

separated into two components: between-treatments variability and within-treatments

variability. This first stage is identical to the single-factor ANOVA introduced in

Chapter 12 with each cell in the two-factor matrix viewed as a separate treatment condi-

tion. The within-treatments variability that is obtained in stage 1 of the analysis is used

to compute the denominator for the F-ratios. As we noted in Chapter 12, within each

treatment, all of the participants are treated exactly the same. Thus, any differences

that exist within the treatments cannot be caused by treatment effects. As a result, the

within-treatments variability provides a measure of the differences that exist when there

are no systematic treatment effects influencing the scores (see Equation 13.12).

The between-treatments variability obtained in stage 1 of the analysis combines all

the mean differences produced by factor A, factor B, and the interaction. The purpose

of the second stage is to partition the differences into three separate components: differ-

ences attributed to factor A, differences attributed to factor B, and any remaining mean

differences that define the interaction. These three components form the numerators for

the three F-ratios in the analysis.

The goal of this analysis is to compute the variance values needed for the three F-ratios.

We need three between-treatments variances (one for factor A, one for factor B, and one for

the interaction), and we need a within-treatments variance. Each of these variances (or mean

squares) is determined by a sum of squares value (SS) and a degrees of freedom value (df):

mean square 5 5MS SS

T H E ST R UC T U R E O F T H E T WO - FAC TO R

A N OVA

Remember that in ANOVA

a variance is called a mean

square, or MS.

SECTION 13.3 / TWO-FACTOR ANOVA (INDEPENDENT MEASURES) 4 1 9

To demonstrate the two-factor ANOVA, we will use a research study based on previous

work by Ackerman and Goldsmith (2011). Their study compared learning performance

by students who studied text either from printed pages or from a computer screen. The

results from the study indicate that students do much better studying from printed pages

if their study time is self-regulated. However, when the researchers fixed the time spent

studying, there was no difference between the two conditions. Apparently, students are

less accurate predicting their learning performance or have trouble regulating study

time when working with a computer screen compared to working with paper. Table 13.8

shows data from a two-factor study replicating the Ackerman and Goldsmith experi-

ment. The two factors are mode of presentation (paper or computer screen) and time

control (self-regulated or fixed). A separate group of n 5 5 students was tested in each of the four conditions. The dependent variable is student performance on a 10-point

quiz covering the text that was studied.

The data are displayed in a matrix with the two levels of time control (factor A)

making up the rows and the two levels of presentation mode (factor B) making up the

columns. Note that the data matrix has a total of four cells or treatment conditions with

a separate sample of n 5 5 participants in each condition. Most of the notation should be familiar from the single-factor ANOVA presented in Chapter 12. Specifically, the

treatment totals are identified by T values, the total number of scores in the entire

study is N 5 20, and the grand total (sum) of all 20 scores is G 5 155. In addition to these familiar values, we have included the totals for each row and for each column

in the matrix. The goal of the ANOVA is to determine whether the mean differences

observed in the data are significantly greater than would be expected if there were no

treatment effects.

The first stage of the two-factor analysis separates the total variability into two compo-

nents: between-treatments and within-treatments. The formulas for this stage are identi-

cal to the formulas used in the single-factor ANOVA in Chapter 12 with the provision

E X A M P L E 1 3 . 4

STAG E 1 O F T H E T WO - FAC TO R A N A LYS I S

Stage 1

Stage 2

Between-treatments variance

Factor A variance

Factor B variance

Interaction variance

Total variance

Within-treatments variance

FIGURE 13.3

Structure for the analysis

for a two-factor ANOVA.

4 2 0 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

that each cell in the two-factor matrix is treated as a separate treatment condition. The

formulas and the calculations for the data in Table 13.8 are as follows:

Total variability

SS G

N total

2 5 2o X

(13.13)

For these data,

SS total

1303 1201.2

5 2

1303 155

101.75 5

This SS value measures the variability for all N 5 20 scores and has degrees of

freedom given by

df total

5 N – 1 (13.14)

For the data in Table 13.8, df total

5 19.

Within-treatments variability To compute the variance within treatments, we first

compute SS and df 5 n – 1 for each of the individual treatment conditions. Then the

within-treatments SS is defined as

SS within treatments

5 oSS each treatment

(13.15)

Factor B: Text Presentation Mode

Paper Computer

Screen

11 4

8 4

9 8

Self-regulated 10 5 T row

5 70

7 4

M 5 9 M 5 5

T 5 45 T 5 25 N 5 20

Factor A: Time Control

SS 5 10 SS 5 12 G 5 155

10 10 oX2 5 1303

7 6

10 10

6 10

Fixed 7 9 T row

5 85

M 5 8 M 5 9

T 5 40 T 5 45

SS 5 14 SS 5 12

T col

5 85 T col

5 70

TABLE 13.8

Data for a two-factor study

comparing two levels of time

control (self-regulated or

fixed by the researchers) and

two levels of text presenta-

tion (paper and computer

screen). The dependent vari-

able is performance on a quiz

covering the text that was

presented. The study involves

four treatment conditions

with n 5 5 participants in

each treatment.

SECTION 13.3 / TWO-FACTOR ANOVA (INDEPENDENT MEASURES) 4 2 1

And the within-treatments df is defined as

df within treatments

5 odf each treatment

(13.16)

For the four treatment conditions in Table 13.8,

SS within treatments

5 10 1 12 1 14 1 12

5 48

df within treatments

5 4 1 4 1 4 1 4

5 16

Between-treatments variability Because the two components in stage 1 must add up

to the total, the easiest way to find SS between treatments

is by subtraction.

SS between treatments

5 SS total

– SS within

(13.17)

For the data in Table 13.8, we obtain

SS between treatments

5 101.75 – 48 5 53.75

However, you can also use the computational formula to calculate SS between treatments

directly.

SS T

n between treatments

5 2o G

(13.18)

For the data in Table 13.8, there are four treatments (four T values), each with

n 5 5 scores, and the between-treatments SS is

SS between treatments

5 1 1 1 45

2 2 2 2

5 1 1

155

405 125 3200 405 1201.25

53.

1 2

5 775

The between-treatments df value is determined by the number of treatments (or the

number of T values) minus one. For a two-factor study, the number of treatments is

equal to the number of cells in the matrix. Thus,

df between treatments

5 number of cells 2 1 (13.19)

For these data, df between

treatments

5 3.

This completes the first stage of the analysis. Note that the two components add to

equal the total for both SS values and df values.

SS between treatments

1 SS within treatments

5 SS total

53.75 1 48 5 101.75

df between treatments

1 df within treatments

5 df total

3 1 16 5 19

4 2 2 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

The second stage of the analysis determines the numerators for the three F-ratios.

Specifically, this stage determines the between-treatments variance for factor A,

factor B, and the interaction.

1. Factor A. The main effect for factor A evaluates the mean differences between

the levels of factor A. For this example, factor A defines the rows of the matrix,

so we are evaluating the mean differences between rows. To compute the SS for

factor A, we calculate a between-treatment SS using the row totals in exactly the

same way as we computed SS between treatments

using the treatment totals (T values)

earlier. For factor A, the row totals are 70 and 85, and each total was obtained

by adding 10 scores.

Therefore,

SS T

N A

ROW

5 2o 2 2

(13.20)

For our data,

SS A

5 1 2

155

2 2 2

490 722.5 12011.25

11.25 5

Factor A involves two treatments (or two rows), easy and difficult, so the

df value is

df A 5 number of rows – 1 (13.21)

5 2 – 1

5 1

2. Factor B. The calculations for factor B follow exactly the same pattern that was

used for factor A, except for substituting columns in place of rows. The main

effect for factor B evaluates the mean differences between the levels of factor B,

which define the columns of the matrix.

SS T

N B

COL

5 2o

2 2

(13.22)

For our data, the column totals are 85 and 70, and each total was obtained by adding

10 scores. Thus,

SS B

5 1 1

5 1 2

155

2 2 2

725.5 490 11201.25

11.25 5

df B 5 number of columns – 1 (13.23)

5 2 – 1

5 1

STAG E 2 O F T H E T WO - FAC TO R A N A LYS I S

SECTION 13.3 / TWO-FACTOR ANOVA (INDEPENDENT MEASURES) 4 2 3

3. The A 3 B Interaction. The A 3 B interaction is defined as the “extra” mean

differences not accounted for by the main effects of the two factors. We use this

definition to find the SS and df values for the interaction by simple subtraction.

Specifically, the between-treatments variability is partitioned into three parts:

the A effect, the B effect, and the interaction (see Figure 13.3). We have already

computed the SS and df values for A and B, so we can find the interaction

values by subtracting to find out how much is left. Thus,

SS A 3 B

5 SS between treatments

– SS A – SS

B (13.24)

For our data,

SS A 3 B

5 53.75 – 11.25 – 11.25

5 31.25

Similarly,

df A 3 B

5 df between treatments

– df A – df

B (13.25)

5 3 – 1 – 1

5 1

An easy to remember alternative formula for df A 3 B

df A 3 B

5 df A 3 df

B (13.26)

5 131 5 1

The two-factor ANOVA consists of three separate hypothesis tests with three sepa-

rate F-ratios. The denominator for each F-ratio is intended to measure the variance

(differences) that would be expected if there are no treatment effects. As we saw

in Chapter 12, the within-treatments variance is the appropriate denominator for an

independent-measures design (see page 352). The within-treatments variance is

called a mean square, or MS, and is computed as follows:

MS SS

df within treatments

within treatments

5 tthin treatments

For the data in Table 13.8,

MS within treatments

5 5 48

16 3

This value forms the denominator for all three F-ratios.

The numerators of the three F-ratios all measured variance or differences between

treatments: differences between levels of factor A, differences between levels of factor

B, and extra differences that are attributed to the A 3 B interaction. These three vari- ances are computed as follows:

MS SS

df MS

df A

B B

A B A B5 5 53

AA B3

For the data in Table 13.9, the three MS values are

MS MS A B

5 5 5 5 11 25

1 11 25

11 25

1 11 25

. .

. . MMS

A B3 5 5

31 25

1 31 25

. .

M E A N S Q UA R E S A N D F - R AT I O S F O R

T H E T WO - FAC TO R A N OVA

4 2 4 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

Finally, the three F-ratios are

F MS

A A

B B

5 5 5

within treatments

11 25

3 3 75

. .

MMS

F MS

M A B

A B

within treatments

5 5

53 3

11 25

3 3 75

. .

SS within treatments

5 5 31 25

3 10 41

. .

To determine the significance of each F-ratio, we must consult the F distribution

table using the df values for each of the individual F-ratios. For this example, all three

F-ratios have df 5 1 for the numerator and df 5 16 for the denominator. Checking the table with df 5 1, 16, we find a critical value of 4.49 for a 5 .05 and a critical value of 8.53 for a 5 .01. For both main effects, we obtained F 5 3.75, so neither of the main effects is significant. For the interaction, we obtained F 5 10.41, which exceeds both of the critical values, so we conclude that there is a significant interaction between the

two factors. That is, the difference between the two modes of presentation depends on

how studying time is controlled.

Table 13.9 is a summary table for the complete two-factor ANOVA from Example

13.4. Although these tables are no longer commonly used in research reports, they pro-

vide a concise format for displaying all of the elements of the analysis.

Source SS df MS F

Between treatments 53.75 3

Factor A (time control) 11.25 1 11.25 F(1, 16) 5 3.75

Factor B (presentation) 11.25 1 11.25 F(1, 16) 5 3.75

A 3 B 31.25 1 31.25 F(1, 16) 5 10.42

Within treatments 48 16 3

Total 101.75 19

TABLE 13.9

A summary table for the

two-factor ANOVA for the

data from Example 13.4.

1. Explain why the within-treatment variability is the appropriate denominator for the

two-factor independent-measures F-ratios.

2. The following data summarize the results from a two-factor independent-measures

experiment:

L E A R N I N G C H E C K

Factor B

B 1

B 2

B 3

Factor A

n 5 5 n 5 5 n 5 5

A 1 T 5 0 T 5 10 T 5 20

SS 5 30 SS 5 40 SS 5 50

n 5 5 n 5 5 n 5 5

A 2 T 5 10 T 5 10 T 5 10

SS 5 60 SS 5 50 SS 5 40

SECTION 13.3 / TWO-FACTOR ANOVA (INDEPENDENT MEASURES) 4 2 5

a. Calculate SS for factor A.

b. Calculate SS for factor B.

c. Given that the between-treatments (or between-cells) SS is equal to 40, what is

the SS for the interaction?

3. The following table summarizes the results from a two-factor ANOVA with 2

levels of factor A, 3 levels of factor B, and n = 6 in each treatment condition. Fill

in all of the blank cells.

Source SS df MS

Between Treat. 75 __

Factor A __ __ __ F 5 __

Factor B __ __ 15 F 5 __

A 3 B __ __ __ F 5 6.00

Within Treat. __ __ __

Total 165 __

1. Within each treatment condition, all individuals are treated in exactly the same way.

Therefore, the within-treatment variability measures the differences that exist between one

score and another when there is no treatment effect causing the scores to be different. This is

exactly the variance that is needed for the denominator of the F-ratios.

2. a. The totals for factor A are 30 and 30, and each total is obtained by adding 15 scores. SS A 5 0.

b. The totals for factor B are 10, 20, and 30, and each total is obtained by adding 10 scores.

SS B 5 20.

c. The interaction is determined by differences that remain after the main effects have been

accounted for. For these data,

SS A 3 B

5 SS between treatments

– SS A – SS

5 40 – 0 – 20

5 20

ANSWERS

Source SS df MS

Between Treat. 75 5

Factor A 9 1 9 F 5 3.00

Factor B 30 2 15 F 5 5.00

A 3 B 36 2 18 F 5 6.00

Within Treat. 90 30 3

Total 165 35

The general technique for measuring effect size with an ANOVA is to compute a value

for h2, the percentage of variance that is explained by the treatment effects. For a two-

factor ANOVA, we compute three separate values for h2: one measuring how much of

the variance is explained by the main effect for factor A, one for factor B, and a third

for the interaction. As we did with the repeated-measures ANOVA (page 404), we

remove any variability that can be explained by other sources before we calculate the

M E AS U R I N G E F F E C T S I Z E F O R T H E

T WO - FAC TO R A N OVA

4 2 6 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

percentage for each of the three specific treatment effects. Thus, for example, before we

compute the h2 for factor A, we remove the variability that is explained by factor B and

the variability explained by the interaction. The resulting equation is,

for factor total

A SS

SS SS SS

B A B

, h 5 2 2

(13.27)

Note that the denominator of Equation 13.27 consists of the variability that is ex-

plained by factor A and the other unexplained variability. Thus, an equivalent version

of the equation is,

for factor within treatments

A SS

SS SS

, h 5 1

(13.28)

Similarly, the h2 formulas for factor B and for the interaction are as follows:

for factor total

B SS

SS SS SS

A A B

B, h 5 2 2

5 3

BB SS1

within treatments

(13.29)

for , total

A B SS

SS SS SS

A B

3 h 5 2 2

5 3 3

11 SS within treatments

(13.30)

Because each of the h2 equations computes a percentage that is not based on the total

variability of the scores, the results are often called partial eta squares. For the data in

Example 13.4, the equations produce the following values:

h 5 2 11

for factor time control)A ( ..

. .

(

11 25 48 0 190

1 5

h for factor presentaB ttion mode) 11.25

11.25 48 5

1 5

0 190

ffor factor interation 31.25

31.25 48 A B3 5

1 5 00 394.

IN THE LITERATURE

REPORTING THE RESULTS OF A TWO-FACTOR ANOVA

The APA format for reporting the results of a two-factor ANOVA follows the same

basic guidelines as the single-factor report. First, the means and standard deviations

are reported. Because a two-factor design typically involves several treatment condi-

tions, these descriptive statistics often are presented in a table or a graph. Next, the

results of all three hypothesis tests (F-ratios) are reported. The results for the study in

Example 13.4 could be reported as follows:

The means and standard deviations for all treatment conditions are shown in

Table 1. The two-factor analysis of variance showed no significant main effect

for time control, F(1, 16) 5 3.75, p . .05, h2 5 0.190, or for presentation mode,

F(1, 16) 5 3.75, p . .05, h2 5 0.190. However, the interaction between factors

was significant, F(1, 16) 5 10.41, p , .01, h2 5 0.394.

SECTION 13.3 / TWO-FACTOR ANOVA (INDEPENDENT MEASURES) 4 2 7

Because the two-factor ANOVA involves three separate tests, you must consider the

overall pattern of results rather than focusing on the individual main effects or the

interaction. In particular, whenever there is a significant interaction, you should be

cautious about accepting the main effects at face value (whether they are significant or

not). Remember, an interaction means that the effect of one factor depends on the level

of the second factor. Because the effect changes from one level to the next, there is no

consistent “main effect.”

Figure 13.4 shows the sample means obtained from the paper versus computer

screen study. Recall that the analysis showed that both main effects were not signifi-

cant but the interaction was significant. Although both main effects were too small to

be significant, it would be incorrect to conclude that neither factor influenced behavior.

For this example, the difference between studying text presented on paper versus on a

computer screen depends on how studying time is controlled. Specifically, there is little

or no difference between paper and a computer screen when the time spent studying is

fixed by the researchers. However, studying text from paper produces much higher quiz

scores when participants regulate their own study time. Thus, the difference between

studying from paper and studying from a computer screen depends on how the time

spent studying is controlled. This interdependence between factors is the source of the

significant interaction.

The validity of the ANOVA presented in this chapter depends on the same three

assumptions that we have encountered with other hypothesis tests for independent-

measures designs (the t test in Chapter 10 and the single-factor ANOVA in Chapter 12):

1. The observations within each sample must be independent (see page 220).

2. The populations from which the samples are selected must be normal.

3. The populations from which the samples are selected must have equal variances

(homogeneity of variance).

As before, the assumption of normality generally is not a cause for concern,

especially when the sample size is relatively large. The homogeneity of variance

I N T E R P R E T I N G T H E R E S U LT S F R O M

A T WO - FAC TO R A N OVA

AS S U M P T I O N S F O R T H E T WO - FAC TO R

A N OVA

Paper

Mode of Text Presentation

Computer Screen

fixed time

self-regulated time

Mean

Quiz

Score

FIGURE 13.4

Sample means for the

data in Example 13.4.

The data are quiz scores

from a two-factor study

examining the effect of

studying text on paper

versus on a computer

screen for either a

fixed time or a self-

regulated time.

4 2 8 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

assumption is more important, and if it appears that your data fail to satisfy this

requirement, you should conduct a test for homogeneity before you attempt the

ANOVA. Hartley’s F-max test (see page 301) allows you to use the sample variances

from your data to determine whether there is evidence for any differences among

the population variances. Remember, for the two-factor ANOVA, there is a separate

sample for each cell in the data matrix. The test for homogeneity applies to all these

samples and the populations they represent.

SUMMARY

1. The repeated-measures ANOVA is used to evaluate the mean differences obtained in a research study comparing two or more treatment conditions using the same sample of individuals in each condition. The test statistic is an F-ratio, in which the numerator measures the variance (differences) between treat- ments and the denominator measures the variance (differences) that is expected without any treatment effects or individual differences.

F MS

MS 5 between treatments

error

2. The first stage of the repeated-measures ANOVA is identical to the independent-measures ANOVA and separates the total variability into two components: between-treatments and within-treatments. Because a repeated-measures design uses the same subjects in every treatment condition, the differences between treatments cannot be caused by individual differences. Thus, individual differences are automatically eliminated from the between-treatments variance in the numerator of the F-ratio.

3. In the second stage of the repeated-measures analysis, individual differences are computed and removed from the denominator of the F-ratio. To remove the individual differences, you first compute the variability between subjects (SS and df) and then subtract these values from the corresponding within-treatments values. The residual provides a measure of error excluding individual differences, which is the appropriate denominator for the repeated-measures F-ratio.

4. Effect size for the repeated-measures ANOVA is measured by computing eta squared, the percentage of variance accounted for by the treatment effect. For the repeated-measures ANOVA

h 5 2

2 SS

SS SS

between treatments

total between subbjects

5 1

SS S

between treatments

between treatments SS

error

Because part of the variability (the SS caused by individual differences) is removed before computing h2, this measure of effect size is often called a partial eta squared.

5. A research study with two independent variables is called a two-factor design. Such a design can be diagramed as a matrix with the levels of one factor defining the rows and the levels of the other factor defining the columns. Each cell in the matrix corre- sponds to a specific combination of the two factors.

6. Traditionally, the two factors are identified as factor A and factor B. The purpose of the ANOVA is to deter- mine whether there are any significant mean differ- ences among the treatment conditions or cells in the experimental matrix. These treatment effects are classified as follows:

a. The A-effect: Overall mean differences among the levels of factor A.

b. The B-effect: Overall mean differences among the levels of factor B.

c. The A 3 B interaction: Extra mean differences that are not accounted for by the main effects.

7. The two-factor ANOVA produces three F-ratios: one for factor A, one for factor B, and one for the A 3 B inter- action. Each F-ratio has the same basic structure:

F MS A B A B

M 5

3 treatment effect

either or or( ) SS

within treatments

The formulas for the SS, df, and MS values for the two-factor ANOVA are presented in Figure 13.5.

RESOURCES 4 2 9

KEY TERMS

Total

SS = Σ X 2 – G 2 — N

df = N – 1

Between treatments

SS = Σ T 2

– G2

— — n N

df = (number of cell) – 1

Factor A (rows)

SS = Σ T 2

ROW

– G2 — Nn

df = (levels of A) – 1

Factor B (columns)

SS = Σ T 2

COL

– G2 — Nn

df = (levels of B) – 1

SS Interaction is found by subtraction

df is found by subtration

Within treatments

SS = ΣSSeach cell

Σdfeach celldf =

= SS for the factor

for the factordf MS factor =

SS within treatments within treatmentsdf

MS within

FIGURE 13.5

The ANOVA for an

independent-measures

two-factor design.

individual differences (396)

between-treatments variance (398)

error variance (398)

between-subjects variance (398)

two-factor design (410)

matrix (411)

cell (411)

main effect (411)

interaction (413)

RESOURCES

Go to CengageBrain.com to access Psychology CourseMate, where you will find an

interactive eBook, glossaries, flashcards, quizzes, statistics workshops, and more.

If your professor has assigned Aplia:

1. Sign in to your account.

2. Complete the corresponding exercises as required by your professor.

3. When finished, click “Grade It Now” to see which areas you have mastered, which

areas need more work, and detailed explanations of every answer.

4 3 0 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

General instructions for using SPSS are presented in Appendix D. Following are

detailed instructions for using SPSS to perform the Single-Factor, Repeated-Measures

Analysis of Variance (ANOVA) presented in this chapter.

Data Entry

Enter the scores for each treatment condition in a separate column, with the scores

for each individual in the same row. All of the scores for the first treatment go in the

VAR00001 column, the second treatment scores go in the VAR00002 column, and

so on.

Data Analysis

1. Click Analyze on the tool bar, select General Linear Model, and click on

Repeated-Measures.

2. SPSS presents a box entitled Repeated-Measures Define Factors. Within the

box, the Within-Subjects Factor Name should already contain Factor 1. If not,

type in Factor 1.

3. Enter the Number of levels (number of different treatment conditions) in the

next box.

4. Click Add.

5. Click Define.

6. One by one, move the column labels for your treatment conditions into the

Within Subjects Variables box. (Highlight the column label on the left and

click the arrow to move it into the box.)

7. If you want descriptive statistics for each treatment, click on the Options box,

select Descriptives, and click Continue.

8. Click OK.

SPSS Output

We used the SPSS program to analyze the data from the study comparing different

strategies for studying that was presented in Example 13.1. Portions of the program

output are shown in Figure 13.6. Note that large portions of the SPSS output are not

relevant for our purposes and are not included in Figure 13.6. The first item of interest

is the table of Descriptive Statistics, which presents the mean, standard deviation,

and number of scores for each treatment. Next, we skip to the table showing Tests

of Within-Subjects Effects. The top line of the factor1 box (Sphericity Assumed)

shows the between-treatments sum of squares, degrees of freedom, and mean square

that form the numerator of the F-ratio. The same line reports the value of the F-ratio

and the level of significance (the p value or alpha level). Similarly, the top line of

the Error (factor1) box shows the sum of squares, the degrees of freedom, and the

mean square for the error term (the denominator of the F-ratio). The final box in

the output (not shown in Figure 13.6) is labeled Tests of Between-Subjects Effects

and the bottom line (Error) reports the between-subjects sum of squares and degrees

of freedom (ignore the mean square and F-ratio, which are not part of the repeated-

measures ANOVA).

Following are detailed instructions for using SPSS to perform the Two-Factor,

Independent-Measures Analysis of Variance (ANOVA) presented in this chapter.

RESOURCES 4 3 1

Data Entry

1. The scores are entered into the SPSS data editor in a stacked format, which

means that all of the scores from all of the different treatment conditions are

entered in a single column (VAR00001).

2. In a second column (VAR00002) enter a code number to identify the level of

factor A for each score. If factor A defines the rows of the data matrix, enter a 1

beside each score from the first row, enter a 2 beside each score from the sec-

ond row, and so on.

3. In a third column (VAR00003) enter a code number to identify the level of fac-

tor B for each score. If factor B defines the columns of the data matrix, enter a

1 beside each score from the first column, enter a 2 beside each score from the

second column, and so on.

Descriptive Statistics

Tests of Within-Subjects Effects

Measure:MEASURE_1

Source

factor 1

Type III Sum of Squares

df Mean Square

F Sig.

Mean

VAR00001

VAR00002

VAR00003

VAR00004

Sphericity Assumed

Greenhouse-Geisser

Huynh-Feldt

Lower-bound

60.000

2.178

3.000

1.000

20.000

27.551

20.000

60.000

14.000

10.889

15.000

5.000

.933

1.286

.933

2.800

21.429

.000

.006

Error (factor 1) Sphericity Assumed

Greenhouse-Geisser

Huynh-Feldt

Lower-bound

5.0000

6.0000

8.0000

9.0000

2.00000

1.67332

1.26491

Std. Deviation N

FIGURE 13.6

Portions of the SPSS output for the repeated-measures ANOVA for the study evaluating different

strategies for studying in Example 13.1.

4 3 2 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

Thus, each row of the SPSS data editor will have one score and two code numbers,

with the score in the first column, the code for factor A in the second column, and the

code for factor B in the third column.

Data Analysis

1. Click Analyze on the tool bar, select General Linear Model, and click on

Univariant.

2. Highlight the column label for the set of scores (VAR0001) in the left box and

click the arrow to move it into the Dependent Variable box.

3. One by one, highlight the column labels for the two factor codes and click the

arrow to move them into the Fixed Factors box.

4. If you want descriptive statistics for each treatment, click on the Options box,

select Descriptives, and click Continue.

5. Click OK.

SPSS Output

We used the SPSS program to analyze the data from the study in Example 13.4, and part

of the program output is shown in Figure 13.7. The output begins with a table listing the

factors (not shown in Figure 13.7), followed by a table showing descriptive statistics,

including the mean and standard deviation for each cell or treatment condition. The

results of the ANOVA are shown in the table labeled Tests of Between-Subjects

Effects. The top row (Corrected Model) presents the between-treatments SS and

df values. The second row (Intercept) is not relevant for our purposes. The next three

rows present the two main effects and the interaction (the SS, df, and MS values,

as well as the F-ratio and the level of significance), with each factor identified by its

column number from the SPSS data editor. The next row (Error) describes the error

term (denominator of the F-ratio), and the final row (Corrected Total) describes the

total variability for the entire set of scores. (Ignore the row labeled Total.)

FOCUS ON PROBLEM SOLVING

1. Before you begin a repeated-measures ANOVA, complete all of the prelimi-

nary calculations needed for the ANOVA formulas. This requires that you

find the total for each treatment (Ts), the total for each person (Ps), the grand

total (G), the SS for each treatment condition, and oX2 for the entire set of N scores. As a partial check on these calculations, be sure that the T values

add up to G and that the P values have a sum of G.

2. To help remember the structure of repeated-measures ANOVA, keep in mind

that a repeated-measures experiment eliminates the contribution of individual

differences. There are no individual differences contributing to the numera-

tor of the F-ratio (MS between treatments

) because the same individuals are used for

all treatments. Therefore, you must also eliminate individual differences in

the denominator. This is accomplished in the second stage of the analysis by

subtracting the between-subjects variance from the within subjects variance

to produce an error term for the F-ratio.

