Essentials of statistics for the behavioral sciences_ -- Frederick J_ Gravetter; Larry B_ Wallnau --

Instructor’s Manual with Test Bank for Gravetter and Wallnau’s

Essentials of Statistics for the

Behavioral Sciences

Frederick |]Gravetter

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Digitized by the Internet Archive in 2021 with funding from Kahle/Austin Foundation

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Instructors Manual with Test Bank for Gravetter and Wallnau’s

ESSENTIALS OF STATISTICS FOR THE BEHAVIORAL SCIENCES FOURTH

EDITION

FREDERICK

J GRAVETTER

State University of New York College at Brockport

VVADSVVORTH THOMSON

Australia

LEARNING

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* United States

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THOMSON

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COPYRIGHT © 2002 Wadsworth Group. Wadsworth is an imprint of the Wadsworth Group, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

For more information about this or any other Wadsworth product, contact: WADSWORTH 511 Forest Lodge Road Pacific Grove, CA 93950 USA

www.wadsworth.com 1-800-423-0563 (Thomson Learning Academic Resource Center)

ALL RIGHTS RESERVED. Instructors ofclasses using Essentials of Statistics for the Behavioral Sciences, 4th Ed.,

by Gravetter and Wallnau, as a textbook may reproduce material from this publication for classroom use. Otherwise, no part of this work covered by the copyright hereon may be reproduced or used in any form or by any means— graphic, electronic, or mechanical, including photocopying, recording, taping, Web distribution, or information storage and retrieval systems—without the written permission of the publisher.

For permission to use material from this work, contact us by Web:

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Printed in the United States of America

sae, ae ae

|

ISBN 0-534-58619-8

CONTENTS

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Section I Test Bank

The first section of this Instructor’s Manual consists of a test bank of exam items to accompany the Statistics textbook. There are approximately 25 multiplechoice questions, 25 true/false questions, and a handful of other text questions for each of the 19 chapters in the book. Answers are provided for all of the test items, and the multiple-choice and true/false questions are cross-referenced to specific pages in the textbook. Finally, you should note that some of the exam items are marked with a www to indicate that these items are available to students on the Study Center web site at www.wadsworth.com.

Section I: Test Bank - page |!

Chapter | Introduction to Statistics Multiple-Choice Questions

Answer a page 3 www

1. A researcher is interested in the eating behavior of rats and selects a group of 25 rats to be tested in a research study. The group of 25 rats isa a. sample b. statistic c. population d. parameter

Answer b page 4 www

2. A researcher is interested in the effect of St. Johns Wort on memory. A group of 25 college students is selected to participate in a research study. The average memory score obtained for the students is a a. sample b. statistic c. population d. parameter

Answer ¢ page 3

3. A researcher is curious about the average IQ of registered voters in the state of Florida. The entire group of registered voters in Florida is an example of a a. sample b. statistic c. population d. parameter

Answer d

4. A researcher is curious about the average IQ of registered voters in the state of

page 4

Florida.

If this average could be obtained, it would be an example of a

a. sample b. statistic c. population d. parameter

Chapter 1: Exam Items - page 2

Answer c

page 3

5. Although research questions typically concern a typically examines a a. sample, population

, aresearch study

b. statistic, sample

c. population, sample d. parameter, population Answer a page 4

6. The relation between a statistic and a parameter is the same as the relation between a. a sample and a population b. a dependent variable and an independent variable c. descriptive statistics and inferential statistics d. an operational definition and a hypothetical construct

Answer d

7. Statistical methods that use sample data to answer general questions about a population are called a. parameters b. statistics ¢c. descriptive statistics d. inferential statistics

page 5

Answer c

page 5 WWWw

Answer a

page 4

8. Statistical techniques that summarize, organize, and simplify data are classified as . population statistics . sample statistics . descriptive statistics . inferential statistics aaa 9. A characteristic, usually a numerical value, that describes an entire population of scores is a a. parameter b. statistic c. variable d. constant

Answer b page 4

10. A characteristic, usually a numerical value, that describes a sample of scores is a parameter

statistic variable . constant aoe

Chapter 1: Exam Items - page 3

Answer b page 15

11.

Using letter grades (A, B, C, D, and F) to classify student performance on an

exam is an example of measurement on a(n)

scale of measurement.

a. nominal b. ordinal c. interval d. ratio Answer d

page 15

12.

Determining a person's age (in years) would involve measurement on a(n) scale of measurement. nominal . ordinal . interval a=6. Tatio

13. Determining a person's occupation would involve measurement on a(n) scale of measurement. a. nominal b. ordinal c. interval d. ratio Answer d

page 16

Answer c

page 11

14. After measuring a set of individuals, a researcher finds that Bob's score is three times greater than Jane's score. These measurements must come from a(n) scale. . nominal . ordinal . interval oo (oF man) . ratio 15. In an experiment, the researcher manipulates the or measures the variable.

variable and observes

a. population, sample b. sample, population c. independent, dependent d. dependent, independent 16. A researcher observed that preschool children who were placed in a red room for their play period showed more aggression than children who were placed in a blue room. For this study, what is the independent variable? a. preschool children b. color of the room c. amount of aggression d. the children's play

Chapter 1: Exam Items - page 4

17. Ina study evaluating the effectiveness of a new medication designed to reduce cholesterol, one sample of individuals is given the medicine and a second sample is given an inactive placebo. Cholesterol level is measured for each individual. For this study, what is the dependent variable? the medication . the placebo cholesterol level . the individuals given the medication paon Answer d page 17

18. Which of the following is not a continuous variable? a. time to solve a problem b. temperature c. height d. number of children in a family

Answer d page 17

19. A variable that has an infinite number of possible values between any two specific measurements is called a(n) variable. a. independent b. dependent c. d.

Answer c

page 17 WWW

discrete continuous

20. A variable that consists of indivisible categories with no other scores existing between neighboring categories is called a(n) variable. a. independent b. dependent c. discrete d. continuous

Answer C

21. What is the first step to be performed in the following mathematical

page 21

expression?

L(X +2)

a. square each value b. sum the squared values c. add 2 points to each value d. square the sum of the values

22. What is the final step to be performed in the following mathematical expression? L(X +2) a. square each value b. sum the squared values c. add 2 points to each value d. square the sum of the values

Chapter 1: Exam Items - page 5

Answer b

page 21

Answer a

page 23

23. For the following scores, what is ©X*? a. 16 b. 24 c. 64 d. (24) = 576

Scores: 2, 0, 4, 2

24. For the following scores, what is (ZX)*? a. 16 b. 24 c. 64 d. (24) = 576

Scores: 2, 0, 4, 2

25. For the following scores, what is 2X +1? a. 9 bo

Scores: 2, 0, 4, 2

C72

d. 36 Answer c page 22

26. For the following scores, what is £(X + 1)? a. 9 oa ream We d. 3(9)= 36

Scores: 2, 0, 4, 2

Answer b page 21

27. How would the following mathematical operation be expressed in summation notation? "Subtract one point from each score and find the sum ofthe resulting values." oe 2) e | : = 1) 1 - xX One! Ba . U1 - X) 28. How would the following mathematical operation be expressed in summation notation? "Add two points to each score, square the resulting value, then find the sum of the squared numbers."

re

be,te es

b. GF 2y 6 2K +2)

re

ae

Chapter 1: Exam Items - page 6

True/False Questions False page 4

1. A summary value, usually numerical, that describes a sample is called a parameter.

True page 3

2. A researcher is interested in the vocabulary skills of 3-year-old children. For this researcher, the entire set of 3-year-old children would be the population.

False

page 3

3. A researcher interested in human memory selects a group of 30 students to participate in a research study. The group of 30 students would be considered a population.

True page 5

4. The goal of descriptive statistics is to simplify, summarize, and organize data.

True page 10

5. To establish a cause-and-effect relation between two variables, a researcher

should use the experimental research method.

page 9

6. The correlation research method requires manipulation and control of variables to establish a relationship.

False

7. In the experimental method, the researcher manipulates the dependent variable.

False

page 11 True page 11

8. A recent report concluded that subjects assigned to a regular exercise program had fewer problems with depression than subjects who were not allowed to exercise regularly. For this study, depression is the dependent variable.

False page 14

9. The reading ability of 2nd-grade students is classified as high, medium, or low.

False

10. Last week the three major networks were ranked as follows: CBS first, NBC second, and ABC third. This is an example of measurement on a nominal scale.

page 15 True page 14 True

page 15

This classification involves measurement on a nominal scale.

11. Determining the favorite color for each child in a pre-school class would be an example of measurement on a nominal scale.

Measurements from a nominal scale can be used to determine whether two individuals are the same or different, but cannot determine the direction or the size 12.

of the difference.

False

page 15

13. Ifa researcher measures two individuals on an ordinal scale, it is impossible to determine which individual has the larger score.

Chapter 1: Exam Items - page 7

True page 17

14. Determining the favorite color for each child in a pre-school class would be an example of measurement of a discrete variable.

True page 17

15. Reaction time is an example of a continuous variable.

True page 17

16. A researcher records the sex of each child born in the county hospital during the month of June. This researcher is measuring a discrete variable.

False page 20

17. In statistical notation, a lower-case letter n is used to identify the number of scores in a population.

False page 21

18. To compute 2X + 3, the first step is to add 3 points to each of the original scores.

False page 21

19. To compute DX’, you first sum the scores and then square the total.

False

20. For any set of scores, DX? = (ZX).

page 21

False page 22

21. To compute U(X + 2)’, you first add 2 points to each score, then sum the resulting values, and finally you square the sum.

False page 23

22. For the following scores, 2X +1=10.

False page 22

23. For the following scores, U(X + 1)?=100.

False page 21

24. For the following scores, (2X)’=30.

True

25. For the following scores, UX? = 30.

Scores: 1, 0, 2,3

Scores: 1, 0, 2,3

Scores: 1, 2, 3, 4

Scores: 1, 2, 3, 4

page 21

Chapter 1: Exam Items - page 8

Other Exam Items 1. Statistical techniques are classified into two major categories: descriptive and inferential.

Describe the general purpose of each category.

2. Define the concept of "sampling error." Note: Your definition should include the concepts of sample, population, statistic, and parameter.

3. Describe the sequence of mathematical operations that would be used to evaluate each of the following expressions. a. LX°

b. (aX) Go Ao 1 d. BCX + 1)

e. U(X +1) 4.

Calculate each value requested for the following set of scores.

Scores:

1, 2, 0, 4

a. UX be 2X?

6c (Ky 5.

Calculate each value requested for the following set of scores. a. Us 2

By

tia

e.

3(X+2y

Scores:

0, 1, 4, 2

2)

6. Calculate each value requested for the following set of scores. ee ve.8 ee b. -2Y 1 3 c. BADY 3 ] da. LAY i az 2 -4

Answers for Other Exam Items 1. The purpose of descriptive statistics is to simplify the organization and presentation of data.

The purpose of inferential statistics is to use the limited data from a sample as the basis for making general conclusions about the population.

Chapter 1: Exam Items - page 9

2. A parameter is a value that is obtained from a population of scores and is used to describe the population. A statistic is a value obtained from a sample and used to describe the sample.

Typically it is impossible to obtain measurements for an entire population, so researchers must rely on information from samples; that is, researchers use statistics to obtain information about

unknown parameters. However, samples provide only limited information about their populations. Thus, sample statistics are usually not identical to their corresponding population parameters. The error or discrepancy between a statistic and the corresponding parameter is called sampling error.

3. a. To compute 2X’, you first square each score, then sum the squared values.

b. To compute (ZX)’, you first sum the scores, then square the sum. c. To compute £X + 1, you first sum the scores, then add 1 to the sum. d. To compute X(X + 1) you first add 1 to each score, then sum the resulting values. e. To compute X(X + 1)’, you first add 1 to each score, then square the resulting values, then sum the squared numbers.

4. a. 7

b21 c. (7% =49 Siew

eae,

b. 15 c. 65 6. a. 6

b. -2 c. -12

d. -2

Chapter 1: Exam Items - page 10

Chapter 2 Frequency Distributions

Multiple-Choice Questions

Questions 1 through 3 concern the following table. x t 3 1 4 2 3 4 2 ] ] 2 1. For these data, N = Ce b. 10 c. 29 d. cannot be determined from the table

2. For these data, XX is a. 10

bes ce) 29 d. cannot be determined from the table

3. The proportion associated with X = 3 is a. 0.10 b. 0.20 ¢.. 0:30 d. 0.40 4. What is the shape of the distribution for the following set of data?

Seorex:

1. 1, 1,2. 2.3.3.3

a. symmetrical b. positively skewed c. negatively skewed d. cumulative

Chapter 2: Exam Items - page 1]

5. What is the shape of the distribution for the following set of data?

Scores:

1, 2, 2.2, 2, 3; 3, 4, 5,6

.

a. symmetrical b. positively skewed c. negatively skewed d. cumulative Answer c

page 37

6, Ina grouped frequency distribution one interval is listed as 50-54. What are the real limits of this interval? a. 50 and 54 b. 50.5 and 54.5 c. 49.5 and 54.5 d. 50.5 and 53.5 7. Ina grouped frequency distribution one interval is listed as 50-54. What is the width of this interval? a. 3 points b. 4 points c. 5 points d. 54 points 8. Frequency distribution polygons are intended for use with a. either interval or ratio scales of measurement b. only ratio scales c. either nominal or ordinal scales d. only nominal scales

9. A distribution of scores is being organized in a grouped frequency distribution table with an interval width of 2 points. If the lowest score in the distribution is X = 41, then the bottom interval in the table should be

a. 40-42

b. 41-43 c. 40-41 d. 41-42

10. Which of the following statements is false regarding grouped frequency distribution tables? a. An interval width should be used that yields about 10 intervals. b. Intervals are listed in descending order, starting with the highest value at the top of the X column. c. The bottom score for each interval is a multiple of the interval width. d. The value for N can be determined by counting the number of intervals in the X column.

Chapter 2: Exam Items - page 12

Answer c page 36

11. Ina frequency distribution graph, frequencies are presented on the ___ and the scores (categories) are listed on the __.

X axis, Y axis horizontal line, vertical line Y axis, X axis oP class interval, horizontal line Be

Answer b page 42

12. Ina distribution with positive skew, scores with the highest frequencies are a. on the right side of the distribution b. on the left side of the distribution c. in the middle of the distribution d. represented at two distinct peaks Questions 13 through 15 refer to the following table showing a frequency distribution of quiz scores.

» ae i

me N hh WOW NO DN Wan

Answer c page 31

13. How many individuals took the quiz? as Beas ona

d. cannot be determined from the information given Answer C page 31

14. Ifa score of X = 3 or higher is needed for a passing grade, how many individuals passed? a. 3 b. 11 c 16 d. cannot be determined from the information given

Answer b page 31

15. How many individuals had a score of X = 2? a. | b, 3 eS d. cannot be determined from the information given

Chapter 2: Exam Items - page 13

Questions 16 through 18 refer to the following table showing a frequency distribution of exam scores.

xX 70-74 65-69 60-64 55-59 50-54 45-49 Answer b page 35

f 3 4 8 2 1 1

16. This is a grouped frequency distribution table where the scores have been grouped into class intervals using an interval width of a. 4 points b. 5 points c. 9 points d. 10 points

Answer d page 35

17. The lowest score on the exam is a. X =45 b. X= 49 CX =475 d. cannot be determined from the information given

Answer b page 35

18. For this set of exam scores, how many students had scores less than X = 60? rgd b. 4 bu 8 d. cannot be determined from the information given

Questions 19 through 22 refer to the following graph showing a frequency distribution of quiz scores.

PNW

Chapter 2: Exam Items - page 14

Answer b page 37

19. This graph is an example of a a. bar graph b. histogram c. polygon d. none of the above

Answer b

page 37

20. For this distribution, how many individuals had a score of X = 4? an

bo 2 c. 4

d. cannot be determined from the information given Answer b page 37

21. How many individuals took the quiz? a> b. 10 6. 30 d. cannot be determined from the information given

Answer d

22. What is the value of ©X for the entire set of scores shown in this distribution? a. 5 iy [5 =) aes

page 37

Questions 23 through 25 refer to the following graph showing a frequency distribution of quiz scores.

4

|

ee wi, ih x

A

Answer b

page 42

Z

3

23. Thisisanexample ofa —_ a. symmetrical

4

5

6

distribution.

b. positively skewed c. negatively skewed d. normal

Chapter 2: Exam [tems - page 15

Answer ¢ page 37

24. How many individual scores are represented in this graph? (What is the value of N?) a. 5

b. 6 er 3

d. cannot be determined from the information given Answer a

25. The lowest score in the class was X =

page 37

this score is

and the number of people with

True/False Questions True page 30

1. The purpose for constructing a frequency distribution table is to get an organized view of the entire set of scores.

True page 30

2. A professor recorded the academic major for each student in an introductory psychology class. If these data were organized in a frequency distribution table, the first column would be a list of academic majors.

True page 31

3. When listing scores in a frequency distribution table, you should not skip any

False page 31

4. Ina frequency distribution table, N can be obtained by counting the number of values listed in the X column.

True page 31

5. For any frequency distribution table, N = Uf.

False page 31

6. Ina frequency distribution table, the 2X may be obtained by simply adding the values in the column labeled X.

True page 32

7. Proportions can be obtained for each interval in a frequency distribution by computing p = f/N.

True page 33

8. In a sample of n = 30 scores, the lowest score is K = 15 and the highest score is X = 44. Ifthe scores are put in a frequency distribution table, then a grouped

scores, even if one has a frequency of zero.

table should be used.

Chapter 2: Exam Items ~ page 16

False page 33

9. As a general rule, a grouped frequency distribution table should have approximately 15 to 20 class intervals.

True page 34

10. Ina grouped frequency distribution table, all of the class intervals should have exactly the same width.

False page 33

11. You should use a grouped frequency distribution table whenever there are more than 15 or 20 scores in the distribution.

True page 35

12. Ina grouped frequency distribution, a class interval listed as 10-19 has an interval width of 10 points.

True page 31

13. There is enough information in a regular frequency distribution table to obtain a complete listing of the original set of scores.

False page 35

14. There is enough information in a grouped frequency distribution table to obtain a complete listing of the original set of scores.

True page 34

15. Ina grouped frequency distribution table, the bottom value in each class interval should be a multiple of the interval width.

False page 34

16. For a grouped frequency distribution table, a class interval should be omitted from the table if there are no scores in the interval.

True page 36

17. Ina frequency distribution graph, the frequency values are presented on the vertical axis and the scores (or measurement categories) are presented on the horizontal axis.

True page 37

18. Bar graphs may be used for data that are measured on a nominal or an ordinal scale.

True page 39

19. A professor records the academic major for each student in a class. If these data are presented in a frequency distribution graph, the graph should not be a polygon.

True page 39

20. A professor records the height (in inches) for each student in a class. If these data are presented in a frequency distribution graph, it would be acceptable to use a polygon.

False

page 37

21. A frequency distribution histogram is best suited for data measured on a nominal scale.

True page 39

22. You should not use a polygon to display a distribution of scores that have been measured on a nominal scale.

Chapter 2: Exam Items - page 17

True

23. It would be acceptable for the data in a histogram to be re-drawn in a polygon.

page 39

False

24. A histogram is drawn so that there are spaces between the bars.

page 37

True

25. A negatively skewed distribution has a tail on the left side of the graph.

page 42

True page 42

26. Ina positively skewed distribution, the scores tend to pile up on the left-hand side of the graph.

False

27. When the scores pile up on the right-hand side of a distribution and taper off

page 42

to the left, then the distribution is positively skewed.