FOCUS ON PROBLEM SOLVING 4 3 3

3. Before you begin a two-factor ANOVA, take time to organize and

summarize the data. It is best if you summarize the data in a matrix with

rows corresponding to the levels of one factor and columns corresponding

to the levels of the other factor. In each cell of the matrix, show the

number of scores (n), the total and mean for the cell, and the SS within

the cell. Also compute the row totals and column totals that are needed

to calculate main effects.

4. For a two-factor ANOVA, there are three separate F-ratios. Although the

three F-ratios use the same error term in the denominator (MS within

), they have

different numerators, which may have different df values.

Descriptive Statistics

Dependent Variable: VAR00001

VAR00003VAR00002 Mean

1.00

2.00

Total

1.00

2.00

Total

1.00

2.00

Total

1.00

2.00

Total

9.0000

5.0000

7.0000

8.0000

9.0000

8.5000

7.0000

7.7500

1.58114

1.73205

2.62467

1.87083

1.73205

1.77951

1.71594

2.66667

2.31414

Std. Deviation N

Tests of Between-Subjects Effects

Dependent Variable: VAR00001

Source

Corrected Model

Intercept

VAR00002

VAR00003

VAR00002 * VAR00003

Error

Total

Corrected Total

Type III Sum of Squares

df Mean Square

F Sig.

53.750 a

1201.250

11.250

31.250

48.000

1303.000

101.750

17.917

1201.250

11.250

31.250

3.000

5.972

400.417

3.750

10.417

.006

.000

.071

.005

FIGURE 13.7

Portions of the SPSS

output for the two-factor

ANOVA for the study in

Example 13.4.

4 3 4 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

DEMONSTRATION 13.1

REPEATED-MEASURES ANOVA

The following data were obtained from a research study examining the effect of sleep

deprivation on motor-skills performance. A sample of five participants was tested on

a motor-skills task after 24 hours of sleep deprivation, tested again after 36 hours, and

tested once more after 48 hours. The dependent variable is the number of errors made

on the motor-skills task. Do these data indicate that the number of hours of sleep

deprivation has a significant effect on motor skills performance?

Participant 24 Hours 36 Hours 48 Hours P totals

A 0 0 6 6 N 5 15

B 1 3 5 9 G 5 45

C 0 1 5 6 oX2 5 245

D 4 5 9 18

E 0 1 5 6

T 5 5 T 5 10 T 5 30

SS 5 12 SS 5 16 SS 5 12

State the hypotheses, and specify alpha. The null hypothesis states that, for the gen- eral population, there are no differences among the three deprivation conditions. Any differences that exist among the samples are simply the result of chance or error. In symbols,

H 0 : �

1 5 �

2 5 �

The alternative hypothesis states that there are differences among the conditions.

H 1 : At least one of the treatment means is different.

We use a 5 .05.

Locate the critical region. Rather than compute the df values and look for a critical value for F at this time, we proceed directly to the ANOVA.

The first stage of the analysis is identical to the independent-measures ANOVA presented

in Chapter 12.

SS G

N total

2 5 2 5 2 5o X

2 2

245 45

15 110

SS within

5 ∑SS inside each treatment

5 12 1 16 1 12 5 40

SS n

N between

75 2 5 1 1 2 5o T

2 2 2 2 2 2 5

15 0

and the corresponding degrees of freedom are

df total

5 N –1 5 14

df within

5 odf 5 4 1 4 1 4 5 12

df between

5 k – 1 5 2

S T E P 1

S T E P 2

S T A g E 1

DEMONSTRATION 13.2 4 3 5

The second stage of the repeated-measures analysis measures and removes the individual

differences from the denominator of the F-ratio.

SS k

N between subjects

5 2

o P

2 2

5 1 1 1 1 2

2 2 2 2 2 2

5 36

SS error

5 SS within

5 SS between subjects

5 40 – 36

5 4

and the corresponding df values are

df between subjects

5 n – 1 5 4

df error

5 df within

– df between subjects

5 12 – 4

5 8

The mean square values that form the F-ratio are as follows:

MS SS

between between

between

error

5 5 5

2 35

SSS

error

5 5 4

8 0 50.

Finally, the F-ratio is

F MS

MS 5 5 5

between

error

0 50 70 00

. .

Make a decision and state a conclusion. With df 5 2, 8 and a 5 .05, the critical value is F 5 4.46. Our obtained F-ratio (F 5 70.00) is well into the critical region, so our decision is to reject the null hypothesis and conclude that there are significant differ- ences among the three levels of sleep deprivation.

DEMONSTRATION 13.2

TWO-FACTOR ANOVA

The following data are from a two-factor study comparing performance in three

treatment conditions for males and females. There is a separate sample of n 5 10

participants in each of the six groups. We use a two-factor ANOVA to evaluate the

significance of the mean differences.

S T A g E 2

S T E P 3

4 3 6 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

State the hypotheses, and select alpha.

For a two-factor study, there are three separate hypotheses, the two main effects and

the interaction.

For factor A, the null hypothesis states that there is no difference in performance for

males versus females. In symbols,

H 0 : �

males 5 �

females

For factor B, the null hypothesis states that there is no difference in performance

among the three treatment conditions. In symbols,

H 0 1 2 3

: � � � B B B

5 5

For the A 3 B interaction, the null hypothesis states that the effect of either factor does not depend on the levels of the other factor.

We will use a 5 .05 for all tests.

Locate the critical region.

Rather than compute the df values and look up critical values for F at this time, we

proceed directly to the ANOVA.

The first stage of the analysis is identical to the independent-measures ANOVA presented in

Chapter 12, where each cell in the data matrix is considered a separate treatment condition.

SS X G

N total

5 2

5 2 5

o 2

2312 240

60 1352

SS within treatments

5 oSS each treatment

5 195 1 275 1 220 1 237 1 240 1 225 5 972

SS T

N between treatments

5 2

2 2

5 1 1 1 1 1 10

9 2 2 2 2 2

240

380

2 2

S T E P 1

S T E P 2

S T A g E 1

Factor B (Treatment Condition)

B 1

B 2

B 3

T 5 10 T 5 50 T 5 30

Male M 5 1 M 5 5 M 5 3

Factor A SS 5 195 SS 5 235 SS 5 220

T 5 30 T 5 30 T 5 90

Female M 5 3 M 5 3 M 5 9

SS 5 237 SS 5 240 SS 5 225

N 5 60 G 5240 oX2 5 2312

DEMONSTRATION 13.2 4 3 7

The corresponding degrees of freedom are

df total

5 N – 1 5 59

df within treatments

5 odf each treatment

5 9 1 9 1 9 1 9 1 9 1 9 5 54

df between treatments

5 number of cells – 1 5 5

The second stage of the analysis partitions the between-treatments variability into

three components: the main effect for factor A, the main effect for factor B, and the

A 3 B interaction.

a. Factor A. The main effect for factor A evaluates the mean differences between the

levels of factor A. For this example, factor A defines the rows of the matrix, so we

are evaluating the mean differences between rows. To compute the SS for factor A,

we calculate a between-treatment SS using the row totals exactly the same as we

computed SS between treatments

using the treatment totals (T values) earlier. For factor A,

the row totals are 90 and 150, and each total was obtained by adding 30 scores.

For factor A (male, female)

SS T

ROW

A 5 2

5 1 2

150

2 2

240

For factor B (treatments)

SS T

COL

B 5 2

5 1 2

2 2

120

240

160

2 2

For the A 3 B interaction

SS A3B

5 SS between treatments

– SS A – SS

5 380 – 60 – 160

5 160

The corresponding degrees of freedom are

df A 5 number of rows – 1 5 1

df B 5 number of columns – 1 5 2

df A3B

5 df between treatments

– df A – df

B 5 5 – 1 – 2 5 2

S T A g E 2

4 3 8 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

The MS value needed for the F-ratios are

A B

5 5

5 53

1 60

160

2 80

160

2 80

within ttreatments 5 5

972

54 18

Finally, the three F-ratios are

F MS

A A

B B

5 5 5

within treatments

18 3 33.

wwithin treatments

wit

5 5

53 3

18 4 44.

F MS

MS A B

A B

hhin treatments

5 5 80

18 4 44.

Make a decision and state a conclusion.

The F-ratio for factor A has df 5 1, 54. With a 5 .05, the critical F value is 4.03 (using df 5 1, 50 from the table). For these data, factor A has no significant effect. Statistically, there is no difference in performance between males and females.

Factor B and the interaction both have df 5 2, 54. With a 5 .05, the critical F value is 3.18 (using df 5 1, 50 from the table). For these data, there are significant differences among the levels of factor B and there is a significant interaction. The in-

teraction indicates that the differences among treatments are not the same for males

as they are for females.

S T E P 3

PROBLEMS

1. How does the denominator of the F-ratio (the error term) differ for a repeated-measures ANOVA com- pared to an independent-measures ANOVA?

2. The repeated-measures ANOVA can be viewed as a two-stage process. What is the purpose of the second stage?

3. A researcher conducts an experiment comparing three treatment conditions with n 5 10 scores in each condition.

a. If the researcher uses an independent-measures design, how many individuals are needed for the study and what are the df values for the F-ratio?

b. If the researcher uses a repeated-measures design, how many individuals are needed for the study and what are the df values for the F-ratio?

4. A researcher conducts a repeated-measures experi- ment using a sample of n 5 8 subjects to evaluate the differences among four treatment conditions. If

the results are examined with an ANOVA, what are the df values for the F-ratio?

5. A researcher uses a repeated-measures ANOVA to evaluate the results from a research study and reports an F-ratio with df 5 2, 30.

a. How many treatment conditions were compared in the study?

b. How many individuals participated in the study?

6. A published report of a repeated-measures research study includes the following description of the statisti- cal analysis. “The results show significant differences among the treatment conditions, F(3, 21) 5 6.10, p , .01.”

a. How many treatment conditions were compared in the study?

b. How many individuals participated in the study?

7. The following data were obtained from a repeated- measures study comparing two treatment conditions.

a. Use a repeated-measures t test with a 5 .05 to determine whether there are significant mean differences between the two treatments. Note that

PROBLEMS 4 3 9

you will need to find the difference score (D) for each person.

b. Use a repeated-measures ANOVA with a 5 .05 to determine whether there are significant mean differences between the two treatments. Within rounding error, you should find that F 5 t2.

Treatments

Person I II Person Totals

A 3 5 P 5 8

B 5 9 P 5 14 N 5 16

C 1 5 P 5 6 G 5 80

D 1 7 P 5 8 oX2 5 500

E 5 9 P 5 14

F 3 7 P 5 10

G 2 6 P 5 8

H 4 8 P 5 12

M 5 3 M 5 7

T 5 24 T 5 56

SS 5 18 SS 5 18

8. A recent study examined how applicants with a facial blemish such as a scar or birthmark fared in job inter- views (Madera & Hebl, 2011). The results indicate that interviewers recalled less information and gave lower ratings to applicants with a blemish. In a similar study, participants conducted computer-simulated interviews with a series of applicants including one with a facial scar and one with a facial birthmark. The following data represent the ratings given to each applicant.

a. Use a repeated-measures ANOVA with a 5 .05 to determine whether there are significant mean differences among the three conditions.

b. Compute h2, the percentage of variance accounted for by the mean differences, to measure the size of the treatment effects.

c. Write a sentence demonstrating how a research report would present the results of the hypothesis test and the measure of effect size.

Applicant

Participant Scar Birthmark No

Blemish Person Total

A 1 1 4 P 5 6

B 3 4 8 P 5 15 N 515

C 0 2 7 P 5 9 G 5 45

D 0 0 6 P 5 6 oX2 5 231

E 1 3 5 P 5 9

M 5 1 M 5 2 M 5 6

T 5 5 T 5 10 T 5 30

SS 5 6 SS 5 10 SS 5 10

9. One of the primary advantages of a repeated-measures design, compared to an independent-measures design, is that it reduces the overall variability by removing variance caused by individual differences. The follow- ing data are from a research study comparing three treatment conditions.

a. Assume that the data are from an independent- measures study using three separate samples, each with n 5 6 participants. Ignore the column of P totals and use an independent-measures ANOVA with a 5 .05 to test the significance of the mean differences.

b. Now assume that the data are from a repeated- measures study using the same sample of n 5 6 participants in all three treatment conditions. Use a repeated-measures ANOVA with a 5 .05 to test the significance of the mean differences.

c. Explain why the two analyses lead to different conclusions.

Treatment 1 Treatment 2 Treatment 3 P

6 9 12 27

8 8 8 24 N 5 18

5 7 9 21 G 5 108

0 4 8 12 oX2 5 800

2 3 4 9

3 5 7 15

M 5 4 M 5 6 M 5 8

T 5 24 T 5 36 T 5 48

SS 5 42 SS 5 28 SS 5 34

10. The following data are from an experiment compar- ing three different treatment conditions:

A B C

0 1 2 N 5 15

2 5 5 oX2 5 354

1 2 6

5 4 9

2 8 8

T 5 10 T 5 20 T 5 30

SS 5 14 SS 5 30 SS 5 30

a. If the experiment uses an independent-measures design, can the researcher conclude that the treat- ments are significantly different? Test at the .05 level of significance.

b. If the experiment is done with a repeated-measures design, should the researcher conclude that the treatments are significantly different? Set alpha at .05 again.

c. Explain why the analyses in parts a and b lead to different conclusions.

4 4 0 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

11. A researcher is evaluating customer satisfaction with the service and coverage of two phone carri- ers. Each individual in a sample of n 5 25 uses one carrier for two weeks and then switches to the other. Each participant then rates the two carriers. The following table presents the results from the repeated-measures ANOVA comparing the aver- age ratings. Fill in the missing values in the table. (Hint: Start with the df values.)

Source SS df MS

Between treatments _____ _____ 2 F 5 _____

Within treatments _____ _____

Between subjects _____ _____

Error 12 _____ _____

Total 23 _____

12. The following summary table presents the results from a repeated-measures ANOVA comparing three treatment conditions with a sample of n 5 11 sub- jects. Fill in the missing values in the table. (Hint: Start with the df values.)

Source SS df MS

Between treatments _____ _____ _____ F 5 5.00

Within treatments 80 _____

Between subjects _____ _____

Error 60 _____ _____

Total _____ _____

13. A recent study indicates that simply giving college students a pedometer can result in increased walking (Jackson & Howton, 2008). Students were given pe- dometers for a 12-week period, and asked to record the average number of steps per day during weeks 1, 6, and 12. The following data are similar to the re- sults obtained in the study.

Number of steps (31000)

Week

Participant 1 6 12 P

A 6 8 10 24

B 4 5 6 15

C 5 5 5 15 G 5 72

D 1 2 3 6 oX2 5 400

E 0 1 2 3

F 2 3 4 9

T 5 18 T 5 24 T 5 30

SS 5 28 SS 5 32 SS 5 40

a. Use a repeated-measures ANOVA with a 5 .05 to determine whether the mean number of steps changes significantly from one week to another.

b. Compute h2 to measure the size of the treatment effect.

c. Write a sentence demonstrating how a research report would present the results of the hypothesis test and the measure of effect size.

14. The following data represent the typical results from a delayed discounting study. The participants are asked how much they would take today instead of waiting for a specific delay period to receive $1000. Each participant responds to all 5 of the delay peri- ods. Use a repeated-measures ANOVA with a 5 .01 to determine whether there are significant differences among the 5 delay periods for the following data:

Participant 1

month 6

months 1

year 2

years 5

years

A 950 850 800 700 550

B 800 800 750 700 600

C 850 750 650 600 500

D 750 700 700 650 550

E 950 900 850 800 650

F 900 900 850 750 650

15. The endorphins released by the brain act as natural painkillers. For example, Gintzler (1980) monitored endorphin activity and pain thresholds in pregnant rats during the days before they gave birth. The data showed an increase in pain threshold as the preg- nancy progressed. The change was gradual until 1 or 2 days before birth, at which point there was an abrupt increase in pain threshold. Apparently a natural painkilling mechanism was preparing the animals for the stress of giving birth. The following data represent pain-threshold scores similar to the results obtained by Gintzler. Do these data indicate a significant change in pain threshold? Use a repeated- measures ANOVA with a 5.01.

Days Before Giving Birth

Subject 7 5 3 1

A 39 40 49 52

B 38 39 44 55

C 44 46 50 60

D 40 42 46 56

E 34 33 41 52

16. The structure of a two-factor study can be presented as a matrix with the levels of one factor determining the rows and the levels of the second factor deter- mining the columns. With this structure in mind,

PROBLEMS 4 4 1

describe the mean differences that are evaluated by each of the three hypothesis tests that make up a two-factor ANOVA.

17. Briefly explain what happens during the second stage of the two-factor ANOVA.

18. The following matrix presents the results from an independent-measures, two-factor study with a sam- ple of n 5 10 participants in each treatment condi- tion. Note that one treatment mean is missing.

Factor B

B 1

B 2

Factor A

A 1 M 5 20 M 5 30

A 2 M 5 40

a. What value for the missing mean would result in no main effect for factor A?

b. What value for the missing mean would result in no main effect for factor B?

c. What value for the missing mean would result in no interaction?

19. The following matrix presents the results of a two- factor study with n 5 10 scores in each of the six treatment conditions. Note that one of the treatment means is missing.

Factor B

B 1

B 2

B 3

Factor A

A 1 M 5 10 M 5 20 M 5 40

A 2 M 5 20 M 5 30

a. What value for the missing mean would result in no main effect for factor A?

b. What value for the missing mean would result in no interaction?

20. A researcher conducts an independent-measures, two-factor study using a separate sample of n 5 15 participants in each treatment condition. The results are evaluated using an ANOVA and the researcher reports an F-ratio with df 5 1, 84 for factor A, and an F-ratio with df 5 2, 84 for factor B.

a. How many levels of factor A were used in the study?

b. How many levels of factor B were used in the study?

c. What are the df values for the F-ratio evaluating the interaction?

21. A researcher conducts an independent-measures, two-factor study with two levels of factor A and three levels of factor B, using a sample of n 5 12 partici- pants in each treatment condition.

a. What are the df values for the F-ratio evaluating the main effect of factor A?

b. What are the df values for the F-ratio evaluating the main effect of factor B?

c. What are the df values for the F-ratio evaluating the interaction?

22. Some people like to pour beer gently down the side of the glass to preserve bubbles. Others, splash it down the center to release the bubbles into a foamy head and free the aromas. Champagne, however is best when the bubbles remain concentrated in the wine. According to an article in the Journal of Agricultural and Food Chemistry, a group of French scientists recently verified the difference between the two pouring methods by measuring the amount of bubbles in each glass of champagne poured two different ways and at three different temperatures (Journal of Agricultural and Food Chemistry, 2010). The following data present the pattern of results obtained in the study.

Champagne Temperature (°F)

40° 46° 52°

n 5 10 n 5 10 n 5 10

Gentle Pour M 5 7 M 5 3 M 5 2

SS 5 64 SS 5 57 SS 5 47

n 5 10 n 5 10 n 5 10

Splashing Pour M 5 5 M 5 1 M 5 0

SS 5 56 SS 5 54 SS 5 46

a. Use a two-factor ANOVA with a 5 .05 to evalu- ate the mean differences.

b. Briefly explain how temperature and pouring influence the bubbles in champagne according to this pattern of results.

23. Example 13.1 in this chapter described a two- factor study examining performance under two audience conditions (factor B) for high and low self-esteem participants (factor A). The following summary table presents possible results from the analysis of that study. Assuming that the study used a separate sample of n 5 15 participants in each treatment condition (each cell), fill in the missing values in the table. (Hint: Start with the df values.)

4 4 2 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

Source SS df MS

Between treatments 67 _____

Audience _____ _____ _____ F 5 ____

Self-esteem 29 _____ _____ F 5

Interaction _____ _____ _____ F 5 5.50

Within treatments _____ _____ 4

Total _____ _____

24. The following table summarizes the results from a two-factor study with 2 levels of factor A and 3 levels of factor B using a separate sample of n 5 11 partici- pants in each treatment condition. Fill in the missing values. (Hint: Start with the df values.)

Source SS df MS

Between treatments ___ ___

Factor A ___ ___ ___ F 5 7

Factor B ___ ___ ___ F 5 8

A 3 B Interaction ___ ___ ___ F 5 3

Within treatments 240 ___ ___

Total ___ ___

25. The following data are from a two-factor study ex- amining the effects of two treatment conditions on males and females.

a. Use an ANOVA with a 5 .05 for all tests to evaluate the significance of the main effects and the interaction.

b. Compute h2 to measure the size of the effect for each main effect and the interaction.

Factor B: Treatment

B 1 B

2 B

Male 3 1 10

1 4 10

1 8 14

6 6 7 T ROW1

5 90

4 6 9

M 5 3 M 5 5 M 5 10

T 5 15 T 5 25 T 5 50

SS 5 18 SS 5 28 SS 5 26 N 5 30

Factor A: Gender

0 2 1 G 5 120

2 7 1 oX2 5 860

0 2 1

0 2 6 T ROW2

5 30

Female 3 2 1

M 5 1 M 5 3 M 5 2

T 5 5 T 5 15 T 5 10

SS 5 8 SS 5 20 SS 5 20

T COL1

5 20 T COL2

5 40 T COL3

5 60

26. Research indicates that paying students to improve their grades simply does not work (Fryer, 2011). However, paying students for specific tasks such as reading books, attending class, or doing homework does have a significant effect. Apparently, students on their own do not understand how to get good grades. If they are told exactly what to do, however, the incentives work. The following data represent a two-factor study attempting to replicate this result.

Paid for Homework

Not Paid for Homework

Paid for Grades

14 2

7 7

10 5

9 7

11 3

9 6

Not Paid for Grades

13 7

7 2

9 4

7 2

11 6

7 3

a. Use a two-factor ANOVA with a 5 .05 to evalu- ate the significance of the main effects and the interaction.

b. Calculate the h2 values to measure the effect size for the two main effects and the interaction.

c. Describe the pattern of results. (How does pay- ing for grades influence performance? How does paying for homework influence performance? Does the effect of paying for homework depend on whether you also pay for grades?)

27. In Chapter 12 (page 390), we described a study reporting that college students who are on Facebook (or have it running in the background) while study- ing had lower grades than students who did not use the social network (Kirschner & Karpinski, 2010). A researcher would like to know if the same re- sult extends to students in lower grade levels. The researcher planned a two-factor study comparing Facebook users with non-users for middle school students, high school students, and college students. For consistency across groups, grades were con- verted into six categories, numbered 0 to 5 from low to high. The results are presented in the following matrix.

a. Use a two-factor ANOVA with a 5 .05 to evalu- ate the mean differences.

b. Describe the pattern of results.

PROBLEMS 4 4 3

Middle School High School College

Non-user

3 5 5

5 5 4

5 2 2

3 4 5

User

5 1 1

3 2 0

2 3 0

2 2 3

28. In Chapter 11, we described a research study in which the color red appeared to increase men’s at- traction to women (Elliot & Niesta, 2008). The same researchers have published other results showing that red also increases women’s attraction to men but does not appear to affect judgments of same-sex indi- viduals (Elliot et al., 2010). Combining these results into one study produces a two-factor design in which men judge photographs of both women and men, which are shown on both red and white backgrounds.

The dependent variable is a rating of attractiveness for the person shown in the photograph. The study uses a separate group of participants for each condi- tion. The following table presents data similar to the results from previous research.

Person Shown in Photograph

Female Male

Background Color for Photograph

n 5 10 n 5 10

White M 5 4.5 M 5 4.4

SS 5 6 SS 5 7

n 5 10 n 5 10

Red M 5 7.5 M 5 4.6

SS 5 9 SS 5 8

a. Use a two-factor ANOVA with a 5 .05 to evaluate the main effects and the interaction.

b. Describe the effect of background color on judg- ments of males and females.

REVIEW

P A R T IV

After completing this part, you should be able to perform an ANOVA to evaluate the significance

of mean differences in three research situations. These include:

4 4 5

1. The single-factor independent-measures design intro- duced in Chapter 12.

2. The single-factor repeated-measures design introduced in Chapter 13.

3. The two-factor independent-measures design intro- duced in Chapter 13.

In this part, we introduce three applications of ANOVA

that use an F-ratio statistic to evaluate the mean differences

among two or more populations. In each case, the F-ratio

has the following structure:

F 5 variance between treatments

variance from rrandom unsystematic sources

The numerator of the F-ratio measures the mean differences

that exist from one treatment condition to another, including

any systematic differences caused by the treatments. The

denominator measures the differences that exist when there

are no systematic factors that cause one score to be different

from another. The F-ratio is structured so that the numerator

and denominator are measuring exactly the same variance

when the null hypothesis is true and there are no system-

atic treatment effects. In this case, the F-ratio should have

a value near 1.00. Thus, an F-ratio near 1.00 is evidence

that the null hypothesis is true. Similarly, an F-ratio that is

much larger than 1.00 provides evidence that a systematic

treatment effect does exist and the null hypothesis should

be rejected.

For independent-measures designs, either single-factor

or two-factor, the denominator of the F-ratio is obtained

by computing the variance within treatments. Inside each

treatment condition, all participants are treated exactly the

same so there are no systematic treatment effects that cause

the scores to vary.

For a repeated-measures design, the same individuals

are used in every treatment condition, so any differences

between treatments cannot be caused by individual differ-

ences. Thus, the numerator of the F-ratio does not include

any individual differences. Therefore, individual differences

must also be eliminated from the denominator to balance

the F-ratio. As a result, the repeated-measures ANOVA is

a two-stage process. The first stage separates the between-

treatments variance (numerator) and the within-treatments

variance. The second stage removes the systematic indi-

vidual differences from the within-treatments variance to

produce the appropriate denominator for the F-ratio.

For a two-factor design, the mean differences between

treatments can be caused by either of the two factors or by

specific combinations of factors. The goal of the ANOVA

is to separate these possible treatment effects so that each

can be evaluated independent of the others. To accomplish

this, the two-factor ANOVA is a two-stage process. The

first stage separates the between-treatments variance and

the within-treatments variance (denominator). The second

stage analyzes the between-treatments variance into three

components: the main effect from the first factor, the main

effect from the second factor, and the interaction.

Note that the repeated-measures ANOVA and the two-

factor ANOVA are both two-stage processes. Both begin

by separating the between-treatments variance and the

within-treatments variance. However, the second stages

of these two ANOVAs serve different purposes and focus

on different components. The repeated-measures ANOVA

focuses on the within-treatments variance and is intended to

remove the individual differences. The two-factor ANOVA

focuses on the between-treatments variance and is intended

to separate the main effects and the interaction.

REVIEW EXERCISES

1. A researcher examining the jet lag that people experi- ence when flying long distances obtained the following data measuring the number of days required to adjust after a long flight. Do the data indicate significant dif- ferences in jet lag depending on the direction of travel? Test with a 5 .05.

Westbound Eastbound Same Time Zone

2 6 1

1 4 0

3 6 1

3 8 1

2 5 0

4 7 0

M 5 2.5 M 5 6 M 5 0.5

SS 5 5.5 SS 5 10 SS 5 1.5

2. The following data were obtained from a repeated- measures study comparing three treatment conditions.

4 4 6 CHAPTER 13 REPEATED-MEASURES AND TWO-FACTOR ANALYSIS OF VARIANCE

Treatments

Person I II III Person Totals

A 0 4 2 P 5 6

B 1 5 6 P 5 12 N 5 18

C 3 3 3 P 5 9 G 5 48

D 0 1 5 P 5 6 oX2 5 184

E 0 2 4 P 5 6

F 2 3 4 P 5 9

M 5 1 M 5 3 M 5 4

T 5 6 T 5 18 T 5 24

SS 5 8 SS 5 10 SS 5 10

a. Do the data indicate significant differences among the three treatments? Test at the .05 level of significance.

b. Calculate h2 to measure the size of the effect. c. Write a sentence demonstrating how the outcome

of the hypothesis test and the measure of effect size would appear in a research report.

3. Briefly describe what is meant by an interaction be- tween factors in a two-factor research study.

4. Most sports injuries are immediate and obvious, like a broken leg. However, some can be more subtle, like the neurological damage that may occur when soccer

players repeatedly head a soccer ball. To examine long- term effects of repeated heading, Downs and Abwender (2002) examined two different age groups of soccer players and swimmers. The dependent variable was performance on a conceptual thinking task. Following are hypothetical data, similar to the research results.

a. Use a two-factor ANOVA with a 5 .05 to evaluate the main effects and interaction.

b. Calculate the effects size (h2) for the main effects and the interaction.

c. Briefly describe the outcome of the study.

Factor B: Age

College Older

Factor A: Sport

Soccer

n 5 20 n 5 20

M 5 9 M 5 4

T 5 180 T 5 80

SS 5 380 SS 5 390

Swimming

n 5 20 n 5 20

M 5 9 M 5 8

T 5 180 T 5 160

SS 5 350 SS 5 400

oX2 5 6360

447

Chapter 14 Correlation 449

Chapter 15 The Chi-Square Statistic: Tests for Goodness of Fit and Independence 509

B ack in Chapter 1, we stated that the primary goal of science is

to establish relationships between variables. Until this point, the

statistics we have presented all attempt to accomplish this goal

by comparing groups of scores using means and variances as the basic

statistical measures. Typically, one variable is used to define the groups,

and a second variable is measured to obtain a set of scores within each

group. Means and variances are then computed for the scores, and the

sample means are used to test hypotheses about population means. If

the hypothesis test indicates a significant mean difference, then we

conclude that there is a relationship between the variables.

However, many research situations do not involve comparing

groups, and many do not produce data that allow you to calculate

means and variances. For example, a researcher can investigate the

relationship between two variables (for example, IQ and creativity)

by measuring both variables within a single group of individuals.

Also, the measurement procedure may not produce numerical scores.

For example, participants can indicate their color preferences by sim-

ply picking a favorite color or by ranking several choices. Without

numerical scores, it is impossible to calculate means and variances.

Instead, the data consist of proportions or frequencies. For example,

a research study may investigate what proportion of people select

red as their favorite color and whether this proportion is different for

introverted people compared with extroverted people.

Notice that these new research situations are still asking questions

about the relationships between variables, and they are still using

sample data to make inferences about populations. However, they are

no longer comparing groups and they are no longer based on means

and variances. In this part, we introduce the statistical methods that

have been developed for these other kinds of research.

Correlations and Nonparametric Tests

P A R T

Aplia for Essentials of Statistics for the Behavioral Sciences

After reading, go to “Resources” at the end of this chapter for

an introduction on how to use Aplia’s homework and learning

resources.

Correlation

14.1 Introduction

14.2 The Pearson Correlation

14.3 Using and Interpreting the Pearson Correlation

14.4 Hypothesis Tests with the Pearson Correlation

14.5 Alternatives to the Pearson Correlation

14.6 Introduction to Linear Equations and Regression

Summary

Focus on Problem Solving

Demonstration 14.1

Problems

C H A P T E R

14 Tools You Will Need The following items are considered essential background material for this chapter. If you doubt your knowledge of any of these items, you should review the appropriate chapter or section before proceeding.

• Sum of squares (SS) (Chapter 4) • Computational formula • Deinitional formula

• z-scores (Chapter 5) • Hypothesis testing (Chapter 8) • Analysis of Variance (Chapter 12)

• MS values and F-ratios

4 5 0 CHAPTER 14 CORRELATION

INTRODUCTION

Correlation is a statistical technique that is used to measure and describe the relation-

ship between two variables. Usually the two variables are simply observed as they exist

naturally in the environment—there is no attempt to control or manipulate the variables.

For example, a researcher could check high school records (with permission) to obtain a

measure of each student’s academic performance, and then survey each family to obtain

a measure of income. The resulting data could be used to determine whether there is a

relationship between high school grades and family income. Notice that the researcher

is not manipulating any student’s grade or any family’s income, but is simply observing

what occurs naturally. You also should notice that a correlation requires two scores for

each individual (one score from each of the two variables). These scores normally are

identified as X and Y. The pairs of scores can be listed in a table, or they can be pre-

sented graphically in a scatter plot (Figure 14.1). In the scatter plot, the values for the X

variable are listed on the horizontal axis and the Y values are listed on the vertical axis.

Each individual is then represented by a single point in the graph so that the horizontal

position corresponds to the individual’s X value and the vertical position corresponds to

the Y value. The value of a scatter plot is that it allows you to see any patterns or trends

that exist in the data. The scores in Figure 14.1, for example, show a clear relationship

between family income and student grades; as income increases, grades also increase.

A correlation is a numerical value that describes and measures three characteristics of

the relationship between X and Y. These three characteristics are as follows:

1. The Direction of the Relationship. The sign of the correlation, positive or

negative, describes the direction of the relationship.

14.1

T H E C H A R AC T E R I ST I C S O F A R E L AT I O N S H I P

S tu

d e

n t’

s a

v e

ra g

e g

ra d

Family income (in $1000)

Person

A B C D E F G H I J K L M N

31 38 42 44 49 56 58 65 70 90 92

106 135 174

Student’s Average Grade

72 86 81 78 85 80 91 89 94 83 90 97 89 95

Family Income

(in $1000)

100

30 55 70 90 110 130 150 170 190

FIGURE 14.1

Correlational data showing the relationship between family income (X) and student grades (Y) for a sample of n 5 14 high school students. The scores are listed in order from lowest to highest family income and are shown in a scatter plot.

SECTION 14.1 / INTRODUCTION 4 5 1

In a positive correlation, the two variables tend to change in the same direction: As

the value of the X variable increases from one individual to another, the Y variable

also tends to increase; when the X variable decreases, the Y variable also decreases.

In a negative correlation, the two variables tend to go in opposite directions. As

the X variable increases, the Y variable decreases. That is, it is an inverse relationship.

The following examples illustrate positive and negative relationships.

Suppose you run the drink concession at the football stadium. After several seasons,

you begin to notice a relationship between the temperature at game time and the

beverages you sell. Specifically, you have noted that when the temperature is low,

you sell relatively little beer. However, as the temperature goes up, beer sales also

go up (Figure 14.2). This is an example of a positive correlation. You also have

noted a relationship between temperature and coffee sales: On cold days, you sell

a lot of coffee, but coffee sales go down as the temperature goes up. This is an ex-

ample of a negative relationship.

2. The Form of the Relationship. In the preceding coffee and beer examples, the

relationships tend to have a linear form; that is, the points in the scatter plot tend

to cluster around a straight line. We have drawn a line through the middle of the

data points in each figure to help show the relationship. The most common use

of correlation is to measure straight-line relationships. However, other forms of

D E F I N I T I O N S

E X A M P L E 1 4 . 1

20 30 40

A m

o u

n t

o f

b e

e r

so ld

50 60 8070

Temperature (in degrees F)

(a) (b)

Relationship between beer sales and temperature

20 30 40

A m

o u

n t

o f

c o

ff e

e s

o ld

50 60 8070

Temperature (in degrees F)

Relationship between coffee sales and temperature

FIGURE 14.2

Examples of positive and negative relationships. (a) Beer sales are positively related to temperature. (b) Coffee sales are

negatively related to temperature.

4 5 2 CHAPTER 14 CORRELATION

relationships do exist and there are special correlations used to measure them.

(We examine alternatives in Section 14.5.)

3. The Strength or Consistency of the Relationship. Finally, the correlation

measures the consistency of the relationship. For a linear relationship, for

example, the data points could fit perfectly on a straight line. Every time

X increases by one point, the value of Y also changes by a consistent and

predictable amount. Figure 14.3(a) shows an example of a perfect linear

relationship. However, relationships are usually not perfect. Although there

may be a tendency for the value of Y to increase whenever X increases, the

amount that Y changes is not always the same, and occasionally, Y decreases

when X increases. In this situation, the data points do not fall perfectly on

a straight line. The consistency of the relationship is measured by the nu-

merical value of the correlation. A perfect correlation always is identified

by a correlation of 1.00 and indicates a perfectly consistent relationship.

For a correlation of 1.00 (or –1.00), each change in X is accompanied by

a perfectly predictable change in Y. At the other extreme, a correlation of

0 indicates no consistency at all. For a correlation of 0, the data points are

scattered randomly with no clear trend [see Figure 14.3(b)]. Intermediate

values between 0 and 1 indicate the degree of consistency.

Examples of different values for linear correlations are shown in Figure 14.3. In

each example we have sketched a line around the data points. This line, called an

envelope because it encloses the data, often helps you to see the overall trend in the

data. As a rule of thumb, when the envelope is shaped roughly like a football, the

correlation is around 0.7. Envelopes that are fatter than a football indicate correla-

tions closer to 0, and narrower shapes indicate correlations closer to 1.00.

You should also note that the sign (� or 2) and the strength of a correlation

are independent. For example, a correlation of 1.00 indicates a perfectly consis-

tent relationship whether it is positive (�1.00) or negative (21.00). Similarly,

correlations of �0.80 and 20.80 are equally consistent relationships. Finally, you

should notice that a correlation can never be greater than �1.00 or less than 21.00.

X (c)

X (a)

X (d)

X (b)

FIGURE 14.3

Examples of different

values for linear correla-

tions: (a) a perfect nega-

tive correlation, –1.00;

(b) no linear trend, 0.00;

tionship, approximately

�0.90; (d) a relatively

weak negative correlation,

approximately –0.40.

SECTION 14.2 / THE PEARSON CORRELATION 4 5 3

THE PEARSON CORRELATION

By far the most common correlation is the Pearson correlation (or the Pearson product–

moment correlation) which measures the degree of straight-line relationship.

The Pearson correlation measures the degree and the direction of the linear

relationship between two variables.

The Pearson correlation is identified by the letter r. Conceptually, this correlation is

computed by

r X Y

5 degree to which and vary together

degreee to which and vary separately

covari

X Y

5 aability of and

variability of and

X Y

X Y sseparately

When there is a perfect linear relationship, every change in the X variable is accom-

panied by a corresponding change in the Y variable. In Figure 14.3(a), for example,

every time the value of X increases, there is a perfectly predictable decrease in the value

of Y. The result is a perfect linear relationship, with X and Y always varying together.

In this case, the covariability (X and Y together) is identical to the variability of X and

Y separately, and the formula produces a correlation with a magnitude of 1.00 or –1.00.

At the other extreme, when there is no linear relationship, a change in the X variable

14.2

D E F I N I T I O N

1. For each of the following, indicate whether you would expect a positive or a nega-

tive correlation.

a. Model year and price for a used Honda

b. IQ and grade point average for high school students

c. Daily high temperature and daily energy consumption for 30 winter days in

New York City

2. The data points would be clustered more closely around a straight line for a

correlation of –0.80 than for a correlation of �0.05. (True or false?)

3. If the data points are clustered close to a line that slopes up from left to right, then

a good estimate of the correlation would be �0.90. (True or false?)

4. If a scatter plot shows a set of data points that form a circular pattern, the correla-

tion should be near zero. (True or false?)

1. a. Positive: Higher model years tend to have higher prices.

b. Positive: More intelligent students tend to get higher grades.

c. Negative: Higher temperature tends to decrease the need for heating.

2. True. The numerical value indicates the strength of the relationship. The sign only indicates

direction.

3. True.

4. True.

L E A R N I N G C H E C K

ANSWERS

4 5 4 CHAPTER 14 CORRELATION

does not correspond to any predictable change in the Y variable. In this case, there is no

covariability, and the resulting correlation is zero.

To calculate the Pearson correlation, it is necessary to introduce one new concept: the

sum of products of deviations, or SP. This new value is similar to SS (the sum of squared

deviations), which is used to measure variability for a single variable. Now, we use SP

to measure the amount of covariability between two variables. The value for SP can be

calculated with either a definitional formula or a computational formula.

The definitional formula for the sum of products is

SP 5 o(X 2 M X )(Y 2 M

Y ) (14.1)

where M X is the mean for the X scores and M

Y is the mean for the Ys.

The definitional formula instructs you to perform the following sequence of opera-

tions:

1. Find the X deviation and the Y deviation for each individual.

2. Find the product of the deviations for each individual.

3. Add the products.

Notice that this process “defines” the value being calculated: the sum of the products

of the deviations.

The computational formula for the sum of products of deviations is

SP XY X Y

n 5 2o

o o

(14.2)

Because the computational formula uses the original scores (X and Y values), it

usually results in easier calculations than those required with the definitional formula,

especially if M X or M

Y is not a whole number. However, both formulas always produce

the same value for SP.

You may have noted that the formulas for SP are similar to the formulas you have learned

for SS (sum of squares). The relationship between the two sets of formulas is described in

Box 14.1. The following example demonstrates the calculation of SP with both formulas.

The same set of n 5 4 pairs of scores is used to calculate SP, first using the definitional

formula and then using the computational formula.

For the definitional formula, you need deviation scores for each of the X values and

each of the Y values. Note that the mean for the Xs is M X 5 3 and the mean for the Ys is

M Y 5 5. The deviations and the products of deviations are shown in the following table:

Scores Deviations Products

X Y X 2 M X

Y 2 M Y

(X 2 M X )(Y 2 M

Y )

1 3 22 22 �4

2 6 21 �1 21

4 4 �1 21 21

5 7 �2 �2 �4

�6 5 SP

T H E S U M O F P R O D UC T S O F

D E V I AT I O N S

E X A M P L E 1 4 . 2

Caution: The n in this

formula refers to the number

of pairs of scores.

SECTION 14.2 / THE PEARSON CORRELATION 4 5 5

For these scores, the sum of the products of the deviations is SP 5 �6. For the computational formula, you need the X value, the Y value, and the XY product

for each individual. Then you find the sum of the Xs, the sum of the Ys, and the sum of

the XY products. These values are as follows:

X Y XY

1 3 3

2 6 12

4 4 16

5 7 35

12 20 66 Totals

Substituting the totals in the formula gives

SP XY X Y

n 5 2o

o o

5 266 12 20

( )

566–60

Both formulas produce the same result, SP 5 6.

Caution: The signs (� and 2)

are critical in determining the

sum of products, SP.

BOX

14.1 COMPARING THE SP AND SS FORMULAS

It will help you to learn the formulas for SP if you

note the similarity between the two SP formulas and

the corresponding formulas for SS that were presented

in Chapter 4. The definitional formula for SS is

SS 5 o(X – M)2

In this formula, you must square each deviation,

which is equivalent to multiplying it by itself. With

this in mind, the formula can be rewritten as

SS 5 o(X – M)(X – M)

The similarity between the SS formula and the

SP formula should be obvious—the SS formula uses

squares and the SP formula uses products. This same

relationship exists for the computational formulas.

For SS, the computational formula is

SS X X

n 5 2o

o2 2

( )

As before, each squared value can be rewritten so

that the formula becomes

SP XX X X

n 5 2o

o o

Again, note the similarity in structure between the SS

formula and the SP formula. If you remember that SS

uses squares and SP uses products, the two new formu-

las for the sum of products should be easy to learn.

4 5 6 CHAPTER 14 CORRELATION

As noted earlier, the Pearson correlation consists of a ratio comparing the covariability

of X and Y (the numerator) with the variability of X and Y separately (the denominator).

In the formula for the Pearson r, we use SP to measure the covariability of X and Y. The

variability of X is measured by computing SS for the X scores and the variability of Y

is measured by SS for the Y scores. With these definitions, the formula for the Pearson

correlation becomes

r SP

SS SS X Y

(14.3)

The following example demonstrates the use of this formula with a simple set of scores.

The Pearson correlation is computed for the following set of n 5 5 pairs of scores.

X Y

0 2

10 6

4 2

8 4

8 6

Before starting any calculations, it is useful to put the data in a scatter plot and make

a preliminary estimate of the correlation. These data have been graphed in Figure 14.4.

Looking at the scatter plot, it appears that there is a very good (but not perfect) positive

correlation. You should expect an approximate value of r 5 �0.8 or �0.9. To find the Pearson correlation, we need SP, SS for X, and SS for Y. The calculations for each of

these values, using the definitional formulas, are presented in Table 14.1. (Note that the

mean for the X values is M X 5 6 and the mean for the Y scores is M

Y 5 4.)

Using the values from Table 14.1, the Pearson correlation is

r SP

SS SS X Y

= = +5 ( ) ( )

= ( ) ( )

28 28

32 0.8

64 16 775

CA L C U L AT I O N O F T H E P E A R S O N CO R R E L AT I O N

E X A M P L E 1 4 . 3

Note that you multiply SS for

X by SS for Y in the denomina-

tor of the Pearson formula.

0 1 2 3 4 5 6 7 8 9 10 X

FIGURE 14.4

Scatter plot of the data

from Example 14.3.

SECTION 14.2 / THE PEARSON CORRELATION 4 5 7

Correlation and the pattern of data points Note that the value we obtained for the

correlation in Example 14.3 is perfectly consistent with the pattern formed by the data

points in Figure 14.4. The positive sign for the correlation indicates that the points are

clustered around a line that slopes up to the right. Second, the high value for the cor-

relation (near 1.00) indicates that the points are very tightly clustered close to the line.

Thus, the value of the correlation describes the relationship that exists in the data.

Because the Pearson correlation describes the pattern formed by the data points, any

factor that does not change the pattern also does not change the correlation. For example,

if 5 points were added to each of the X values in Figure 14.4, then each data point would

move to the right. However, because all of the data points shift to the right, the overall

pattern is not changed, it is simply moved to a new location. Similarly, if 5 points were

subtracted from each X value, the pattern would shift to the left. In either case, the overall

pattern stays the same and the correlation is not changed. In the same way, adding a con-

stant to (or subtracting a constant from) each Y value simply shifts the pattern up (or down)

but does not change the pattern and, therefore, does not change the correlation. Multiplying

each X and/or Y value by a constant also does not change the pattern formed by the data

points. For example, if each of the X values in Figure 14.4 were multiplied by 2, then the

same scatter plot could be used to display either the original scores or the new scores. The

current figure shows the original scores, but if the values on the X-axis (0, 1, 2, 3, and so

on) are doubled (0, 2, 4, 6, and so on), then the same figure would show the pattern formed

by the new scores. In summary, adding a constant to (or subtracting a constant from) each

X and/or Y value does not change the pattern of data points and does not change the correla-

tion. Also, multiplying (or dividing) each X and/or Y value by a constant does not change

the pattern and does not change the value of the correlation.

The Pearson correlation measures the relationship between an individual’s location in

the X distribution and his or her location in the Y distribution. For example, a positive

correlation means that individuals who have a high X score also tend to have a high

Y score. Similarly, a negative correlation indicates that individuals with high X scores

tend to have low Y scores.

Recall from Chapter 5 that z-scores identify the exact location of each individual score

within a distribution. With this in mind, each X value can be transformed into a z-score,

z X , using the mean and standard deviation for the set of Xs. Similarly, each Y score can be

transformed into z Y . If the X and Y values are viewed as a sample, then the transformation is

completed using the sample formula for z (Equation 5.3). If the X and Y values form a com-

plete population, then the z-scores are computed using Equation 5.1. After the transforma-

tion, the formula for the Pearson correlation can be expressed entirely in terms of z-scores.

For a sample, 1

r z z

n X y

5 2

( )

(14.4)

T H E P E A R S O N CO R R E L AT I O N A N D

Z - S CO R E S

Scores Deviations Squared Deviations Products

X Y X – M X

Y 2 M Y

(X 2 M X )2 (Y 2 M

Y )2 (X 2 M

X )(Y 2 M

Y )

0 2 26 22 36 4 �12

10 6 �4 �2 16 4 �8

4 2 22 22 4 4 �4

8 4 �2 0 4 0 0

8 6 �2 �2 4 4 �4

SS X 5 64 SS

Y 5 16 SP 5 �28

TABLE 14.1

Calculation of SS X , SS

Y , and

SP for a sample of n 5 5 pairs of scores.

4 5 8 CHAPTER 14 CORRELATION

For a population, r 5 o z z

X y

(14.5)

Note that the population value is identified with a Greek letter, in this case the letter

rho (r), which is the Greek equivalent of the letter r.

1. Can SP ever have a value less than zero?

2. Calculate the sum of products of deviations (SP) for the following set of scores.

Use the definitional formula and then the computational formula. Verify that you

get the same answer with both formulas.

X Y

0 1

4 3

5 3

2 2

4 1

3. For the following data:

a. Sketch a scatter plot and make an estimate of the Pearson correlation.

b. Compute the Pearson correlation.

X Y

2 6

1 5

3 3

0 7

4 4

1. Yes. SP can be positive, negative, or zero depending on the relationship between X and Y.

2. SP 5 5

3. b. r 5 2 8

10 5 20.80

L E A R N I N G C H E C K

ANSWERS

USING AND INTERPRETING THE PEARSON CORRELATION

Although correlations have a number of different applications, a few specific examples

are presented next to give an indication of the value of this statistical measure.

1. Prediction. If two variables are known to be related in a systematic way, then it

is possible to use one of the variables to make predictions about the other. For

example, when you applied for admission to college, you were required to sub-

mit a great deal of personal information, including your scores on the Scholastic

Achievement Test (SAT). College officials want this information because it helps

to predict your chances of success in college. It has been demonstrated over several

years that SAT scores and college grade point averages are correlated. Students

14.3

W H E R E A N D W H Y CO R R E L AT I O N S A R E

U S E D

SECTION 14.3 / USING AND INTERPRETING THE PEARSON CORRELATION 4 5 9

who do well on the SAT tend to do well in college; students who have difficulty

with the SAT tend to have difficulty in college. Based on this relationship, college

admissions officers can predict the potential success of each applicant. You should

note that this prediction is not perfectly accurate. Not everyone who does poorly

on the SAT has trouble in college. That is why you also submit letters of recom-

mendation, high school grades, and other information with your application.

2. Validity. Suppose that a psychologist develops a new test for measuring intel-

ligence. How could you show that this test truly measures what it claims; that

is, how could you demonstrate the validity of the test? One common technique

for demonstrating validity is to use a correlation. If the test actually measures

intelligence, then the scores on the test should be related to other measures of

intelligence—for example, standardized IQ tests, performance on learning tasks,

problem-solving ability, and so on. The psychologist could measure the correla-

tion between the new test and each of these other measures of intelligence to

demonstrate that the new test is valid.

3. Reliability. In addition to evaluating the validity of a measurement procedure,

correlations are used to determine reliability. A measurement procedure is con-

sidered reliable to the extent that it produces stable, consistent measurements.

That is, a reliable measurement procedure produces the same (or nearly the

same) scores when the same individuals are measured twice under the same

conditions. For example, if your IQ were measured as 113 last week, you would

expect to obtain nearly the same score if your IQ were measured again this

week. One way to evaluate reliability is to use correlations to determine the

relationship between two sets of measurements. When reliability is high, the

correlation between two measurements should be strong and positive.

4. Theory Verification. Many psychological theories make specific predictions

about the relationship between two variables. For example, a theory may predict

a relationship between brain size and learning ability; a developmental theory

may predict a relationship between the parents’ IQs and the child’s IQ; a social

psychologist may have a theory predicting a relationship between personality

type and behavior in a social situation. In each case, the prediction of the theory

could be tested by determining the correlation between the two variables.

When you encounter correlations, there are four additional considerations that you

should bear in mind:

1. Correlation simply describes a relationship between two variables. It does not

explain why the two variables are related. Specifically, a correlation should not

and cannot be interpreted as proof of a cause-and-effect relationship between

the two variables.

2. The value of a correlation can be affected greatly by the range of scores repre-

sented in the data.

3. One or two extreme data points, often called outliers, can have a dramatic effect

on the value of a correlation.

4. When judging how “good” a relationship is, it is tempting to focus on the numeri-

cal value of the correlation. For example, a correlation of �0.5 is halfway between

0 and 1.00 and, therefore, appears to represent a moderate degree of relationship.

However, a correlation should not be interpreted as a proportion. Although a cor-

relation of 1.00 does mean that there is a 100% perfectly predictable relationship

I N T E R P R E T I N G CO R R E L AT I O N S

4 6 0 CHAPTER 14 CORRELATION

between X and Y, a correlation of 0.5 does not mean that you can make predictions

with 50% accuracy. To describe how accurately one variable predicts the other, you

must square the correlation. Thus, a correlation of r 5 0.5 means that one variable partially predicts the other, but the predictable portion is only r2 5 0.52 5 0.25 (or 25%) of the total variability.

We now discuss each of these four points in detail.

One of the most common errors in interpreting correlations is to assume that a cor-

relation necessarily implies a cause-and-effect relationship between the two variables.

(Even Pearson blundered by asserting causation from correlational data [Blum, 1978].)

We constantly are bombarded with reports of relationships: Cigarette smoking is related

to heart disease; alcohol consumption is related to birth defects; carrot consumption is

related to good eyesight. Do these relationships mean that cigarettes cause heart disease

or carrots cause good eyesight? The answer is no. Although there may be a causal re-

lationship, the simple existence of a correlation does not prove it. Earlier, for example,

we discussed a study showing a relationship between high school grades and family

income. However, this result does not mean that having a higher family income causes

students to get better grades. For example, if mom gets an unexpected bonus at work,

it is unlikely that her child’s grades will also show a sudden increase. To establish a

cause-and-effect relationship, it is necessary to conduct a true experiment (see p. 13)

in which one variable is manipulated by a researcher and other variables are rigorously

controlled. The fact that a correlation does not establish causation is demonstrated in

the following example.

Suppose we select a variety of different cities and towns throughout the United

States and measure the number of churches (X variable) and the number of serious

crimes (Y variable) for each. A scatter plot showing hypothetical data for this study

is presented in Figure 14.5. Notice that this scatter plot shows a strong, positive cor-

relation between churches and crime. You also should note that these are realistic

data. It is reasonable that small towns would have less crime and fewer churches

and that large cities would have large values for both variables. Does this relation-

ship mean that churches cause crime? Does it mean that crime causes churches?

It should be clear that both answers are no. Although a strong correlation exists

between number of churches and crime, the real cause of the relationship is the size

of the population.

Whenever a correlation is computed from scores that do not represent the full range

of possible values, you should be cautious in interpreting the correlation. Suppose, for

example, that you are interested in the relationship between IQ and creativity. If you

select a sample of your fellow college students, your data probably will represent only

a limited range of IQ scores (most likely from 110 to 130). The correlation within this

restricted range could be completely different from the correlation that would be ob-

tained from a full range of IQ scores. For example, Figure 14.6 shows a strong positive

relationship between X and Y when the entire range of scores is considered. However,

this relationship is obscured when the data are limited to a restricted range.

To be safe, you should not generalize any correlation beyond the range of data repre-

sented in the sample. For a correlation to provide an accurate description for the general

population, there should be a wide range of X and Y values in the data.

CO R R E L AT I O N A N D CAU SAT I O N

E X A M P L E 1 4 . 4

CO R R E L AT I O N A N D R E ST R I C T E D R A N G E

SECTION 14.3 / USING AND INTERPRETING THE PEARSON CORRELATION 4 6 1

An outlier is an individual with X and/or Y value that is substantially different (larger or

smaller) from the values obtained for the other individuals in the data set. The data point of

a single outlier can have a dramatic influence on the value obtained for the correlation. This

effect is illustrated in Figure 14.7. Figure 14.7(a) shows a set of n 5 5 data points for which the correlation between the X and Y variables is nearly zero (actually r 5 –0.08). In Figure 14.7(b), one extreme data point (14, 12) has been added to the original data set. When this

outlier is included in the analysis, a strong, positive correlation emerges (now r 5 � 0.85). Note that the single outlier drastically alters the value for the correlation and, thereby, can

affect one’s interpretation of the relationship between variables X and Y. Without the outlier,

one would conclude there is no relationship between the two variables. With the extreme

data point, r 5 �0.85, which implies a strong relationship with Y increasing consistently as X increases. The problem of outliers is a good reason for looking at a scatter plot, instead

of simply basing your interpretation on the numerical value of the correlation. If you only

O U T L I E R S

0 10 20

Number of churches

N u

m b

e r

o f

se ri o

u s

c ri m

e s

30 40 50 60 70

FIGURE 14.5

Hypothetical data showing

the logical relationship

between the number of

churches and the number

of serious crimes for a

sample of U.S. cities.

X values

X values restricted to a limited range

v a

lu e

FIGURE 14.6

In this example, the

full range of X and Y

values shows a strong,

positive correlation, but

the restricted range of

scores produces a

correlation near zero.

4 6 2 CHAPTER 14 CORRELATION

X values

2 4 6 8 12 1410

Y v a

lu e

2 4 6 8 12 1410

Y v a

lu e

r = 0.85r = –0.08

X values

Data with Outlier Included

Subject X Y

A B C D E F

1 3 6 4 5

3 5 4 1 2

Original Data

Subject X Y

A B C D E

1 3 6 4 5

3 5 4 1 2

(a) (b)

FIGURE 14.7

A demonstration of how one extreme data point (an outlier) can influence the value of a correlation.