True page 42

28. The scores for a very easy exam would probably form a negatively skewed distribution.

Other Exam Items 1. Briefly explain when and why you should use a grouped frequency distribution table instead of a regular table.

2. Briefly explain what information is available in a regular frequency distribution table that is not available in a grouped table.

3. For the following data, construct a grouped frequency distribution table. Include columns for proportion, and percent.

41 32 67 44 46 47 36 53 56 45 31 43 25 33 25 38 48 57 42 49 4. For the following quiz scores, sketch a histogram of the frequency distribution.

5 2 10°9 7 ©: 10 1837 34-19 OP atT a Ben

Chapter 2: Exam Items - page 18

5. Answer each question for the following data.

Zo 37 Ais td) a 0 13 P9521 49 37) 29 20 16-17 34 16 35 24 16 12 29 30 a. Which is more appropriate, a regular or a grouped frequency distribution table?

Explain why. b. Construct the appropriate table. c. Based on the table, what is the shape of the distribution?

6. For the following data, construct a grouped frequency distribution table using an interval width of 10. Include columns for proportion and percent.

30 22 58 28 35 67 27 42 46 35 OL -3a 18.33 20 38 48.30.31 39

Answers for Other Exam Items 1. You should use a grouped table whenever the range of scores is too wide to list each value in a regular table. Usually, a range of more than 15 points will require a grouped table.

2. A regular table identifies each individual score exactly. However, in a grouped table, you simply know that an individual score is located in a particular interval; you do not know its exact value. a

xX 6-69 60-64 s-59 50-54 45-49 40-44 35-39 30-34 23-29

f 1 0 2 1 5 4 2 3 2

Pp 0.05 0 0410 0.05 0.25 0.20 0.10 O15 0.10...

% 5% 0% . 10% 5% 25% 20% 10% 15% 10%

Chapter 2: Exam Items - page 19

Fh

PNM oOo We

5. a. A grouped frequency distribution table would be better. The lowest score is X = 12 and the highest is X = 49. A regular table would require 38 rows. b.

x 45 - 49 40 - 44 20 oo 30 - 34 Za 22 20 - 24 15-19 10-14

WHA WLR BW S| W

c. The distribution is positively skewed.

a

nak 60-69 S059 40-49 30-39 20-29 i=1o

Hf 1 924 3 9 4 1

Pp 0.05 010, O15 045 0.20 9:05

% 5% 10% 15% 45% 20% 5%

Chapter 2: Exam Items - page 20

Chapter 3 Central Tendency Multiple-Choice Questions Answer d

1. What is the value of the mean for the following set of scores?

page 53

eoores: 1) 3,5, 0, 1

a, 10 3 or 2 a2

www

Answer C page 53 WWW

Answer a

page 53

Answer a

page 57

2.

A sample of n=5 scores has a mean of X = 9. What is UX for this sample? a. 9/5 = 1.80 b. 5/9 =0.555 c. 3(9)=45 d. cannot be determined from the information given

3. A population of scores has 2X = 60 and a mean of = 12. How many scores are in this population? ra) b 12 c. 60 d. cannot be determined from the information given

4. A sample has a mean of X = 42. Ifa new score with a value of X = 50 is added to the sample, what effect will it have on the sample mean? a. The sample mean will increase. b. The sample mean will decrease. c. The sample mean will remain the same. d. cannot be determined from the information given 5. A sample has a mean ofX = 30. If one score with a value of X = 10 is removed from the sample, what effect will it have on the sample mean? a. The sample mean will increase. b. The sample mean will decrease. c. The sample mean will remain the same. d. cannot be determined from the information given

Chapter 3: Exam Items - page 21

6. A sample has a mean of X = 25. Ifa new score with a value ofX = 25 is added to the sample, what effect will it have on the sample mean? a. The sample mean will increase. b. The sample mean will decrease. c. The sample mean will remain the same. d. cannot be determined from the information given 7. Asample of n= 25 scores has a mean of X = 11. Ifeach score in the sample is multiplied by 6, the new mean would be

a. 6 b. 11 Cruly d. 66 Answer a

page 59

8. After 5 points are added to every score in a distribution, the mean is calculated and found to be » = 30. What was the value of the mean for the original distribution?

Fee)

b. 30 C35 d. cannot be determined from the information given

Answer d page 59

9. Which of the following is a property of the mean? a. Changing the value of a score will change the value of the mean. b. Adding a constant to each score will add the same constant to the mean. c. Multiplying each score by a constant will multiply the mean by the same constant.

d. all of the above Answer b

page 57

10. A set of N = 20 exam scores has a mean of p = 50. The instructor discovered that one student cheated on the exam so 20 points are subtracted from that student's score. What is the new mean for the class? a. 50 b. 49 oe: 30 d. cannot be determined from the information given

Answer a

11.

page 57

Which measure(s) of central tendency is (are) certain to be changed? a. the mean ; b. the median c. the mean and the median d. the mode

The value of one score in a distribution is changed from X = 20 to X = 30.

Chapter 3: Exam Items - page 22

12. One sample of n= 4 scores has a mean of X = 10, and a second sample of n = 8 scores has a mean of X = 20. Ifthe two samples are combined, the mean for the

combined sample will be

;

a. equal to 15 b. greater than 15 but less than 20 c. less than 15 but more than 10 d. none ofthe above Answer a

page 57

Answer c

page 54

Answer c

page 54

13. A sample of n = 20 scores has a mean of X = 55. After one score is removed from the sample, the mean for the remaining scores is found to be X = 51. From this information you can conclude that the removed score was a. greater than 55 b. less than 55 c. It is impossible to estimate the magnitude of the score.

14. Ina population of N = 6, five of the individuals all have scores that are exactly 1 point above the mean. From this information you can determine that the score for the sixth individual must be a. also above the mean by | point b. below the mean by one point c. below the mean by 5 points d. not enough information to describe the 6th score 15. Ina sample of n= 3 scores, the first person has a score that is below the mean by 2 points and the second person is below the mean by 4 points. Which of the following provides the most accurate description of the score for the third person? a. It is above the mean. b. It is above the mean by an average of 3 points. c. It is above the mean by 6 points. d. The score cannot be described based on the information given.

16. What is the median for the following set of scores?

Scores:

1, 3, 9, 10, 22

Scores:

1, 4, 6, 17

a. 6

b. 9 C.2

ae

Answer b page 61

17. What is the median for the following set of scores? a. 4

ae hn QR SI

Chapter 3: Exam Items - page 23

18. A teacher gave a reading test to a class of Sth-grade students and computed the mean, median, and mode for the test scores.

Which of the following statements

cannot be an accurate description of the scores? a. The majority of the students had scores above the mean. b. The majority of the students had scores above the median. c. The majority of the students had scores above the mode. d. All of the above must be false statements.

Answer C

page 63

Answer a

page 70

Answer a

page 70

Answer c

page 71

Answer c

page 67 www

19. A distribution can have more than one mean . median mode . none of the above of ao

20. For a distribution of scores, the mean is equal to the median. This distribution is most likely to be a. symmetrical b. positively skewed c. negatively skewed d. impossible to determine the shape 21. Fora perfectly symmetrical distribution with p = 30, the median would have a value : . equal to 30 . greater than 30 . less than 30 . cannot be determined from the information given fm aot

22. For a positively skewed distribution with = 30, the median would have a value ; equal to 30 . greater than 30 . less than 30 . cannot be determined from the information given of uo 23. A set of individuals is measured on a nominal scale. To determine the central tendency for the resulting measurements, a researcher should use the a. mean b. median c. mode d. It is impossible to determine central tendency for nominal measurements.

Chapter 3: Exam Items - page 24

Answer b page 71

24. A distribution is positively skewed. three measures of central tendency? a. mean b. mean c. mean d. mean

= 40, median = 60, median = 40, median = 50, median

= 50, mode = 50, mode = 60, mode = 50, mode

Which is the most probable order for the = 60 = 40 = 50 = 50

25. Ina negatively skewed distribution of exam scores, Tom scored at the mean, Mary scored at the median, and Jane scored at the mode. Who had the highest score? a. Tom b. Mary c. Jane

d. cannot be determined from the information given Answer c page 62

26. For any distribution, you can be sure that at least one individual has a score equal to the a. mean b. median c. mode d. all of the above 27. For an extremely skewed distribution of scores the best measure of central tendency would be the a. mean b. median c. mode d. Central tendency cannot be determined for a skewed distribution.

True/False Questions True page 51

1. One purpose for central tendency is to find a single score that can serve as a representative value for an entire distribution.

True page 5]

2. One purpose for central tendency is to describe the location of the center of an entire distribution of scores.

False page 53

3. Last summer I had three monthly phone bills of $20, $30, and $100. My mean phone bill for the summer was $75.

Chapter 3: Exam Items - page 25

True page 53

4. A sample of n= 6 scores has a mean of X =8. The scores in the sample must sum to £X = 48.

True

page 59

5. A population has p = 40. If 5 points are added to every score, the mean will be . changed to p = 45.

False page 59

6. Adding a constant amount to every score in a population will not change the value of the mean.

True page 59

7. After each score in a population is multiplied by 3 the mean is found to be 1 = 90. Based on this information, you can conclude that the original mean was i= 30;

False page 71

8. A sample of n= 10 scores will always have five scores below the mean and five scores above the mean.

False page 54

9. The mean is considered to be the “balance point” for a distribution because exactly have the scores exactly half of the scores are located above the mean and exactly half are below the mean.

True page 71

10. It is possible for more than half of the scores in a distribution to have values greater than the mean.

False

page 60

11. It is possible for more than half of the scores in a distribution to have values greater than the median.

False page 71

12. A distribution with a mean p: = 50 and a median of 70 probably is positively skewed.

True page 65

13. Inaskewed distribution, the mean is often displaced toward the tail so that it is not a good, representative measure of central tendency.

True

14. Ifa negatively skewed distribution has a mean of 50, then the mode is probably greater than 50.

page 71 True page 63

15. A distribution can have more than one mode.

True page 67

16. It is impossible to compute the mean for data that have been measured on a nominal scale. :

False page 67

17. It is impossible to measure central tendency for data that have been measured on a nominal scale.

Chapter 3: Exam Items - page 26

True page 70

18. In a symmetrical distribution, the mean is equal to the median.

False page 73

19. Ona 50-point exam Tom has a score of X = 23. This means that Tom scored below the median.

False page 63

20. In addition to having two modes, a bimodal distribution typically has two medians.

True page 71

21. For a positively skewed distribution, the mean tends to have a larger value than either the median or the mode.

True page 65

22. Extreme scores in a distribution are more likely to affect the value of the mean than the value of the median.

True page 55

23. If one sample with a mean of X = 10 is combined with a second sample with a mean of X = 20, then the mean for the combined sample will be between 10 and 20.

True page 67

24. The mode is the best way to measure central tendency for data from a nominal scale.

True page 68

25. Ina published report of research results, a sample mean is typically identified by the letter M.

True page 68

26. Ina graph showing the relation between an independent variable and a dependent variable, the values of the independent variable are displayed on the horizontal axis.

Other Exam Items

1. What is the purpose for obtaining a measure of central tendency?

2. Explain why it is necessary to have more than one standard procedure for defining and measuring central tendency.

Chapter 3: Exam Items - page 27

3. Describe two circumstances where the median would provide a better measure of central tendency than the mean.

4. Although it was not discussed in the textbook, describe what happens to the median and the mode when a constant is added to every score in a distribution.

5. Find the mean, the median, and the mode for the set of scores in the frequency distribution table below. xX 5 4 3 2 1 6.

f 2 3 2 2 1

Mean

=

Median

=

Mode =

Find the mean, the median, and the mode for the set of scores in the frequency distribution

histogram below.

4

Mean

=

3

Median

=

2

Mode

=

e

7. A-set ofn=7 scores has a mean of X = 10. Another set of scores has n = 3 and X = 20. If these two sets of scores are combined, what is the mean for the combined group?

8. A sample of n= 5 scores has a mean of X = 8. If one score with a value ofX = 4 is removed from the sample, what is the mean for the remaining scores?

9. A sample of n= 5 scores has a mean of X = 10. One new score is added to the sample and the new mean is computed to be X = 11. What is the value of the score that was added to the sample?

Chapter 3: Exam Items - page 28

Answers for Other Exam Items

1. The purpose of a measure of central tendency is to find the single score that identifies the middle of the distribution and best represents the entire distribution.

2. No single method for measuring central tendency will produce a good, representative value in every situation.

3. a. When there are extreme scores (skewed distribution) b. When there are undetermined scores. c. When there is an open-ended distribution d. When the data are measured on an ordinal scale

4. When you add a constant to every score, the entire distribution is shifted to a new location.

All of the individual scores are shifted, and the mean, median, and mode move along with the whole distribution.

5. Mean = 33/10 = 3.30.

Median=3.5.

Mode =4.

6. Mean = 27/10 =2.70.

Median=2.5.

Mode =2.

7. For the combined sample, n = 10, 2X = 130, and Gees

8. The original sample had n = 5 scores with 2X = 40. When the score is removed, there are n= 4 scores with 2X = 36. The new mean is X = 9.

9. The original sample had n= 5 scores with YX = 50. After the new score is added, there are n= 6 scores with YX = 66. Because the sum increased by 16 points, the new score must be m= 16.

Chapter 3: Exam Items - page 29

Chapter 4 Variability

Multiple-Choice Questions Answer c

page 78

Answer d page 81

1. Ina population of N = 10 scores, the smallest score is X = 8 and the largest score is X = 20. The range for the population is art b. 12 e. 13 d. cannot be determined without more information

2. Which of the following deviation scores corresponds to the score that is farthest away from the mean? a. 0 by 5 (a)

d. -10 Answer b page 81

3. Ina population with a mean of p = 50, a score of X = 45 would have a deviation score of a. 5 b. -5 c. 45

d. cannot be determined without more information Answer b

page 83

4. The symbol SS stands for the a. sum of squared scores b. sum of squared deviations c. sum of scores, squared d. sum of the deviations, squared

5. A population of N = 5 scores has SS = 40. What is the variance for this

population? a. 40/5 =8 b. 8° = 64 c. 40/4 = 10 d. 10° = 100

Chapter 4: Exam Items - page 30

6. A sample of n=5 a. 40/5 =8 b. 8° = 64 c. 40/4 = 10 d. 10?=100

scores has SS = 40. What is the variance for this sample?

7. Sample variance is identified by the symbol Ass

as Go

ds Answer ¢

page 85

8. Population standard deviation is identified by the symbol a. S

b. s c. o

dog Answer a

page 85

9. A population has SS = 30 and co”? = 6. What is the number of scores in the population? a N=9d

by M6 ¢. W180 d. cannot be determined without additional information

10. What is the relationship between the standard deviation and variance? a. Standard deviation equals the variance divided by N. b. Standard deviation equals the variance divided by n - 1. c. Variance is the square root of standard deviation. d. Standard deviation is the square root of variance.

11. A population of scores has yp = 50 and o = 12. If you subtract five points from every score in the population, then the value for the new standard deviation will be a7 b. 12 Ge 24 d. insufficient information, cannot be determined Answer Cc

12. A population has p = 40 and o = 8. If each score is multiplied by 2, the new

page 97

standard a. b. & d.

deviation will be 4 8 16 insufficient information, cannot be determined

Chapter 4: Exam Items - page 31

Answer b

page 96

13. A population of scores has p = 20 and o = 5. If two points are added to every score in the population, then the new values for the mean and standard deviation would be a. w= 20 ando=5 b, w= 22 ando=5 ¢. p=20ando=7 d. w=22 ando=7 14. A population of scores has p, = 20 and o = 5. If every score in the population is multiplied by 2, then the new values for the mean and standard deviation would a b. p=40 ando=5 c. p= 20 ando=10 d. p= 40 ando=10

Answer c

15. To compute sample variance, the value of SS is divided by

page 89

Answer a

page 88

the variability of 16. The variability of the scores in a sample tends to be the scores in the population from which the sample was obtained. a. smaller than b. larger than c. exactly the same as 17. Sample variance provides an unbiased estimate of the population variance.

The term unbiased means _ : a. for any sample, the sample variance will equal the population variance b. for large samples, the sample variance will equal the population variance c. if sample variance is computed for many samples, the average of all the sample variances will equal the population variance d. none of the above

Answer c

page 94

18. Which of the following sample statistics does not provide an unbiased estimate of the corresponding population parameter? a. the sample mean b. the sample variance c. the sample standard deviation

Chapter 4: Exam Items - page 32

Answer b

page 80

19. You must determine Q1 and Q3 to compute the ___.

a. sum of squared deviations b. semi-interquartile range c. variance d. standard deviation

Answer a

20. Ifa population has a mean of

page 82

avalueof a. 0 b. 16

= 24 with o = 4 and N = 10, then X(X - 1) has

op 25

d. 400 Answer a

page 88

21. A sample of n= 8 scores has SS = 50. If these same scores were a population, then the SS value for the population would be a. 50 b. greater than 50 c. less than 50 d. impossible to determine without additional information 22. What is the value of SS for the following set of scores? a. 144 bi. 62 c. 14 d. none of the above

S2O.0h:

23. For the population of scores 5 2 5 4, the variance equals a. 6 be Z

Answer e

page 82

24. Which of the following samples would have the largest value for sample variance? a.

1.3.0

&, 11, 13,15 & DLS 559 ad) 101, 103, 105 e. All the samples would have exactly the same variance.

Chapter 4: Exam Items - page 33

25. Which set of scores has the least amount of variability?

GO U es poe: SAG eee RLF 143, 145, 145, 147 . 40, 27, 805105 fa0 26. Which set of parameters would make X = 85 an extreme score in the distribution? a. w= 80 ando=5 b. up = 80 and o = 20 c. p= 70ando=5 d. p= 70 and o = 20 Answer ¢ page 94

27. The smallest score in a population is X = 5 and the largest score is X = 10.

Based on this information, you can conclude that a. the population mean is somewhere between 5 and 10. b. the population standard deviation is smaller than 6 c. all of the above d. none of the above

Answer b

28. For a particular sample, the largest distance (deviation) between a score and the mean is 11 points. The smallest distance between a score and the mean is 4 points. Therefore, the standard deviation . must be less than 4 must be somewhere between 4 and 11 . must be greater than 11 & . It is impossible to say anything about the standard deviation. aow

page 90

29. If you have a score of X = 75 on an exam, which set of parameters would give you the highest position within the class? a. p= 70ando=5 b. p= 70 ando= 15 c. p= 60 ando=5 d. w= 60 ando=15

True/False Questions True

page 79

1. The range is generally considered to be a relatively crude way to measure variability.

Chapter 4: Exam Items - page 34

True page 81

2. A negative deviation always indicates a score that is less than the mean.

True page 82

3. For any population of scores, the sum of the deviations will be equal to zero.

False page 83

4. SS is the sum of the squared scores.

True page 83

5.