“go by the numbers,” you might overlook the fact that one extreme data point inflated the

size of the correlation.

A correlation measures the degree of relationship between two variables on a scale from

0 to 1.00. Although this number provides a measure of the degree of relationship, many

researchers prefer to square the correlation and use the resulting value to measure the

strength of the relationship.

One of the common uses of correlation is for prediction. If two variables are cor-

related, you can use the value of one variable to predict the other. For example, college

admissions officers do not just guess which applicants are likely to do well; they use

other variables (SAT scores, high school grades, and so on) to predict which students are

most likely to be successful. These predictions are based on correlations. By using cor-

relations, the admissions officers expect to make more accurate predictions than would

be obtained by chance. In general, the squared correlation (r2) measures the gain in ac-

curacy that is obtained from using the correlation for prediction. The squared correlation

measures the proportion of variability in the data that is explained by the relationship

between X and Y. It is sometimes called the coefficient of determination.

The value r2 is called the coefficient of determination because it measures the

proportion of variability in one variable that can be determined from the relation-

ship with the other variable. A correlation of r 5 0.80 (or –0.80), for example, means that r2 5 0.64 (or 64%) of the variability in the Y scores can be predicted from the relationship with X.

CO R R E L AT I O N A N D T H E ST R E N G T H O F T H E

R E L AT I O N S H I P

D E F I N I T I O N

SECTION 14.3 / USING AND INTERPRETING THE PEARSON CORRELATION 4 6 3

In earlier chapters (see pp. 262, 291, and 322), we introduced r2 as a method for

measuring effect size for research studies where mean differences were used to com-

pare treatments. Specifically, we measured how much of the variance in the scores was

accounted for by the differences between treatments. In experimental terminology, r2

measures how much of the variance in the dependent variable is accounted for by the

independent variable. Now we are doing the same thing, except that there is no inde-

pendent or dependent variable. Instead, we simply have two variables, X and Y, and we

use r2 to measure how much of the variance in one variable can be determined from its

relationship with the other variable. The following example demonstrates this concept.

Figure 14.8 shows three sets of data representing different degrees of linear relation-

ship. The first set of data [Figure 14.8(a)] shows the relationship between IQ and shoe

size. In this case, the correlation is r 5 0 (and r2 5 0), and you have no ability to predict a person’s IQ based on his or her shoe size. Knowing a person’s shoe size provides no

information (0%) about the person’s IQ. In this case, shoe size provides no help ex-

plaining why different people have different IQs.

Now consider the data in Figure 14.8(b). These data show a moderate, positive

correlation, r 5 �0.60, between IQ scores and college grade point averages (GPA). Students with high IQs tend to have higher grades than students with low IQs. From this

relationship, it is possible to predict a student’s GPA based on his or her IQ. However,

you should realize that the prediction is not perfect. Although students with high IQs

tend to have high GPAs, this is not always true. Thus, knowing a student’s IQ provides

some information about the student’s grades, or knowing a student’s grades provides

some information about the student’s IQ. In this case, IQ scores help explain the fact

that different students have different GPAs. Specifically, you can say that part of the

differences in GPA are accounted for by IQ. With a correlation of r 5 �0.60, we obtain r2 5 0.36, which means that 36% of the variance in GPA can be explained by IQ.

Finally, consider the data in Figure 14.8(c). This time we show a perfect linear

relationship (r 5 �1.00) between monthly salary and yearly salary for a group of college employees. With r 5 1.00 and r2 5 1.00, there is 100% predictability. If you know a person’s monthly salary, you can predict perfectly the person’s annual salary. If

two people have different annual salaries, the difference can be completely explained

(100%) by the difference in their monthly salaries.

E X A M P L E 1 4 . 5

Shoe size IQ Annual salary

C o

lle g

e G

P A

M o

n th

ly s

a la

FIGURE 14.8

Three sets of data showing three different degrees of linear relationship.

(a) (b) (c)

4 6 4 CHAPTER 14 CORRELATION

Just as r2 was used to evaluate effect size for mean differences in Chapters 9, 10,

and 11, r2 can now be used to evaluate the size or strength of the correlation. The same

standards that were introduced in Table 9.3 (p. 264) apply to both uses of the r2 mea-

sure. Specifically, an r2 value of 0.01 indicates a small effect or a small correlation, an

r2 value of 0.09 indicates a medium correlation, and an r2 of 0.25 or larger indicates a

large correlation.

More information about the coefficient of determination (r2) is presented

in Section 14.5. For now, you should realize that whenever two variables are

consistently related, it is possible to use one variable to predict values for the

second variable.

1. A researcher finds a correlation of r 5 20.71 between the time spent playing video games each week and grade point average for a group of high school boys. This

means that playing video games causes students to get lower grades. (True or false?)

2. A researcher finds a correlation of r 5 0.60 between salary and the number of

years of education for a group of 40-year-old men. How much of the variance in

salary is explained by the years of education?

1. False. You cannot conclude that there is a cause-and-effect relationship based on a correlation.

2. r2 5 0.36, or 36%

L E A R N I N G C H E C K

ANSWERS

HYPOTHESIS TESTS WITH THE PEARSON CORRELATION

The Pearson correlation is generally computed for sample data. As with most

sample statistics, however, a sample correlation is often used to answer questions

about the corresponding population correlation. For example, a psychologist would

like to know whether there is a relationship between IQ and creativity. This is a

general question concerning a population. To answer the question, a sample would

be selected, and the sample data would be used to compute the correlation value.

You should recognize this process as an example of inferential statistics: using

samples to draw inferences about populations. In the past, we have been concerned

primarily with using sample means as the basis for answering questions about

population means. In this section, we examine the procedures for using a sample

correlation as the basis for testing hypotheses about the corresponding population

correlation.

The basic question for this hypothesis test is whether a correlation exists in the

population. The null hypothesis is “No. There is no correlation in the population,”

or “The population correlation is zero.” The alternative hypothesis is “Yes. There is

a real, nonzero correlation in the population.” Because the population correlation

is traditionally represented by r (the Greek letter rho), these hypotheses would be stated in symbols as

H 0 : r 5 0 (There is no population correlation.)

H 1 : r  0 (There is a real correlation.)

14.4

T H E H Y P OT H E S E S

SECTION 14.4 / HYPOTHESIS TESTS WITH THE PEARSON CORRELATION 4 6 5

When there is a specific prediction about the direction of the correlation, it is pos-

sible to do a directional, or one-tailed, test. For example, if a researcher is predicting a

positive relationship, the hypotheses would be

H 0 : r  0 (The population correlation is not positive.)

H 1 : r . 0 (The population correlation is positive.)

The correlation from the sample data is used to evaluate the hypotheses. For the

regular, nondirectional test, a sample correlation near zero provides support for H 0 and a

sample value far from zero tends to refute H 0 . For a directional test, a positive value for

the sample correlation would tend to refute a null hypothesis stating that the population

correlation is not positive.

Although sample correlations are used to test hypotheses about population

correlations, you should keep in mind that samples are not expected to be identi-

cal to the populations from which they come; there is some discrepancy (sampling

error) between a sample statistic and the corresponding population parameter.

Specifically, you should always expect some error between a sample correlation

and the population correlation it represents. One implication of this fact is that,

even when there is no correlation in the population (r 5 0), you are still likely to obtain a nonzero value for the sample correlation. This is particularly true for

small samples. Figure 14.9 illustrates how a small sample from a population with

a near-zero correlation could result in a correlation that deviates from zero. The

colored dots in the figure represent the entire population and the three circled

dots represent a random sample. Note that the three sample points show a rela-

tively good, positive correlation even through there is no linear trend (r 5 0) for the population.

X values

Y v

a lu

e s

FIGURE 14.9

Scatter plot of a popula-

tion of X and Y values

with near-zero correla-

tion. However, a small

sample of n 5 3 data points from this popula-

tion shows a relatively

strong, positive correla-

tion. Data points in the

sample are circled.

4 6 6 CHAPTER 14 CORRELATION

When you obtain a nonzero correlation for a sample, the purpose of the hypothesis

test is to decide between the following two interpretations:

1. There is no correlation in the population (r 5 0), and the sample value is the result of sampling error. Remember, a sample is not expected to be identical to

the population. There always is some error between a sample statistic and the

corresponding population parameter. This is the situation specified by H 0 .

2. The nonzero sample correlation accurately represents a real, nonzero correlation

in the population. This is the alternative stated in H 1 .

The correlation from the sample helps to determine which of these two interpretations

is more likely. A sample correlation near zero supports the conclusion that the population

correlation is also zero. A sample correlation that is substantially different from zero sup-

ports the conclusion that there is a real, nonzero correlation in the population.

The hypothesis test evaluating the significance of a correlation can be conducted using

either a t statistic or an F-ratio. The F-ratio is discussed later (pp. 493–496), and we

focus on the t statistic here. The t statistic for a correlation has the same general struc-

ture as t statistics introduced in Chapters 9, 10, and 11.

t 5 2sample statistic population parameter

sstandard error

In this case, the sample statistic is the sample correlation (r) and the corresponding

parameter is the population correlation (r). The null hypothesis specifies that the popu- lation correlation is r 5 0. The final part of the equation is the standard error, which is determined by

standard error for 2

s 1 r

n r

r 5 5 2

2 (14.6)

Thus, the complete t statistic is

t r

1 r

2 5

( ) ( )2

(14.7)

The t statistic has degrees of freedom defined by df 5 n – 2. An intuitive explanation for this value is that a sample with only n 5 2 data points has no degrees of freedom. Specifically, if there are only two points, they will fit perfectly on a straight line, and

the sample produces a perfect correlation of r 5 11.00 or r 5 –1.00. Because the first two points always produce a perfect correlation, the sample correlation is free to vary

only when the data set contains more than two points. Thus, df 5 n – 2. The following examples demonstrate the hypothesis test.

A researcher is using a regular, two-tailed test with a 5 .05 to determine whether a nonzero correlation exists in the population. A sample of n 5 30 individuals is obtained and produces a correlation of r 5 0.35. The null hypothesis states that there is no cor- relation in the population.

H 0 : r 5 0

T H E H Y P OT H E S I S T E ST

E X A M P L E 1 4 . 6

SECTION 14.4 / HYPOTHESIS TESTS WITH THE PEARSON CORRELATION 4 6 7

For this example, df 5 28 and the critical values are t 5 ±2.048. With r2 5 0.352 5 0.1225, the data produce

t = =5 2

0.35 0

0.1225

0.35

0.177 1

1 28( ) / ..97

The t value is not in the critical region so we fail to reject the null hypothesis. The

sample correlation is not large enough to reject the null hypothesis.

With a sample of n 5 30 and a correlation of r 5 0.35, this time we use a direc-

tional, one-tailed test to determine whether there is a positive correlation in the

population.

H 0 : r  0 (There is not a positive correlation.)

H 1 : r . 0 (There is a positive correlation.)

The sample correlation is positive, as predicted, so we simply need to determine

whether it is large enough to be significant. For a one-tailed test with df 5 28 and a 5 .05, the critical value is t 5 1.701. In the previous example, we found that this sample produces t 5 1.97, which is beyond the critical boundary. For the one-tailed test, we reject the null hypothesis and conclude that there is a significant positive

correlation in the population.

As with most hypothesis tests, if other factors are held constant, the likelihood of

finding a significant correlation increases as the sample size increases. For example, a

sample correlation of r 5 0.50 produces a nonsignificant t(8) 5 1.63 for a sample of n 5 10, but the same correlation produces a significant t(18) 5 2.45 if the sample size is increased to n 5 20.

E X A M P L E 1 4 . 7

IN THE LITERATURE

REPORTING CORRELATIONS

Correlations are typically reported using APA format. The statement should include the

sample size, the calculated value for the correlation, whether it is a statistically signifi-

cant relationship, the probability level, and the type of test used (one- or two-tailed).

For example, a correlation might be reported as follows:

A correlation for the data revealed a significant relationship between amount

of education and annual income, r 5 10.65, n 5 30, p , .01, two tails.

Sometimes a study might look at several variables, and correlations between all pos-

sible variable pairings are computed. Suppose, for example, that a study measured people’s

annual income, amount of education, age, and intelligence. With four variables, there are

six possible pairings leading to six different correlations. The results from multiple cor-

relations are most easily reported in a table called a correlation matrix, using footnotes to

indicate which correlations are significant. For example, the report might state:

4 6 8 CHAPTER 14 CORRELATION

Occasionally a researcher may suspect that the relationship between two variables

is being distorted by the influence of a third variable. Earlier in the chapter, for

example, we found a strong positive relationship between the number of churches

and the number of serious crimes for a sample of different towns and cities (see

Example 14.4, p 460). However, it is unlikely that there is a direct relationship

between churches and crime. Instead, both variables are influenced by population:

Large cities have a lot of churches and high crime rates compared to smaller towns,

which have fewer churches and less crime. If population were controlled, there

probably would be no real correlation between churches and crime.

Fortunately, there is a statistical technique, known as partial correlation, that allows

a researcher to measure the relationship between two variables while controlling or

holding constant the influence of a third variable. Thus, a researcher could use a partial

correlation to examine the relationship between churches and crime without the risk

that the relationship is distorted by the size of the population.

PA R T I A L CO R R E L AT I O N S

1. A researcher obtains a correlation of r 5 20.39 for a sample of n 5 25 individuals. Does this sample provide sufficient evidence to conclude that there is a significant,

nonzero correlation in the population? Assume a two-tailed test with a 5 .05.

2. As sample size gets smaller, what happens to the magnitude of the correlation

necessary for significance? Explain why this occurs.

1. No. The sample correlation produces t 5 22.03. With df 5 23 and a 5 .05, the critical value is 2.069. The sample value is not in the critical region.

2. As the sample size gets smaller, the magnitude of the correlation needed for significance

gets larger. With a small sample, it is easy to get a relatively large correlation just by chance.

Therefore, a small sample requires a very large correlation before you can be confident there

is a real (nonzero) relationship in the population.

L E A R N I N G C H E C K

ANSWERS

The analysis examined the relationships among income, amount of education,

age, and intelligence for n 5 30 participants. The correlations between pairs of variables are reported in Table 1. Significant correlations are noted in the table.

TABLE 1

Correlation matrix for income, amount of education, age, and intelligence

Education Age IQ

Income �.65* �.41** �.27

Education �.11 �.38**

Age 2.02

n 5 30

*p , .01, two tails

**p , .05, two tails

SECTION 14.4 / HYPOTHESIS TESTS WITH THE PEARSON CORRELATION 4 6 9

A partial correlation measures the relationship between two variables while

controlling the influence of a third variable by holding it constant.

In a situation with three variables, X, Y, and Z, it is possible to compute three individual

Pearson correlations:

1. r XY

, measuring the correlation between X and Y

2. r XZ

, measuring the correlation between X and Z

3. r YZ

, measuring the correlation between Y and Z

These three individual correlations can then be used to compute a partial correlation.

For example, the partial correlation between X and Y, holding Z constant, is determined

by the formula

r r r r

r r XY Z

XY XZ YZ

XZ YZ

− 5 2

2 2

( )

( )( )1 12 2

(14.8)

The following example demonstrates the calculation and interpretation of a partial

correlation.

We begin with the hypothetical data shown in Table 14.2. These scores have been con-

structed to simulate the church/crime/population situation for a sample of n 5 15 cities.

The X variable represents the number of churches, Y represents the number of crimes,

and Z represents the population for each city. Note that there are three categories for the

size of the population (three values for Z) corresponding to small, medium, and large

cities. For these scores, the individual Pearson correlations are all large and positive:

a. The correlation between churches and crime is r XY

5 0.923.

b. The correlation between churches and population is r XZ

5 0.961.

c. The correlation between crime and population is r YZ

5 0.961.

D E F I N I T I O N

E X A M P L E 1 4 . 8

Number of Churches (X) Number of Crimes (Y) Population (Z)

1 4 1

2 3 1

3 1 1

4 2 1

5 5 1

7 8 2

8 11 2

9 9 2

10 7 2

11 10 2

13 15 3

14 14 3

15 16 3

16 17 3

17 13 3

TABLE 14.2

Hypothetical data showing

the relationship between

the number of churches, the

number of crimes, and the

population of a set of n 5 15

cities.

4 7 0 CHAPTER 14 CORRELATION

The data points for the 15 cities are shown in the scatter plot in Figure 14.10. Notice that

the population variable, Z, separates the scores into three distinct groups: When Z 5 1, the population is low and churches and crime (X and Y) are also low; when Z 5 2, the popula- tion is moderate and churches and crime (X and Y) are also moderate; and when Z 5 3, the population is large and churches and crime are both high. Thus, as the population increases

from one city to another, the number of churches and crimes also increase, and the result is

a strong positive correlation between churches and crime.

Within each of the three population categories, however, there is no linear relation-

ship between churches and crime. Specifically, within each group, the population vari-

able is constant and the five data points for X and Y form a circular pattern, indicating

no consistent linear relationship. The partial correlation allows us to hold population

1 2 3 4 5 6 7 8 9

Number of Churches

Z = 1 Small Cities

Z = 2 Medium Cities

Z = 3 Large Cities

N u

m b

e r

o f

C ri m

e s

10 11 12 13 14 15 16 17

FIGURE 14.10

Hypothetical data showing the relationship between the number of churches and the number of crimes for three

groups of cities. Those with small populations (Z 5 1), those with medium populations (Z 5 2), and those with large populations (Z 5 3).

SECTION 14.4 / HYPOTHESIS TESTS WITH THE PEARSON CORRELATION 4 7 1

constant across the entire sample and measure the underlying relationship between

churches and crime without any influence from population. For these data, the partial

correlation is

r XY Z− 5

2 2

0 923 0 961 0 961

1 0 961 1 0 961 2 2

. . .

. .

( )

( ) (( )

5 0

0 076.

5 0

Thus, when the population differences are eliminated, there is no correlation remain-

ing between churches and crime (r 5 0).

In Example 14.8, the population differences, which correspond to the different val-

ues of the Z variable, were eliminated mathematically in the calculation of the partial

correlation. However, it is possible to visualize how these differences are eliminated

in the actual data. Looking at Figure 14.10, focus on the five points in the bottom left

corner. These are the five cities with small populations, few churches, and little crime.

The five points in the upper right corner represent the five cities with large populations,

many churches, and a lot of crime. The partial correlation controls population size by

mathematically equalizing the populations for all 15 cities. Population is increased for

the five small cities. However, increasing the population also increases churches and

crime. Similarly, population is decreased for the five large cities, which also decreases

churches and crime. In Figure 14.10, imagine the five points in the bottom left moving

up and to the right so that they overlap with the points in the center. At the same time,

the five points in the upper right move down and to the left so that they also overlap

the points in the center. When population is equalized, all 15 data points are clustered

in the center of the figure with no clear positive or negative trend. In other words, the

remaining correlation between churches and crime is r 5 0.

In Example 14.8 we used a partial correlation to demonstrate that an apparent

relationship between churches and crime was actually caused by the influence of

a third variable, population. It also is possible to use partial correlations to dem-

onstrate that a relationship is not caused by the influence of a third variable. As an

example, consider research examining the relationship between exposure to sexual

content on television and sexual behavior among adolescents (Collins et al., 2004).

The study consisted of a survey of 1,792 adolescents, 12 to 17 years old, who re-

ported their television viewing habits and their sexual behaviors. The results showed

a clear relationship between television viewing and behaviors. Specifically, the more

sexual content the adolescents watched on television, the more likely they were to

engage in sexual behaviors. One concern for the researchers was that the observed

relationship may be influenced by the age of the participants. For example, the

older adolescents (age 17) probably watch more programs with sexual content and

engage in more sexual behaviors than the younger adolescents (age 12) do. Although

the viewing of sexual content on television and the participants’ sexual behaviors

increase together, the observed relationship may simply be the result of age differ-

ences. To address this problem, the researcher used a partial correlation technique to

control or hold constant the age variable. The results clearly showed that a relation-

ship still exists between television sexual content and sexual behavior even after the

influence of the participants’ ages was accounted for.

4 7 2 CHAPTER 14 CORRELATION

ALTERNATIVES TO THE PEARSON CORRELATION

The Pearson correlation measures the degree of linear relationship between two vari-

ables when the data (X and Y values) consist of numerical scores from an interval

or ratio scale of measurement. However, other correlations have been developed for

nonlinear relationships and for other types of data. In this section, we examine three

additional correlations: the Spearman correlation, the point-biserial correlation, and the

phi-coefficient. As you will see, all three can be viewed as special applications of the

Pearson correlation.

When the Pearson correlation formula is used with data from an ordinal scale (ranks),

the result is called the Spearman correlation. The Spearman correlation is used in two

situations.

First, the Spearman correlation is used to measure the relationship between X and Y

when both variables are measured on ordinal scales. Recall from Chapter 1 that an ordinal

scale typically involves ranking individuals rather than obtaining numerical scores. Rank-

order data are fairly common because they are often easier to obtain than interval or ratio

scale data. For example, a teacher may feel confident about rank-ordering students’ leader-

ship abilities but would find it difficult to measure leadership on some other scale.

In addition to measuring relationships for ordinal data, the Spearman correlation

can be used as a valuable alternative to the Pearson correlation, even when the original

raw scores are on an interval or a ratio scale. As we have noted, the Pearson correlation

measures the degree of linear relationship between two variables—that is, how well

the data points fit on a straight line. However, a researcher often expects the data to

show a consistently one-directional relationship but not necessarily a linear relation-

ship. For example, Figure 14.11 shows the typical relationship between practice and

performance. For nearly any skill, increasing amounts of practice tend to be associ-

ated with improvements in performance (the more you practice, the better you get).

However, it is not a straight-line relationship. When you are first learning a new skill,

practice produces large improvements in performance. After you have been performing

a skill for several years, however, additional practice produces only minor changes in

performance. Although there is a consistent relationship between the amount of practice

and the quality of performance, it clearly is not linear. If the Pearson correlation were

computed for these data, it would not produce a correlation of 1.00 because the data do

not fit perfectly on a straight line. In a situation like this, the Spearman correlation can

be used to measure the consistency of the relationship, independent of its form.

14.5

T H E S P E A R M A N CO R R E L AT I O N

1. Sales figures show a positive relationship between temperature and ice cream

consumption; as temperature increases, ice cream consumption also increases.

Other research shows a positive relationship between temperature and crime

rate (Cohn & Rotton, 2000). When the temperature increases, both ice cream

consumption and crime rates tend to increase. As a result, there is a positive

correlation between ice cream consumption and crime rate. However, what do

you think is the true relationship between ice cream consumption and crime

rate? Specifically, what value would you predict for the partial correlation be-

tween the two variables if temperature were held constant?

1. There should be no systematic relationship between ice cream consumption and crime rate.

The partial correlation should be near zero.

L E A R N I N G C H E C K

ANSWER

SECTION 14.5 / ALTERNATIVES TO THE PEARSON CORRELATION 4 7 3

The reason that the Spearman correlation measures consistency, rather than form,

comes from a simple observation: When two variables are consistently related, their

ranks are linearly related. For example, a perfectly consistent positive relationship

means that every time the X variable increases, the Y variable also increases. Thus, the

smallest value of X is paired with the smallest value of Y, the second-smallest value

of X is paired with the second smallest value of Y, and so on. Every time the rank for

X goes up by 1 point, the rank for Y also goes up by 1 point. As a result, the ranks fit

perfectly on a straight line. This phenomenon is demonstrated in the following example.

Table 14.3 presents X and Y scores for a sample of n 5 4 people. Note that the data show a perfectly consistent relationship. Each increase in X is accompanied by an increase

in Y. However, the relationship is not linear, as can be seen in the graph of the data in

Figure 14.12(a).

Next, we convert the scores to ranks. The lowest X is assigned a rank of 1, the next

lowest a rank of 2, and so on. The Y scores are then ranked in the same way. The ranks

are listed in Table 14.3 and shown in Figure 14.12(b). Note that the perfect consistency

for the scores produces a perfect linear relationship for the ranks.

The preceding example demonstrates that a consistent relationship among scores

produces a linear relationship when the scores are converted to ranks. Thus, if you want

to measure the consistency of a relationship for a set of scores, you can simply convert

the scores to ranks and then use the Pearson correlation formula to measure the linear

relationship for the ranked data. The degree of linear relationship for the ranks provides

a measure of the degree of consistency for the original scores.

E X A M P L E 1 4 . 9

Amount of practice (X )

L e

v e

l o

f p

e rf

o rm

a n

c e

( Y

)

FIGURE 14.11

Hypothetical data showing

the relationship between

practice and performance.

Although this relationship

is not linear, there is a

consistent positive relation-

ship. An increase in perfor-

mance tends to accompany

an increase in practice.

Person X Y X-Rank Y-Rank

A 4 9 3 3

B 2 2 1 1

C 10 10 4 4

D 3 8 2 2

TABLE 14.3

Scores and ranks for

Example 14.9.

4 7 4 CHAPTER 14 CORRELATION

To summarize, the Spearman correlation measures the relationship between two

variables when both are measured on ordinal scales (ranks). There are two general situ-

ations in which the Spearman correlation is used:

1. Spearman is used when the original data are ordinal. In this case, you rank the

ordinal scores and apply the Pearson correlation formula to the set of ranks.

2. Spearman is used when the original scores are numerical values from an

interval or ratio scale and the goal is to measure the consistency of the rela-

tionship between X and Y, independent of the specific form of the relation-

ship. In this case, the original scores are first converted to ranks, and then

the Pearson correlation formula is used with the ranks. Because the Pearson

formula measures the degree to which the ranks fit on a straight line, it

also measures the degree of consistency in the relationship for the original

scores. Incidentally, when there is a consistently one-directional relationship

between two variables, the relationship is said to be monotonic. Thus, the

Spearman correlation measures the degree of monotonic relationship

between two variables.

In either case, the Spearman correlation is identified by the symbol r S to differentiate

it from the Pearson correlation. The complete process of computing the Spearman cor-

relation, including ranking scores, is demonstrated in Example 14.10.

The following data show a nearly perfect monotonic relationship between X and Y.

When X increases, Y tends to decrease, and there is only one reversal in this general

trend. To compute the Spearman correlation, we first rank the X and Y values, and we

then compute the Pearson correlation for the ranks.

E X A M P L E 1 4 . 1 0

1 3 5 7 9

Y s

c o

re s

X scores

Scores

2 4 6 8 100

A C

1 2 3 4

Y r

a n

k s

X ranks

Ranks

(a) (b) FIGURE 14.12

Scatter plots showing (a) the scores and (b) the ranks for the data in Example 14.9. Notice that there is a consistent,

positive relationship between the X and Y scores, although it is not a linear relationship. Also notice that the scatter plot

of the ranks shows a perfect linear relationship.

The word monotonic describes

a sequence that is consistently

increasing (or decreasing). Like

the word monotonous, it means

constant and unchanging.

SECTION 14.5 / ALTERNATIVES TO THE PEARSON CORRELATION 4 7 5

To compute the correlation, we need SS for X, SS for Y, and SP. Remember that all

of these values are computed with the ranks, not the original scores. The X ranks are

simply the integers 1, 2, 3, 4, and 5. These values have oX 5 15 and oX2 5 55. The SS for the X ranks is

SS X X

n X

5  2 

5 2 5 2

2 2

55 15

5 10

( ) ( )

Note that the ranks for Y are identical to the ranks for X; that is, they are the integers

1, 2, 3, 4, and 5. Therefore, the SS for Y is identical to the SS for X:

SS Y 5 10

To compute the SP value, we need oX, oY, and oXY for the ranks. The XY values are listed in the table with the ranks, and we already have found that both the Xs and

the Ys have a sum of 15. Using these values, we obtain

SP XY Y

n 5  2

  5 5 2

X( ) ( ) ( )( ) 36

15 15

5 9−

Finally, the Spearman correlation simply uses the Pearson formula for the ranks.

r SP

SS SS s

X Y

5 5 2

5 2 ( )( ) ( )

10 10 0 9.

The Spearman correlation indicates that the data show a consistent (nearly perfect)

negative trend.

When you are converting scores into ranks for the Spearman correlation, you may

encounter two (or more) identical scores. Whenever two scores have exactly the same

value, their ranks should also be the same. This is accomplished by the following

procedure:

1. List the scores in order from smallest to largest. Include tied values in the list.

2. Assign a rank (first, second, etc.) to each position in the ordered list.

3. When two (or more) scores are tied, compute the mean of their ranked positions,

and assign this mean value as the final rank for each score.

R A N K I N G T I E D S CO R E S

Original Data

X Y

3 12

4 10

10 11

11 9

12 2

Ranks

X Y XY

1 5 5

2 3 6

3 4 12

4 2 8

5 1 5

36 5 oXY

We have listed the X values

in order so that the trend is

easier to recognize.

4 7 6 CHAPTER 14 CORRELATION

The process of finding ranks for tied scores is demonstrated here. These scores have

been listed in order from smallest to largest.

Scores Rank Position Final Rank

3 1 1.5 Mean of 1 and 2

3 2 1.5

5 3 3

6 4 5 Mean of 4, 5, and 6

6 5 5

6 6 5

12 7 7

Note that this example has seven scores and uses all seven ranks. For X 5 12, the largest score, the appropriate rank is 7. It cannot be given a rank of 6 because that rank

has been used for the tied scores.

If the original scores are ordinal values, then they are ranked using exactly the same

process for ranking tied scores. For example, suppose a researcher has the following

letter grades for a sample of n 5 7 students: A, A, B, C, C, D, and F. The grades would receive the following ranks: 1.5, 1.5, 3, 4.5, 4.5, 6, and 7.

After the original X values and Y values have been ranked, the calculations necessary

for SS and SP can be greatly simplified. First, you should note that the X ranks and the

Y ranks are really just a set of integers: 1, 2, 3, 4, … , n. To compute the mean for these

integers, you can locate the midpoint of the series by M 5 (n � 1)/2. Similarly, the SS for this series of integers can be computed by

SS n n

5 2

2 1

( ) ( )Try it out.

Also, because the X ranks and the Y ranks are the same values, the SS for X is identical

to the SS for Y.

Because calculations with ranks can be simplified and because the Spearman

correlation uses ranked data, these simplifications can be incorporated into the

final calculations for the Spearman correlation. Instead of using the Pearson

formula after ranking the data, you can put the ranks directly into a simplified

formula:

r D

n n s

5 2 2

1 6

( )

(14.9)

where D is the difference between the X rank and the Y rank for each individual.

This formula produces the same result that would be obtained from the Pearson

formula. However, note that this special formula can be used only after the scores

have been converted to ranks and only when there are no ties among the ranks. If

there are relatively few tied ranks, the formula still may be used, but it loses accuracy

as the number of ties increases. The application of this formula is demonstrated in

the following example.

S P E C I A L F O R M U L A F O R T H E S P E A R M A N

CO R R E L AT I O N

Caution: In this formula,

you compute the value of the

fraction and then subtract

from 1. The 1 is not part of

the fraction.

SECTION 14.5 / ALTERNATIVES TO THE PEARSON CORRELATION 4 7 7

To demonstrate the special formula for the Spearman correlation, we use the same data

that were presented in Example 14.10. The ranks for these data are shown again here:

Ranks Difference

X Y D D2

1 5 4 16

2 3 1 1

3 4 1 1

4 2 22 4

5 1 24 16

38 5 oD 2

Using the special formula for the Spearman correlation, we obtain

r D

n n s 5 2

2 5 2

2 5 2 5 21

1 1

6 38

5 25 1 1

228

120 1

( ) ( )

( ) 11 90 0 90. .52

This is exactly the same answer that we obtained in Example 14.10, using the

Pearson formula on the ranks.

E X A M P L E 1 4 . 1 1

1. Describe what is measured by a Spearman correlation, and explain how this correlation

is different from the Pearson correlation.

2. If the following scores are converted into ranks, what rank will be assigned to the

individuals who have scores of X 5 7?

Scores: 1, 1, 1, 3, 6, 7, 7, 8, 10

3. Rank the following scores and compute the Spearman correlation:

X Y

2 7

12 3

9 16

14 5

1. The Spearman correlation measures the consistency of the direction of the relationship between

two variables. The Spearman correlation does not depend on the form of the relationship,

whereas the Pearson correlation measures how well the data fit a linear form.

2. Both scores get a rank of 6.5 (the average of 6 and 7).

3. r S 5 20.60

L E A R N I N G C H E C K

ANSWERS

In Chapters 9, 10, and 11, we introduced r2 as a measure of effect size that often accompa-

nies a hypothesis test using the t statistic. The r2 used to measure effect size and the r used

to measure a correlation are directly related, and we now have an opportunity to demonstrate

the relationship. Specifically, we compare the independent-measures t test (Chapter 10) and

a special version of the Pearson correlation known as the point-biserial correlation.

T H E P O I N T- B I S E R I A L CO R R E L AT I O N A N D

M E AS U R I N G E F F E C T S I Z E W I T H R 2

4 7 8 CHAPTER 14 CORRELATION

The point-biserial correlation is used to measure the relationship between two vari-

ables in situations in which one variable consists of regular, numerical scores, but the

second variable has only two values. A variable with only two values is called a dichot-

omous variable or a binomial variable. Some examples of dichotomous variables are

1. Male versus female

2. College graduate versus not a college graduate

3. First-born child versus later-born child

4. Success versus failure on a particular task

5. Older than 30 years versus younger than 30 years

To compute the point-biserial correlation, the dichotomous variable is first converted

to numerical values by assigning a value of zero (0) to one category and a value of one

(1) to the other category. Then, the regular Pearson correlation formula is used with the

converted data.

To demonstrate the point-biserial correlation and its association with the r2 measure

of effect size, we use the data from Example 10.1 (p. 289). The original example com-

pared high school grades for two groups of students: one group who regularly watched

Sesame Street as 5-year-old children and one who did not watch the program. The

data from the independent-measures study are presented on the left side of Table 14.4.

Notice that the data consist of two separate samples and the independent-measures t

was used to determine whether there was a significant mean difference between the two

populations represented by the samples.

It is customary to use the

numerical values 0 and 1, but

any two different numbers

would work equally well and

would not affect the value of

the correlation.

TABLE 14.4

The same data are organized

in two different formats.

On the left-hand side, the

data appear as two separate

samples appropriate for an

independent-measures t

hypothesis test. On the right-

hand side, the same data are

shown as a single sample,

with two scores for each

individual: the original high

school grade and a dichoto-

mous score (Y) that identifies

the group in which the par-

ticipant is located (Sesame

Street 5 1 and No-Sesame Street 5 0). The data on the right are appropriate for a

point-biserial correlation.

Data for the independent-measures

t test. Two separate samples, each

with n 5 10 scores.

Average High School Grade

Watched Sesame Street

Did Not Watch Sesame Street

86 99 90 79

87 97 89 83

91 94 82 86

97 89 83 81

98 92 85 92

n 5 10

M 5 93

SS 5 200

n 5 10

M 5 85

SS 5 160

Data for the point-biserial correlation. Two scores,

X and Y, for each of the n 5 20 participants.

Participant Grade X Group Y

A 86 1

B 87 1

C 91 1

D 97 1

E 98 1

F 99 1

G 97 1

H 94 1

I 89 1

J 92 1

K 90 0

L 89 0

M 82 0

N 83 0

O 85 0

P 79 0

Q 83 0

R 86 0

S 81 0

T 92 0

SECTION 14.5 / ALTERNATIVES TO THE PEARSON CORRELATION 4 7 9

On the right-hand side of Table 14.4 we have reorganized the data into a form that is

suitable for a point-biserial correlation. Specifically, we used each student’s high school

grade as the X value and we have created a new variable, Y, to represent the group, or

condition, for each student. In this case, we have used Y 5 1 for students who watched Sesame Street and Y 5 0 for students who did not watch the program.

When the data in Table 14.4 were originally presented in Chapter 10, we conducted

an independent-measures t hypothesis test and obtained t 5 4.00 with df 5 18. We measured the size of the treatment effect by calculating r2, the percentage of variance

accounted for, and obtained r2 5 0.47. Calculating the point-biserial correlation for these data also produces a value for r.

Specifically, the X scores produce SS 5 680; the Y values produce SS 5 5.00, and the sum of the products of the X and Y deviations produces SP 5 40. The point-biserial correlation is

r SP

SS SS X Y

5 5 5 5 ( )( ) ( ) ( )

680 5

58 31 0 686

. .

Notice that squaring the value of the point-biserial correlation produces r2 5 (0.686)2 5 0.47, which is exactly the value of r2 we obtained measuring effect size.

In some respects, the point-biserial correlation and the independent-measures

hypothesis test are evaluating the same thing. Specifically, both are examining the

relationship between the TV-viewing habits of 5-year-old children and their future aca-

demic performance in high school.

1. The correlation is measuring the strength of the relationship between the two

variables. A large correlation (near 1.00 or 21.00) would indicate that there is

a consistent, predictable relationship between high school grades and watch-

ing Sesame Street as a 5-year-old child. In particular, the value of r2 measures

how much of the variability in grades can be predicted by knowing whether the

participants watched Sesame Street.

2. The t test evaluates the significance of the relationship. The hypothesis test

determines whether the mean difference in grades between the two groups is

greater than can be reasonably explained by chance alone.

As we noted in Chapter 10 (p. 291–295), the outcome of the hypothesis test and

the value of r2 are often reported together. The t value measures statistical signifi-

cance and r2 measures the effect size. Also, as we noted in Chapter 10, the values

for t and r2 are directly related. In fact, either can be calculated from the other by

the equations

r t

t df t

r df

2 2

2 1

5 �

5 2

and ( ) /

where df is the degrees of freedom for the t statistic.

However, you should note that r2 is determined entirely by the size of the correla-

tion, whereas t is influenced by the size of the correlation and the size of the sample.

For example, a correlation of r 5 0.30 produces r2 5 0.09 (9%) no matter how large

the sample may be. On the other hand, a point-biserial correlation of r 5 0.30 for a

total sample of 10 people (n 5 5 in each group) produces a nonsignificant value of

t 5 0.889. If the sample is increased to 50 people (n 5 25 in each group), the same

correlation produces a significant t value of t 5 2.18. Although t and r are related, they

are measuring different things.

4 8 0 CHAPTER 14 CORRELATION

When both variables (X and Y) measured for each individual are dichotomous, the cor-

relation between the two variables is called the phi-coefficient. To compute phi (), you

follow a two-step procedure:

1. Convert each of the dichotomous variables to numerical values by assigning a 0

to one category and a 1 to the other category for each of the variables.

2. Use the regular Pearson formula with the converted scores.

This process is demonstrated in the following example.

A researcher is interested in examining the relationship between birth-order position and

personality for individuals who have at least one sibling. A random sample of n 5 8 participants is obtained, and each individual is classified in terms of birth-order position

as first-born versus later-born. Then, each individual’s personality is classified as either

introvert or extrovert.

The original measurements are then converted to numerical values by the following

assignments:

Birth Order Personality

First-born child 5 0 Introvert 5 0

Later-born child 5 1 Extrovert 5 1

The original data and the converted scores are as follows:

Original Data Converted Scores

Birth Order (X)

Personality (Y)

Birth Order (X)

Personality (Y)

1st Introvert 0 0

3rd Extrovert 1 1

1st Extrovert 0 1

2nd Extrovert 1 1

4th Extrovert 1 1

2nd Introvert 1 0

1st Introvert 0 0

3rd Extrovert 1 1

The Pearson correlation formula is then used with the converted data to compute the

phi-coefficient.

Because the assignment of numerical values is arbitrary (either category

could be designated 0 or 1), the sign of the resulting correlation is meaningless.

As with most correlations, the strength of the relationship is best described by the

value of r2, the coefficient of determination, which measures how much of the vari-

ability in one variable is predicted or determined by the association with the second

variable.

We also should note that although the phi-coefficient can be used to assess the rela-

tionship between two dichotomous variables, the more common statistical procedure is

a chi-square statistic, which is examined in Chapter 15.

T H E P H I - CO E F F I C I E N T

E X A M P L E 1 4 . 1 2

SECTION 14.6 / INTRODUCTION TO LINEAR EQUATIONS AND REGRESSION 4 8 1

INTRODUCTION TO LINEAR EQUATIONS AND REGRESSION

Earlier in the chapter, we introduced the Pearson correlation as a technique for describ-

ing and measuring the linear relationship between two variables. Figure 14.13 presents

hypothetical data showing the relationship between SAT scores and college grade point

average (GPA). Note that the figure shows a good, but not perfect, positive relationship.

Also note that we have drawn a line through the middle of the data points. This line

serves several purposes:

1. The line makes the relationship between SAT scores and GPA easier to see.

2. The line identifies the center, or central tendency, of the relationship, just as the

mean describes central tendency for a set of scores. Thus, the line provides a

simplified description of the relationship. For example, if the data points were

removed, the straight line would still give a general picture of the relationship

between SAT scores and GPA.

3. Finally, the line can be used for prediction. The line establishes a precise,

one-to-one relationship between each X value (SAT score) and a correspond-

ing Y value (GPA). For example, an SAT score of 620 corresponds to a GPA

of 3.25 (see Figure 14.13). Thus, the college admissions officers could use the

straight-line relationship to predict that a student entering college with an SAT

score of 620 should achieve a college GPA of approximately 3.25.

14.6

1. The following data represent job-related stress scores for a sample of n 5 8 individuals. These people also are classified by salary level.

a. Convert the data into a form suitable for the point-biserial correlation.

b. Compute the point-biserial correlation for these data.

Salary More than $40,000

Salary Less than $40,000

8 4

6 2

5 1

3 3

2. A researcher would like to know whether there is a relationship between gender and

manual dexterity for 3-year-old children. A sample of n 5 10 boys and n 5 10 girls is obtained and each child is given a manual-dexterity test. Five of the girls failed the test

and only two of the boys failed. Describe how these data could be coded into a form

suitable for computing a phi-coefficient to measure the strength of the relationship.

1. a. Salary level is a dichotomous variable and can be coded as Y 5 1 for individuals with salary more than $40,000 and Y 5 0 for salary less than $40,000. The stress scores produce SS

X 5 36, the salary codes produce SS

Y 5 2, and SP 5 6.

b. The point-biserial correlation is 0.71.

2. Gender could be coded with male 5 0 and female 5 1. Manual dexterity could be coded with failure 5 0 and success 5 1. Eight boys would have scores of 0 and 1 and two would have scores of 0 and 0. Five girls would have scores of 1 and 1 and five would have scores of 1 and 0.

L E A R N I N G C H E C K

ANSWERS

4 8 2 CHAPTER 14 CORRELATION

Our goal in this section is to develop a procedure that identifies and defines the

straight line that provides the best fit for any specific set of data. This straight line does

not have to be drawn on a graph; it can be presented in a simple equation. Thus, our

goal is to find the equation for the line that best describes the relationship for a set of

X and Y data.

In general, a linear relationship between two variables X and Y can be expressed by

the equation

Y 5 bX � a (14.10)

where a and b are fixed constants.

For example, a local video store charges an annual membership fee of $5,

which allows you to rent videos and games for $2 each. With this information, the

total cost for 1 year can be computed using a linear equation that describes the

relationship between the total cost (Y) and the number of videos and games

rented (X).

Y 5 2X � 5

In the general linear equation, the value of b is called the slope. The slope

determines how much the Y variable changes when X is increased by 1 point. For the

video store example, the slope is b 5 2 and indicates that your total cost increases by $2 for each video you rent. The value of a in the general equation is called

the Y-intercept because it determines the value of Y when X 5 0. (On a graph, the a value identifies the point where the line intercepts the Y-axis.) For the video-

store example, a 5 5; there is a $5 membership charge even if you never rent a video.

L I N E A R E Q UAT I O N S

2.00 G

ra d

e p

o in

t a

v e

ra g

2.50

1.50

4.00

3.50

3.00

1.00

0.50

420 460 500 540 620580

SAT scores

660 700

FIGURE 14.13

Hypothetical data showing

the relationship between

SAT scores and GPA with

a regression line drawn

through the data points.

The regression line de-

fines a precise, one-to-one

relationship between each

X value (SAT score) and

its corresponding Y value

(GPA).

Note that a positive slope

means that Y increases

when X is increased, and

a negative slope indicates

that Y decreases when X is

increased.

SECTION 14.6 / INTRODUCTION TO LINEAR EQUATIONS AND REGRESSION 4 8 3

Figure 14.14 shows the general relationship between the annual cost and number

of videos for the video store example. Notice that the relationship results in a straight

line. To obtain this graph, we picked any two values of X and then used the equation to

compute the corresponding values for Y. For example,

When X 5 3: When X 5 8:

Y 5 bX � a Y 5 bX � a

5 $2(3) � $5 5 $2(8) � $5

5 $6 � $5 5 $16 � $5

5 $11 5 $21

Next, these two points are plotted on the graph: one point at X 5 3 and Y 5 11, the other point at X 5 8 and Y 5 21. Because two points completely determine a straight line, we simply drew the line so that it passed through these

two points.

When drawing a graph of

a linear equation, it is wise

to compute and plot at least

three points to be certain that

you have not made a mistake.

1 2 3 4 5 6 7 8 9 10

To ta

l c

o st

( Y

)

Number of Videos Rented (X)

FIGURE 14.14

The relationship between

total cost and number of

videos rented each month.

The video store charges a

$5 monthly membership

fee and $2 for each video

or game rented. The rela-

tionship is described by a

linear equation, Y 5 2X �5, where Y is the total

cost and X is the number

of videos.

4 8 4 CHAPTER 14 CORRELATION

Because a straight line can be extremely useful for describing a relationship between

two variables, a statistical technique has been developed that provides a standardized

method for determining the best-fitting straight line for any set of data. The statistical

procedure is regression, and the resulting straight line is called the regression line.

The statistical technique for finding the best-fitting straight line for a set of data

is called regression, and the resulting straight line is called the regression line.

The goal for regression is to find the best-fitting straight line for a set of data. To

accomplish this goal, however, it is first necessary to define precisely what is meant by

“best fit.” For any particular set of data, it is possible to draw lots of different straight

lines that all appear to pass through the center of the data points. Each of these lines can

be defined by a linear equation of the form Y 5 bX � a where b and a are constants that determine the slope and Y-intercept of the line, respectively. Each individual line has

its own unique values for b and a. The problem is to find the specific line that provides

the best fit to the actual data points.

To determine how well a line fits the data points, the first step is to define mathematically the

distance between the line and each data point. For every X value in the data, the linear equa-

tion determines a Y value on the line. This value is the predicted Y and is called Ŷ (“Y hat”).

The distance between this predicted value and the actual Y value in the data is determined by

distance 5 Y 2 Ŷ

R E G R E S S I O N

D E F I N I T I O N

T H E L E AST- S Q UA R E S S O L U T I O N

1. A local gym charges a $25 monthly membership fee plus $2 per hour for aerobics

classes. What is the linear equation that describes the relationship between the total

monthly cost (Y) and the number of class hours each month (X)?

2. For the linear equation, Y 5 23X � 7, what happens to the value of Y each time

X is increased by 1 point?

3. Use the linear equation Y 5 2X 2 7 to determine the value of Y for each of the following values of X: 1, 3, 5, 10.

4. If the slope constant (b) in a linear equation is positive, then a graph of the equation

is a line tilted from lower left to upper right. (True or false?)

1. Y 5 2X � 25

2. The slope is 23, so Y decreases by 3 points each time X increases by 1 point.

X Y

1 –5

3 –1

5 3

10 13

4. True. A positive slope indicates that Y increases (goes up in the graph) when X increases

(goes to the right in the graph).

L E A R N I N G C H E C K

ANSWERS

SECTION 14.6 / INTRODUCTION TO LINEAR EQUATIONS AND REGRESSION 4 8 5

Note that we simply are measuring the vertical distance between the actual data

point (Y) and the predicted point on the line. This distance measures the error between

the line and the actual data (Figure 14.15).

Because some of these distances are positive and some are negative, the next step is to

square each distance to obtain a uniformly positive measure of error. Finally, to determine

the total error between the line and the data, we add the squared errors for all of the data

points. The result is a measure of overall squared error between the line and the data:

total squared error 5 o(Y 2 Ŷ)2

Now we can define the best-fitting line as the one that has the smallest total squared

error. For obvious reasons, the resulting line is commonly called the least-squared-

error solution. In symbols, we are looking for a linear equation of the form

Ŷ 5 bX � a

For each value of X in the data, this equation determines the point on the line (Ŷ)

that gives the best prediction of Y. The problem is to find the specific values for a and

b that make this the best-fitting line.

The calculations that are needed to find this equation require calculus and some

sophisticated algebra, so we do not present the details of the solution. The results, how-

ever, are relatively straightforward, and the solutions for b and a are as follows:

b SP

SS X

(14.11)

where SP is the sum of products and SS X is the sum of squares for the X scores.

A commonly used alternative formula for the slope is based on the standard devia-

tions for X and Y. The alternative formula is

b s

s y

(14.12)

X Values

X, Y data point

Y V

a lu

e s

Y = bX + aˆ

Distance = Y – Ŷ

FIGURE 14.15

The distance between the

actual data point (Y) and

the predicted point on the

line (Ŷ) is defined as

Y 2 Ŷ. The goal of

regression is to find the

equation for the line that

minimizes these distances.

4 8 6 CHAPTER 14 CORRELATION

where s Y is the standard deviation for the Y scores, s

X is the standard deviation for the

X scores, and r is the Pearson correlation for X and Y. The value of the constant a in the

equation is determined by

a 5 M Y 2 bM

X (14.13)

Note that these formulas determine the linear equation that provides the best predic-

tion of Y values. This equation is called the regression equation for Y.

The regression equation for Y is the linear equation.

Ŷ 5 bX � a (14.14)

where the constant b is determined by Equation 14.11, or 14.12 and the constant

a is determined by Equation 14.13. This equation results in the least squared

error between the data points and the line.

The scores in the following table are used to demonstrate the calculation and use of the

regression equation for predicting Y.

X Y X – M X

Y – M Y

(X – M X )2 (Y – M

X )2 (X – M

X ) (Y – M

Y )

5 10 1 3 1 9 3

1 4 –3 –3 9 9 9

4 5 0 –2 0 4 0

7 11 3 4 9 16 12

6 15 2 8 4 64 16

4 6 0 –1 0 1 0

3 5 –1 –2 1 4 2

2 0 –2 –7 4 49 14

SS X 5 28 SS

Y 5 156 SP 5 56

For these data, oX 5 32, so M X 5 4. Also, oY 5 56, so M

Y 5 7. These values have

been used to compute the deviation scores for each X and Y value. The final three col-

umns show the squared deviations for X and for Y, and the products of the deviation

scores.

Our goal is to find the values for b and a in the regression equation. Using Equations

14.11 and 14.13, the solutions for b and a are

b SP

SS X

5 5 5 56

28 2

a 5 M Y 2 bM

X 5 7 2 2(4) 5 21

The resulting equation is

Ŷ 5 2X 2 1

The original data and the regression line are shown in Figure 14.16.

D E F I N I T I O N

E X A M P L E 1 4 . 1 3

SECTION 14.6 / INTRODUCTION TO LINEAR EQUATIONS AND REGRESSION 4 8 7

The regression line shown in Figure 14.16 demonstrates some simple and very pre-

dictable facts about regression. First, the calculation of the Y-intercept (Equation 14.13)

ensures that the regression line passes through the point defined by the mean for X and

the mean for Y. That is, the point identified by the coordinates M X , M

Y will always be

on the line. We have included the two means in Figure 14.16 to show that the point they

define is on the regression line. Second, the sign of the correlation (� or –) is the same

as the sign of the slope of the regression line. Specifically, if the correlation is positive,

then the slope is also positive and the regression line slopes up to the right. On the other

hand, if the correlation is negative, then the slope is negative and the line slopes down

to the right. A correlation of zero means that the slope is also zero and the regression

equation produces a horizontal line that passes through the data at a level equal to the

mean for the Y values. Note that the regression line in Figure 14.16 has a positive slope.

One consequence of this fact is that all of the points on the line that are above the mean

for X are also above the mean for Y. Similarly, all of the points below the mean for X

are also below the mean for Y. Thus, every individual with a positive deviation for X is

predicted to have a positive deviation for Y, and everyone with a negative deviation for

X is predicted to have a negative deviation for Y.

As we noted at the beginning of this section, one common use of regression equations is

for prediction. For any given value of X, we can use the equation to compute a predicted

value for Y. For the equation from Example 14.13, an individual with a score of X 5 3 would be predicted to have a Y score of

Ŷ 5 2X 2 1 5 6 2 1 5 5

U S I N G T H E R E G R E S S I O N E Q UAT I O N

F O R P R E D I C T I O N

1 2 3 4 5 6 7 8

M = 4x

M = 7 y

Y = X – ˆ 12

FIGURE 14.16

The X and Y data points

and the regression line for

the n 5 8 pairs of scores

in Example 14.13.

4 8 8 CHAPTER 14 CORRELATION

Although regression equations can be used for prediction, a few cautions should be

considered whenever you are interpreting the predicted values:

1. The predicted value is not perfect (unless r 5 �1.00 or 21.00). If you examine Figure 14.16, it should be clear that the data points do not fit perfectly on the

line. In general, there is some error between the predicted Y values (on the line)

and the actual data. Although the amount of error varies from point to point,

on average the errors are directly related to the magnitude of the correlation.

With a correlation near 1.00 (or 21.00), the data points generally are clustered

close to the line and the error is small. As the correlation gets nearer to zero, the

points move away from the line and the magnitude of the error increases.

2. The regression equation should not be used to make predictions for X values that fall

outside of the range of values covered by the original data. For Example 14.13, the

X values ranged from X 5 1 to X 5 7, and the regression equation was calculated

as the best-fitting line within this range. Because you have no information about

the X-Y relationship outside this range, the equation should not be used to predict Y

for any X value lower than 1 or greater than 7.

So far, we have presented the regression equation in terms of the original values, or

raw scores, for X and Y. Occasionally, however, researchers standardize the scores by

transforming the X and Y values into z-scores before finding the regression equation.

The resulting equation is often called the standardized form of the regression equation

and is greatly simplified compared to the raw-score version. The simplification comes

from the fact that z-scores have standardized characteristics. Specifically, the mean for

a set of z-scores is always zero and the standard deviation is always 1. As a result, the

standardized form of the regression equation becomes

ẑ z Y X

5 (beta)

(14.15)

First notice that we are now using the z-score for each X (z X ) to predict the z-score for

the corresponding Y (z Y ). Also, note that the slope constant that was identified as b in the

raw-score formula is now identified as beta. Because both sets of z-scores have a mean of

zero, the constant a disappears from the regression equation. Finally, when one variable, X, is

being used to predict a second variable, Y, the value of beta is equal to the Pearson correlation

for X and Y. Thus, the standardized form of the regression equation can also be written as

ẑ Y

5 rz X

(14.16)

Because the process of transforming all of the original scores into z-scores can be

tedious, researchers usually compute the raw-score version of the regression equation

(Equation 14.14) instead of the standardized form. However, most computer programs

report the value of beta as part of the output from linear regression, and you should

understand what this value represents.

STA N DA R D I Z E D F O R M O F T H E R E G R E S S I O N

E Q UAT I O N S

1. Sketch a scatter plot for the following data—that is, a graph showing the X, Y data

points:

X Y

4 13

2 5

5 12

1 6

L E A R N I N G C H E C K

SECTION 14.6 / INTRODUCTION TO LINEAR EQUATIONS AND REGRESSION 4 8 9

It is possible to determine a regression equation for any set of data by simply using

the formulas already presented. The linear equation you obtain is then used to gener-

ate predicted Y values for any known value of X. However, it should be clear that the

accuracy of this prediction depends on how well the points on the line correspond

to the actual data points—that is, the amount of error between the predicted values,

Ŷ , and the actual scores, Y values. Figure 14.17 shows two different sets of data that

have exactly the same regression equation. In one case, there is a perfect correla-

tion (r 5 �1) between X and Y, so the linear equation fits the data perfectly. For the second set of data, the predicted Y values on the line only approximate the real

data points.

A regression equation, by itself, allows you to make predictions, but it

does not provide any information about the accuracy of the predictions. To mea-

sure the precision of the regression, it is customary to compute a standard error

of estimate.

The standard error of estimate gives a measure of the standard distance between

the predicted Y values on the regression line and the actual Y values in the data.