True page 81

6. A deviation of zero corresponds to a score exactly equal to the mean.

True page 89

7. Ifa sample has a variance of 16, then the sample standard deviation is 4.

True

page 96

8. Ifaconstant is added to every score in a distribution, then the standard deviation will remain unchanged.

False page 96

9. A population has p = 20 and o = 5. If five points are added to every score in the population, the new set of scores will have o = 10

True page 97

10. If every score in a population is multiplied by 3, then the population standard deviation will also be multiplied by 3.

True

11. The scores in a sample tend to be less variable than the scores in the population from which the sample was obtained.

page 88 True

page 86

When calculating SS, it is impossible to obtain a value less than zero unless a

mistake is made.

12. The value for the standard deviation cannot be larger than the value for the range.

False page 83

13.

For any distribution, L(X - 1)’ will always equal zero.

False

14. A sample of n= 5 scores has SS = 20. The variance for this sample is 4.

page 89

15. A population of N = 5 scores has SS = 20. The variance for this population

True page 85

is 4.

True page 94

16. Ina distribution with p = 40 and o = 2, a score of X = 46 would be considered an extreme value (far out in the tail).

Chapter 4: Exam Items - page 35

distribution with p = 40 and o = 12, a score of X = 46 would be

False

17. Ina

page 94

considered an extreme value (far out in the tail).

True

18. Ina

page 86

is X = 35. For this distribution, the standard deviation cannot be greater than 5.

True page 94

19. Ifa distribution has p = 20 and o = 1, you can be confident that the majority of the scores are located between 18 and 22.

True page 81

20. The standard deviation provides a description of the average distance from the mean.

True page 88

21. For a specific set of scores, the value you obtain for SS will be the same whether the set of scores is a sample or a population.

False page 85

22.

True page 89

23. By using n- | in the formula for sample variance, you will get an unbiased estimate of population variance.

False page 92

24. When a sample statistic is said to be “unbiased” it means that the value of the statistic will always be equal to the corresponding population parameter.

‘rue page 85

25. For a population, variance is the mean of the squared deviation scores.

False page 96

26. Large sample variance makes it easier to see means and mean differences in sample data.

distribution with p = 40, the largest score is X = 45 and the smallest score

When computing the population variance, SS is divided by n-

1.

Other Exam Items 1. For the following set of data, find the range and the semi-interquartile range. scores:

3,.5, 3,.2, 604, 6, 7,1, 4; 3,2

2. For the following set of data, compute the value forSS.

Scores:

Chapter 4: Exam Items - page 36

5 2 2 7 9

3. For the following population of scores, use the definitional formula (the definition) to compute SS. Then compute the variance and the standard deviation for the population.

Scores: 9 1 8 6 4. For the following sample, use the computational formula to calculate SS. Scores: 1, 7, 1, 1 sample variance and standard deviation.

Then, compute the

5. Calculate the variance and standard deviation for the following sample data. mcoress, 2°93) 2-457 (5-30 6 4

6. A distribution of N = 10 scores has p = 50 and o = 10. If another five individuals, all with scores of X = 50, are added to this distribution, what will happen to the standard deviation?

Will

it increase, decrease, or stay the same? Explain your answer.

7. Without some correction, sample variability is said to be "biased." Define the term biased, and explain how this bias is corrected in the formula for sample variance.

Answers for Other Exam Items 1. The range is 8 points and the semi-interquartile range is (5.5 - 2.5)/2 = 1.5 points.

2. so = 38 3. SS=38;

0° =9.5;

o=3.08

4, §S=27;

# =9: s=3

5. SS=24;

s*=3;s=1.73

6. The variability will decrease. Adding new scores at the mean will not change the mean, but it does result in more scores being clustered at the mean. As a result the standard deviation (average distance from the mean) will decrease.

7. On average, sample variability is less than population variability. Whenever a statistic consistently underestimates (or overestimates) the corresponding population parameter, the statistic is said to be biased. The bias in sample variability is corrected by dividing by n - 1 (instead of n) in the formula for sample variance.

Chapter 4: Exam Items - page 37

Chapter 5 z-Scores

Multiple-Choice Questions

Answer d page 106 www

1. A z-score of z = -2.00 indicates a position in a distribution a. above the mean by 2 points b. above the mean by a distance equal to 2 standard deviations c. below the mean by 2 points d. below the mean by a distance equal to 2 standard deviations

Answerd page 107

2. Of the following z-score values, which one represents the most extreme location on the left-hand side of the distribution? a. z=+1.00 b. z= +2.00 c. z=-1.00 d. z= -2.00

Answer a page 108

3. For a population with p = 80 and o = 10, the z-score corresponding to X = 85 : would be +0.50 +1.00 +2.00 aos or o.00

Answerb page 108

4. For a population with : = 80 and o = 10, the z-score corresponding to X = 70 would be : a. -10 b. -1.00 go 00 ds +10

Chapter 5: Exam Items - page 38

Answer b page 109

5. For a population with 1 = 60 and o = 8, the X value corresponding to z= -0.50 would be a. -4

b. 56 c. 64 dP S9S Answer c

page 109 www

Answer c page 106 Www

6. Fora population with p = 60 and o = 8, the X value corresponding to z= 1.50 would be H.. TZ age oy We 6. 72 d. 90

7. A population of scores has pp = 44. In this population, an X value of 40 corresponds to z= -0.50. What is the population standard deviation? ae 2 ni:2 cs d.6

Answer b

page 106

8. A population of scores has p = 50. In this population, an X value of 58 corresponds to z= 2.00. What is the population standard deviation?

And b. 4 c: 8 d. 16

9. A population of scores has o = 20. In this population, a score of X = 80 corresponds to z = +0.25. What is the population mean? a. 70

Bb. 75 e. B35 d. 90 Answer c

page 106

10. A population of scores has o = 10. In this population, a score of X = 60 corresponds to z = -1.50. What is the population mean? a. -30 b. 45 ei)

d. 90

Chapter 5: Exam Items - page 39

11. Under what circumstances would a score that is 15 points above the mean be considered an extreme score? a. when the population mean is much larger than 15 b. when the population standard deviation is much larger than 15 c. when the population mean is much smaller than 15 d. when the population standard deviation is much smaller than 15 Answer c

page 106

Answer c

12. A very bright student is described as having an IQ that is three standard deviations above the mean. If this student's IQ is reported as a z-score, the z-score would be a ut 3 b. ut+3o0 rete) d. cannot be determined from the information given

13.

A z-score of z = +3.00 indicates a location that is a. near the center of the distribution b. slightly above the mean c. far above the mean in the extreme right-hand tail of the distribution d. The location depends on the mean and standard deviation for the distribution.

14.

A z-score of z = -0.25 indicates a location that is a. at the center of the distribution

page 107

Answer b

page 107

b. slightly below the mean c. far below the mean in the extreme left-hand tail of the distribution

d. The location depends on the mean and standard deviation for the distribution.

Answer c

page 106

15. A population distribution has » = 80 and o = 6. In this distribution a z-score of z = +2.00 identifies a location a. two points above the mean b. two points below the mean c. twelve points above the mean d. twelve points below the mean

Answer c

page 106

16. For a population with o = 4, an individual with a deviation score of +2 would have a z-score of a. +2.00 b. +8.00 C.F -30 d. cannot be determined without knowing the population mean

Chapter 5: Exam Items - page 40

Answer a

page 112 www

Answer b

page 112

17. Suppose you earned a score of X = 54 onan exam. Which set of parameters would give you the highest grade? a. p =50 ando=2 b. up = 50 ando =8 c. h=58 ando=2 d. p=58 ando=8

18. Suppose you earned a score of X = 45 on an exam. would give you the highest grade?

Which set of parameters

a. w=50ando=2 b. p= 50 ando=10 GC. = S53 ando=2Z d. nu=55 ando= 10 Answer b page 112

19. Last week Tom had exams in statistics and in English. He scored 10 points above the mean on both exams. From this information, you can conclude that a. Tom has identical z-scores for the two exams b. Both of Tom's z-scores are positive c. Tom will have a higher z-score for the exam with the lower mean d. none of the above

Answer c

20. A population with p = 85 and o = 12 is transformed into z-scores. After the transformation, the population of z-scores will have a mean of

page 110

a. p= 85 b. un= 1.00 cj p= 0

d. cannot be determined from the information given Answer b

page 111

21. A population with p = 85 and o = 12 is transformed into z-scores. After the transformation, the population of z-scores will have a standard deviation of a. o= 85 b. o = 1.00 c. 6 =0

d. cannot be determined from the information given Answer d

page 106

22. For a symmetrical population with p = 100 the z-score corresponding to X = 120 would be a. 1.20 b. 2.00 e. 1.00

d. cannot be determined from the information given

Chapter 5: Exam Items - page 41

Answer d page 111 www

23. A population has p = 50 and o = 10. If these scores are transformed into z-scores, the population of z-scores will have ameanof___ and a standard deviationof _.. a. 50, 10

bo0et c. 0, 10

d. 0, 1 24. One advantage of transforming X values to z-scores is a. all negative numbers are eliminated b. the distribution is transformed to a normal shape c. all scores are moved closer to the mean d. all of the above e. none of the above

25. A distribution with p = 35 and o = 8 is being standardized so that the new mean and standard deviation will be 1 = 50 and o = 10. When the distribution is standardized, what value will be obtained for a score of X = 39 from the original distribution? a. X = 54 b. X = 55 c. X=1.10

d. impossible to determine without more information Answer b page 114

26. A distribution with p = 35 and o = 8 is being standardized so that the new mean and standard deviation will be p = 50 and o = 10. In the new, standardized distribution your score is X = 60. What was your score in the original distribution? a. b. c.

X =45 X = 43 X = 1.00

d. impossible to determine without more information Answer c

page 114

27. Using z-scores, a population with p = 37 and o = 8 is standardized so that the new mean is = 50 and o = 10. How does an individual's z-score in the new distribution compare with his/her z-score in the original population? a. new z= old z+ 13 b. new z= (10/6)(old z) c. newz=oldz d. cannot be determined with the information given

Chapter 5: Exam Items - page 42

True/False Questions True

1. A positive z-score always corresponds to a score greater than the mean.

page 106 True page 106

2. For any distribution, a z-score of z = -0.50 corresponds to a location below the

False page 110

3. Fora positively skewed distribution, a z-score of zero will be located below (to the left of) the mean.

True page 107

4. Within any distribution of scores, the location specified by z = +1 and the location specified by z = -1 are exactly the same distance from the mean.

True

5. For any population, a z-score of -2.00 indicates a more extreme location (farther from the mean) than a z-score of +1.00.

page 107 True

page 108

mean.

6. Ina distribution with p = 50 and o = 10, a score of X = 55 corresponds to a z-score of z= +0.50.

page 108

7. Ina distribution with p = 50 and o = 10, a score of X = 30 corresponds to a z-score of z= -3.00.

False page 108

8. Ina distribution with p = 80 and o = 20, a score of X = 85 corresponds to a zscore of z = 1.50.

True

9. Ina distribution with 1: = 80 and o = 20, a score of X = 100 corresponds to a zscore of z = 1.00.

False

page 108 False

page 109

10. Ina distribution with p = 40 and o = 12, a z-score of z= -0.50 corresponds to a score ofX = 46.

page 109

11. Ina distribution with p = 40 and o = 12, a z-score ofz = 2.00 corresponds to an X value of X = 64.

False page 106

12. For any population, a z-score of +1.00 corresponds to a location exactly 1 point above the mean.

True page 106

13. For any population, a z-score of +1.00 corresponds to a location above the mean by one standard deviation.

True page 106

14. Ina distribution with p = 80, a score of X = 90 corresponds to z = +2.00.

The standard deviation for this population is o = 5S.

True

Chapter 5: Exam Items - page 43

False page 106

15. Ina distribution with 1: = 80, a score of X = 70 corresponds to z = -0.50. The standard deviation for this population is o = 5.

False

16. Ina distribution with o = 8, a score of X = 64 corresponds to z = -0.50. The mean for this population is 1 = 60.

page 106

page 106

17. Ina distribution with o = 4, a score of X = 48 corresponds to z= 2.00 The mean for this population is p = 40.

True page 106

18. On an exam, Tom scored 8 points above the mean and had a z-score of +2.00.

The standard deviation for the set of exam scores must be o = 4.

True page 112

19. On an exam with p = 70, you have a score of X = 76. In this situation you should expect a better grade if o = 3 than if o = 12.

False page 110

20. Transforming X values into z-scores makes it possible to describe an entire distribution with a single number.

False page 106

21. In any population of scores, at least one individual will have a z-score of zero.

False

22. Ifa population of N = 10 scores is transformed into z-scores, there will be five positive z-scores and five negative z-scores.

True

page 106

page 110

23. Ifa distribution of scores is transformed into z-scores then the sum of the positive z-scores will be exactly equal to the sum of the negative z-scores (ignoring the signs).

True

24. The population mean always corresponds to a z-score of zero.

True

page 110 True page 106

25. Iftwo individuals in a population have identical X scores, they also will have

True page 110

26. Transforming X values into z-scores will not change the shape of the distribution.

False page 110

27. One advantage of transforming X values into z-scores is that the transformation always creates a normal shaped distribution.

True

28. Whenever a population is transformed into z-scores, £z = 0.

identical z-scores.

page 110

Chapter 5: Exam Items - page 44

Other Exam Items

1. Describe the general purpose of a z-score.

2. Describe what happens to the mean, the standard deviation, and the shape of a distribution when all of the scores are transformed into z-scores.

3. Fora population with p = 70 and o = 8, find the z-score that corresponds to each of the following X values.

X=74 z= X=68

z=

X=78 X=70

z= z=

4. For a population with » = 50 and o = 4, find the X value that corresponds to each of the following z-scores.

5. Ona

Ze ol Z= Alo z= 2.00

5 Gee Does x=

Z==]025

)

psychology exam with p = 76 and o = 12, Tom scored at the mean, Mary had a score of

X = 73, and Bill had a z-score of z = -0.50. Place these three students in order from lowest to

highest score.

Student with lowest score Student with middle score

Student with highest score 6. A set of exam scores are reported as X values and z-scores.

On this exam a score of X = 75

corresponds to a z-score of z = +2.00, and a score of X = 60 corresponds to a z-score of z =

-1.00. What are the values for the mean and standard deviation for this exam? (Hint: Sketch a distribution and locate each of the z-score positions. )

7. A population of scores with p = 38 and o = 6 is standardized to create a new population with u = 50 and o = 10. What is the new value for each of the following scores from the original population? original score new value

X= 26 K= 35 X = 44 X= 47

es

Chapter 5: Exam Items - page 45

Answers for Other Exam Items 1. The purpose of a z-score is to describe a location within a distribution using a single number.

2. When an entire distribution of scores is transformed into z-scores, the resulting distribution will have a mean of zero, a standard deviation of one, and the same shape as the original distribution.

3.

z=+0.50 z= -0.25 z= +1.00 z= 0

4. X=44 » ae) | X= 58 X= 45 5. Bill: X =70, z=-0.50 Mary: X = 73, z=-0.25 Tom:

X=76,

z=0

6.

w=65 ando=5

7.

X=26 => z=-2.00 X=35 = z=-0.50

=> X=30 => X=45

X=44 X=47

> =

= z=+1.00 = 2=+1,50

X=60 X=65

Chapter 5: Exam Items - page 46

Chapter 6 Probability

Multiple-Choice Questions 1. Probability values are always fev). greater than or equal to zero b. less than or equal to one Cc. positive numbers d. all of the above

2. A random sample requires that a. every individual has an equal chance of being selected b. the probabilities cannot change during a series of selections c. there must be sampling with replacement d. all of the above Answer b

page 123

Answer d page 123

3. A colony of laboratory rats contains 7 albino rats and 23 hooded rats. What is the probability of randomly selecting an albino rat from this colony? a. 7/23 b. 7/30 Cc. 23/30 d. 23/7 4. An introductory psychology class has 9 freshman males, 11 freshman females, 8 sophomore males, and 12 sophomore females. What is the probability of randomly selecting a male from this group? a. 920 b. 8/20 e. 17/20 d. 17/40

5. A jar contains 40 red marbles and 10 black marbles. If you take a random sample of a marble from this jar, what is the probability that the marble will be red? a. 10/50 b. 40/50 c. 10/40 d. cannot be determined with the information given

Chapter 6: Exam Items - page 47

Answer a

6. A jar contains 40 red marbles and 10 black marbles. If you take a random

page 125

sample of n= 3 marbles from this jar, and the first two marbles are both red, what

is the probability that the third marble will be black? a. 10/50 b. 8/48 c. 9/48 c. 10/48 7. For a normal distribution, the proportion in the tail beyond z = 1.50 is p = 0.0668. Based on this information, what is the proportion in the tail beyond z= -1.50? a. 0.0668 b. -0.0668 C.1029332 d. -0.9332 Answer c

page 130

Answer b

page 129 www

8. The proportion of a normal distribution that corresponds to values less than z= 1.00 is p = 0.8413. Base on this information, what is the proportion that corresponds to values less than z = -1.00? a. 0.8413 b. -0.8413 €: 01587 d. -0.1587 9. A vertical line is drawn through a normal distribution at z = +0.25. The line separates the distribution into two sections and the smaller section corresponds to _____ of the whole distribution. a. 25%

b. 40.13% c. 59.87% d. 75% Answer b page 129 www

10. A vertical line is drawn through a normal distribution at z = -1.00. The line separates the distribution into two sections and the larger section corresponds to of the whole distribution. a. 75% b. 84.13% c. -84.13% gd. 99%

Chapter 6: Exam Items - page 48

Answer c

page 132

11. What proportion of a normal distribution is located in the tail beyond a z-score of z= -1.50? a. -0.668

b. -0.9332 c. 0.0668 d. 0.9332 Answer ¢

page 131 www

Answer b

page 131

Answer c

page 131

Answer a

page 132

12. What proportion of a normal distribution is located in the tail beyond z = 2.00? a. 1.14% b. 2% c. 2.28% d.97 12%

13. What proportion of the scores in a normal distribution correspond to z-scores greater than +1.04?. a. 0.3508 b. 0.1492 c. 0.6492 d. 0.8508

14. What proportion of the scores in a normal distribution have z-scores less than Z = 0.86? a. 0.3051 b. 0.1949 c. 0.8051 d. 0.6949 15. What proportion of the scores in a normal distribution have z-scores less than Zim = soe

a. 0.0934 b. 0.4066 c. 0.5934 d. 0.9066 Answer Cc

page 132

16. What proportion ofthe scores in a normal distribution have z-scores greater than z = -1.25? a. 0.1056 b. -0.1056 c. 0.8944 d. -0.8944

Chapter 6: Exam Items - page 49

Answer d

page 132 www

Answer c

page 132

17. For a normal distribution, what z-score value separates the lowest 20% of the distribution from the highest 80%? a. z= 0.20 b. z= 0.80 c.