Conceptually, the standard error of estimate is very much like a standard deviation:

Both provide a measure of standard distance. Also, the calculation of the standard error

of estimate is very similar to the calculation of standard deviation.

To calculate the standard error of estimate, we first find the sum of squared devia-

tions (SS). Each deviation measures the distance between the actual Y value (from the

data) and the predicted Y value (from the regression line). This sum of squares is com-

monly called SS residual

because it is based on the remaining distance between the actual

Y scores and the predicted values on the line.

SS residual

5 o(Y 2 Ŷ)2 (14.17)

The obtained SS value is then divided by its degrees of freedom to obtain a measure

of variance. This procedure should be very familiar:

Variance 5 SS

The degrees of freedom for the standard error of estimate are df 5 n – 2. The reason

for having n – 2 degrees of freedom, rather than the customary n – 1, is that we now are

measuring deviations from a line rather than deviations from a mean. To find the equa-

tion for the regression line, you must know the means for both the X and the Y scores.

Specifying these two means places two restrictions on the variability of the data, with

the result that the scores have only n – 2 degrees of freedom. (Note: the df 5 n – 2 for

SS residual

is the same df 5 n – 2 that we encountered when testing the significance of the

Pearson correlation on page 466.)

T H E STA N DA R D E R R O R O F E ST I M AT E

D E F I N I T I O N

a. Find the regression equation for predicting Y from X. Draw this line on your

graph. Does it look like the best-fitting line?

b. Use the regression equation to find the predicted Y value corresponding to each X

in the data.

1. a. SS X 5 10, SP 5 20, b 5 2, a 5 3. The equation is Ŷ 5 2X � 3.

b. The predicted Y values are 11, 7, 13, and 5.

ANSWERS

4 9 0 CHAPTER 14 CORRELATION

The final step in the calculation of the standard error of estimate is to take the square

root of the variance to obtain a measure of standard distance. The final equation is

standard error of estimate residual5 5 SS

o((Y Y ^

) 2

(14.18)

The following example demonstrates the calculation of this standard error.

Y = X – ˆ 12

X2 3 4 5 6 7 81

(a)

(b)

FIGURE 14.17

(a) A scatter plot showing

data points that perfectly fit

the regression line defined

by the equation Ŷ 5 2X – 1. Note that the correlation is

r 5 11.00. (b) A scatter plot for the data in Example

14.13. Notice that there is

error between the actual data

points and the predicted Y

values on the regression line.

Recall that variance measures

the average squared distance.

SECTION 14.6 / INTRODUCTION TO LINEAR EQUATIONS AND REGRESSION 4 9 1

The same data that were used in Example 14.13 are used here to demonstrate the calcu-

lation of the standard error of estimate. These data have the regression equation

Ŷ 5 2X 2 1

Using this regression equation, we have computed the predicted Y value, the re-

sidual, and the squared residual for each individual, using the data from Example 14.13.

Data Predicted Y value Residual Squared Residual

X Y Ŷ 5 2X 2 1 Y 2 Ŷ (Y 2 Ŷ)2

5 10 9 1 1

1 4 1 3 9

4 5 7 22 4

7 11 13 22 4

6 15 11 4 16

4 6 7 21 1

3 5 5 0 0

2 0 3 23 9

0 SS residual

5 44

First note that the sum of the residuals is equal to zero. In other words, the sum of the

distances above the line is equal to the sum of the distances below the line. This is true

for any set of data and provides a way to check the accuracy of your calculations. The

squared residuals are listed in the final column. For these data, the sum of the squared

residuals is SS residual

5 244. With n 5 8, the data have df 5 n – 2 5 6, so the standard

error of estimate is

standard error of estimate residual5 5 SS

444

6 2 7085 .

Remember: The standard error of estimate provides a measure of how accurately

the regression equation predicts the Y values. In this case, the standard distance

between the actual data points and the regression line is measured by standard error

of estimate 5 2.708.

It should be clear from Example 14.14 that the standard error of estimate is directly re-

lated to the magnitude of the correlation between X and Y. If the correlation is near 1.00

(or –1.00), then the data points are clustered close to the line, and the standard error of

estimate is small. As the correlation gets nearer to zero, the data points become more

widely scattered, the line provides less accurate predictions, and the standard error of

estimate grows larger.

Earlier (p. 462), we observed that squaring the correlation provides a measure

of the accuracy of prediction. The squared correlation, r2, is called the coefficient

of determination because it determines what proportion of the variability in Y is

predicted by the relationship with X. Because r2 measures the predicted portion of

the variability in the Y scores, we can use the expression (1 – r2) to measure the

unpredicted portion. Thus,

E X A M P L E 1 4 . 1 4

R E L AT I O N S H I P B E T W E E N T H E

STA N DA R D E R R O R A N D T H E CO R R E L AT I O N

4 9 2 CHAPTER 14 CORRELATION

predicted variability 5 SS regression

5 r2SS Y (14.19)

unpredicted variability 5 SS residual

5 (1 – r2)SS Y (14.20)

For example, if r 5 0.80, then the predicted variability is r2 5 0.64 (or 64%) of the total variability for the Y scores and the remaining 36% (1 – r2) is the unpredicted vari-

ability. Note that when r 5 1.00, the prediction is perfect and there are no residuals. As the correlation approaches zero, the data points move farther off the line and the

residuals grow larger. Using Equation 14.20 to compute SS residual

, the standard error of

estimate can be computed as

standard error of estimate residual5 5 SS

(11

2 2

r SS

Y )

(14.21)

Because it is usually much easier to compute the Pearson correlation than to

compute the individual (Y – Ŷ )2 values, Equation 14.20 is usually the easiest way

to compute SS residual

, and Equation 14.21 is usually the easiest way to compute the

standard error of estimate for a regression equation. The following example demon-

strates this new formula.

We use the same data used in Examples 14.13 and 14.14, which produced SS X 5 28,

SS Y 5 156, and SP 5 56. For these data, the Pearson correlation is

r 5 5 5 56

28(156)

66.09 0.847

With SS Y 5 156 and a correlation of r 5 0.847, the predicted variability from the

regression equation is

SS regression

5 r2SS Y 5 (0.8472)(156) 5 0.718(156) 5 112.01

Similarly, the unpredicted variability is

SS residual

5 (1 2 r2)SS Y 5 (1 2 0.8472)(156) 5 0.282(156) 5 43.99

Notice that the new formula for SS residual

produces the same value, within rounding

error, that we obtained by adding the squared residuals in Example 14.14. Also note

that this new formula is generally much easier to use because it requires only the cor-

relation value (r) and the SS for Y. The primary point of this example, however, is that

SS residual

and the standard error of estimate are closely related to the value of the correla-

tion. With a large correlation (near �1.00 or 21.00), the data points are close to the

regression line, and the standard error of estimate is small. As a correlation gets smaller

(near zero), the data points move away from the regression line, and the standard error

of estimate gets larger.

Because it is possible to have the same regression equation for several different

sets of data, it is also important to consider r2 and the standard error of estimate. The

regression equation simply describes the best-fitting line and is used for making pre-

dictions. However, r2 and the standard error of estimate indicate how accurate these

predictions are.

E X A M P L E 1 4 . 1 5

SECTION 14.6 / INTRODUCTION TO LINEAR EQUATIONS AND REGRESSION 4 9 3

As we noted earlier in the chapter, a sample correlation is expected to be representative

of its population correlation. For example, if the population correlation is zero, then the

sample correlation is expected to be near zero. Note that we do not expect the sample

correlation to be exactly equal to zero. This is the general concept of sampling error that

was introduced in Chapter 1 (p. 8). The principle of sampling error is that there is typi-

cally some discrepancy or error between the value obtained for a sample statistic and the

corresponding population parameter. Thus, when there is no relationship whatsoever in

the population, a correlation of r 5 0, you are still likely to obtain a nonzero value for the sample correlation. In this situation, however, the sample correlation is caused by chance

and a hypothesis test usually demonstrates that the correlation is not significant.

Whenever you obtain a nonzero value for a sample correlation, you also obtain

real, numerical values for the regression equation. However, if there is no real rela-

tionship in the population, both the sample correlation and the regression equation are

meaningless—they are simply the result of sampling error and should not be viewed

as an indication of any relationship between X and Y. In the same way that we tested

the significance of a Pearson correlation, we can test the significance of the regression

equation. In fact, when a single variable, X, is being used to predict a single variable,

Y, the two tests are equivalent. In each case, the purpose for the test is to determine

whether the sample correlation represents a real relationship or is simply the result of

sampling error. For both tests, the null hypothesis states that there is no relationship

between the two variables in the population. For a correlation,

H 0 : the population correlation is r 5 0

For the regression equation,

H 0 : the slope of the regression equation (b or beta) is zero

The process of testing the significance of a regression equation is called analysis

of regression and is very similar to the analysis of variance (ANOVA) presented in

Chapter 12. As with ANOVA, the regression analysis uses an F-ratio to determine

whether the variance predicted by the regression equation is significantly greater than

would be expected if there were no relationship between X and Y. The F-ratio is a ratio

of two variances, or mean square (MS) values, and each variance is obtained by dividing

an SS value by its corresponding degrees of freedom. The numerator of the F-ratio is

MS regression

, which is the variance in the Y scores that is predicted by the regression equa-

tion. This variance measures the systematic changes in Y that occur when the value of X

A N A LYS I S O F R E G R E S S I O N : T E ST I N G

T H E S I G N I F I CA N C E O F T H E R E G R E S S I O N

E Q UAT I O N

1. Describe what is measured by the standard error of estimate for a regression equation.

2. As the numerical value of a correlation increases, what happens to the standard

error of estimate?

3. A sample of n 5 6 pairs of X and Y scores produces a correlation of r 5 0.80 and SS

Y 5 100. What is the standard error of estimate for the regression equation?

1. The standard error of estimate measures the average, or standard, distance between the

predicted Y values on the regression line and the actual Y values in the data.

2. A larger correlation means that the data points are clustered closer to the line, which means

the standard error of estimate is smaller.

3. The standard error of estimate 5 36

4 5 3.

L E A R N I N G C H E C K

ANSWERS

4 9 4 CHAPTER 14 CORRELATION

increases or decreases. The denominator is MS residual

, which is the unpredicted variance

in the Y scores. This variance measures the changes in Y that are independent of changes

in X. The two MS value are defined as

MS SS

df regression

regression

with5 df MS5 5 1 and residuals

residual

resi

df ddual

with 2df n5 2

The F-ratio is

F MS

MS 5 5 2df n

regression

residual

with 1, 2

(14.22)

The complete analysis of SS and degrees of freedom is diagrammed in Figure 14.18.

The analysis of regression procedure is demonstrated in the following example, using

the same data that we used in Examples 14.13, 14.14, and 14.15.

The data consist of n 5 8 pairs of scores with a correlation of r 5 0.847 and SS Y 5 156.

The null hypothesis states either that there is no relationship between X and Y in the

population, or that the regression equation does not account for a significant portion of

the variance for the Y scores. The F-ratio for the analysis of regression has df 5 1, n 2 2. For these data, df 5 1, 6. With a 5 .05, the critical value is 5.99.

As noted in the previous section, the SS for the Y scores can be separated into two

components: the predicted portion corresponding to r2 and the unpredicted, or residual,

portion corresponding to (1 2 r2). With r 5 0.847, we obtain r2 5 0.718 and

predicted variability 5 SS regression

5 0.718(156) 5 112.01

unpredicted variability 5 SS residual

5 (1 2 0.718)(156) 5 0.282(156) 5 43.99

Using these SS values and the corresponding df values, we calculate a variance, or

MS, for each component. For these data the MS values are

MS SS

df regression

regression

5 5 1112.01

1 112.015

MS SS

df residual

residual

43.99

6 5 5 5 7.33

E X A M P L E 1 4 . 1 6

SSregression

r2SSY

SSresidual

(1 � r2)SSY

SSY

dfregression 5 1 dfresidual 5 n � 2

dfY 5 n � 1

FIGURE 14.18

The partitioning of SS and df for analysis of regression. The variability in the original Y scores (both SS Y and df

Y ) is partitioned

into two components: (1) the variability that is explained by the regression equation, and (2) the residual variability.

SECTION 14.6 / INTRODUCTION TO LINEAR EQUATIONS AND REGRESSION 4 9 5

Finally, the F-ratio for evaluating the significance of the regression equation is

F MS

MS 5 5 5

regression

residual

112.01

7.33 15.28

The F-ratio is in the critical region, so we reject the null hypothesis and conclude

that the regression equation does account for a significant portion of the variance for

the Y scores. The complete analysis of regression is summarized in Table 14.5 which is

a common format for computer printouts of regression analysis.

In a situation with a single X variable and a single Y variable, testing the significance of

the regression equation is equivalent to testing the significance of the Pearson correla-

tion. Therefore, whenever the correlation between two variables is significant, you can

conclude that the regression equation is also significant. Similarly, if a correlation is

not significant, then the regression equation is also not significant. Earlier in the chapter

(p. 466), we introduced a hypothesis test using the following t statistic to evaluate the

significance of a correlation.

t 5 2r

(1 )

( 2)

We now demonstrate that this t statistic is equivalent to the F-ratio used for analysis

of regression. Specifically, the two test statistics are related by the basic equality F 5 t2 that was introduced in Chapters 12 and 13 (pp. 397 and 407). Recall that a t statistic with

df 5 n 2 1 can be squared to produce an equivalent F-ratio with df 5 1, n 2 1. We begin the demonstration of equivalence by removing the population correlation,

r, from the t equation. This value is always zero, as specified by the null hypothesis, and its removal does not affect the equation. Next, we square the t statistic to produce

the corresponding F-ratio.

t 2

5 5 2

F r

2 (1 )

( 2)

Finally, multiply the numerator and the denominator by SS Y to produce

t 2

5 5 2

F r SS

r SS

( )

(1 )( )

( 2)

You should recognize the numerator of the new F-ratio as SS regression

(see Equation 14.19).

Dividing by df 5 1, which does not change the value, produces MS regression

. Similarly, the

S I G N I F I CA N C E O F R E G R E S S I O N A N D

S I G N I F I CA N C E O F T H E CO R R E L AT I O N

Source SS df MS F

Regression 112.01 1 112.01 15.28

Residual 43.99 6 7.33

Total 156.00 7

TABLE 14.5

A summary table showing

the results of the analysis of

regression in Example 14.16.

4 9 6 CHAPTER 14 CORRELATION

denominator of the F-ratio is SS residual

(see Equation 14.20) divided by df 5 n – 2, which is MS

residual . Thus, the squared t value is identical to the F-ratio used in analysis of regression

(Equation 14.22) to evaluate the significance of the regression equation.

1. A set of n 5 18 pairs of scores produces a Pearson correlation of r 5 0.60 with SS

Y 5 100. Find SS

regression and SS

residual and compute the F-ratio to evaluate the

significance of the regression equation of predicting Y.

1. SS regression

5 36 with df 5 1. SS residual

5 64 with df 5 16. F 5 9.00. With df 5 1, 16, the F-ratio is significant with either a 5 .05 or a 5 .01.

L E A R N I N G C H E C K

ANSWER

SUMMARY

1. A correlation measures the relationship between two variables, X and Y. The relationship is described by three characteristics:

a. Direction. The sign of the correlation (� or 2) specifies the direction.

b. Form. The Pearson correlation measures the degree of straight line relationship, but other correlations measure the consistency or strength of the relation- ship, independent of any specific form.

c. Strength or consistency. The numerical value of the correlation measures the strength or consistency of the relationship from 0 (not consistent) to 1.00 (perfect).

2. The most commonly used correlation is the Pearson correlation, which measures the degree of linear rela- tionship. The Pearson correlation is identified by the letter r and is computed by

r SS SS

X Y

SP 5

In this formula, SP is the sum of products of deviations.

definitional formula: SP 5 o(X 2 M X )(Y 2 M

Y )

computational formula: SP XY X Y

n 5 2o

o o

3. A correlation between two variables should not be interpreted as implying a causal relationship. Simply because X and Y are related does not mean that X causes Y or that Y causes X.

4. To evaluate the strength of a relationship, you square the value of the correlation. The resulting value, r2, is called the coefficient of determination because it measures the portion of the variability in one variable that can be predicted using the relationship with the second variable.

5. A sample correlation, r, can be used to evaluate the significance of the corresponding population correlation, r, using a t statistic with df 5 n 2 2.

t 5 2r

(1 )

( 2)

6. A partial correlation measures the linear relationship between two variables by eliminating the influence of a third variable by holding it constant.

7. The Spearman correlation (r S ) measures the consistency

of direction in the relationship between X and Y—that is, the degree to which the relationship is one-directional, or monotonic. The Spearman correlation is computed by a two-stage process:

a. Rank the X scores and the Y scores separately. b. Compute the Pearson correlation using the ranks.

8. The point-biserial correlation is used to measure the strength of the relationship when one of the two variables is dichotomous. The dichotomous variable is coded using values of 0 and 1, and the regular Pearson formula is applied. Squaring the point- biserial correlation produces the same r2 value that is obtained to measure effect size for the independent- measures t test. When both variables, X and Y, are dichotomous, the phi-coefficient can be used to mea- sure the strength of the relationship. Both variables are coded 0 and 1, and the Pearson formula is used to compute the correlation.

9. When there is a general linear relationship between two variables, X and Y, it is possible to construct a linear equation that allows you to predict the Y value corresponding to any known value of X.

predicted Y value 5 Ŷ5 bX 1 a

RESOURCES 4 9 7

The technique for determining this equation is called regression. By using a least-squares method to mini- mize the error between the predicted Y values and the actual Y values, the best-fitting line is achieved when the linear equation has

b SS

5 5 5 2 SP

r s

s a M bMY

Y X and

10. The linear equation generated by regression (called the regression equation) can be used to compute a predicted Y value for any value of X. However, the prediction is not perfect, so for each Y value, there is a predicted portion and an unpredicted, or residual, portion. Overall, the predicted portion of the Y score variability is measured by r2, and the residual portion is measured by 1 2 r2.

predicted variability 5 SS regression

5 r2SS Y

unpredicted variability 5 SS residual

5 (1 2 r2)SS Y

11. The residual variability can be used to compute the standard error of estimate, which provides a measure of

the standard distance (or error) between the predicted Y values on the line and the actual data points. The stan- dard error of estimate is computed by

standard error of estimate 2

residual SS

n 5

2 MS5

residual

It is also possible to compute an F-ratio to evaluate the significance of the regression equation. The process is called analysis of regression and determines whether the equation predicts a significant portion of the variance for the Y scores. First a variance, or MS, value is computed for the predicted variability and the residual variability,

MS SS

df regression

regression

5 MS rresidual

residual

5 SS

where df regression

5 1 and df residual

5 n 22.

F MS

MS 5 5 2df n

regression

residual

with 1, 2

The F-ratio for analysis of regression is equivalent to the t statistic evaluating the significance of the correlation.

KEY TERMS

correlation (450)

positive correlation (451)

negative correlation (451)

perfect correlation (452)

Pearson correlation (453)

sum of products (SP) (454)

restricted range (460)

coefficient of determination (462)

correlation matrix (467)

partial correlation (463)

Spearman correlation (472)

linear relationship (477)

point-biserial correlation (478)

dichotomous variable (478)

phi-coefficient (480)

linear equation (482)

slope (482)

Y-intercept (482)

regression (484)

regression line (484)

least-squared-error solution (485)

regression equation for Y (486)

standard error of estimate (489)

predicted variability (SS regression

) (492)

unpredicted variability (SS residual

) (492)

analysis of regression (493)

RESOURCES

Go to CengageBrain.com to access Psychology CourseMate, where you will find an

interactive eBook, glossaries, flashcards, quizzes, statistics workshops, and more.

4 9 8 CHAPTER 14 CORRELATION

If your professor has assigned Aplia:

1. Sign in to your account.

2. Complete the corresponding exercises as required by your professor.

3. When finished, click “Grade It Now” to see which areas you have mastered, which

areas need more work, and detailed explanations of every answer.

General instructions for using SPSS are presented in Appendix D. Following are de-

tailed instructions for using SPSS to perform The Pearson, Spearman, point-biserial,

and partial correlations. Note: We focus on the Pearson correlation and then describe

how slight modifications to this procedure can be made to compute the Spearman,

point-biserial, and partial correlations. Separate instructions for the phi-coefficient are

presented at the end of this section.

Data Entry

The data are entered into two columns in the data editor, one for the X values

(VAR00001) and one for the Y values (VAR00002), with the two scores for each indi-

vidual in the same row.

Data Analysis

1. Click Analyze on the tool bar, select Correlate, and click on Bivariate.

2. One by one, move the labels for the two data columns into the Variables box.

(Highlight each label and click the arrow to move it into the box.)

3. The Pearson box should be checked but, at this point, you can switch to the

Spearman correlation by clicking the appropriate box.

4. Click OK.

SPSS Output

We used SPSS to compute the correlation for the data in Example 14.13, and the output

is shown in Figure 14.19. The program produces a correlation matrix showing all of

the possible correlations, including the correlation of X with X and the correlation of

Y with Y (both are perfect correlations). You want the correlation of X and Y, which is

contained in the upper right corner (or the lower left). The output includes the signifi-

cance level (p value or alpha level) for the correlation.

To compute a partial correlation, click Analyze on the tool bar, select Correlate, and

click on Partial. Move the column labels for the two variables to be correlated into the

Variables box and move the column label for the variable to be held constant into the

Controlling for box and click OK.

To compute the Spearman correlation, enter either the X and Y ranks or the X and

Y scores into the first two columns. Then follow the same Data Analysis instructions

that were presented for the Pearson correlation. At step 3 in the instructions, click on

the Spearman box before the final OK. (Note: If you enter X and Y scores into the data

editor, SPSS converts the scores to ranks before computing the Spearman correlation.)

RESOURCES 4 9 9

To compute the point-biserial correlation, enter the scores (X values) in the first

column and enter the numerical values (usually 0 and 1) for the dichotomous variable

in the second column. Then, follow the same Data Analysis instructions that were pre-

sented for the Pearson correlation.

The phi-coefficient can also be computed by entering the complete string of

0s and 1s into two columns of the SPSS data editor, then following the same Data

Analysis instructions that were presented for the Pearson correlation. However, this

can be tedious, especially with a large set of scores. The following is an alternative

procedure for computing the phi-coefficient with large data sets.

Data Entry

1. Enter the values 0, 0, 1, 1 (in order) into the first column of the SPSS data editor.

2. Enter the values 0, 1, 0, 1 (in order) into the second column.

3. Count the number of individuals in the sample who are classified with X 5 0 and Y 5 0. Enter this frequency in the top box in the third column of the data editor. Then, count how many have X 5 0 and Y 5 1 and enter the frequency in the second box of the third column. Continue with the number who have X 5 1 and Y 5 0, and finally the number who have X 5 1 and Y 5 1. You should end up with 4 values in column three.

4. Click Data on the Tool Bar at the top of the SPSS Data Editor page and select

Weight Cases at the bottom of the list.

5. Click the circle labeled Weight cases by, and then highlight the label for the

column containing your frequencies (VAR00003) on the left and move it into

the Frequency Variable box by clicking on the arrow.

6. Click OK.

Correlations

**. Correlation is significant at the 0.01 level (2-tailed).

VAR00001

VAR00001 1 .847

.008

8 8

.847**

.008

Pearson Correlation

Sig. (2-tailed)

VAR00002 Pearson Correlation

Sig. (2-tailed)

VAR00002

FIGURE 14.19

The SPSS output showing

the correlation for the data

in Example 14.13.

5 0 0 CHAPTER 14 CORRELATION

7. Click Analyze on the tool bar, select Correlate, and click on Bivariate.

8. One by one, move the labels for the two data columns containing the 0s and 1s

(probably VAR00001 and VAR00002) into the Variables box. (Highlight each

label and click the arrow to move it into the box.)

9. Verify that the Pearson box is checked.

10. Click OK.

SPSS Output

The program produces the same correlation matrix that was described for the Pearson

correlation. Again, you want the correlation between X and Y, which is in the upper

right corner (or lower left). Remember, with the phi-coefficient, the sign of the correla-

tion is meaningless.

Following are detailed instructions for using SPSS to perform the Linear Regression

presented in this chapter.

Data Entry

Enter the X values in one column and the Y values in a second column of the SPSS

data editor.

Data Analysis

1. Click Analyze on the tool bar, select Regression, and click on Linear.

2. In the left-hand box, highlight the column label for the Y values, then click the

arrow to move the column label into the Dependent Variable box.

3. Highlight the column label for the X values and click the arrow to move it into

the Independent Variable(s) box.

4. Click OK.

SPSS Output

We used SPSS to perform regression for the data in Examples 14.13, 14.14,

and 14.15 and the output is shown in Figure 14.20. The Model Summary table

presents the values for R, R2, and the standard error of estimate. (Note: R is simply the

Pearson correlation between X and Y.) The ANOVA table presents the analysis of regres-

sion evaluating the significance of the regression equation, including the F-ratio and the

level of significance (the p value or alpha level for the test). The Coefficients table summa-

rizes both the unstandardized and the standardized coefficients for the regression equation.

The table shows the values for the constant (a) and the coefficient (b). The standardized

coefficient is the beta values. Again, beta is simply the Pearson correlation between X and

Y. Finally, the table uses a t statistic to evaluate the significance of the predictor variable.

This is identical to the significance of the regression equation and you should find that t is

equal to the square root of the F-ratio from the analysis of regression.

FOCUS ON PROBLEM SOLVING

1. A correlation always has a value from �1.00 to –1.00. If you obtain a correlation

outside this range, then you have made a computational error.

FOCUS ON PROBLEM SOLVING 5 0 1

2. When interpreting a correlation, the sign and the numerical value must be

considered separately. Remember that the sign indicates the direction of the

relationship between X and Y. The numerical value reflects the strength of

the relationship. Therefore, a correlation of –0.90 is as strong as a correlation

of �0.90.

3. Before you begin to calculate a correlation, sketch a scatter plot of the data and

make an estimate of the correlation. (Is it positive or negative? Is it near 1 or

near 0?) After computing the correlation, compare your final answer with your

original estimate.

4. The F-ratio for analysis of regression is usually calculated using the actual

SS regression

and SS residual

. However, you can simply use r2 in place of SS regression

and

you can use 1 – r2 in place of SS residual

. Note: You must still use the correct df

values for the numerator and the denominator.

FIGURE 14.20

Portions of the SPSS output from the analysis of regression for the data in Examples 14.13, 14.14, and 14.15.

Model Summary

Model

.847 a

.718 .671 2.70801

R Square Adjusted R

Square Std. Error of

the Estimate

Coefficients a

Model

1 (Constant)

VAR00001

-1.000

2.000

2.260

.512

–.442

3.908

.674

.008.847

Unstandardized Coefficients

B Std. Error Beta t Sig.

Standardized Coefficients

ANOVA b

Model

1 Regression

Residual

Total

112.000

44.000

156.000

112.000 15.273 .008 a

7.333

Sum of Squares

df F Sig.Mean Square

5 0 2 CHAPTER 14 CORRELATION

DEMONSTRATION 14.1

CORRELATION AND REGRESSION

Calculate the Pearson correlation for the following data:

Person X Y

A 0 4 M X 5 4 with SS

X 5 40

B 2 1 M Y 5 6 with SS

Y 5 54

C 8 10 SP 5 40

D 6 9

E 4 6

Sketch a scatter plot. We have constructed a scatter plot for the data (Figure 14.21) and placed an envelope around the data points to make a preliminary estimate of the correlation. Note that the envelope is narrow and elongated. This indicates that the correlation is large—perhaps 0.80 to 0.90. Also, the correlation is positive because increases in X are generally accompanied by increases in Y. We also have sketched a straight line through the middle of the data points, roughly approximating the slope and Y-intercept of the regression line.

Compute the Pearson correlation. For these data, the Pearson correlation is

r SP

SS SS X Y

5 5 5 5 5 40

40(54)

2160

40 0.861

46 48.

In step 1, our preliminary estimate for the correlation was between �0.80 and �0.90.

The calculated correlation is consistent with this estimate.

Compute the values for the regression equation. The general form of the regression equation is

^ Y bX a b

SS a M

Y 5 1 5 5 2where and bM

S T E P 1

S T E P 2

S T E P 3

0 0 1 2 3 4 5 6 7 8 9 10

YFIGURE 14.21

The scatter plot for the

data of Demonstration

14.1. An envelope is

drawn around the points

to estimate the magnitude

of the correlation. A line

is drawn through the

middle of the envelope.