Z=0.84 d. z=-0.84

18. For a normal distribution, what z-score value separates the highest 10% of the distribution from the lowest 90%? a. Z= 0.90 b. z= -0.90 c. Z=1.28 d. z=-1.28

Answer a page 135

19. What proportion of a normal distribution is located between z = 1.00 and Z= 1.50? a. 0.0919 b. 0.7745 c. 0.5000 d02255

Answer d page 135

20. What proportion of a normal distribution is located between z= -1.16 and Z=+1,16? a. 0.3770 b. 0.6230 c. 0.2460 d. 0.7540

Answer a page 134

21. For a normal distribution with 1 = 60 and o = 8, the probability of selecting a

www

score greater than X = 64 is equal to a. the proportion of the distribution with z-scores greater than 0.50 b. the proportion of the distribution with z-scores greater than 1.00

c. the proportion of the distribution with z-scores greater than 2.00 d. the proportion of the distribution with z-scores greater than 4.00 Answer a page 134

22. For a normal distribution with a mean of 1 = 40 and o = 4, what is the probability of sampling an individual with a score greater than 46? a. 0.0668 b. 0.4452 R09 332 d. 0.0548

Chapter 6: Exam Items - page 50

Answer b

page 134

23. For a normal distribution with a mean of 1 = 60 and o = 8, what is the probability of selecting an individual with a score greater than 54? a. 0.2266 b. 0.7734 c. 0.7266 d. 0.2734

Answer c

24. For a normal distribution with a mean of u = 40 and o = 4, what is the

page 134

probability of sampling an individual with a score less than 46? a. 0.0668 b. 0.4452 ¢, 0.9332 d. 0.0548

www

Answer a

page 134

Answer c

page 137 Www

25. For a normal distribution with a mean of ,1 = 60 and o = 8, what is the probability of selecting an individual with a score less than 54? a. 0.2266 by 0.7734 c. 0.7266 d. 0.2734 26. Scores on the SAT form a normal distribution with p = 500 and o = 100. The State College accepts students whose SAT scores are in the top 60% of the distribution. What is the minimum SAT score needed for admission to the State College? a. 440

b. 460 c. 475

dc. 525 Answer c

27. A normal distribution has a mean of tp = 500 and o = 100. What score is

page 137

needed to place in the top 20% of the distribution? ay 220 b. 580 c. 584 d. 700

Answer b

28. A normal distribution has a mean of p = 36 and o = 4. What proportion ofthe distribution is located between scores of X = 30 and X = 38? a. 0.3753 b. 0.6247 @.. 0;2583 d. 0.7417

page 135

Chapter 6: Exam Items - page 51

True/False Questions True page 124

1. Probability values are always greater than or equal to zero.

True page 124

2. The value for a probability can never exceed 1.00, unless you have made a computational error.

True page 124

3. As defined in the text, random sampling requires that each individual in the population has an equal chance of being selected.

True page 125

4. As defined in the text, random sampling requires sampling with replacement.

True page 125

5. Sampling with replacement means that it is possible for the same individual to be selected more than once for a single sample.

False page 124

6. If there are 10 seniors and 40 juniors in a class, then the probability of randomly selecting one individual who is a senior is p(senior) = 1/4.

True page 124

7. If there are 50 students in a class, then the probability of randomly selecting any particular individual is p = 1/50.

True page 124

8. A jar contains 15 red marbles and 75 blue marbles. If you randomly select a marble from this jar, the probability of obtaining a red marble is p = 15/90.

True

9. A jar contains 10 red marbles and 20 blue marbles. If you take a random

page 125

sample of two marbles from this jar, then the probability that the second marble is blue is exactly equal to the probability that the first marble is blue.

False

page 125

10. A jar contains 10 red marbles and 20 blue marbles. If you take a random sample of two marbles from this jar, then the probability that the second marble is blue depends on the color of the first marble.

False page 127

11. One major advantage of transforming X values into z-scores is that the zscores always form a normal distribution.

Tre page 130

12. For any normal distribution, the mean and the median will have the same value.

Chapter 6: Exam Items - page 52

True

13. The Unit Normal Table can be used for any normal distribution, no matter

page 134

what the values are for the mean and standard deviation.

False page 134

14. The Unit Normal Table lists proportions for any distribution, provided that the distribution has been transformed into z-scores.

True

15. When the z-score value in a normal distribution is positive, the tail is on the

page 130

right-hand side of the distribution.

True page 130

distribution is on the right-hand side.

False page 130

17. When determining the probability of selecting a score that is below the mean, you will get a negative value for probability.

True

18. For any normal distribution, the proportion corresponding to scores greater than z = +1.00 is exactly equal to the proportion corresponding to scores less than

page 130

16. When the z-score value in a normal distribution is negative, the body of the

B= = 100:

True page 130

19. For any normal distribution, the portion in the tail beyond z = 1.00 and the portion in the tail beyond z = -1.00 are each exactly 15.87% of the total distribution.

True

page 132

20. A vertical line is drawn through a normal distribution dividing the distribution into two parts. If the smaller part equals 0.0668 of the distribution, then the line was drawn at z = 1.50 or at z = -1.50.

True page 132

21. A vertical line is drawn through a normal distribution dividing the distribution into two parts. If the larger part equals 0.9750 of the distribution, then the line was drawn at z = 1.96 or at z= -1.96.

False page 131

22. For any normal distribution, the proportion in the tail beyond z = -2.00 is p = -0.0228.

True page 231

23. For any normal distribution, exactly 2.5% ofthe scores are located in the tail beyond z = 1.96.

True page 131

24. For any normal distribution, a z-score ofz = -0.25 separates the highest 59.87% ofthe scores from the rest of the distribution.

False page 231

25. For any normal distribution, a z-score of z= 1.28 separates the highest 89.97% ofthe scores from the rest of the distribution.

Chapter 6: Exam Items - page 53

Other Exam Items 1. Describe the two conditions that must be satisfied for random sampling.

2. Ifa vertical line is drawn through a normal distribution at each of the following z-score locations, then determine whether the tail is on the left side or the right side of the line and determine what proportion of the distribution is located in the tail. a. z=+1.00 b. z= +0.50 c. Z=-1.50 d. z=-2.00 3.

For anormal distribution, find the z-score location of a line that would separate the

distribution as described in each of the following: a. Where should the line be drawn to separate the top 10% from the rest? © b. Where should the line be drawn to separate the top 20% from the rest? c. Where should the line be drawn to separate the bottom 40% from the rest?

4. A normal distribution has a mean of 1 = 61 with o = 8. Find the following probabilities:

p(X > 66) p(X < 55)

> px < 70) op

pol =X = 73)

[ow

5. A normal distribution has a mean of p = 28 and o = 5. Find the scores associated with the following regions: a. the score needed to be in the top 41% of the distribution b. the score needed to be in the top 72% of the distribution c. the scores that mark off the middle 60% of the distribution

Answers for Other Exam Items

1. The two conditions for random sampling are a. every individual has an equal chance of being selected, and b. if more than one individual is selected, the probabilities must remain constant after each selection.

Chapter 6: Exam Items - page 54

. The tail is on the right. p = 0.1587 . The tail is on the right. p = 0.3085 . The tail is on the left. p = 0.0668 iy . The tail is on the left. fen Te) eh p = 0.0228 z= 1.28 .

Z= 0.84 . z=-0.25

. p= 0.2643 p = 0.2266 p = 0.8708 p = 0.8276 Bia 5 EEi) WahWe . 23.80 < X < 32.20

Chapter 6: Exam Items - page 55

Chapter 7 The Distribution of Sample Means

Multiple-Choice Questions 1. The distribution of sample means consists of a. all the scores contained in a particular sample b. all the scores contained in a specific population c. the sample means for all the possible random samples (of a specific size) from a particular population d. the sample means for all the possible random samples (for all sample sizes) from a particular population 2. The mean of the distribution of sample means is called a. the expected value of X b. the standard error of X c. the sample mean d. the central limit mean 3. The standard deviation of the distribution of sample means is called a. the expected value of X b. the standard error of X c. the sample standard deviation d. the central limit standard deviation Answer a

page 149

4. The symbol that corresponds to the standard error of X is a.

OX

aos >) x Answer a

page 149

5. When a random sample is selected from a population, the sample mean is not expected to be exactly equal to the population mean. On average, the distance between a sample mean and the population mean is predicted by a. the standard error b. the expected value c. the mean of the population d. the standard deviation of the population Chapter 7: Exam Items - page 56

6. When the sample size is greater than n = 30 ; a. the distribution of sample means will be approximately normal b. the sample mean will be equal to the population mean c. all of the above d. none of the above 7. The standard error of X provides a measure of a. the maximum possible discrepancy between X and pt b. the minimum possible discrepancy between X and p c. the exact discrepancy between each specific X and 1 d. none of the above Answer c page 149

8. Iftwo samples of exactly the same size are selected from the same population, then the two sample means will have a. exactly the same expected value b. exactly the same standard error c. both of the above d. neither of the above

Answer c page 149

9. As sample size increases, the expected value of X

Answer b page 151

10. As sample size increases, the standard error of ye a. also increases b. decreases

a. also increases b. decreases c. stays constant

c. stays constant

Answer d page 148

11. The distribution of sample means

Answer d page 151

12. Fora population with uy = 80 and o = 20, the distribution of sample means based on n= 16 will have an expected value of ae i a bs: c. 20 d. 80

a. is always normal b. is normal only if the population distribution is normal c. is normal only if the sample size is greater than 30 d. none of the above

Chapter 7: Exam Items - page 57

Answer b page 151 www

13. Fora population with p = 80 and o = 20, the distribution of sample means based on n = 4 will have a standard error of aa

b. 10 C20

d. 80 Answer d page 151

14. Which of the following samples would have a standard error of 2 points? a. n=5 scores from a population with o = 10 b. n=5 scores from a population with o = 20 c. n= 100 scores from a population with o = 10 d. n= 100 scores from a population with o = 20

Answer c page 151

15. If you select a random sample of n = 16 scores from a population with p = 70 and o = 20, how much error would you expect, on average, between the sample mean and the population mean?

1.25 points . 4 points . 5 points 6 . 20 points CG 16. Which of the following samples would have the largest standard error? a. n= 25 scores from a population with o = 10 b. n= 25 scores from a population with o = 20 c. n= 100 scores from a population with o = 10 d. n= 100 scores from a population with o = 20

Answer c page 151

17. In general, the standard error of X gets smaller as a. sample size and standard deviation both increase b. sample size and standard deviation both decrease c. sample size increases and standard deviation decreases d. sample size decreases and standard deviation increases

Answer b page 151 www

18. For a particular population a sample of n = 4 scores has a standard error of 10.

For the same population, a sample of n = 16 scores would have a standard error of a. 2 Ee bird c. 10 d.

Chapter 7: Exam Items - page 58

Answer b

page 153

19. A positively skewed population has p = 50 and o = 20. A random sample of n= 4 scores obtained from this population has a mean of X = 55. What is the z-score corresponding to this sample mean? a, O25

b. 0.50 c. 1.00

d. 2.00 20. A normal population has p = 50 and o = 8. A random sample of n = 4 scores from this population has a mean of 54. What is the z-score for this sample mean? a. +0.50 b. +1.00 ce, +2.00 d. +4.00

Answer b page 153

21. A random sample of n= 4 scores is obtained from a population with o = 10.

If the sample mean is 10 points greater than the population mean, then the sample mean would have a z-score of a. +10.00 b. +2.00 1000) d. cannot be determined without knowing the population mean 22. A random sample of n = 4 scores is obtained from a normal population with yu = 20 and o = 4. What is the probability of obtaining a mean greater than xX =22 for this sample? a. 0.50 b. 1.00 c. 0.1587 d. 0.3085

Answer d

page 153

23. A random sample is obtained from a population with » = 80 and o = 10.

Which of the following samples would have the largest z-score? a. n= 25 with X = 82

b. n= 25 with X = 84 c. n= 100 with X = 82 d. n= 100 withX = 84 Answer c

24. A sample of n= 25 scores is determined to have a standard error of 2 points.

page 151

What is the standard deviation for the population from which the sample was

www

obtained? a

b. 2/5 ¢. 10

d. 50 Chapter 7: Exam Items - page 59

Answer c

25. A random sample is selected from a population with p = 80 and o = 10. To

page 151

ensure a standard error of 2 points or less, the sample size should be at least ; a n=s5 bei = i0 ce. n=25 d. It is impossible to obtain a standard error less than 2 for any sized sample.

Answer a

26. If sample size (n) is held constant, the standard error will population variance increases. a. increase b. decrease

page 151

as the

c. stay constant

d. cannot answer with the information given Answer a

page 153

27. Ifa sample is selected from a normal population, then the probability that the sample mean will have a z-score greater than z = 2.00 is a. p= 0.0228 bap = 09772 c. p=0 .0456 d. cannot determine without knowing the sample size

True/False Questions True page 145

1. Two samples probably will have different means even if they are both the same size and they are both selected from the same population.

False page 155

2. Ifa sample of at least 30 scores is randomly selected from a normal population, the sample mean will be equal to the population mean.

False page 148

3. Ifa sample contains at least n = 30 scores, then the scores in the sample will

form a normal shaped distribution.

False page 148

4. According to the Central Limit Theorem, the expected value for a sample mean approaches zero as the sample size approaches infinity.

True

page 148

5. It is possible for the distribution of sample means to be normal even if it is based on samples with less than n = 30.

False

6. The mean of the distribution of sample means is called the standard error of X.

page 149

Chapter 7: Exam Items - page 60

True page 149

7. The standard error of X provides a measure of the distance, on average, between a sample mean and the population mean.

False page 151

8. To compute standard error for a specific sample mean, you simply find the difference between the sample mean and the population mean; ox = (X - 11).

True page 151

9. The standard error will always be less than or equal to the standard deviation.

True page 150

10. Ifa sample consists of a single score (n = 1), then the sample mean will be equal to the score, X = X, and the standard error will be equal to the population standard deviation.

True page 151

11. The standard error of X can never be greater than the standard deviation of the population from which the sample is selected.

True page 152

12. A sample mean with a z-score value between -0.50 and +0.50 would be considered a fairly typical, representative sample.

True page 152

13. A sample mean with a z-score value greater than z = 3.00 would be considered an extreme, unrepresentative sample.

True page 149

14. The mean of the distribution of sample means is always equal to the population mean.

True page 151

15. As sample size increases, the magnitude of the standard error decreases.

False page 149

16. As sample size increases, the magnitude of the expected value decreases.

False

page 151

17. On average, the difference between a sample mean and the population mean increases as the sample size increases.

True page 149

18. The standard deviation of the distribution of sample means is called the standard error of X.

True page 151

19. On average, a sample of n= 100 scores will provide a better estimate of the population mean than a sample of n = 50 scores.

True

20. Assuming that all other factors are held constant, as the population variability increases, the standard error also increases.

page 151 False

page 152

21. For a population with 1: = 100, you are more likely to obtain a sample mean greater than 110 with a sample of n = 100 than with a sample of n= 50.

Chapter 7: Exam Items - page 6]

False

page 151 True

22. The distribution of sample means based on samples of n = 4 selected from a population with p: = 80 and o = 20 will have a standard error of 5.

23. The distribution of sample means based on samples of n= 16 selected from a

page 149

population with 1: = 80 and o = 20 will have an expected value of 80.

False page 151

24. A sample of n = 25 scores has a standard error of 2. This sample was selected from a population with o = 50.

False page 151

25. Ifa random sample of n = 16 scores is obtained from a population with 0 = 20, the sample mean is guaranteed to be within five points of the population mean.

False page 153

26. A population has pp = 50 ando=10. Fora sample of n= 4 scores from this population, a sample mean of X = 55 would be considered an extreme value.

True page 153

27. A population has p = 50 and o=10. Fora sample of n= 100 scores from this population, a sample mean of X = 55 would be considered an extreme value.

True

28. A researcher obtained X = 33 for a sample of n= 16 scores selected from a population with 1, = 30 and o = 12. This sample mean corresponds to a z-score of Z= 1.00;

page 153

page 153

29. A sample of n = 4 scores is randomly selected from a population with p = 80 and o = 16. If the sample mean is X = 84, then the corresponding z-score is Z= 1.00;

True page 153

30. Ifa sample mean corresponds to a z-score value near zero, then you can conclude that the sample mean is relatively close to the population mean.

True

31. If you continue selecting samples from a normal population, you will find that

False

page 153

most of the samples, more than 50%, will have sample means with z-scores between z = -1.00 and z = +1.00.

Other Exam Items 1. Define the distribution of sample means.

2. Describe what is measured by the standard error of X.

Chapter 7: Exam Items - page 62

3. Describe the major differences between the following two distributions. a. A population of scores with pp = 50 and o = 6. b. The distribution of sample means based on samples of n = 36 selected from a population

with pp = 50 and o = 6.

4. For a population with p = 80 and o = 10, a. how close, on the average, would you expect a single score (n = 1) to be to the population mean? b. how close, on the average, would you expect the mean for a sample of n = 4 scores to be to the population mean? c. how close, on the average, would you expect the mean for a sample of n = 25 scores to be to the population mean?

5. Each of the following samples was obtained from a population with pp = 80 and 6 = 12. Find the z-score corresponding to each sample mean. a. X = 86 for a sample of n= 4 b. X =77 for a sample of n= 16 c. X = 84 for a sample of n = 36 6. A normal distribution has p = 40 and o = 8. a. Describe the distribution of sample means based on samples of n = 16 selected from this population. b. Ofall the possible samples of n= 16, what proportion will have sample means greater than 42? c. Of all the possible samples of n = 16, what proportion will have sample means less than 39°?

7. For anormal population with p = 100 and o = 12, a. what is the probability of obtaining a sample mean greater than 106 for a sample of n= 4 scores? b. what is the probability of obtaining a sample mean greater than 106 for a sample of n= 16 scores?

Answers for Other Exam Items 1. The distribution of sample means is the set of sample means obtained from all the possible random samples of a specified size (n) taken from a particular population.

2. The standard error of X is the standard deviation for the distribution of sample means, and it measures the standard distance (or deviation) between a sample mean X and the population mean LL.

Chapter 7: Exam Items - page 63

3. The population consists of scores and the distribution of sample means consists of means (statistics). Although both distributions have the same mean, 1 = 50, the distribution of sample means has a standard deviation given by the standard error, 6g = 1. The distribution of sample means will be normal because the sample size (n) is greater than 30.

4.a.

o=10

b.

Gz=5

C.

Ors

5. a. : c.

z=+1.00 =-1.00 z=+2.00

6. a. The distribution will be normal with an expected value of 1: = 40 and a standard error of 2 points. b. z= 1.00, p = 0.1587 c Z=-0.50, p= .3085 7,4, a

£2=6/6=1.00.

Z= 6/3. =2.00.

p=0,1587 p=0,0228

Chapter 7: Exam Items - page 64

Chapter 8 Introduction to Hypothesis Testing

Multiple-Choice Questions

Answer c

page 167 Www

Answer a

page 169

. A hypothesis test 1s a. a descriptive technique that allows researcher to describe a sample b. a descriptive technique that allows researcher to describe a population a an inferential technique that uses the data from a sample to draw inferences about a population d. an inferential technique that uses information about a population to make predictions about a sample No .

The null hypothesis a.

states that the treatment has no effect

b . is denoted by the symbol H, e is always stated in terms of sample statistics d . all of the above Answer d page 169

Answer b

page 173

The null hypothesis a. concerns a population with an unknown mean b. is denoted by the symbol Hy e. generally states that there is no effect, no change, or no difference d. all of the above . The final step of hypothesis testing is to as locate the values associated with the critical region b . make a statistical decision about Hy c. collect the sample data and compute the test statistic d . state the hypotheses and select an alpha level . Ina hypothesis test, the “critical region” consists of a. sample values that are highly likely to be obtained if Hy is true b. sample values that are very unlikely to be obtained if Hy is true €. sample values that provide proof that Hy is true d. sample values that provide proof that Hy is false Chapter 8: Exam Items - page 65

Answer a page 171

6. The critical region for a hypothesis test consists of a. outcomes that have a very low probability if the null hypothesis is true b. outcomes that have a high probability if the null hypothesis is true c. outcomes that have a very low probability whether or not the null hypothesis is true d. outcomes that have a high probability whether or not the null hypothesis

is true Answer d page 179

7. Increasing the alpha level (for example from a = .01 to a = .05) . increases the probability of a Type I error . increases the size of the critical region . increases the probability that the sample will fall into the critical region . all of the above . none of the above oa0gdgp

Answer a

8. In general, increasing the alpha level (for example from .01 to .05) will a. increase the likelihood of rejecting the null hypothesis b. decrease the likelihood of rejecting the null hypothesis c. not affect the likelihood of rejecting the null hypothesis

page 179

Answer c

page 179 www

9. Which combination of factors will increase the chances of rejecting the null hypothesis? a. a large standard error and a large alpha level b. a large standard error and a small alpha level c. a small standard error and a large alpha level d. a small standard error and a small alpha level 10. Ina hypothesis test, a z-score value near zero . is probably in the critical region . means that you should probably reject the null hypothesis . is strong evidence of a statistically significant effect . all of the above ff oaade . none of the above

Answer d page 173

11. Ina hypothesis test, an extreme z-score value, like z = +3 or z= +4,

is probably in the critical region . means that you should probably reject the null hypothesis is strong evidence of a statistically significant effect . all of the above ono sp none of the above

Chapter 8: Exam Items - page 66

Answer a page 173

12. A sample of n= 25 individuals is selected from a population with p = 80 and a treatment is administered to the sample. If the treatment has a large effect, then a. the sample mean should be very different from 80 and should lead you to reject the null hypothesis b. the sample mean should be very different from 80 and should lead you to fail to reject the null hypothesis c. the sample mean should be close to 80 and should lead you to reject the null hypothesis d. the sample mean should be close 80 and should lead you to fail to reject the null hypothesis

13. A sample of n = 25 individuals is selected from a population with 1 = 80 and a treatment is administered to the sample. If the treatment has no effect, then a. the sample mean should be very different from 80 and should lead you to reject the null hypothesis b. the sample mean should be very different from 80 and should lead you to fail to reject the null hypothesis c. the sample mean should be close to 80 and should lead you to reject the null hypothesis d. the sample mean should be close 80 and should lead you to fail to reject the null hypothesis Answer c

page 176

By definition, a Type I error is a. rejecting a false H, b. rejecting a false Hy c. rejecting a true Hy d. failing to reject a false Hy OS: A Type I error means that a researcher has a. falsely concluded that a treatment has an effect b. correctly concluded that a treatment has no effect c. falsely concluded that a treatment has no effect d. correctly concluded that a treatment has an effect 16. A Type II error means that a researcher has a. falsely concluded that a treatment has an effect b. correctly concluded that a treatment has no effect c. falsely concluded that a treatment has no effect d. correctly concluded that a treatment has an effect

The probability of committing a Type I error a. is determined solely by the size of the treatment effect b. cannot be controlled by the experimenter c. is determined by the level of significance that one chooses d. is determined by the value for beta (B) one selects Chapter 8: Exam Items - page 67

Answer a

page 177

18. A researcher risks a Type | error . anytime Hy, is rejected . anytime H, is rejected . anytime the decision is "fail to reyect Hy" ao . all of the above

Answer b

19. Ina hypothesis test, an extreme, low-probability value for the z-score is an

page 171

indication that | a. the null hypothesis is probably correct b. the null hypothesis is probably wrong c. the null hypothesis is certainly correct d. the null hypothesis is certainly wrong

Answer b page 185

20. One assumption for a hypothesis test with z-scores is that the distribution of sample means ; a. is based on less than 30 means b. is normal c. has a pu the value of which is predicted by H, d. none of the above

Answer d page 175

21. In hypothesis tests, standard error measures a. the size of the treatment effect b. variability among population means c. the size of the critical region d. the amount of difference between the sample mean and the population mean that is reasonable to expect just by chance 22. The critical boundaries for a hypothesis test are z = +1.96 and -1.96. If the z-score for the sample data is z = -1.90, then the correct statistical decision is

a. fail to reject H, b. fail to reject Hy, c. reject Hi, d.. reject Hi, Answer a

page 188

23. A researcher expects a treatment to produce an increase in the population mean. Assuming a normal distribution, what is the critical z-score for a one-tailed test witha = .01? a, 12.83 by £2.58 C2 t1.65 a. 42.33

Chapter 8: Exam Items - page 68

Answer c

24. By selecting a smaller alpha level, a researcher is

page 177

a. attempting to make it easier to reject Hy b. better able to detect a treatment effect c. reducing the risk of a Type I error d. all of the above

25. Ifa treatment has a very small effect, then a hypothesis test evaluating the treatment effect is likely to a. result ina Type | error b. result in a Type II error c. correctly reject the null hypothesis d. correctly fail to reject the null hypothesis Answer a page 178

26. The probability of a Type II error is expressed as a. B b. 1-8 eae d.l-a

27. A researcher is predicting that a treatment will increase scores. If this treatment is evaluated using a directional hypothesis test, then the critical region for the test a. would be entirely in the right-hand tail of the distribution b. would be entirely in the left-hand tail of the distribution c. would be divided equally between the two tails of the distribution d. cannot answer without knowing the value of the a level Answer a

page 186

28. A population is known to have a mean of p = 45. Ifa researcher predicts that the experimental treatment will produce a decrease in the population mean, then the null hypothesis for a one-tailed test would state a. p> 45 b. p< 45 Cc. p<45

d. X > 45 Answer a

page 186

29. A population is known to have a mean of p = 50. A treatment is expected to lower scores for individuals in this population. Ifthe treatment is evaluated using a one tailed hypothesis, then the null hypothesis would state a.

p> 50

b. p > 50 e

p< 50

d. wp<50

Chapter 8: Exam Items - page 69

Answer c

page 182

30. A researcher reports that there is a significant difference between two treatments at the .05 level of significance. This means that a. if the researcher repeated the experiment, there is a 95% probability that the same decision would be reached b. the difference obtained in the experiment is at least 5% larger than the standard error fo). there is a 5% probability (or less) that the difference is due to chance d. the average score for one treatment is at least 5% higher than the average score for the other treatment

True/False Questions True page 169

1. The null hypothesis always concerns a population even though the data come from a sample.

True

2. In most research situations, the goal is to reject the null hypothesis.

page 173

page 173

3. A decision to reject the null hypothesis means that the data do not provide evidence for a real treatment effect.

True page 173

4. Ifthe data provide convincing evidence that the treatment does have a real effect, then the statistical decision is to reject the null hypothesis.

True page 171

5. For a hypothesis test, the critical region is defined as the set of sample outcomes that is very unlikely to be obtained if the null hypothesis is true.

False page 171

6. Ifthe obtained sample data are in the critical region, then we conclude that the data provide support for the null hypothesis.

True page 171

7. When the z-score for sample data falls within the critical region, the decision is to reject Hy.

True page 174

8. If the obtained sample data are not in the critical region, then the correct statistical decision is “fail to reject the null hypothesis.”

False

Chapter 8: Exam Items - page 70

page 174

9. Ifyou fail to reject the null hypothesis, it means that the data do not provide sufficient evidence to say that the treatment has a real effect.

True page 176

10. Whenever the statistical decision is to reject the null hypothesis, there is a risk of a Type I error.

True page 179

11. Ifa specific sample leads to rejecting the null hypothesis with a = .01, then the same sample would also lead to rejecting the null hypothesis with a = .05.

False page 179

12. Ifa specific sample leads to rejecting the null hypothesis with a = .05, then the same sample would also lead to rejecting the null hypothesis with a = .01.

True page 177

13. A Type II error occurs when a treatment actually does have an effect on the scores but the effect fails to show up in a research study.

True page 177

14. One way to reduce the risk of a Type | error is to lower the alpha level.

False page 177

15. One way to reduce the risk of a Type I error is to use a larger sample.

False page 177

16. You are more likely to make a Type I error with a sample of n= 4 than with a sample of n= 100 even if you use a = .05 for both tests.

True page 182

17. Ina research report, the term statistically significant is used to indicate that the null hypothesis was rejected.

True page 183

18. Ina research report, the notation p < .05 is used when the null hypothesis is rejected and the treatment is shown to have a significant effect using an alpha level ofa =.05.

True page 177

19. Alpha (a) is the probability of committing a Type I error.

False

20. Ina Type II error, the experimenter concludes there is evidence for an effect

page 177

when in fact an effect does not exist.

True page 177

21. Changing the level of significance from .01 to .05 increases the risk of aType I error.

False page 179

22. When a research report demonstrates a significant treatment effect at the .05

True

level, you can be more confident that the effect is real than if the researcher had

reported a significant effect with an alpha level of only .01.

Chapter 8: Exam Items - page 71

True page 176

23. There is always a possibility that the decision reached in a hypothesis test is incorrect.

False page 179

24. To be certain that a treatment actually does have a real effect, a researcher should use a large value for alpha.

False page 189

25. If the null hypothesis is rejected using a one-tailed test, then it certainly would be rejected if the researcher had used a two-tailed test.

True page 188

26. In a directional hypothesis test, the entire critical region is located in one tail of the distribution.

True page 188

27. If a researcher is predicting that a treatment will increase scores, then the critical region for a directional test will be in the right-hand tail.

Other Exam Items 1. Define the critical region for a hypothesis test, and explain how the critical region is related to the alpha level.

2. The term error is used in two different ways in hypothesis testing: a. Type I error (or Type I) b. standard error What can a researcher do to influence the size of the standard error? Does this action have any effect on the probability of a Type I error? What can a researcher do to influence the probability of a Type I error? Does this action have any effect on the size of the standard error?

3. A researcher investigates whether or not a new cold medication disrupts mental alertness. It is known that scores on a standardized test containing a variety of problem-solving tasks are normally distributed with p = 64 ando=8. A random sample of n= 16 subjects are given the

drug and then tested.

For this sample, the mean is K = 58. Are the data sufficient to conclude

that the medication affects performance?

Test with a = .01, two tails.

4. Suppose reaction time for recognition of common words like red or green is normally distributed with pp, = 220 milliseconds and o = 32. A psychologist presents a sample-of n = 16 subjects with words that include names of colors. However, the color names are printed ina different color ink. That is, the word red might be printed in blue ink, or green might be printed in red. For these words, the sample had a mean reaction time of X = 240. Does presentation of color names in another color alter reaction time? Use a two-tailed test with a = .05.

Chapter 8: Exam Items - page 72

5. A study examines self-esteem and depression in children. A sample of 25 children with low self-esteem is given a standardized test for depression. The average score for the group is X = 93.3. For the general population, scores on the depression test form a normal distribution with a mean of 1: = 90 and a standard deviation of 6 =15. Do children with low self-esteem show significantly more depression? Use a one-tailed test with a = .05.

6. A researcher knows that the weights of 6-year olds are normally distributed with p = 20.9 kg and o = 3.2. She suspects that children in poverty-stricken regions are undernourished and therefore underweight. With a sample of n= 16 children, the researcher obtains a sample mean of X = 18.3. Use a one-tailed test and the .01 of significance to determine if the weights for this sample are significantly lower than what would be expected for the regular population of 6-yearolds.

Answers for Other Exam Items 1. The critical region consists of data that would be very unlikely if the null hypothesis is true.

The value of alpha is used to define a precise probability for the term very unlikely.

2. The standard error can be decreased by increasing the sample size. Changes in sample size have no effect on the probability of a Type I error. However, the probability of a Type I error is under the direct control of the researcher who selects an alpha level for the test. The choice of an alpha level does not affect the standard error.

3. ox = 2.00 and z = -3.00. Reject Hy and conclude that the medication has a significant effect.

4. oz = 8.00 and z= 2.50. Reject Hy and conclude that changing the color of ink has a significant effect.

5. Hy: u < 90. Hj: p> 90. The critical region is in the tail beyond z = 1.65. The standard error is 3.00 and z= 1.10. Fail to reject H, and conclude that there is no evidence that the children with low self-esteem are more depressed.

6. Hy: 1 > 20.9. Hy: p < 20.9. The critical region is in the tail beyond -2.33. ox = 0.8 and z = -3.25. Reject H, and conclude that children from poverty stricken areas weigh significantly less than other children.

Chapter 8: Exam Items - page 73

Chapter 9 Introduction to the t Statistic

Multiple-Choice Questions

Answer c

page 203

1. In what circumstances is the t statistic used instead of a z-score for a hypothesis test? a. The t statistic is used when the sample size is n = 30 or larger. b. The t statistic is used when the population mean is unknown. c. The t statistic is used when the population variance (or standard deviation) is unknown. d. The t statistic is used if you are not sure that the population distribution is normal.

2. The major difference between the t statistic formula and the z-score formula is a. the t statistic uses the sample variance in place of the population variance b. the t statistic uses the sample mean in place of the population mean c. the t statistic computes standard error by dividing the standard deviation by df=n - | instead of dividing by n d. all of the above Answer b page 204

3. Which of the following terms is not required when using the t statistic? a. n b. o cae

d. sors’ or SS Answer c

page 205 www

4. If two samples are selected from the same population, under what circumstances will the two sample have exactly the same t statistic? a. if the sample size (n) is the same for both samples. b. if the samples are the same size and have the same mean c. if the samples are the same size and have the same mean and have the same sample variance d. none of the above

Chapter 9: Exam Items - page 74

Answer b

page 204

5. To calculate a t statistic for a specific sample, what information is needed about the population from which the sample was obtained? . the value for N . the value for p

. the value for o or o° . All of the above are needed to compute t. . No population values are needed to compute t.

Ff ono 6. To calculate a t statistic, what information is needed from the sample? the value for n_ the value for X

the value for s or s’ All of the above are needed to compute t.

No sample values are needed to compute t.

FP GOO

7, On average, what value is expected for the t statistic when the null hypothesis is true? a. 0 bak Cas76 dat 1.96 Answer b page 202

8. The estimated standard error, sz, provides a measure of the "average" or standard discrepancy between j a. ascore and the population mean (X and 1) b. a sample mean and the population mean (X and 1)

c. ascore and the sample mean (X and X) d. none of the above 9. The magnitude of the estimated standard error is a. directly related to sample variance and directly ict to sample size b. directly related to sample variance and inversely related to sample size c. inversely related to sample variance and directly related to sample size d. inversely related to sample variance and inversely related to sample size 10. Holding everything else constant, increasing sample size a. decreases standard error b. increases the magnitude of the t statistic c. increases degrees of freedom d. all of the above

Chapter 9: Exam Items - page 75

Answer d

page 202 www

Answer a

page 202

11. What is the sample variance and the estimated standard error for a sample of n= 4 scores with SS = 300?

. § = 10 and sy =5 . S = 100 and sg = 20 . S = 10 and sz = 20 DD aA0C«wWwCU . § = 100 and sy = 5 12. What is the estimated standard error for a sample of n= 9 scores with s? = 36? a. V36/9

b. V36/8

c. 36/V9 d. 36/V8 Answer c

page 204

13. Ifa researcher reports a t statistic with df = 18, how many individual subjects participated in the experiment? aon—d7 b. n=18 C= 19 d. cannot be determined from the information given

14. Ifa sample consists of 30 individuals, then df for the t statistic for this sample would be ; ao29 b. 30 ee

|

d. cannot be determined from the information given Answer b

page 205 www

Answer c

page 205

15. When n is small (less than 30), the t distribution a. is identical in shape to the normal z distribution b. is flatter and more spread out than the normal z distribution c. is taller and narrower than the normal z distribution d. cannot be specified, making hypothesis tests impossible

16. With a = .05 the two-tailed critical region for a sample of n = 10 subjects would have boundaries of a. t=+1.96 b. t=+1.833 c. t=+2.262 d. t=+2.228

Chapter 9: Exam Items - page 76

Answer d page 205

17. With a = .01 the two-tailed critical region for a sample of n = 20 subjects would have boundaries of a, t=+2.845 b. t =+2.528 c. t=+2.539 d. t= +2.861

Answer a

18. With a = .01, the one-tailed critical region for a sample of n = 30 would have a boundary of a, t=2.462 b. t= 2.756 ¢. t=2.457 d, t=2.750

page 205

Answer b

page 206

19. With a = .05, the critical values for a two-tailed z-score test are corresponding values for a two-tailed t test with a = .05 tend to be a. less than +1.96 (closer to zero) b. greater than +1.96 (farther from zero) c. also equal to +1.96

1.96. The

8, the critical value for a one-tailed t test is t = 1.860.

Answer b

20. With a = .05 and df=

page 206

Assuming all other factors are held constant, if the df value were increased to i= 20, the critical value of t would a. increase b. decrease c. stay the same d. not enough information to answer

Answer c

21. A sample has a mean of X = 39.5 and a standard deviation of s= 4.3. Ina two-tailed hypothesis test with a = .05, this sample produces a t statistic of t = 2.14. Based on this information, the correct statistical decision is a. reject the null hypothesis b. fail to reject the null hypothesis ¢. cannot answer without additional information

page 206

Answer b

page 206

22. A sample of n= 25 scores produces at statistic of t= -2.05. Ifthe researcher is using a two-tailed test with a = .05, the correct statistical decision is a. reject the null hypothesis b. fail to reject the null hypothesis c. cannot answer without additional information

Chapter 9: Exam Items - page 77

23. For a sample of n = 15, what t values determine the two-tailed critical region for a = .05? +1.761, -1.761 . +2.145, -2.145 ene ame be SPIS35 =1.792 oS ac Answer a

page 214

Answer c

page 204 Www

24. An important assumption for the t test is that a. the population distribution is normal b. the sample size be less than 30 c. the value for o is known d. the t distribution is normal 25. Two samples from the same population both have n = 10 scores with X = 45.

If t statistics are computed for these two samples, then a. the two t statistics will be identical b. the sample with the larger variance will produce the larger t statistic c. the sample with the smaller variance will produce the larger t statistic

Answer a

26. Two different samples, each with n= 10 scores, are selected from the same

page 205

population. Ifthe two sample means are identical, then the two sample means will produce a. identical z-scores b. identicalt statistics c. all of the above d. none of the above

True/False Questions

page 201

1. When the value of o° (or o) is not known, it is impossible to use a z-score for a hypothesis test.

True page 201

2. The t statistic is used for hypothesis tests in situations where the population standard deviation (or variance) is unknown.

True page 203

3. Compared to a z-score, a hypothesis test with a t statistic requires less information about the population.

True

Chapter 9: Exam Items - page 78

False page 212

4. In general, a hypothesis test with a t statistic has a greater risk of a Type | error than a hypothesis test with a z-score. Assume a = .05 for both tests.

True page 204

5. The larger the sample size, the larger the df value for the t statistic.

True page 202

6. Inat statistic, the estimated standard error provides a measure of how much difference is reasonable to expect between a sample mean and the population mean.

False page 202

7. Estimated standard error is computed from o°.

True page 202

8. As sample size increases, the estimated standard error tends to decrease.

True page 202

9. As the sample variance increases, the estimated standard error also increases.

False page 202

10. Increasing the alpha level from a = .01 to a = .05 will produce an increase in the estimated standard error.

False page 204

11. The t statistic uses oy in the denominator instead of o.

True

12. To calculate a t statistic, you must first compute the sample mean and the

page 210

sample variance (or sample standard deviation).