PROBLEMS 5 0 3

For these data, 40

40 1.00 andb a5 5 55 2 5 16 1(4) 2.00

Thus, the regression equation is Ŷ 5 (1)X 1 2.00 or simply, Ŷ 5 X 1 2.

Evaluate the significance of the correlation and the regression equation. The null hy- pothesis states that, for the population, there is no linear relationship between X and Y, and that the values obtained for the sample correlation and the regression equation are simply the result of sampling error. In terms of the correlation, H

0 says that the popula-

tion correlation is zero (r 5 0). In terms of the regression equation, H 0 says that the

equation does not predict a significant portion of the variance, or that the beta value is zero. The test can be conducted using either the t statistic for a correlation or the F-ratio for analysis of regression. Using the F-ratio, we obtain

SS regression

5 r2(SS Y ) 5 (0.861)2(54) 5 40.03 with df 5 1

SS residual

5 (1 2 r2)(SS Y ) 5 (1 2 0.8612)(54) 5 13.97 with df 5 n 5 2 5 3

F MS

MS 5 5

regression

residual

40.03/1

13.97/33 .605 8

With df 5 1, 3 and a 5 .05, the critical value is 10.13. Fail to reject the null hypoth- esis. The correlation and the regression equation are both not significant.

S T E P 4

PROBLEMS

1. a. What information is provided by the sign (1 or 2) of the Pearson correlation?

b. What information is provided by the numerical value of the Pearson correlation?

2. Calculate SP (the sum of products of deviations) for the following scores. Note: Both means are whole numbers, so the definitional formula works well.

X Y

0 2

1 4

4 5

3 3

7 6

3. Calculate SP (the sum of products of deviations) for the following scores. Note: Both means are decimal values, so the computational formula works well.

X Y

0 2

0 1

1 0

2 1

1 2

0 3

4. For the following scores,

X Y

1 3

3 5

2 1

2 3

a. Sketch a scatter plot and estimate the Pearson correlation.

b. Compute the Pearson correlation.

5. For the following scores,

X Y

1 7

4 2

1 3

1 6

2 0

0 6

2 3

1 5

a. Sketch a scatter plot and estimate the Pearson correlation.

b. Compute the Pearson correlation.

6. For the following scores,

5 0 4 CHAPTER 14 CORRELATION

X Y

1 6

4 1

1 4

1 3

3 1

a. Sketch a scatter plot and estimate the value of the Pearson correlation.

b. Compute the Pearson correlation.

7. With a small sample, a single point can have a large effect on the magnitude of the correlation. To create the following data, we started with the scores from problem 8 and changed the first X value from X 5 1 to X 5 6.

X Y

6 6

4 1

1 4

1 3

3 1

a. Sketch a scatter plot and estimate the value of the Pearson correlation.

b. Compute the Pearson correlation.

8. For the following set of scores,

X Y

6 4

3 1

5 0

6 7

4 2

6 4

a. Compute the Pearson correlation. b. Add 2 points to each X value and compute the

correlation for the modified scores. How does adding a constant to every score affect the value of the correlation?

c. Multiply each of the original X values by 2 and compute the correlation for the modified scores. How does multiplying each score by a constant affect the value of the correlation?

9. Judge and Cable (2010) report the results of a study demonstrating a negative relationship be- tween weight and income for a group of women professionals. Following are data similar to those obtained in the study. To simplify the weight vari- able, the women are classified into five categories that measure actual weight relative to height, from

1 5 thinnest to 5 5 heaviest. Income figures are annual income (in thousands), rounded to the nearest $1,000.

a. Calculate the Pearson correlation for these data. b. Is the correlation statistically significant? Use a

two-tailed test with a 5 .05.

Weight (X) Income (Y)

1 125

2 78

4 49

3 63

5 35

2 84

5 38

3 51

1 93

4 44

10. The researchers cited in the previous problem also examined the weight/salary relationship for men and found a positive relationship, suggesting that we have very different standards for men than for women (Judge & Cable, 2010). The following are data similar to those obtained for working men. Again, weight relative to height is coded in five categories from 1 5 thinnest to 5 5 heaviest. Income is recorded as thousands earned annually.

a. Calculate the Pearson correlation for these data. b. Is the correlation statistically significant? Use a

two-tailed test with a 5 .05.

Weight (X) Income (Y)

4 156

3 88

5 49

2 73

1 45

3 92

1 53

5 148

11. Identifying individuals with a high risk of Alzheimer’s disease usually involves a long series of cognitive tests. However, researchers have developed a 7-Minute Screen, which is a quick and easy way to accomplish the same goal. The question is whether the 7-Minute Screen is as effective as the complete series of tests. To address this ques- tion, Ijuin et al. (2008) administered both tests to a group of patients and compared the results. The following data represent results similar to those obtained in the study.

PROBLEMS 5 0 5

Patient 7-Minute Screen Cognitive Series

A 3 11

B 8 19

C 10 22

D 8 20

E 4 14

F 7 13

G 4 9

H 5 20

I 14 25

a. Compute the Pearson correlation to measure the degree of relationship between the two test scores.

b. Is the correlation statistically significant? Use a two-tailed test with a 5 .01.

c. What percentage of variance for the cognitive scores is predicted from the 7-Minute Screen scores? (Compute the value of r2.)

12. As we have noted in previous chapters, even a very small effect can be significant if the sample is large enough. Suppose, for example, that a researcher obtains a correlation of r 5 0.60 for a sample of n 5 10 participants.

a. Is this sample sufficient to conclude that a signifi- cant correlation exists in the population? Use a two-tailed test with a 5 .05.

b. If the sample had n 5 25 participants, is the cor- relation significant? Again, use a two-tailed test with a 5 .05.

13. A researcher measures three variables, X, Y, and Z, for each individual in a sample of n 5 25. The Pearson correlations for this sample are r

XY 5 0.8,

r XZ

5 0.6, and r YZ

5 0.7. a. Find the partial correlation between X and Y,

holding Z constant. b. Find the partial correlation between X and Z,

holding Y constant. (Hint: Simply switch the labels for the variables Y and Z to correspond with the labels in the equation.)

14. Problem 9 presented data showing a negative rela- tionship between weight and income for a sample of working women. However, weight was coded in five categories, which could be viewed as an ordinal scale rather than an interval or ratio scale. If so, a Spearman correlation is more appropriate than a Pearson correlation. Convert the weights and the incomes into ranks and compute the Spearman correlation for the scores in problem 9.

15. Problem 23 in Chapter 10 presented data showing that mature soccer players, who have a history of hitting soccer balls with their heads, had significantly

lower cognitive scores than mature swimmers, who do not suffer repeated blows to the head. The independent-measures t test produced t 5 2.11 with df 5 11 and a value of r2 5 0.288 (28.8%).

a. Convert the data from this problem into a form suitable for the point-biserial correlation (use 1 for the swimmers and 0 for the soccer players), and then compute the correlation.

b. Square the value of the point-biserial correlation to verify that you obtain the same r2 value that was computed in Chapter 10.

16. Studies have shown that people with high intel- ligence are generally more likely to volunteer as participants in research, but not for research that involves unusual experiences such as hypnosis. To examine this phenomenon, a researcher ad- ministers a questionnaire to a sample of college students. The survey asks for the student’s grade point average (as a measure of intelligence) and whether the student would like to take part in a future study in which participants would be hyp- notized. The results showed that 7 of the 10 lower- intelligence people were willing to participant but only 2 of the 10 higher-intelligence people were willing.

a. Convert the data to a form suitable for comput- ing the phi-coefficient. (Code the two intelligence categories as 0 and 1 for the X variable, and code the willingness to participate as 0 and 1 for the Y variable.)

b. Compute the phi-coefficient for the data.

17. Sketch a graph showing the line for the equation Y 5 –2X � 4. On the same graph, show the line for Y 5 X 24.

18. The regression equation is intended to be the “best fitting” straight line for a set of data. What is the criterion for “best fitting”?

19. A set of n 5 20 pairs of scores (X and Y values) has SS

X 5 16, SS

Y 5 100, and SP 5 32. If the mean for

the X values is M X 5 6 and the mean for the Y values

is M Y 5 20,

a. Calculate the Pearson correlation for the scores. b. Find the regression equation for predicting Y from

the X values.

20. A set of n 5 25 pairs of scores (X and Y values) pro- duces a regression equation of Ŷ5 3X – 2. Find the predicted Y value for each of the following X scores: 0, 1, 3, –2.

21. For the following set of data, a. Find the linear regression equation for predicting

Y from X. b. Calculate the standard error of estimate for the

regression equation.

5 0 6 CHAPTER 14 CORRELATION

25. Problem 9 examined the relationship between weight and income for a sample of n 5 10 women. Weights were classified in five categories and had a mean of M 5 3 with SS 5 20. Income, measured in thousands, had a mean score of M 5 66 with SS 5 7430, and SP 5 2359.

a. Find the regression equation for predicting income from weight. (Identify the income scores as X values and the weight scores as Y values.)

b. What percentage of the variance in the income is accounted for by the regression equation? (Compute the correlation, r, then find r2.)

c. Does the regression equation account for a significant portion of the variance in income? Use a5 .05 to evaluate the F-ratio.

26. There appears to be some evidence suggesting that earlier retirement may lead to memory decline (Rohwedder & Willis, 2010). The researchers gave a memory test to men and women aged 60 to 64 years in several countries that have different retirement ages. For each country, the research- ers recorded the average memory score and the percentage of individuals in the 60 to 64 age range who were retired. Note that a higher percentage retired indicates a younger retirement age for that country. The following data are similar to the results from the study. Use the data to find the regression equation for predicting memory scores from the percentage of people aged 60 to 64 who are retired.

Country % Retired (X) Memory Score (Y)

Sweden 39 9.3

U.S.A. 48 10.9

England 59 10.7

Germany 70 9.1

Spain 74 6.4

Netherlands 78 9.1

Italy 81 7.2

France 87 7.9

Belgium 88 8.5

Austria 91 9.0

27. The regression equation is computed for a set of n 5 18 pairs of X and Y values with a correlation of r 5 180 and SS

Y 5 100.

a. Find the standard error of estimate for the regres- sion equation.

b. How big would the standard error be if the sample size were n 5 38?

X Y

7 6

9 6

6 3

12 5

9 6

5 4

22. Does the regression equation from problem 21 ac- count for a significant portion of the variance in the Y scores? Use a 5 .05 to evaluate the F-ratio.

23. For the following scores,

X Y

3 6

6 1

3 4

3 3

5 1

a. Find the regression equation for predicting Y from X. b. Calculate the predicted Y value for each X.

24. Although you might suspect that dissatisfied people would be the most likely individuals to participate in political activities in an attempt to change things, research tends to show just the opposite. Flavin and Keane (2011) found a positive relationship between life satisfaction and political participation, which included activities such as attending rallies, contributing to can- didates, and displaying a yard sign. Following are data similar to those obtained in the study.

Life Satisfaction Political Participation

5 4

8 7

3 2

6 9

3 5

1 3

4 6

2 4

a. Find the regression equation for predicting politi- cal participation from life satisfaction.

b. Using a 5 .05, test the significance of the regression equation.

PROBLEMS 5 0 7

28. a. One set of 20 pairs of scores, X and Y values, produces a correlation of r 5 0.70. If SS

Y 5

150, find the standard error of estimate for the regression line.

b. A second set of 20 pairs of X and Y values produces a correlation of r 5 0.30. If SS

Y 5 150,

find the standard error of estimate for the regression line.

29. a. A researcher computes the regression equation for a sample of n 5 25 pairs of scores, X and Y values. If an analysis of regression is used to test the signifi- cance of the equation, what are the df values for the F-ratio?

b. A researcher evaluating the significance of a regression equation obtains an F-ratio with df 5 1, 18. How many pairs of scores, X and Y values, are in the sample?

Aplia for Essentials of Statistics for the Behavioral Sciences

After reading, go to “Resources” at the end of this chapter for

an introduction on how to use Aplia’s homework and learning

resources.

The Chi-Square Statistic: Tests for Goodness of Fit and Independence 15.1 Parametric and Nonparametric

Statistical Tests

15.2 The Chi-Square Test for Goodness of Fit

15.3 The Chi-Square Test for Independence

15.4 Measuring Effect Size for the Chi-Square Test for Independence

15.5 Assumptions and Restrictions for Chi-Square Tests

Summary

Focus on Problem Solving

Demonstrations 15.1 and 15.2

Problems

C H A P T E R

15 Tools You Will Need The following items are considered essential background material for this chapter. If you doubt your knowledge of any of these items, you should review the appropriate chapter or section before proceeding.

• Proportions (math review, Appendix A) • Frequency distributions (Chapter 2)

5 1 0 CHAPTER 15 THE CHI-SQUARE STATISTIC: TESTS FOR GOODNESS OF FIT AND INDEPENDENCE

PARAMETRIC AND NONPARAMETRIC STATISTICAL TESTS

All of the statistical tests that we have examined thus far are designed to test hypotheses

about specific population parameters. For example, we used t tests to assess hypotheses

about a population mean (µ) or mean difference (µ 1 – µ

2 ). In addition, these tests typi-

cally make assumptions about other population parameters. Recall that, for analysis of

variance (ANOVA), the population distributions are assumed to be normal and homo-

geneity of variance is required. Because these tests all concern parameters and require

assumptions about parameters, they are called parametric tests.

Another general characteristic of parametric tests is that they require a numerical

score for each individual in the sample. The scores then are added, squared, averaged,

and otherwise manipulated using basic arithmetic. In terms of measurement scales,

parametric tests require data from an interval or a ratio scale (see Chapter 1).

Often, researchers are confronted with experimental situations that do not conform

to the requirements of parametric tests. In these situations, it may not be appropriate to

use a parametric test. Remember that, when the assumptions of a test are violated, the

test may lead to an erroneous interpretation of the data. Fortunately, there are several

hypothesis-testing techniques that provide alternatives to parametric tests. These alter-

natives are called nonparametric tests.

In this chapter, we introduce two commonly used examples of nonparametric tests.

Both tests are based on a statistic known as chi-square and both tests use sample data

to evaluate hypotheses about the proportions or relationships that exist within popula-

tions. Note that the two chi-square tests, like most nonparametric tests, do not state

hypotheses in terms of a specific parameter and they make few (if any) assumptions

about the population distribution. For the latter reason, nonparametric tests sometimes

are called distribution-free tests.

One of the most obvious differences between parametric and nonparametric tests

is the type of data they use. All of the parametric tests that we have examined so far

require numerical scores. For nonparametric tests, on the other hand, the participants

are usually just classified into categories such as Democrat and Republican, or High,

Medium, and Low IQ. Note that these classifications involve measurement on nominal

or ordinal scales, and they do not produce numerical values that can be used to calcu-

late means and variances. Instead, the data for many nonparametric tests are simply

frequencies—for example, the number of Democrats and the number of Republicans in

a sample of n 5 100 registered voters. Occasionally, you have a choice between using a parametric and a nonparametric

test. Changing to a nonparametric test usually involves transforming the data from nu-

merical scores to nonnumerical categories. For example, you could start with numerical

scores measuring self-esteem and create three categories consisting of high, medium,

and low self-esteem. In most situations, the parametric test is preferred because it is

more likely to detect a real difference or a real relationship. However, there are situa-

tions for which transforming scores into categories might be a better choice.

1. Occasionally, it is simpler to obtain category measurements. For example, it is

easier to classify students as high, medium, or low in leadership ability than to

obtain a numerical score measuring each student’s ability.

2. The original scores may violate some of the basic assumptions that underlie

certain statistical procedures. For example, the t tests and ANOVA assume that

the data come from normal distributions. Also, the independent-measures tests

assume that the different populations all have the same variance (the homogeneity-

of-variance assumption). If a researcher suspects that the data do not satisfy these

15.1

SECTION 15.2 / THE CHI-SQUARE TEST FOR GOODNESS OF FIT 5 1 1

assumptions, it may be safer to transform the scores into categories and use a

nonparametric test to evaluate the data.

3. The original scores may have unusually high variance. Variance is a major compo-

nent of the standard error in the denominator of t statistics and the error term in the

denominator of F-ratios. Thus, large variance can greatly reduce the likelihood that

these parametric tests will find significant differences. Converting the scores to cat-

egories essentially eliminates the variance. For example, all individuals fit into three

categories (high, medium, and low) no matter how variable the original scores are.

4. Occasionally, an experiment produces an undetermined, or infinite, score. For

example, a rat may show no sign of solving a particular maze after hundreds of

trials. This animal has an infinite, or undetermined, score. Although there is no

absolute number that can be assigned, you can say that this rat is in the highest

category, and then classify the other scores according to their numerical values.

THE CHI-SQUARE TEST FOR GOODNESS OF FIT

Parameters such as the mean and the standard deviation are the most common way to

describe a population, but there are situations in which a researcher has questions about

the proportions or relative frequencies for a distribution. For example,

How does the number of women lawyers compare with the number of men in the

profession?

Of the two leading brands of cola, which is preferred by most Americans?

In the past 10 years, has there been a significant change in the proportion of college

students who declare a business major?

Note that each of the preceding examples asks a question about proportions in the

population. In particular, we are not measuring a numerical score for each individual.

Instead, the individuals are simply classified into categories and we want to know what

proportion of the population is in each category. The chi-square test for goodness of

fit is specifically designed to answer this type of question. In general terms, this chi-

square test uses the proportions obtained for sample data to test hypotheses about the

corresponding proportions in the population.

The chi-square test for goodness of fit uses sample data to test hypotheses about

the shape or proportions of a population distribution. The test determines how

well the obtained sample proportions fit the population proportions specified by

the null hypothesis.

Recall from Chapter 2 that a frequency distribution is defined as a tabulation of

the number of individuals located in each category of the scale of measurement. In

a frequency distribution graph, the categories that make up the scale of measurement

are listed on the X-axis. In a frequency distribution table, the categories are listed in

the first column. With chi-square tests, however, it is customary to present the scale of

measurement as a series of boxes, with each box corresponding to a separate category

on the scale. The frequency corresponding to each category is simply presented as a

number written inside the box. Figure 15.1 shows how the grade distribution for a class

of n 5 40 students can be presented as a graph, a table, or a series of boxes. The scale of measurement for this example consists of five grade categories (A, B, C, D, and F).

15.2

D E F I N I T I O N

The name of the test comes

from the Greek letter x (chi, pronounced “kye”),

which is used to identify

the test statistic.

5 1 2 CHAPTER 15 THE CHI-SQUARE STATISTIC: TESTS FOR GOODNESS OF FIT AND INDEPENDENCE

For the chi-square test of goodness of fit, the null hypothesis specifies the proportion (or

percentage) of the population in each category. For example, a hypothesis might state

that 50% of all lawyers are men and 50% are women. The simplest way of presenting

this hypothesis is to put the hypothesized proportions in the series of boxes representing

the scale of measurement:

H 0 :

Men Women

50% 50%

Although it is conceivable that a researcher could choose any proportions for the null

hypothesis, there usually is some well-defined rationale for stating a null hypothesis.

Generally H 0 falls into one of the following categories:

1. No Preference, Equal Proportions. The null hypothesis often states that there

is no preference among the different categories. In this case, H 0 states that the

population is divided equally among the categories. For example, a hypothesis

stating that there is no preference among the three leading brands of soft drinks

would specify a population distribution as follows:

H 0 :

Brand X Brand Y Brand Z

The no-preference hypothesis is used in situations in which a researcher wants

to determine whether there are any preferences among the categories, or

whether the proportions differ from one category to another.

Because the null hypothesis for the goodness-of-fit test specifies an exact dis-

tribution for the population, the alternative hypothesis (H 1 ) simply states that

the population distribution has a different shape from that specified in H 0 . If

T H E N U L L H Y P OT H E S I S F O R T H E G O O D N E S S -

O F- F I T T E ST

(Preferences in the population

are equally divided among the

three soft drinks.)

FIGURE 15.1

A distribution of grades for a sample of n 5 40 students. The same frequency distribution is shown as a bar graph, as a table, and with the frequencies written in a series of boxes.

A B C D F

A B C D E f

Grade

X f

A B C D F

5 11 16 6 25

11 16 6 2

SECTION 15.2 / THE CHI-SQUARE TEST FOR GOODNESS OF FIT 5 1 3

the null hypothesis states that the population is equally divided among three

categories, then the alternative hypothesis says that the population is not divided

equally.

2. No Difference from a Known Population. The null hypothesis can state

that the proportions for one population are not different from the propor-

tions that are known to exist for another population. For example, suppose

it is known that 28% of the licensed drivers in the state are younger than

30 years old and 72% are 30 or older. A researcher might wonder whether

this same proportion holds for the distribution of speeding tickets. The null

hypothesis would state that tickets are handed out equally across the popu-

lation of drivers, so there is no difference between the age distribution for

drivers and the age distribution for speeding tickets. Specifically, the null

hypothesis would be

H 0 :

Tickets Given to Drivers

Younger than 30

Tickets Given to Drivers

30 or Older

28% 72%

The no-difference hypothesis is used when a specific population distribution is

already known. For example, you may have a known distribution from an earlier

time, and the question is whether there has been any change in the proportions. Or,

you may have a known distribution for one population (drivers) and the question is

whether a second population (speeding tickets) has the same proportions.

Again, the alternative hypothesis (H 1 ) simply states that the population proportions

are not equal to the values specified by the null hypothesis. For this example, H 1

would state that the number of speeding tickets is disproportionately high for one

age group and disproportionately low for the other.

The data for a chi-square test are remarkably simple. There is no need to calculate

a sample mean or SS; you just select a sample of n individuals and count how

many are in each category. The resulting values are called observed frequencies.

The symbol for observed frequency is f o . For example, the following data represent

observed frequencies for a sample of 40 college students. The students were clas-

sified into three categories based on the number of times they reported exercising

each week.

No Exercise 1 Time a Week More Than

Once a Week

n 5 4015 19 6

Notice that each individual in the sample is classified into one and only one of the

categories. Thus, the frequencies in this example represent three completely separate

groups of students: 15 who do not exercise regularly, 19 who average once a week, and

6 who exercise more than once a week. Also note that the observed frequencies add up

to the total sample size: of o 5 n. Finally, you should realize that we are not assigning

individuals to categories. Instead, we are simply measuring individuals to determine the

category in which they belong.

T H E DATA F O R T H E G O O D N E S S - O F- F I T T E ST

(Proportions for the population

of tickets are not different from

proportions for drivers.)

5 1 4 CHAPTER 15 THE CHI-SQUARE STATISTIC: TESTS FOR GOODNESS OF FIT AND INDEPENDENCE

The observed frequency is the number of individuals from the sample who are classi-

fied in a particular category. Each individual is counted in one and only one category.

The general goal of the chi-square test for goodness of fit is to compare the data (the

observed frequencies) with the null hypothesis. The problem is to determine how well

the data fit the distribution specified in H 0 —hence the name goodness of fit.

The first step in the chi-square test is to construct a hypothetical sample that represents

how the sample distribution would look if it were in perfect agreement with the propor-

tions stated in the null hypothesis. Suppose, for example, the null hypothesis states that the

population is distributed in three categories with the following proportions:

H 0 :

Category A Category B Category C

25% 50% 25%

If this hypothesis is correct, how would you expect a random sample of n 5 40 in- dividuals to be distributed among the three categories? It should be clear that your best

strategy is to predict that 25% of the sample would be in category A, 50% would be in

category B, and 25% would be in category C. To find the exact frequency expected for

each category, multiply the sample size (n) by the proportion (or percentage) from the

null hypothesis. For this example, you would expect

25% of 40 5 0.25(40) 5 10 individuals in category A

50% of 40 5 0.50(40) 5 20 individuals in category B

25% of 40 5 0.25(40) 5 10 individuals in category C

The frequency values predicted from the null hypothesis are called expected fre-

quencies. The symbol for expected frequency is f e , and the expected frequency for each

category is computed by

expected frequency 5 f e 5 pn (15.1)

where p is the proportion stated in the null hypothesis and n is the sample size.

The expected frequency for each category is the frequency value that is pre-

dicted from the proportions in the null hypothesis and the sample size (n). The

expected frequencies define an ideal, hypothetical sample distribution that would

be obtained if the sample proportions were in perfect agreement with the propor-

tions specified in the null hypothesis.

Note that the no-preference null hypothesis always produces equal f e values

for all categories because the proportions (p) are the same for all categories. On

the other hand, the no-difference null hypothesis typically does not produce equal

values for the expected frequencies because the hypothesized proportions typically

vary from one category to another. You also should note that the expected frequen-

cies are calculated, hypothetical values and the numbers that you obtain may be

decimals or fractions. The observed frequencies, on the other hand, always repre-

sent real individuals and always are whole numbers.

D E F I N I T I O N

E X P E C T E D F R E Q U E N C I E S

D E F I N I T I O N

(The population is distributed

across the three categories with

25% in category A, 50% in cate-

gory B, and 25% in category C.)

SECTION 15.2 / THE CHI-SQUARE TEST FOR GOODNESS OF FIT 5 1 5

The general purpose of any hypothesis test is to determine whether the sample data

support or refute a hypothesis about the population. In the chi-square test for goodness

of fit, the sample is expressed as a set of observed frequencies (f o values), and the null

hypothesis is represented by a set of expected frequencies (f e values). The chi-square

statistic simply measures how well the data (f o ) fit the hypothesis (f

e ). The symbol for

the chi-square statistic is x2. The formula for the chi-square statistic is

chi-square 5 x 5  2

2 f f

o e

( )

(15.2)

As the formula indicates, the value of chi-square is computed by the following steps:

1. Find the difference between f o (the data) and f

e (the hypothesis) for each category.

2. Square the difference. This ensures that all values are positive.

3. Next, divide the squared difference by f e .

4. Finally, add the values from all of the categories.

The first two steps determine the numerator of the chi-square statistic and should

be easy to understand. Specifically, the numerator measures how much difference

there is between the data (the f o values) and the hypothesis (represented by the

f e values). The final step is also reasonable: we add the values to obtain the total

discrepancy between the data and the hypothesis. Thus, a large value for chi-square

indicates that the data do not fit the hypothesis, and leads us to reject the null

hypothesis.

However, the third step, which determines the denominator of the chi-square statistic,

is not so obvious. Why must we divide by f e before we add the category values? The

answer to this question is that the obtained discrepancy between f o and f

e is viewed as

relatively large or relatively small depending on the size of the expected frequency. This

point is demonstrated in the following analogy.

Suppose that you were going to throw a party and you expected 1,000 people to

show up. However, at the party you counted the number of guests and observed that

1,040 actually showed up. Forty more guests than expected are no major problem

when all along you were planning for 1,000. There probably will be enough beer

and potato chips for everyone. On the other hand, suppose you had a party and you

expected 10 people to attend but instead 50 actually showed up. Forty more guests

in this case spell big trouble. How “significant” the discrepancy is depends in part

on what you were originally expecting. With very large expected frequencies, al-

lowances are made for more error between f o and f

e . This is accomplished in the

chi-square formula by dividing the squared discrepancy for each category, (f o 2 f

e )2,

by its expected frequency.

It should be clear from the chi-square formula that the numerical value of chi-square is

a measure of the discrepancy between the observed frequencies (data) and the expected

frequencies (H 0 ). As usual, the sample data are not expected to provide a perfectly

accurate representation of the population. In this case, the proportions, or observed

frequencies, in the sample are not expected to be exactly equal to the proportions in

the population. Thus, if there are small discrepancies between the f o and f

e values, we

obtain a small value for chi-square and we conclude that there is a good fit between the

data and the hypothesis (fail to reject H 0 ). However, when there are large discrepancies

T H E C H I - S Q UA R E STAT I ST I C

T H E C H I - S Q UA R E D I ST R I B U T I O N A N D

D E G R E E S O F F R E E D O M

5 1 6 CHAPTER 15 THE CHI-SQUARE STATISTIC: TESTS FOR GOODNESS OF FIT AND INDEPENDENCE

between f o and f

e , we obtain a large value for chi-square and conclude that the data do

not fit the hypothesis (reject H 0 ). To decide whether a particular chi-square value is

“large” or “small,” we must refer to a chi-square distribution. This distribution is the set

of chi-square values for all of the possible random samples when H 0 is true. Much like

other distributions that we have examined (t distribution, F distribution), the chi-square

distribution is a theoretical distribution with well-defined characteristics. Some of these

characteristics are easy to infer from the chi-square formula.

1. The formula for chi-square involves adding squared values, so you can

never obtain a negative value. Thus, all chi-square values are zero or

larger.

2. When H 0 is true, you expect the data (f

o values) to be close to the hypothesis

(f e values). Thus, we expect chi-square values to be small when H

0 is true.

These two factors suggest that the typical chi-square distribution is positively

skewed (Figure 15.2). Note that small values, near zero, are expected when H 0 is true

and large values (in the right-hand tail) are very unlikely. Thus, unusually large values

of chi-square form the critical region for the hypothesis test.

Although the typical chi-square distribution is positively skewed, there is one

other factor that plays a role in the exact shape of the chi-square distribution—the

number of categories. Recall that the chi-square formula requires that you add values

from every category. The more categories you have, the more likely it is that you will

obtain a large sum for the chi-square value. On average, chi-square is larger when

you are adding values from 10 categories than when you are adding values from only

3 categories. As a result, there is a whole family of chi-square distributions, with the

exact shape of each distribution determined by the number of categories used in the

study. Technically, each specific chi-square distribution is identified by degrees of

freedom (df), which are determined by the number of categories. For the goodness-

of-fit test, the degrees of freedom are

df 5 C 2 1 (15.3)

where C is the number of categories. A brief discussion of this df formula is pre-

sented in Box 15.1. Figure 15.3 shows the general relationship between df and the

shape of the chi-square distribution. Note that the chi-square values tend to get

larger (shift to the right) as the number of categories and the degrees of freedom

increase.

χ20

Critical region

FIGURE 15.2

Chi-square distributions

are positively skewed. The

critical region is placed

in the extreme tail, which

reflects large chi-square

values.

Caution: The df for a chi-

square test is not related to

sample size (n), as it is in

most other tests.

SECTION 15.2 / THE CHI-SQUARE TEST FOR GOODNESS OF FIT 5 1 7

Recall that a large value for the chi-square statistic indicates a big discrepancy between

the data and the hypothesis, and suggests that we reject H 0 . To determine whether a

particular chi-square value is significantly large, you must consult the table entitled The

Chi-Square Distribution (Appendix B). A portion of the chi-square table is shown in

Table 15.1. The first column lists df values for the chi-square test, and the other column

heads are proportions (alpha levels) in the extreme right-hand tail of the distribution.

The numbers in the body of the table are the critical values of chi-square. The table

shows, for example, that when the null hypothesis is true and df 5 3, only 5% (a 5 .05) of the chi-square values are greater than 7.81, and only 1% (a 5 .01) are greater than 11.34. Thus, with df 5 3, any chi-square value greater than 7.81 has a probability of p , .05, and any value greater than 11.34 has a probability of p , .01.

LO CAT I N G T H E C R I T I CA L R E G I O N F O R

A C H I - S Q UA R E T E ST

FIGURE 15.3

The shape of the chi-square

distribution for different

values of df. As the number

of categories increases,

the peak (mode) of the

distribution has a larger

chi-square value.

χ20

df = 1

df = 5

df = 9

BOX

15.1 A CLOSER LOOK AT DEGREES OF FREEDOM

Degrees of freedom for the chi-square test literally

measure the number of free choices that exist when

you are determining the null hypothesis or the expected

frequencies. For example, when you are classifying

individuals into three categories, you have exactly two

free choices in stating the null hypothesis. You may

select any two proportions for the first two categories,

but then the third proportion is determined. If you

hypothesize 25% in the first category and 50% in the

second category, then the third category must be 25%

to account for 100% of the population.

Category A Category B Category C

25% 50% ?

In general, you are free to select proportions for all

but one of the categories, but then the final proportion

is determined by the fact that the entire set must total

100%. Thus, you have C – 1 free choices, where C

is the number of categories: degrees of freedom, df,

equal C 2 1.

5 1 8 CHAPTER 15 THE CHI-SQUARE STATISTIC: TESTS FOR GOODNESS OF FIT AND INDEPENDENCE

The following example demonstrates the complete process of hypothesis testing with

the goodness-of-fit test.

A psychologist examining art appreciation selected an abstract painting that had no obvi-

ous top or bottom. Hangers were placed on the painting so that it could be hung with any

one of the four sides at the top. The painting was shown to a sample of n 5 50 participants, and each was asked to hang the painting in the orientation that looked correct. The follow-

ing data indicate how many people chose each of the four sides to be placed at the top:

Top up (correct) Bottom up Left side up Right side up

18 17 7 8

The question for the hypothesis test is whether there are any preferences among the

four possible orientations. Are any of the orientations selected more (or less) often than

would be expected simply by chance?

State the hypotheses and select an alpha level. The hypotheses can be stated as follows:

H 0 : In the general population, there is no preference for any specific orientation.

Thus, the four possible orientations are selected equally often, and the popu-

lation distribution has the following proportions:

Top up (correct) Bottom up Left side up Right side up

25% 25% 25% 25%

H 1 : In the general population, one or more of the orientations is preferred

over the others.

We use a 5 .05.

Locate the critical region. For this example, the value for degrees of freedom is

df 5 C 2 1 5 4 2 1 5 3

E X A M P L E O F T H E C H I - S Q UA R E T E ST F O R G O O D N E S S O F F I T

E X A M P L E 1 5 . 1

S T E P 1

S T E P 2

Proportion in Critical Region

0.10 0.05 0.025 0.01 0.005

1 2.71 3.84 5.02 6.63 7.88

2 4.61 5.99 7.38 9.21 10.60

3 6.25 7.81 9.35 11.34 12.84

4 7.78 9.49 11.14 13.28 14.86

5 9.24 11.07 12.83 15.09 16.75

6 10.64 12.59 14.45 16.81 18.55

7 12.02 14.07 16.01 18.48 20.28

8 13.36 15.51 17.53 20.09 21.96

9 14.68 16.92 19.02 21.67 23.59

TABLE 15.1

A portion of the table of

critical values for the chi-

square distribution.

SECTION 15.2 / THE CHI-SQUARE TEST FOR GOODNESS OF FIT 5 1 9

For df 5 3 and a 5 .05, the table of critical values for chi-square indicates that the critical x2 has a value of 7.81. The critical region is sketched in Figure 15.4.

Calculate the chi-square statistic. The calculation of chi-square is actually a two-stage

process. First, you must compute the expected frequencies from H 0 and then calculate

the value of the chi-square statistic. For this example, the null hypothesis specifies

that one-quarter of the population (p 5 25%) will be in each of the four categories. According to this hypothesis, we should expect one-quarter of the sample to be in

each category. With a sample of n 5 50 individuals, the expected frequency for each category is

f pn e

5 5 5 1

4 50 12 5( ) .

The observed frequencies and the expected frequencies are presented in

Table 15.2.

Using these values, the chi-square statistic may now be calculated.

x 2 2

2 2 18 12 5

12 5

17 12 5

5  2

5 2

1 2

f f

o e

( )

( ) ( ). .

.55

7 12 5

12 5

8 12 5

12 5

30 25

12 5

2 2

1 2

5 1

( ) ( )

. . .

12 5

30 25

12 5

20 25

12 5

2 42 1 62 2 42

1 1

5 1 1 11

1 62

8 08

State a decision and a conclusion. The obtained chi-square value is in the critical region.

Therefore, H 0 is rejected, and the researcher may conclude that the four orientations are

not equally likely to be preferred. Instead, there are significant differences among the

four orientations, with some selected more often and others less often than would be

expected by chance.

S T E P 3

S T E P 4

Expected frequencies are

computed and may be decimal

values. Observed frequencies

are always whole numbers.

7.810

df = 3 α = .05

FIGURE 15.4

For Example 15.1, the

critical region begins at a

chi-square value of 7.81.

5 2 0 CHAPTER 15 THE CHI-SQUARE STATISTIC: TESTS FOR GOODNESS OF FIT AND INDEPENDENCE

IN THE LITERATURE

REPORTING THE RESULTS FOR CHI-SQUARE

APA style specifies the format for reporting the chi-square statistic in scientific journals.

For the results of Example 15.1, the report might state:

The participants showed significant preferences among the four orientations for

hanging the painting, x2(3, n 5 50) 5 8.08, p , .05.

Note that the form of the report is similar to that of other statistical tests we have

examined. Degrees of freedom are indicated in parentheses following the chi-square

symbol. Also contained in the parentheses is the sample size (n). This additional infor-

mation is important because the degrees of freedom value is based on the number of

categories (C), not sample size. Next, the calculated value of chi-square is presented,

followed by the probability that a Type I error has been committed. Because we ob-

tained an extreme, very unlikely value for the chi-square statistic, the probability is

reported as less than the alpha level. Additionally, the report may provide the observed

frequencies (f o ) for each category. This information may be presented in a simple sen-

tence or in a table.

We began this chapter with a general discussion of the difference between parametric

tests and nonparametric tests. In this context, the chi-square test for goodness of fit is

an example of a nonparametric test; that is, it makes no assumptions about the param-

eters of the population distribution, and it does not require data from an interval or ratio

scale. In contrast, the single-sample t test introduced in Chapter 9 is an example of a

parametric test: It assumes a normal population, it tests hypotheses about the popula-

tion mean (a parameter), and it requires numerical scores that can be added, squared,

divided, and so on.

Although the chi-square test and the single-sample t are clearly distinct, they are also

very similar. In particular, both tests are intended to use the data from a single sample

to test hypotheses about a single population.

The primary factor that determines whether you should use the chi-square test or the

t test is the type of measurement that is obtained for each participant. If the sample data

consist of numerical scores (from an interval or ratio scale), then it is appropriate to

compute a sample mean and use a t test to evaluate a hypothesis about the population

mean. For example, a researcher could measure the IQ for each individual in a sample

of registered voters. A t test could then be used to evaluate a hypothesis about the mean

IQ for the entire population of registered voters. On the other hand, if the individu-

als in the sample are classified into nonnumerical categories (on a nominal or ordinal

scale), then the researcher would use a chi-square test to evaluate a hypothesis about the

population proportions. For example, a researcher could classify people according to

gender by simply counting the number of males and females in a sample of registered

voters. A chi-square test would then be appropriate to evaluate a hypothesis about the

population proportions.

G O O D N E S S O F F I T A N D T H E S I N G L E - SA M P L E

t T E ST

TABLE 15.2

The observed frequencies

and the expected frequen-

cies for the chi-square test in

Example 15.1.

Observed Frequencies

Top up (Correct)

Bottom up

Left Side up

Right Side up

18 17 7 8

Expected Frequencies 12.5 12.5 12.5 12.5

SECTION 15.3 / THE CHI-SQUARE TEST FOR INDEPENDENCE 5 2 1

THE CHI-SQUARE TEST FOR INDEPENDENCE

The chi-square statistic may also be used to test whether there is a relationship between

two variables. In this situation, each individual in the sample is measured or classified

on two separate variables. For example, Table 15.3 shows the data from a classic experi-

ment demonstrating how eyewitness memory can be influenced by the questions that

witnesses are asked (Loftus and Palmer, 1974). In the study, a sample of 150 students

watched a film of an automobile accident. After watching the film, the students were

separated into three groups and questioned about the accident. One group was asked,

“About how fast were the cars going when they smashed into each other?” Another

group received the same question except that the verb was changed to “hit” instead of

“smashed into.” A third group served as a control and was not asked any question about

the speed of the two cars. A week later, the participants returned and were asked if they

remembered seeing any broken glass in the accident. (There was no broken glass in

the film.) Notice that the researchers are manipulating the form of the initial question

and then measuring a yes/no response to a follow-up question 1 week later. Table 15.3

shows the structure of this design as a matrix with the independent variable (different

groups) determining the rows of the matrix and the two categories for the dependent

variable (yes/no) determining the columns. The number in each cell of the matrix is

the frequency count showing how many participants are classified in that category.

For example, of the 50 students who heard the word smashed, there were 16 (32%)

who claimed to remember seeing broken glass even though there was none in the film.

By comparison, only 7 of the 50 students (14%) who heard the word hit said they re-

called seeing broken glass. Note that the data consist of frequencies, not scores, from a

15.3

1. For a chi-square test, the observed frequencies are always whole numbers.

(True or false?)

2. For a chi-square test, the expected frequencies are always whole numbers.

(True or false?)

3. A researcher has developed three different designs for a computer keyboard. A sample

of n 5 60 participants is obtained, and each individual tests all three keyboards and identifies his or her favorite. The frequency distribution of preferences is as follows:

Design A Design B Design C

n 5 6023 12 25

a. What is the df value for the chi-square statistic?

b. Assuming that the null hypothesis states that there are no preferences among

the three designs, find the expected frequencies for the chi-square test.

1. True. Observed frequencies are obtained by counting people in the sample.

2. False. Expected frequencies are computed and may be fractions or decimal values.

3. a. df 5 2

b. According to the null hypothesis, one-third of the population would prefer each design. The expected frequencies should show one-third of the sample preferring each design.

The expected frequencies are all 20.

L E A R N I N G C H E C K

ANSWERS

5 2 2 CHAPTER 15 THE CHI-SQUARE STATISTIC: TESTS FOR GOODNESS OF FIT AND INDEPENDENCE

sample. The researchers would like to use the frequencies from the sample to test a hy-

pothesis about the corresponding frequency distribution in the population. Specifically,

the researchers would like to know whether the sample data provide enough evidence

to conclude that there is a significant relationship between eyewitnesses’ memories and

the questions they were asked during the initial interview.

You should recognize that the eyewitness study shown in Table 15.3 is an example

of experimental research (Chapter 1, page 13). The researchers manipulated an inde-

pendent variable and controlled other variables by randomly assigning the participants

to groups or treatment conditions. However, similar data are often obtained from non-

experimental studies. For example, Table 15.4 presents hypothetical data for a sample

of n 5 200 students who have been classified by personality and color preference. The number in each box, or cell, of the matrix indicates the frequency, or number of

individuals in that particular group. In Table 15.4, for example, there are 10 students

who were classified as introverted and who selected red as their preferred color. To

obtain these data, the researcher first selects a random sample of n 5 200 students. Each student is then given a personality test and is asked to select a preferred color

from among the four choices. Note that the classification is based on the measurements

for each student; the researcher does not assign students to categories. Once again, the

goal is to use the frequencies from the sample to test a hypothesis about the population

frequency distribution. Specifically, are these data sufficient to conclude that there is

a significant relationship between personality and color preference in the population

of students?

The procedure for using sample frequencies to evaluate hypotheses concerning rela-

tionships between variables involves another test using the chi-square statistic. In this

situation, however, the test is called the chi-square test for independence.

The chi-square test for independence uses the frequency data from a sample

to evaluate the relationship between two variables in the population. Each

individual in the sample is classified on both of the two variables, creating a

two-dimensional frequency-distribution matrix. The frequency distribution for

the sample is then used to test hypotheses about the corresponding frequency

distribution for the population.

D E F I N I T I O N

TABLE 15.3

A frequency distribution table

showing the number of par-

ticipants who answered either

yes or no when asked whether

they recalled seeing any broken

glass 1 week after witness-

ing a video of an automobile

accident. Immediately after the

video, one group was asked

how fast the cars were going

when they “smashed into” each

other. A second group was

asked how fast the cars were

going when they “hit” each

other. A third group served as

a control and was not asked

about the speed of the cars.

Verb Used

to Ask About

the Speed of

the Cars

Smashed into

Hit

Control (Not Asked)

Response to the

Question: Did You See

Any Broken Glass?

Yes No

16 34

7 43

6 44

Introvert

Red Yellow Green Blue

10 3 15 22 50

Extrovert 90 17 25 18 150

100 20 40 40 n 5 200

TABLE 15.4

Color preferences according

to personality types.

SECTION 15.3 / THE CHI-SQUARE TEST FOR INDEPENDENCE 5 2 3

The null hypothesis for the chi-square test for independence states that the two variables

being measured are independent; that is, for each individual, the value obtained for one

variable is not related to (or influenced by) the value for the second variable. This general

hypothesis can be expressed in two different conceptual forms, each viewing the data

and the test from slightly different perspectives. The data in Table 15.4 describing color

preference and personality are used to present both versions of the null hypothesis.

H 0 version 1 For this version of H

0 , the data are viewed as a single sample with each

individual measured on two variables. The goal of the chi-square test is to evaluate the

relationship between the two variables. For the example we are considering, the goal is

to determine whether there is a consistent, predictable relationship between personality

and color preference. That is, if I know your personality, will it help me to predict your

color preference? The null hypothesis states that there is no relationship. The alternative

hypothesis, H 1 , states that there is a relationship between the two variables.

H 0 : For the general population of students, there is no relationship between

color preference and personality.

This version of H 0 demonstrates the similarity between the chi-square test for in-

dependence and a correlation. In each case, the data consist of two measurements (X

and Y) for each individual, and the goal is to evaluate the relationship between the two

variables. The correlation, however, requires numerical scores for X and Y. The chi-

square test, on the other hand, simply uses frequencies for individuals classified into

categories.

H 0 version 2 For this version of H

0 , the data are viewed as two (or more) separate

samples representing two (or more) populations or treatment conditions. The goal of the

chi-square test is to determine whether there are significant differences between the popu-

lations. For the example we are considering, the data in Table 15.4 would be viewed as a

sample of n 5 50 introverts (top row) and a separate sample of n 5 150 extroverts (bot- tom row). The chi-square test determines whether the distribution of color preferences for

introverts is significantly different from the distribution of color preferences for extroverts.

From this perspective, the null hypothesis is stated as follows:

H 0 : In the population of students, the distribution of color preferences for

introverts has the same shape as the distribution for extroverts. The two

distributions have the same proportions.

Notice that the null hypothesis does not say that the two distributions are identical; it

simply says that they have the same proportions. For example, if 10% of the introverts

prefer yellow, then 10% of the extroverts should also prefer yellow.

The second version of H 0 demonstrates the similarity between the chi-square test

and an independent-measures t test (or ANOVA). In each case, the data consist of two

(or more) separate samples that are being used to test for differences between two

(or more) populations. The t test (or ANOVA) requires numerical scores to compute

means and mean differences. However, the chi-square test simply uses frequencies

for individuals classified into categories. The null hypothesis for the chi-square test

states that the populations have the same proportions (same shape). The alternative

hypothesis, H 1 , simply states that the populations have different proportions. For

the example we are considering, H 1 states that the shape of the distribution of color

preferences for introverts is different from the shape of the distribution of color pref-

erences for extroverts.

T H E N U L L H Y P OT H E S I S F O R T H E C H I -

S Q UA R E T E ST F O R I N D E P E N D E N C E

5 2 4 CHAPTER 15 THE CHI-SQUARE STATISTIC: TESTS FOR GOODNESS OF FIT AND INDEPENDENCE

Equivalence of H 0 version 1 and H

0 version 2 Although we have presented two

different statements of the null hypothesis, these two versions are equivalent. The first

version of H 0 states that color preference is not related to personality. If this hypothesis

is correct, then the distribution of color preferences should not depend on personality.

In other words, the distribution of color preferences should have the same proportions

for introverts and for extroverts, which is the second version of H 0 .

For example, if we found that 60% of the introverts preferred red, then H 0 would

predict that we also should find that 60% of the extroverts prefer red. In this case, know-

ing that an individual prefers red does not help you predict his or her personality. Note

that finding the same proportions indicates no relationship.

On the other hand, if the proportions were different, it would suggest that there is

a relationship. For example, if red is preferred by 60% of the extroverts but only 10%

of the introverts, then there is a clear, predictable relationship between personality

and color preference. (If I know your personality, then I can predict your color prefer-

ence.) Thus, finding different proportions means that there is a relationship between

the two variables.

Two variables are independent when there is no consistent, predictable relation-

ship between them. In this case, the frequency distribution for one variable is

not dependent on the categories of the second variable. As a result, when two

variables are independent, the frequency distribution for one variable has the

same shape (same proportions) for all categories of the second variable.

Thus, stating that there is no relationship between two variables (version 1 of H 0 )

is equivalent to stating that the distributions have equal proportions (version 2 of H 0 ).

The chi-square test for independence uses the same basic logic that was used for

the goodness-of-fit test. First, a sample is selected and each individual is classified

or categorized. Because the test for independence considers two variables, every

individual is classified on both variables, and the resulting frequency distribution is

presented as a two-dimensional matrix (see Table 15.4). As before, the frequencies

in the sample distribution are called observed frequencies and are identified by the

symbol f o .

The next step is to find the expected frequencies, or f e values, for this chi-square test.

As before, the expected frequencies define an ideal hypothetical distribution that is in

perfect agreement with the null hypothesis. Once the expected frequencies are obtained,

we compute a chi-square statistic to determine how well the data (observed frequencies)

fit the null hypothesis (expected frequencies).

The easiest way to find the expected frequencies is to begin with the null hypoth-

esis stated in terms of equal proportions. For the example we are considering, the null

hypothesis states

H 0 : The frequency distribution of color preference has the same shape (same

proportions) for both categories of personality.

To find the expected frequencies, we first determine the overall distribution of color

preferences and then apply this distribution to both categories of personality. Table

15.5 shows an empty matrix corresponding to the data from Table 15.4. Notice that

the empty matrix includes all of the row totals and column totals from the original

sample data. The row totals and column totals are essential for computing the expected

frequencies.

D E F I N I T I O N

O B S E R V E D A N D E X P E C T E D

F R E Q U E N C I E S

SECTION 15.3 / THE CHI-SQUARE TEST FOR INDEPENDENCE 5 2 5

The column totals for the matrix describe the overall distribution of color prefer-

ences. For these data, 100 people selected red as their preferred color. Because the total

sample consists of 200 people, the proportion selecting red is 100 out of 200, or 50%.

The complete set of color preference proportions is as follows:

100 out of 200 5 50% prefer red

20 out of 200 5 10% prefer yellow

40 out of 200 5 20% prefer green

40 out of 200 5 20% prefer blue

The row totals in the matrix define the two samples of personality types. For

example, the matrix in Table 15.5 shows a total of 50 introverts (the top row) and a

sample of 150 extroverts (the bottom row). According to the null hypothesis, both

personality groups should have the same proportions for color preferences. To find the

expected frequencies, we simply apply the overall distribution of color preferences

to each sample. Beginning with the sample of 50 introverts in the top row, we obtain

expected frequencies of

50% prefer red: f e 5 50% of 50 5 0.50(50) 5 25

10% prefer yellow: f e 5 10% of 50 5 0.10(50) 5 5

20% prefer green: f e 5 20% of 50 5 0.20(50) 5 10

20% prefer blue: f e 5 20% of 50 5 0.20(50) 5 10

Using exactly the same proportions for the sample of n 5 150 extroverts in the bot- tom row, we obtain expected frequencies of

50% prefer red: f e 5 50% of 150 5 0.50(50) 5 75

10% prefer yellow: f e 5 10% of 150 5 0.10(50) 5 15

20% prefer green: f e 5 20% of 150 5 0.20(50) 5 30

20% prefer blue: f e 5 20% of 150 5 0.20(50) 5 30

The complete set of expected frequencies is shown in Table 15.6. Notice that the row

totals and the column totals for the expected frequencies are the same as those for the

original data (the observed frequencies) in Table 15.4.

A simple formula for determining expected frequencies Although expected frequen-

cies are derived directly from the null hypothesis and the sample characteristics, it is not

necessary to go through extensive calculations to find f e values. In fact, there is a simple

formula that determines f e for any cell in the frequency distribution matrix:

f f f

n e

c r5

(15.4)

Introvert

Red Yellow Green Blue

Extrovert 150

100 20 40 40

TABLE 15.5

An empty frequency distribu-

tion matrix showing only the

row totals and column totals.

(These numbers describe the

basic characteristics of the

sample from Table 15.4.)

5 2 6 CHAPTER 15 THE CHI-SQUARE STATISTIC: TESTS FOR GOODNESS OF FIT AND INDEPENDENCE

where f c is the frequency total for the column (column total), f

r is the frequency total for

the row (row total), and n is the number of individuals in the entire sample. To demon-

strate this formula, we compute the expected frequency for introverts selecting yellow

in Table 15.6. First, note that this cell is located in the top row and second column in

the table. The column total is f c 5 20, the row total is f

r 5 50, and the sample size is n

5 200. Using these values in formula 15.4, we obtain

f f f

n e

c r5 5 5 20 50

200 5

( )

This is identical to the expected frequency we obtained using percentages from the

overall distribution.

The chi-square test for independence uses exactly the same chi-square formula as the

test for goodness of fit:

x 2 2

5  2f f

o e

( )

As before, the formula measures the discrepancy between the data (f o values) and the

hypothesis (f e values). A large discrepancy produces a large value for chi-square and in-

dicates that H 0 should be rejected. To determine whether a particular chi-square statistic

is significantly large, you must first determine degrees of freedom (df) for the statistic

and then consult the chi-square distribution in Appendix B. For the chi-square test of

independence, degrees of freedom are based on the number of cells for which you can

freely choose expected frequencies. Recall that the f e values are partially determined by the

sample size (n) and by the row totals and column totals from the original data. These vari-

ous totals restrict your freedom in selecting expected frequencies. This point is illustrated

in Table 15.7. Notice that we have filled in only three of the f e values in the table. However,

these three values determine all of the other expected frequencies. For example, the bottom

number in the first column must be 75 to produce a column total of 100. Similarly, the final

value in the first row must be 10 to produce a row total of 50. In general, the entire bottom

T H E C H I - S Q UA R E STAT I ST I C A N D

D E G R E E S O F F R E E D O M

Introvert

Red Yellow Green Blue

25 5 10 10 50

Extrovert 75 15 30 30 150

100 20 40 40

TABLE 15.6

Expected frequencies cor-

responding to the data in

Table 15.4. (This is the

distribution predicted by

the null hypothesis.)

Red Yellow Green Blue

25 5 10 ? 50

? ? ? ? 150

100 20 40 40

TABLE 15.7

Degrees of freedom and

expected frequencies. (Once

three values have been se-

lected, all of the remaining

expected frequencies are

determined by the row totals

and the column totals. This

example has only three free

choices, so df 5 3.)

SECTION 15.3 / THE CHI-SQUARE TEST FOR INDEPENDENCE 5 2 7

row and the entire last column are restricted by the row totals and the column totals. As a

result, we may freely choose all but one f e in each row and all but one f

e in each column. If

R is the number of rows and C is the number of columns, and you remove the last column

and the bottom row from the matrix, you are left with a smaller matrix that has C 2 1

columns and R 2 1 rows. This smaller matrix determines the df value. Specifically, the

total number of f e values that you can freely choose is (R 2 1)(C 2 1), and the degrees of

freedom for the chi-square test of independence are given by the formula

df � (R 2 1)(C 2 1) (15.5)

Also note that once you calculate the expected frequencies to fill the smaller matrix,

the rest of the f e values can be found by subtraction.

The following example demonstrates the complete hypothesis-testing procedure for the

chi-square test for independence.

Many parents allow their underage children to drink alcohol in limited situations when

an adult is present to supervise. The idea is that teens will learn responsible drinking

habits if they first experience alcohol in a controlled environment. Other parents take

a strict no-drinking approach with the idea that they are sending a clear message about

what is right and what is wrong. Recent research, however, suggests that the more

permissive approach may actually result in more negative consequences (McMorris,

et al., 2011). In the study, teens who were allowed to drink with their parents were sig-

nificantly more likely to experience alcohol-related problems than teens who were not

allowed to drink. In an attempt to replicate this study, researchers surveyed a sample of

150 students each year from age 14 to 17. The students were asked about their alcohol

use and about alcohol-related problems such as binge drinking, fights, and blackouts.

The results are shown in Table 15.8. Do the data show a significant relationship be-

tween the parents’ rules about alcohol and subsequent alcohol-related problems?

State the hypotheses, and select a level of significance. According to the null hypoth-

esis, the two variables are independent. This general hypothesis can be stated in two

different ways:

Version 1

H 0 : In the general population, there is no relationship between parents’ rules

for alcohol use and the development of alcohol-related problems.

This version of H 0 emphasizes the similarity between the chi-square test and a

correlation.

A N E X A M P L E O F T H E C H I - S Q UA R E T E ST F O R

I N D E P E N D E N C E

E X A M P L E 1 5 . 2

S T E P 1

Experience with Alcohol-Related Problems

Not Allowed to Drink

No Yes

71 9 80

Allowed to Drink 89 31 120

160 40 n � 200

TABLE 15.8

A frequency distribution show-

ing experience with alcohol-