True page 204

13. The larger the value for df, the more a t distribution resembles a normal distribution.

True page 205

14. Assuming all other factors are held constant, t statistics tend to be more variable than z-scores.

True page 204

15. The t distribution is symmetrical and has a mean of zero.

False page 205

16. As the value ofdf increases, the t distribution tends to become flatter and

more spread out.

True page 206

17. For any fixed value of a, the critical values for the t statistic will move closer to zero as df increases.

True page 206

18. The boundaries for the critical region for a two-tailed test using a t statistic with a = .05 will never be less than +1.96.

Chapter 9: Exam Items - page 79

True

page 206

19. For a hypothesis test using a t statistic, the boundaries for the critical region will change if the sample size is changed. (Assume all other factors are held constant).

False page 206

20. While conducting a two-tailed hypothesis test, a researcher obtains a t statistic of t = 2.10 for a sample of n= 15. The researcher should reject the null hypothesis with a = .05.

False page 205

21. Two separate samples, each with n = 30 scores, from the same population will produce identical t statistics if the two sample means are equal.

True

page 214

22. An important assumption for the t test is that the population distribution is normal.

False page 204

23. In general, an increase in the sample variance will produce an increase in the magnitude of the t statistic.

False page 210

24. In general, a small value (near zero) for a t statistic will result in rejecting the null hypothesis.

False page 209

25. The t distribution can be used for hypothesis tests only ifn > 30.

False page 204

26. A sample of n = 15 scores will produce a t statistic with df= 16.

False page 206

27. For a two-tailed hypothesis test with a = .05 and a sample of n = 25 scores, the boundaries for the critical region would be t = +2.060.

Other Exam Items 1. Although you can compute a z-score for a sample of n = 1, it is impossible to compute a t statistic for a sample that has only one score. Explain why.

2. How does the shape of a t distribution compare with the normal distribution? How does the shape of a t distribution depend on sample size (n)?

Chapter 9: Exam Items - page 80

3. A nutrition store in the mall is selling “Memory Booster” which is a concoction of herbs and minerals that is intended to improve memory performance. To test the effectiveness of the herbal mix, a researcher obtains a sample of n = 16 people and has each person take the suggested dosage each day for 4 weeks.

At the end of the four week period, each individual takes a

standardized memory test. The scores from the sample produced a mean of X = 24 with a sample variance of s* = 64. In the general population, the standardized test is know to have a mean of 1 = 20. Do the sample data support the conclusion that the Memory Booster has a significant effect? Test with a = .05.

4. A normal population has a mean of p = 25. After receiving a treatment, a sample of n = 4 scores yields the following data: 20 28 20 20. Is there evidence for a significant treatment effect? Use an alpha level of .05.

5. A sample of freshmen takes a reading comprehension test and their scores are summarized below. Ifthe mean for the general population on this test is p = 12, can you conclude that this sample is significantly different from the population. Test with a = .05. sample Scores:

16, 8, 8, 6,9, 11, 13,9, 10

Answers for Other Exam Items 1. Before you can find a t statistic, you must first compute the sample variance or standard deviation (s* or s). With only one score, it is impossible to compute variance for the sample.

2. The t distributions tend to approximate a normal shape, but they generally have greater variability (look flatter) than a normal distribution. As sample size increases, the t distribution becomes closer to a normal shape.

3. For these data, t(15) = 2.00. Fail to reject Hy. Based on the sample data, the Memory Booster does not have a significant effect.

4. H,: 1 =25. For these data, X = 22, s’ = 16, and t(3) = -1.50. Fail to reject Hy. 5. For these data X = 10 and SS = 72. The standard error is 1 and t = -2.00. Fail to reject Hy.

Chapter 9: Exam Items - page 81

Chapter 10 The t Test for Two Independent Samples Multiple-Choice Questions Answer d page 223

1. An independent-measures t hypothesis test is appropriate when a. the value for o is known b. the mean for a treated group of subjects is compared to a known population mean c. one sample is used to test a hypothesis about one population d. there are two separate samples containing different subjects

Answer c

2. Which of the following research situations is most likely to use an independentmeasures design? a. Evaluate the effectiveness of a diet program by measuring how much weight is lost during 4 weeks of dieting. b. Evaluate the effectiveness of a cholesterol medication by comparing cholesterol levels before and after the medication. c. Evaluate the difference in verbal skills between 3-year-old girls and 3year-old boys. d. Evaluate the development of verbal skills between age 2 and age 3 for a sample of girls.

page 223 Www

3. The null hypothesis for the independent-measures t test states a. there is no difference between the two sample mean b. there is no difference between the two population means c. the difference between the two sample means is identical to the difference between the two population means d. none of the above

4, What is the pooled variance for the following two samples?

Sample 1: n= 6 and SS = 200.

Sample 2: n= 10 and SS = 300 . 250 , 300 . 500/14 . 500/16 2 SoS aa

Chapter 10: Exam Items - page 82

Answer c

page 227

Answer c

page 227

5. One sample has n = 6 and SS = 20 and a second sample has n = 6 and SS = 30.

What is the pooled variance for the two samples?

Boa) breod) c. 50/10 Gd. 50/12 6. One sample has a variance of s’? = 10 and a second sample has s’ = 20. If the pooled variance is computed for these two samples, then the value obtained will be ; . smaller than 10 . larger than 20 somewhere between 10 and 20 . exactly equal to 15 oS ao

7. One sample of n= 5 scores has a variance of s* = 10 and a second sample of n= 10 scores has s* = 20. If the pooled variance is computed for these two samples, then the value obtained will be a. closer to 10 than to 20 b. closer to 20 than to 10 c. half way between 10 and 20 d. cannot be determined without more information 8. As the sample variance increases, the value of the t statistic a. increases (moves away from zero) b. decreases (moves toward zero) c. the value of t is not influenced by sample variance Answer b

page 228

Answer Cc

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9. Ina hypothesis test, if an independent-measures t statistic has a value zero, then a. the two population means must be equal b. the two sample means must be equal c. the two sample variances must be equal d. none of the above

10. Which of the following sets of data would produce the largest value for an independent-measures t statistic? a. The two sample means are 10 and 12 with sample variances of 20 and 25. b. The two sample means are 10 and 12 with variances of 120 and 125. c. The two sample means are 10 and 20 with variances of 20 and 25. d. The two sample means are 10 and 20 with variances of 120 and 125.

Chapter 10: Exam Items - page 83

Answer a

page 231

Answer b

page 228

11. If the null hypothesis is true, then the t statistic (on average) should have a value of a. 0 bl c. 1:96 d. cannot be determined without additional information 12. What is the estimated standard error for the independent-measures t statistic for the following two samples?

Sample 1: n=6 and SS = 56 Sample 2: n= 4 and SS = 40 a.

b. v5 Cal

dO Answer c

page 227

13. What is the pooled variance for the following two samples?

Sample 1: n= 6 and SS = 56 Sample 2: n= 4 and SS = 40 5

v5 12 7 6 aoo8 Answer c

14. Two samples, each with n = 6 subjects, produce a pooled variance of 20.

page 228

Based on this information, the estimated standard error for the sample mean difference would be a. 20/6

www

b. 20/12 c. v20/6 + 20/6

d V20/5 + 20/5 Answer b page 228

15. An independent-measures research study uses two samples, each with n = 10, to compare two treatments. If the results are evaluated with a t statistic using a two-tailed test with a = .05, then the critical region would have boundaries of a Si eee b. t= +2.101 c. t= +2.093

d. t= +2.086

Chapter 10: Exam Items - page 84

Answer c

page 228 www

16. A researcher reports t(24) = 5 for an independent-measures experiment. How many subjects participated in the entire experiment? a. 24 br 25 C226

d. 12 Answer a page 228

Answer a

page 228

17. An independent-measures experiment uses two samples with n = 7 in each to compare two experimental treatments. have degrees of freedom equal to a: k2 b. 6 c;, 13 d. 9

The t statistic from this experiment will

18. An independent-measures experiment uses one sample with n = 10 and a second sample with n = 15 to compare two experimental treatments. The t statistic from this experiment will have degrees of freedom equal to a.23 b. 24 C25 d. cannot be determined with unequal sample sizes

Answer d page 239

19. The homogeneity of variance assumption states that a. the two sample variances are equal b. the two samples come from the same population c. variance must stay constant for each subject in the experiment d. the samples come from populations with equal variances

Answer ¢

20. The alternative hypothesis for an independent-measures t test states

page 224

a. Hy - HW, = 0 b. X, - X, #0

eee d.

Answer b page 232

Ky

a =e

=

21. The results of an independent-measures research study are reported as "t(5) = -2.12, p> .05, two tails."

For this study, what t values formed the boundaries for the critical region? . +2.015 and -2.015 +2971 and =2:371 , 42,776 and -2.776 & fF ao . cannot be determined from the information given

Chapter 10: Exam Items - page 85

www

22. The results of an independent-measures research study are reported as "t(5) = -2.12, p > .05, two tails." What statistical decision was made? a. The researcher failed to reject Hy b. The null hypothesis was rejected. c. There was a marginally significant effect. d. cannot be determined from the researcher's statement

Answer c

23. The results of an independent-measures research study are reported as "t(5) = -2.12, p > .05, two tails.". How many subjects participated in the study?

Answer a

page 232

page 232

Answer a

page 228

24. A research study uses n = 6 subjects in treatment 1 and a separate sample of n = 6 subjects in treatment 2. For this study the independent-measures t statistic has degrees of freedom equal to a: “10

be cole

deat Answer d page 228

25. An independent-measures research study with n = 6 subjects in each treatment has SS = 70 for treatment 1 and SS = 50 for treatment 2. For these data, the

estimated standard error for sample mean differences equals a. 4

ped? c. 120 d. 2 26. For an independent-measures research study, the value of Cohen’s d or r° helps to describe a. the risk of a Type I error b. the risk of a Type II error c. how much difference there is between the two treatments d. whether the difference between the two treatments is likely to have occurred by chance Answer b page 236

27. For an independent-measures research study, the data show a 10-point difference between the two treatment means and a pooled variance of 4. Given this information, the value of Cohen’s d is a. 10/4 b. 10/2 c. 4/10 d. 2/10 Chapter 10: Exam Items - page 86

True/False Questions True page 224

1. The independent-measures t test uses data from two separate samples to test a hypothesis about the mean difference between two populations.

False page 224

2. The null hypothesis for an independent-measures t test states that the two sample means are equal.

True page 224

3. The null hypothesis for the independent-measures t test states that there is no difference between the two population means.

True

4. For an independent-measures study, H, for a two-tailed test states that

page 224

Hy, - H, #0.

True

page 228

5. For a hypothesis test with an independent-measures t, the larger the difference between the two sample means, the greater the likelihood that you will reject the null hypothesis.

False page 228

6. For a hypothesis test with an independent-measures t, the larger the two sample variances, the greater the likelihood that you will reject the null hypothesis.

False page 228

7. One sample with n = 8 and a second sample with n = 10 will produce an independent-measures t statistic with df= 9.

False

8. For the independent-measures t statistic, df = (n, + 1) + (n, + 1).

page 228 True

page 228

9. Two separate samples, each with n = 9, will produce an independent-measures t statistic with df= 16.

True

10. For an independent-measures t statistic, the magnitude of the estimated

page 228

standard error is directly related to the sample variances (the bigger the variances, the bigger the error).

True page 228

11. For an independent-measures t statistic, the magnitude of the estimated standard error is inversely related to the size of the samples (the bigger the samples, the smaller the error).

Chapter 10: Exam Items - page 87

True page 227 False page 228

12. If two samples each have n= 5 and SS = 40, then the pooled variance is 10.

13. An increase in sample variance will produce an increase in the value for the t statistic (farther from zero).

False page 227

14. Pooled variance is obtained by adding two SS values and dividing by 2.

False page 228

15. If other factors are held constant, the larger the estimated standard error of sample mean differences, the more likely you will obtain a t statistic in the critical region.

True page 228

16. For a two-tailed hypothesis with an independent-measures t statistic using two samples, each with n = 10 scores, the boundaries for the critical region using a= .05 aret = +2.101.

True

page 228

17. If an independent-measures t statistic has df = 19, then it is impossible for the two samples to be the same size.

True page 228

18. Ina hypothesis test, the only way to obtain a value of zero for the independent measures t statistic is when X, = X).

True page 239

19. Homogeneity of variance assumes that the two population variances are equal.

False

20. The F-max test is used to determine if the two population means are equal.

page 240 False page 227

21. When pooling variances, the resulting value will be closer to the variance for the sample with the smaller n.

False page 240

22. The F-max test for homogeneity of variance requires that you find the ratio of the smaller variance over the larger variance.

False page 232

23. Ifa researcher reports that t(6) = 1.98, p > .05, then Hy was rejected.

True

24. The results from an independent-measures t hypothesis test are reported as “4(24) = 3.85, p < .01, two tailed.” For this test, df= 24.

page 232

False page 232

25. The results from an independent-measures t hypothesis test are reported as “t(24) = 3.85, p < .01, two tailed.” The research study for this test used a total of 25 subjects.

Chapter 10: Exam Items - page 88

False page 234

26. When a researcher reports a “significant difference” between two treatments based on the outcome of an independent-measures t hypothesis test, it means that there is a very large difference between the two treatments.

True page 234

27. In an independent-measures t hypothesis test, if the estimated standard error is very small, then it is possible to obtain a significant difference between treatments even if the actual mean difference is small.

False page 236

28. Cohen’s d is computed by first finding the difference between the two sample means and then dividing by the pooled variance.

Other Exam Items 1. Describe the data that are collected for an independent-measures t test and the hypotheses that the test evaluates.

2. Explain why the homogeneity of variance assumption must be satisfied before you can interpret the results of an independent-measures t test.

3. A biopsychologist studies the role of the brain chemical serotonin in aggression. One sample of rats serves as a control group and receives a placebo. A second sample of rats receive a drug that lowers brain levels of serotonin. Then the researcher tests the animals by recording the number of aggressive responses each of the rats display. For the data below, is there a significant effect of the drug on aggression? What, if any, role does serotonin have in aggression? Use an alpha level of .05, two tails.

Low Serotonin n=6

Control n=8

X=22

X=14

SS = 108

SS = 180

Chapter 10: Exam Items - page 89

4. An educational psychologist studies the effect of frequent testing on retention of course material. In one of the professor's sections, students are given quizzes each week. The second section receives only two tests during the semester. At the end of the semester, both sections receive the same final exam, and the scores are summarized below.

Do the data indicate that

testing frequency had a significant effect on performance? Use a two-tailed test at the .01 level of significance.

Frequent Quizzes n=15 Kone SS = 100

Two Exams ya 18) X= 78 SS = 110

5. The following data are from an independent-measures experiment comparing two treatment conditions. Do these data indicate a significant difference between the treatments at the .05 level of significance?

Treatment

|

13 9 fi 1]

Treatment 2

9 5 ®) 9

6. The following data were obtained from an independent measures study. Determine if there is a significant difference between the treatments. Use a = .01 Treatment |

Treatment 2

5 l 2 3 4

6 10 14 12 18

Answers for Other Exam Items 1. An independent measures t statistic requires scores from two separate samples. The null hypothesis states that there is no difference between the means of the two populations (or treatments) from which the samples were obtained.

2. Before you can average (or pool) the two sample variances it is necessary that both samples are estimating the same population variance. Otherwise, the pooled value is meaningless. If the homogeneity assumption is violated, an extreme value for t could be caused by either a. an incorrect value for pu, - 1, from Hy, or b. a meaningless value for the pooled variance. In this situation, you cannot conclude that an extreme t necessarily indicates that Hy is wrong.

Chapter 10: Exam Items - page 90

3. Hg: p,- , =0. For these data the pooled variance is 24, the standard error is 2.65, and the t statistic is (12) = 3.02. Reject Hy. Lower serotonin levels cause an increase in aggression.

4. Pooled variance = 7.5, standard error = 1, t(28) = 4.00. Reject the null hypothesis.

5. For treatment 1, X = 10 and SS = 20. For treatment 2, X = 7 and SS = 16. For these data the pooled variance is 6, the standard error is V3 = 1.73, and the t statistic is (6) = 1.73. Fail to reject Hy.

6. For treatment 1, K = 3 and SS = 10. For treatment 2, K = 12 and SS = 80. The pooled variance = ]1.25, standard error = 2.12, t(8) = -4.25. Reject Hy.

Chapter 10: Exam Items - page 91

Chapter 11 The t Test for Two Related Samples

Multiple-Choice Questions Answer b page 250 www

1. A repeated-measures study would not be appropriate for which of the following situations? a. A researcher would like to study the effect of practice on performance. b. A researcher would like to compare individuals from two different populations. c. The effect of a treatment is studied in a small group of individuals with a rare disease. d. A developmental psychologist examines how behavior unfolds by observing the same group of children at different ages.

Answer c page 250

2. Which of the following studies uses a matched-subjects design? a. A group of twins is tested for visual acuity on one day and then tested again the following day after receiving a drug. b. A sample of children is tested for vocabulary skills at age six and then again at age ten. c. Subjects are assigned to one of two treatment groups so that each subject in one group has the same IQ as a subject in the other group. d. none of the above

Answer b page 252 www

3. For a repeated-measures hypothesis test, the null hypothesis states a D=v b. tp = 0

Answer a

4. What is the value of D for the following set of difference scores?

page 251

C.d. D, Hy = Hp=D,

Scores: 3, -8, 6, -4, -2 a. -l

b. -5 c. 23 d. 23/5 = 4.6

Chapter 11: Exam Items - page 92

Answer b page 254

5. A repeated-measures experiment and a matched-subjects experiment both produce t statistics with df= 20. Which experiment used more subjects? a. repeated measures b. matched subjects c. They both used n = 21 subjects. d. They both used n = 22 subjects.

6. A repeated-measures experiment and an independent-measures experiment both produce t statistics with df = 20. Which experiment used more subjects? a. repeated measures b. independent measures c. They both used n= 21 subjects. d. They both used n = 22 subjects.

Answer a

page 261

7. A researcher conducts an experiment comparing two treatment conditions and obtains 20 scores in each treatment. Which design would require the smallest number of subjects? a. a repeated-measures design b. an independent-measures design c. a matched-subjects design d. bandc 8. A researcher conducts a research study comparing two treatment conditions and obtains 20 scores in each treatment. If the researcher used a repeatedmeasures design, then how many subjects participated in the research study? a. 10 b. 20 ae d. 40

Answer b

page 261

9. For an experiment comparing two treatment conditions, an independentmeasures design would obtain score(s) for each subject and a repeatedmeasures design would obtain score(s) for each subject. a

]

b. <2 c.

9 d. NN

Chapter 11: Exam Items - page 93

Answer b

page 262 www

10. Ina repeated-measures experiment, each subject participates in one treatment condition and then moves on to a second treatment condition. One of the major concerns in this type of study is that the effects of the first treatment may influence the subject's score in the second treatment. This problem is called a. individual differences b. carry-over effect c. homogeneity of variance d. bi-treatment effect 11. A research report describing the results from a repeated-measures study states, “The data showed a significant difference between treatments, t(22) = 4.71, p <.05." From this report you can conclude that a. a total of 22 individuals participated in the study b. a total of 23 individuals participated in the study c. a total of 24 individuals participated in the study d. It is impossible to determine the number of participants from the information given.

Answer a

page 257

Answer d

page 261

Answer d

page 254

Answer d

page 254

12. A research report describing the results from a repeated-measures study states, “The data showed a significant difference between treatments, t(22) = 4.71, p <.05." From this report you can conclude that the outcome of the hypothesis test was a. to reject the null hypothesis. b. to fail to reject the null hypothesis c. It is impossible to determine whether or not the null hypothesis was rejected from the information given.

13. For which of the following situations would a repeated-measures design have the maximum advantage over an independent-measures design? a. when many subjects are available and individual differences are small b. when very few subjects are available and individual differences are small c. when many subjects are available and individual differences are large d. when very few subjects are available and individual differences are large

14.

What is the value of sp for the following set of D-scores? a. 12 ba c. 4 d. ]

Scores:

15. What is the value for sp for a set of n= 9 D-scores with SS = 72? a. 72 b. 9 c3 d. 1 Chapter 11: Exam Items - page 94

4, 8, 4, 4

Answer c

page 254

16. For the repeated-measures t statistic, df=

a n,n,

2

b. (n, - 1) + (, - 1) Cc. tael d. n, +n, - 1 Answer c

page 254

17. A researcher uses a repeated-measures design to compare subjects’ performance before treatment with their performance after treatment. If all of the subjects show improved performance of 8 or 9 points after treatment, then the researcher should find a. a sample mean difference near zero b. at statistic near zero c. the variance of the difference scores is near zero d. none of the above 18. In general, as the sample size is increased in a repeated-measures experiment, . the sample mean difference decreases . the variance of the difference scores decreases . the standard error of the mean difference decreases . all of the above anda

19. A repeated-measures experiment with a sample of n = 16 subjects would produce at statistic withdf= a. 15 16 30 a ao Answer d

page 254

20. With a = .05 and a sample of n = 12 subjects in a repeated-measures experiment, the critical region for the t statistic has boundaries of a. t=42.228 b. t=+1.812 c. t=+1.796 d. t=+2.201

Answer a

page 256

21. Ifthe null hypothesis is true, on average the expected value for the related-samples t statistic is a. 0 Bad e. 1,96 db 180

Chapter 11: Exam Items - page 95

Answer ¢

page 254

22. Which of the following sets of data is most likely to produce a significant t statistic?

a. D=2 and SS = 10 b. D =2 and SS = 100

c. D=10 and SS = 10 d. D= 10 and SS = 100 Answer b page 254 www

23. In general, what characteristics of the data (difference scores) are most likely

to produce a significant t statistic for the repeated-measures hypothesis test? a. a large mean difference and large variance b. a large mean difference and small variance c. a small mean difference and large variance d. a small mean difference and small variance 24. An advantage of a repeated-measured design is that it reduces the contribution of error variability due to

a. D b. degrees of freedom c. the effect of the treatment d. individual differences Answer a page 254

25. A researcher is using a repeated-measures study to evaluate the difference between two treatments. If there is a consistent difference between the treatments then the data should produce a. a small variance for the difference scores and a small standard error b. a small variance for the difference scores and a large standard error c. a large variance for the difference scores and a small standard error d. a large variance for the difference scores and a large standard error

Answer d

26. For a repeated-measures experiment comparing two treatment conditions, the t statistic has df= 10. How many subjects were used in the entire study? a9 be22 e 18 d. 11

page 254

Answer a

page 254

27. A repeated-measures experiment comparing two treatment conditions uses a sample of n= 15 subjects. If the results are evaluated using a t statistic, what would be the value of df? a. 14 b. 16 c. 28

d, 29

Chapter 11: Exam Items - page 96

Answer b page 262

28. A major concern with a repeated-measures study is the possibility of a. negative values for the difference scores b. carry-over effects c. obtaining a mean difference that is due to individual differences rather than treatment differences d. all of the above

Answer ¢

29. A researcher obtains a t statistic of t = 2.00 from a repeated measures study using n= 17 participants. If the effect size is measured using 1’ then the value of % for the study would be ar “© = 4.00 b. r = 4/16 = 0.25 c. r = 4/20 = 0.20 d. r cannot be determined from the information provided.

page 256

Answer b page 256

30. A repeated-measures study produces a sample mean difference of D = 4.5 with SS = 2400 for a sample of n = 25 individuals. Based on this information, the effect size using Cohen’s d would be a. d= 4.5/100 = 0.045 b. d= 4.5/10 = 0.45 ce d=4,5 d. d cannot be determined from the information provided.

True/False Questions False page 250

1. For a repeated-measures design, a separate sample is used for each treatment condition.

True

2. With a repeated-measures design, it is possible to evaluate the difference between two treatments using only one sample of subjects.

page 250

page 251

3. When computing the difference scores (D values) for a repeated-measures study, always subtract the smaller score from the larger score so that the result is positive.

False page 251

4. When computing difference scores (D values) the sign of the difference (+ or -) is not important. You need to record only the size of the difference.

False

Chapter 11: Exam Items - page 97

True page 253

5. The formula and calculations for the repeated-measures t statistic are identical to those for the single-sample t except that you use difference scores (D values) instead of X values.

True

page 254

6. The numerator of the repeated-measures t formula measures how much difference there is between a sample mean and a hypothesized population mean.

True page 254

7. Estimated standard error for a repeated-measures t statistic is based on the variance of the difference scores.

False page 254

8. In general, as the variance of the difference scores increases, the magnitude of the t statistic also increases.

True page 254

9. The larger the variance of the difference scores, the larger the standard error for the repeated-measures t.

True page 254

10. For a repeated-measures study, when the variance for the difference scores is small, it is an indication that the difference between treatment conditions is fairly consistent for all of the subjects.

False

11, For a repeated-measures design, df =n, + n, - 2 for the t statistic.

page 254

page 254

12. A repeated-measures t statistic with df = 20 indicates that a total of 21 subjects participated in the research study.

False page 254

13. A repeated-measures research study comparing two treatments with a sample of 18 subjects will produce a t statistic with df= 19.

True page 252

14. The null hypothesis for a repeated-measures t test states that u, = 0.

True page 252

15. The null hypothesis for a repeated-measures t test concerns the entire population of difference scores, not just the D values in the sample.

True

16. The repeated measures t statistic can be used with either a repeated-measures or a matched-subjects design.

True

page 250 True page 262

17. In general, using a repeated-measures design to evaluate the difference between two treatments will produce a smaller standard error than would be obtained using an independent-measures design to compare the same two treatments.

True page 262

18. Repeated-measures experiments reduce the variance contributed by individual differences.

Chapter 11: Exam Items - page 98

True page 262

19. A general concern for researchers using a repeated-measures design is that the participants’ scores in the second treatment may be influenced by aftereffects from the first treatment.

True page 262

20. A general concern for researchers is that the subjects in one treatment will be noticeably different (smarter, older, etc.) than the subjects in another treatment.

However, this is not a problem for a repeated-measures design.

True page 250

21. A researcher wants to compare performance scores for men versus women on a motor-skills task. It would not be possible to use a repeated-measures design for this study.

False page 250

22. Repeated-measures designs are particularly well-suited to research questions concerning the difference between two distinct populations (for example, males versus females).

True page 261

23. Repeated-measures designs are particularly well-suited to research questions concerning changes that occur over time.

True page 254

24. Ifall other factors are held constant, an increase in the magnitude of D will cause an increase in the size of the t statistic.

False page 254

25. If all other factors are held constant, an increase in the variability of the D scores will cause an increase in the size of the t statistic.

False page 261

26. One disadvantage of a repeated-measures design is that it typically requires more subjects than an independent-measures design.

True page 263

27. If two treatments are both expected to produce a permanent or long-lasting change in the subjects, then a repeated-measures design would not be appropriate for comparing the two treatments.

Chapter 11: Exam Items - page 99

Other Exam Items 1. Briefly explain the advantages and disadvantages of using a repeated-measures design as opposed to an independent-measures design.

2. Calculate the difference scores and find D for the following data from a repeated-measures study.

Subject Treatment 1 Treatment 2 A 34 39 B 4] 48 c 38 35 D Phe 36

3. One of the major advantages of a repeated-measures design is that it removes individual differences from the variance and, therefore, reduces the standard error.

The following two sets

of data demonstrate this fact. The first set of data represents the original results from a repeatedmeasures study. To create the second set of data we started with the original scores but increased the individual differences by adding 100 points to each score for subject B, adding 200 points to each score for subject C, and adding 300 points to each score for subject D. Note that this change produces a huge increase in the variability of the scores within each treatment condition.

Set 1 Subject IU A

Set Z Subject _I

II

12

14

A

12

14

B 1 cc 9 De-10"

wF7 11 *12

BY? Ce D

«107-117 209° 2 310 312

a. Find the difference scores for each set of data and compute the mean and variance for each sample of difference scores. b. You should find that both sets of data produce the same mean difference and the same variance for the difference scores. Explain what happened to the huge individual differences that were added to the second set of data.

4. After a sample of n = 25 high school students took a special training course, their SAT scores averaged D = 18 points higher with SS =9600. On the basis of this sample, can you conclude that the training course has a significant effect on SAT scores? Use a two-tailed test at the .05 level of significance.

Chapter 11: Exam Items - page 100

5. A researcher would like to determine if relaxation training will affect the number of headaches for chronic headache sufferers. For a week prior to training, each subject records the number of headaches suffered. Subjects then receive relaxation training and for the week following training the number of headaches is again measured. The change in number of headaches before and after training is as follows (minus signs indicate a reduction): Difference scores: -5 -1 -4 -3 -6 -8 0 +1 -1 Is there an effect of training? Use a two-tailed test with a = .05 6. A therapist would like to evaluate the effectiveness of a procedure for treating depression. A sample of n = 16 depressed clients is obtained and each person's level of depression 1s measured using a standardized questionnaire before they begin the therapy program. Two weeks after therapy, each client's level of depression is measured again. For this sample, the average level of depression dropped by D = 6.8 points following therapy. The difference scores had a variance of s? = 64. Do these data indicate a significant change in depression? Use a two-tailed test at the .O5 level of significance.

7. A teacher gives a third grade class of n= 16 a vocabulary test at the beginning of the school year. To evaluate the changes that occur during the year, students are tested again at the end of the year. Their test scores revealed an average improvement of D = 5.7 points with s* = 144.

Does this constitute a significant improvement? Use a one-tailed test with an alpha level of .05.

Answers for Other Exam Items 1. A repeated measures design tends to be more precise or more powerful than an independent measures design because it eliminates variability due to individual differences. Also, a repeatedmeasure design uses fewer subjects than an independent-measures design. However, the results from a repeated-measures experiment can be confounded by carry-over effects.

2.

The difference scores are 5, 7, -3, 7. For these data, £D = 16 and D= +4.

3. a. The difference scores for Set 1 are identical to the difference scores for Set 2. For both sets of data, D = +4, SS = 48, and s’ = 16. b. The process of subtracting the first score from the second score for each individual eliminates the individual differences. As a result, the difference scores (D values) show how much change there is for each subject, but do not show the magnitude of the original scores.

Chapter 11: Exam Items - page 101

4. The standard error for these data is 4 and t(24) = 18/4 = 4.50. Reject H) and conclude that the preparatory course has a significant effect.

5. For these data D = -3, s* = 9 and t(8) = -3.00. Reject H, and conclude that the relaxation training has a significant effect.

6. For these data the standard error is s5 = 2 and t(15) = 3.4. Reject Hy. These data are sufficient to conclude that there has been a change in depression with a = .0S5.

7. For these data, D = 5.7, s5 = 3, and t(15) = 1.90. For a one-tailed test the critical value is 1.753. Reject Hy and conclude that there has been a significant change.

Chapter 11: Exam Items - page 102

Chapter 12 Estimation

Multiple-Choice Questions

Answer Cc

page 275 www

1. Which of the following research questions would be an appropriate situation for using estimation instead of a hypothesis test? a. Does a new cholesterol medication have any effect? b. Is there any difference in mathematics skills between adolescent boys and adolescent girls? & How much improvement will result from a special program to teach reading skills? d. Does the amount of background noise have an effect on people’s ability to concentrate?

2. A researcher knows that a special reading program will improve students’ reading scores. To determine how much improvement will occur, the researcher obtains a sample of n = 25 students and measures each individual’s reading performance before and after the special program. The data will be used to estimate the mean amount of improvement. Which estimation equation should the researcher use? a. the single-sample t equation the independent-measures t equation é the repeated-measures t equation

3. A researcher would like to determine the average mathematical skill level for freshmen entering the college. A sample of n = 30 freshmen is obtained and each person is given a standardized mathematics skills test. The researcher plans to use the sample data to estimate the mean score for the entire population of freshmen.

Which estimation equation should the researcher use? a. the single-sample t equation b. the independent-measures t equation c. the repeated-measures t equation

Chapter 12: Exam Items - page 103

4. Which of the following most closely resembles the general form of an interval estimate? a. statistic = parameter + error b. statistic = parameter x error c. parameter = statistic + error d. error = parameter + statistic 5. In general, the goal of estimation is to a. determine whether or not a treatment has any effect b. determine whether the effect of a treatment is to increase or to decrease , GA Answer a

page 279

Answer d

page 281

Answer a

page 280

ores terminehow much effect a treatment has

6. The purpose of a confidence interval is to _ a.use a sample mean or mean difference to estimate the corresponding population mean or mean difference b. use jt to estimate the value of a sample mean c. use a level of confidence to estimate a sample mean d. use the sample mean to determine a level of confidence 7. For a single sample of n = 23, the t values used to determine the 90% confidence interval are a. fii) b. +2.819 ¢. 42.074 Ge 17

8. For a point estimate of a population mean using a single-sample t statistic, the appropriate value(s) for t would be =0 b. t=+1.00 c. t=+1.96 d. cannot answer without additional informatioin

Answer b

page 282

9. The independent-measures t statistic allows researchers to use sample data to estimate a. the value of an unknown population mean b. the difference between two population means c. the mean for the entire population of difference scores

10. Ifall other factors are held constant, increasing the sample size will . increase the standard error . decrease the width of the confidence interval . increase the width of the confidence interval Ff ao f . none of the above

Chapter 12: Exam Items - page 104

11. Which combination of factors would definitely reduce the width of a confidence interval? a. use a larger sample and increase the level of confidence b. use a smaller sample and increase the level of confidence use a larger sample and decrease the level of confidence . use a smaller sample and decrease the level of confidence Answer c

page 288

12. Ifall other factors are held constant, which of the following would produce the smallest width for a confidence interval for 1? a. an 80% confidence interval based on a sample of n = 25 . a 90% confidence interval based on a sample of n = 25 éan 80% confidence interval based on a sample of n = 100 . 290% confidence interval based on a sample of n= 100

Answer b page 280

13. A sample of n= 16 scores is obtained from an unknown population. The sample mean is X = 52 with SS = 20. Using this information, the point estimate of the population mean is a 37 b. 52 G: S220 de Sees

Answer c

14. A sample of n = 25 scores is obtained from a normal population. If the sample

page 281

mean is X = 37 and s* = 100, the 95% confidence interval for 1, would be

a. 37 + 2.064(10) b. 37 =0.95(10)

c. S72: 2.064(2) d. 37+0.95(2) Answer d

page 281

15. The 95% confidence interval for ,: is computed from sample data and the interval estimate ranges from 65 to 75. This indicates that a. 95% of the scores in the population fall between X = 65 and X = 75 . you should reject Hy for any sample value between 65 and 75 the standard error is 10 . there is a 95% probatility that the population mean falls somewhere ans between 65 and 75

Answer a

page 285

16. The correct formula for estimating the value of [Up is a. hy = D = tsp

b. Up =Sz+tD C. Up =D+ts iu,

As

Chapter 12: Exam Items - page 105

Answer c

17. For an independent-measures research study, estimation would be used to

page 282

a. estimate the population mean b. estimate the sample mean & estimate the difference between two population means d. estimate the difference between two sample means Answer a

page 288 www

Answer ¢

page 284

Answer d page 284

18. In general, the larger the value of the standard error, a. the wider the confidence interval b. the more narrow the confidence interval c. The size of the standard error has no relation to the width of a confidence interval.

19. Two separate samples, each with n = 4, are used to construct a 99% confidence interval for the mean difference between two populations. The correct t values for the interval are a. 3.143 b. +2.998 eas 07 d. +2.58 20. Two separate sample, each with n = 4, are used to estimate the mean difference between two populations. If the two sample means are X, = 2] and

X, = 28, then the point estimate for 1, - p1, is a. 21

b. 28 c. 6 d.7 Answer c

21. An 80% confidence interval for the population mean ranges from 14 to 22.

page 281

Based on this information,

a. the sample mean is X = 14 b. the sample mean is X = 22

& the sample mean is X = 18 . none of the above Answer b page 286

22. The results from a repeated-measures research study are being used to estimate how much difference there is between two treatments. If the study used a total of n= 10 subjects, then the t values for a 90% confidence interval would be a. t=+1.812 b. t=+1.833 c. t=+1.383 d. t=+1.372

Chapter 12: Exam Items - page 106

23. The 90% confidence interval for the difference between two population means extends from 4.00 to 8.00. Based on this information you can conclude that the

difference between the two sample means was a. 2 points b. 4 points . 6 points . 8 points 24. Compared to a point estimate, an interval estimate

a. has greater precision but less confidence b. has greater precision and greater confidence c. has less precision and less confidence dq has less precision but greater confidence

25.

Estimation would be appropriate a. after a hypothesis test where the decision was to reject the null hypothesis b. after a hypothesis test where the decision was to fail to reject the null hypothesis c. both of the above d. neither of the above

True/False Questions 1. For confidence intervals, precision increases as the interval gets wider. page 274

False page 275

2. Estimation is used to determine whether or not a treatment effect has occurred.

True —

3. For estimation, you do not need to state a hypothesis about the unknown

page 276

population parameter.

4, The sample mean (or mean difference) is always used as the point estimate of the population mean (or mean difference).

5. The sample mean (or mean difference) is always located exactly in the center of the confidence interval estimate for the population mean (or mean difference).

Chapter 12: Exam Items - page 107

True page 288

True page 280

6. Ifall other factors are held constant, as degrees of freedom increase, the width of a confidence interval decreases.

7. When making a point estimate for p, you should use the value of x.

True page 281

8. If the 90% confidence interval for 1 is from 34 to 50, then there is a 10% chance that p does not fall within that range.

True page 281

9. The sample mean will always be exactly in the middle of a confidence interval estimating the population mean.

False page 281

10. For a 75% confidence interval, the correct t values are t = +0.75 and t = -0.75.

True page 281

11. As the percentage of confidence increases, the width of the confidence interval also increases.

False page 288

12. Holding everything else constant, the width of the confidence interval will

True

page 281

increase as the standard error decreases.

13. A sample of n= 25 scores is being used to construct an 80% confidence interval for the population mean. The correct t values for the estimation equation are +1.318.

True page 284

14. Two samples, one with n = 6 and one with n = 8, are being used to estimate the difference between two population means. The t values for the 95% confidence interval are +2.179.

True page 284

15. A sample in treatment I has a mean of X = 12 with SS = 23. A second sample is given treatment II and has X = 16 with SS = 29. If these data are used to estimate the population mean difference between the two treatments, then the point estimate would be 4 points.

False

16. Ifa researcher wanted to estimate how much difference there is in annual income between college graduates and non-graduates, the appropriate estimation equation would be the repeated-measures t equation.

page 282

True

page 280

17. Iftwo samples have equal means, then they will produce identical point estimates for the population mean.

True

18. Estimation with the independent-measures t statistic is typically used to

page 282

determine how much difference there is between two treatments or two

populations.

Chapter 12: Exam Items - page 108

False page 288

19, In general, as sample size increases, the width of a confidence interval also increases.

True page 280

20. The correct formula for estimating the population mean, when o is not known, is ji = X= ts.

True

page 288

21. In general, a 90% confidence interval will be wider than an 80% confidence interval.

False page 288

22. In general, a confidence interval based on a sample of n = 25 will be wider than an interval based on a sample of n= 10.

False page 275

23. Usually, it would be inappropriate to use estimation after a hypothesis test where the decision was to reject the null hypothesis.

True

24. A point estimate can be viewed as equivalent to a 0% confidence interval.

page 273

25. The 95% confidence interval for the difference between two treatment means extends from -2.50 to +5.50. Based on this information, you can conclude that there is no significant difference between the treatments at the .05 level of significance.

True page 288

Other Exam Items

1. Describe point estimates and interval estimates in terms of their confidence and precision in estimating

w.

2. A sample of part time students are asked how many hours per week they work ona job. For the following sample data, determine the boundaries of the 99% confidence interval for u.

Scores:

a

eae

a

Bs

3. A sample of n= 9 has a mean ofX = 72 with SS = 800. Use the sample to make a point estimate of and to construct the 90% confidence interval for p.

Chapter 12: Exam Items - page 109

4. For a sample of n = 25 high school students, the mean improvement in SAT scores following a preparatory course was D = 44 points with s = 20. On average, how much improvement can be expected for the population of high school students? Use a 90% confidence interval.

5. A researcher would like to estimate how much difference in cigarette smoking results from viewing graphic videos (including an autopsy) on the harmful effects of smoking. A sample of 16 smokers is obtained. Half of the smokers (n = 8) view the video and the other half (n = 8) serves as a control group. The number of packs of cigarettes each person smoked during the following week is recorded. The data are as follows: Controls X = 17.5 SS = 122

Video Viewers X= 8.1 SS = 102

a. Use the sample data to make a point estimate of the mean difference in smoking behavior between the control group and the video viewers. b. Make an interval estimate of the mean difference so that you are 80% confident that the population mean difference is in your interval.

Answers for Other Exam Items 1. Point estimates are perfectly precise but have no confidence. Interval estimates provide confidence (the wider the interval, the more confidence), but they give up precision (the wider the interval, the less precision).

2. w= 14+ 4.604(1.14), df= 4. The confidence interval extends from 8.75 to 19.25.

3. For the point estimate, use extends from 65.81 to 78.19.

X = 72.

= 72 + 1.86(3.33), df= 8. The confidence interval

4. w= 44+ 1.711(4), df= 24. The confidence interval extends from 37.156 to 50.844.

5. a. Use X, - X, = 9.4 for the point estimate of 1, - 1).

Db. Hy - Hy = 9.4 + 1.345(2), df= 14. The confidence interval extends from 6.71 to 12.09.

Chapter 12: Exam Items - page 110

Chapter 13 Introduction to Analysis of Variance

Multiple-Choice Questions

Answer ¢ page 302

1. For an experiment comparing more than two treatment conditions you should use analysis of variance rather than separate t tests because a. you are less likely to make a mistake in the computations of ANOVA b. a test based on variances is more sensitive than a test based on means ANOVA has less risk of a Type I error . ANOVA has less risk of a Type II error

Answer c page 302

2. The distinction between the "testwise" alpha level and the "experimentwise" alpha level is important a. whenever you do an analysis of variance b. when the study is comparing exactly two treatments when the study is comparing more than two treatments d. only when there are fewer than 30 scores in each treatment

Answer a page 300 Www

3. The null hypothesis for an ANOVA states that ; a. there are no differences between any of the population means b. at least one of the population means is different from the others c. all of the population means are different from each other d. none of the above

Answer b page 305

4. When the null hypothesis is true for an ANOVA, the expected value for the F-ratio is

WWw

a. 0

_b.1.00 ge: kel d. N-k

Chapter 13: Exam Items - page 111

Answer a

5. In the notation for ANOVA, the letter n refers to

page 307

refers to

and the letter N

the number of scores in each treatment, the number of scores in the entire

study b. the number of scores in the entire study, the number of scores in each treatment

c. the sum of the scores in each treatment, the sum of the scores for the

entire study d. the sum of the scores for the entire study, the sum of the scores in each treatment Answer b page 301

6. In analysis of variance, the F-ratio is a ratio of a. two (or more) sample means

b. two variances | c. sample means divided by sample variances d. none of the above

Answer a page 321

7. Inan ANOVA, which of the following is most likely to produce a large value for the F-ratio?

www

a, large mean differences and small sample variances_ b. large mean differences and large sample variances ¢. small mean differences and small sample variances d. small mean differences and large sample variances

Answer ¢ page 315 www

8. For an independent-measures experiment comparing two treatment conditions with a sample of n = 10 in each treatment, the F-ratio would have df equal to a. 18

b. 19 aie. 18 4419 Answerd page 311

9.‘ The value for the within-treatments degrees of freedom, dfjiuins IS determined by a. n- |

b. N- 1 c. k- 1

» e= ko Answer a

10.

page 315

a = .05 would be

For an F-ratio with df= 2, 10, the critical value for a hypothesis test using

a 410

-

b. 7.56 19.59 d. 99.40 Chapter 13: Exam Items - page 112

Answer c

page 315

11. A researcher uses analysis of variance to test for mean differences among four treatments with a sample of n = 6 in each treatment. The F-ratio for this analysis would have what df values? a. df= 3, 5 b. df=3, 15

c. df=3,

20

d. df= 4, 24 Answer b page 311 www

12. A researcher reports an F-ratio with df = 3, 36 for an independent-measures experiment. How many treatment conditions were compared in this experiment? a3 b. 4 c, 36

d. 39 Answer c

page 311

13. A researcher reports an F-ratio with df= 3, 36 for an independent-measures experiment. How many individual subjects participated in the experiment?

4136 boa

c. 40

d. cannot be determined from the information given Answer d

page 311

14. An experiment compares two treatment conditions with a sample of n = 20 in each treatment. If the data are analyzed with ANOVA, the analysis would have LO

ai a. 18 b, 19 c 38

— d.39 Answer b page 300

15. In analysis of variance, the term factor refers to a. a dependent variable b. an independent variable c. a treatment mean d. a treatment total

Answer a

16. A treatment effect refers todifferences between scores that are caused by the

page 304

different treatment conditions. The differences (or variability) produced by = treatment effects will contribute to a. the numerator of the F-ratio

WWW

b. the denominator of the F-ratio c. both the numerator and the denominator of the F-ratio d. Treatment effects do not contribute to the F-ratio because they are removed before the F-ratio is computed.

Chapter 13: Exam Items - page 113

Based on

Answer b

17. An analysis of variance produces SSyenween = 40 and SS, jinn = 60.

page 321

this information, the percentage of variance accounted for, r’, is equal to

a. 40/60 = 67% b. 40/100 = 40% | c. 60/40 = 150% d. 60/100 = 60% Answer c page 328

Answer a page 328 www

18. Post hoc tests are necessary after an ANOVA whenever a. H, is rejected b. there are more than two treatments c. both of the above d. You always should do post hoc tests after an ANOVA.

. The purpose of post hoc test is to a. determine which treatments are different b. determine how much difference there is between treatments c. determine whether or not a Type I Error was made in the ANOVA

d. determine whether or not a complete ANOVA is justified Answer b page 328

20. Under what circumstances are post tests necessary? a. reject the null hypothesis with k = 2 treatments b. reject the null hypothesis with k > 2 treatments c. fail to reject the null hypothesis with k = 2 treatments d. fail to reject the null hypothesis with k > 2 treatments 2, In general the distribution of F-ratios is a. symmetrical with a mean of zero b. positively skewed with all values greater than or equal to ZETO c. negatively skewed with all values greater than or equal to zero d. symmetrical with a mean equal to dfhenveen pee In analysis of variance MS provides a measure of a. variance — b. average differences among means c. the total variability for the set of N scores d. the overall mean for the set of N scores 2, SSyetween treatments Provides a measure of

a. differences between the treatment means

_

b. differences between subjects inside the treatments c. the sum of the scores for each of the treatments d. the sum of the treatment totals

Chapter 13: Exam Items - page 114

Answer a ai 24. In an independent-measures experiment with three treatment conditions, all page 322 three treatments have the same mean, X, = X,= X,. For these data SSyemeen treatments equals

a. 0 b. 1.00 Ce o(S.00) d. cannot be determined from the information given

25. A research study compares three treatments with n = 5 in each treatment. If page 309

the SS values for the three treatments are 25, 20, and 15, then the analysis of

variance would produce SS,,.,,, equal to aoe

b. 20

c. 60. d. cannot be determined from the information given Answer a page 312

26. A research study compares three treatments with n = 5 in each treatment. If the T values for the three treatments are 5, 10, and 15, then the between-

treatments degrees of freedom would be equal to @2 b. 3 G12

d. 14 Answer c

page 315

27. A research study compares three treatments with n = 5 in each treatment. If the data were examined using an analysis of variance, the F-ratio would have df=

7 aeBe bedav,

d. 2,14

True/False Questions 1. The main advantage of analysis of variance, compared to t tests, is that ANOVA can be used to evaluate differences between variances as well as differences between means.

2. When a study involves more than two treatment conditions, an analysis of variance will evaluate all of the separate mean differences in a single test.

Chapter 13: Exam Items - page 115

True. page 303

3. The basic "analysis" in ANOVA involves partitioning the total variability into two components: between-treatments variability and within-treatments variability.

False page 313

4. Inan analysis of variance, MS,,. will always equal the sum Of MSyerween and

True

5. Inan analysis of variance, df,,.., will always equal the sum Of dfyerween and df, thin:

page 312 6. When there are more than two treatments in an ANOVA, rejecting the null

False page 327

hypothesis means that all of the treatment means are significantly different from each other.

page 299

7. Analysis of variance is used to test for differences in variance between two or more populations.

False page 305

8. When the null hypothesis is true, the F-ratio for analysis of variance is expected, on average, to be zero.

True.

9. In general, a large value for an F-ratio indicates that the null hypothesis is wrong.

False

page 306

* page 319

10. A research report states that there are “significant differences between treatments with F(2, 27) = 5.36, p < .05.” Based on this report, you can conclude

that the decision from the ANOVA was to reject the null hypothesis.

11. A research report states that there are “significant differences between |page 319

treatments with F(2, 27) = 5.36, p < .05.” Based on this report, you can conclude

that the research study evaluated the mean difference between three treatment conditions.

False_ page 301

’

False page 327

True’ page 305

False page 323

THUG, page 323

12. Withk =3 treatments, the alternative hypothesis for analysis of variance always states that all three treatment means are different.

13. With k = 3 treatments, rejecting the null hypothesis for analysis of variance means you have concluded that all three treatment means are different.

14. When the null hypothesis is true, the F-ratio is balanced so that the numerator and the denominator are both measuring exactly the same sources of variance.

15. As sample variance increases for each of the separate samples, the value of the F-ratio will also increase.

16.

The larger the differences between treatment means, the larger the F-ratio.

Chapter 13: Exam Items - page 116

page 314 _ __page 314

17. The critical region for the F-ratio from an analysis of variance is located

entirely in one tail of the distribution.

18. For an analysis of variance comparing two treatments with n= 5 in each treatment, the F-ratio will have df= 2, 8.

True ——~19. page 314

The distribution of F-ratios is positively skewed.

20. The F-distribution table lists a critical value of F = 4.26 for an F-ratio with page 314

df= 2,9 anda=.05.

Ifthe data produce an F-ratio of F(2,9) = 4.10, the correct

decision would be to reject H, at the .05 level of significance.

page 313

21. For an experiment with k = 3 treatments, MSyeseen = 10. For this analysis, Sy |B

True page 328

22. Post tests would not be needed if H, were rejected in an ANOVA comparing k = 2 treatments.

False page 328

23. Post tests are used after analysis of variance to determine why the null hypothesis was not rejected.

False page 328

24. Post tests are used to determine how much difference there is between treatment conditions.

False page 328

25. Post hoc tests are used to reduce the risk of a Type I error.

True —

page 322

26. In analysis of variance, a value of F = 0 indicates that all of the samples have

exactly the same mean.

27. An F-ratio of F = 1.00 indicates that the data provide no evidence of any True page 305 treatment effect.

Chapter 13: Exam Items - page 117

Other Exam Items 1. Describe the circumstances under which you should use ANOVA instead of t tests, and explain why t tests are inappropriate in these circumstances.

2. What value is expected, on average, for the F-ratio in ANOVA when the null hypothesis is true? Explain why this value is expected.

3. State the alternative hypothesis (H,) for an ANOVA comparing four treatment conditions.

Explain why the hypothesis is stated in this way.

4. Describe the circumstances when post tests are used and explain why these tests are necessary.

5. The data below are from an independent-measures experiment comparing three difference treatment conditions.

Treatment | Treatment 2 Treatment 3 4 0 1

no = 3

55 =9

SS =6

a. Use an analysis of variance with a = .05 to determine whether these data indicate any significant differences among the treatments. b. Use the Scheffe test to determine which of the three treatments are significantly different from each other. Use the .05 level of significance for all tests.

6. A psychologist would like to know whether the change in seasons has any consistent effect on people's mood. In the middle of each of the four seasons the psychologist selects a random sample of n= 10 students. Each individual in these four separate samples is given a standardized mood/depression questionnaire. The data from this study are examined using an analysis of variance and the results are shown in the summary table below. Fill in all missing values in the table.

Source Between Treatments Within Treatments Total

SS 2 132

df j= ee

MS _ eas

Chapter 13: Exam Items - page 118

FF=__

7. A psychologist would like to examine the effects of a new drug on the activity level of animals. Three samples of rats are selected with n = 5 in each sample. The rats in the first sample serve as a control and do not get any of the drug. The rats in the second group receive a small dose, and the rats in the third group each get a large dose of the drug. The psychologist records the activity level for each animal. The data from this experiment are presented below.

Do these data indicate any significant differences among the three groups? Test with a = .05.

No drug

Smalldose

Large dose

0 l 3 0 l

] 3 4 ] |

3 8 6 4 if

c=2 SS =6

Can So=s

eo) no 10

G=45 2X*=229

Answers for Other Exam Items

1. ANOVA must be used when there are more than two treatment conditions. In this situation it would require several t tests to compare all the treatments, and the more tests that are done the more risk there is for a Type I error.

2. When the null hypothesis is true, the F-ratio is expected to be near 1.00. The structure of the F-ratio is balanced so that the numerator and denominator are measuring identical sources of variance if the treatment effect is zero (H, is true).

3. With k = 4 treatments the alternative hypothesis would simply state, "At least one mean is different from the others." This general statement is used because there are several different alternatives to H, and it usually is not productive to try to list them all.

4. Post tests are done after an analysis of variance where the decision is to reject the null hypothesis and when the test is comparing three or more treatments. The decision to reject Hy simply indicates that at least one of the treatment means is different from the others. The

purpose of the post tests is to determine exactly which treatment means are different.

Chapter 13: Exam Items - page 119

> a.

Source

Ss)

cdr

Between Within Total

26 18 44

2 9 11

With df = 2, 9 the critical value is 4.26. Reject H, and conclude that there are significant differences among the treatments. b. Start with the largest mean difference, Treatment 1 versus 3: SSyerween = 24-5, MSperween = 12.25, F(2,9) = 6.125 (significant). Treatment 2 versus 3: SSyeween = 12-5, MSpetween = 6.25, F(2,9) = 3.125 (not significant). No other treatment differences are significant.

Source Between Treatments Within Treatments Total

SS

df

F2 13Z

36-2 39

Source

SS

df

MS

Between Treatments Within Treatments Total

70 24 94

_ be 14

2

MS

ay

Wa

With a = .05 and df = 2, 12 the critical value is 3.88. Reject H, and conclude that the drug has a significant effect on activity level.

Chapter 13: Exam Items - page 120

Chapter 14 More Advanced Analysis of Variance: Repeated-Measures and Two-Factor Designs

Multiple-Choice Questions Answer b page 347 www

1. A repeated-measures analysis of variance can be viewed as a two-stage process in which a. the first stage is identical to the independent-measures analysis and the second stage removes individual differences from the numerator of the Fratio b. the first stage is identical to the independent-measures analysis and the second stage removes individual differences from the denominator of the F-ratio

c. the first stage removes individual differences from the numerator of the Fratio and the second stage is identical to the independent-measures analysis d. the first stage removes individual differences from the denominator of the F-ratio and the second stage is identical to the independent-measures analysis

Answer a page 346

2. In the F-ratio for a repeated-measures ANOVA, variability due to individual differences , a. is automatically eliminated from the numerator but must be computed and subtracted out of the denominator b. is automatically eliminated from the denominator but must be computed and subtracted out of the numerator c. is automatically eliminated from both the numerator and the denominator d. must be computed and subtracted out of the numerator and the denominator

Chapter 14: Exam Items - page 121

Answer d page 352

3. Ifk =3 and n=5, then the df values for a repeated-measures F are cae . 2,4 pe 2,8 . none of the above sp eno

Answer a

page 345

4. The sources of variability that contribute to the between-treatments SS for a repeated-measures ANOVA are a. treatment effect and chance/error b. treatment effect and individual differences c. treatment effect, individual differences, and chance/error d. individual differences and chance/error

5. Ina repeated-measures ANOVA, a P value is obtained by a. finding the sum of the scores for a treatment b. summing the T values of all treatments c. finding the sum of the scores for a participant in the study d.

N-k Answer a

6. When using Tukey's HSD or the Scheffé post hoc test for a repeated-measures

page 353

ANOVA, the denominator for the F ratio should be

a. MS error

b. MS between subjects c. MS within d. MS between treatment Answer d

page 353

Answer d

page 353 www

7. The results of a repeated-measures ANOVA are reported as F(3, 27)=1.12, p> .01. What alpha level was used?

Ba5 B27 rela hayis ceo 8. The results of a repeated-measures ANOVA are reported as F(3, 27) = 1.12, p > .01. How many treatment conditions were used in the study?

Anis b. 10 ria d. none of the above

Chapter 14: Exam Items - page 122

Answer a

page 350

9. The results of a repeated-measures ANOVA indicate that the within-treatment SS is 12 and the between-subject SS is 4. For this analysis, the error SS is a. 8 b. 3 c. 48 d. insufficient information to find error SS

Answer c

10. A repeated-measures research study uses n= 6 subjects. If the within-

page 351

treatment df for the ANOVA is 15, what is the value for the error df?

a. 3

2 eebe erg

Pa Answer b

page 345

11. An important advantage of a repeated-measures study is a. a different sample of subjects serves in each treatment, making it possible to compare different populations b. individual differences do not contribute to the analysis, thereby reducing unsystematic and unpredicted error c. a larger number of subjects generally is required, which provides a better representation of the population d. all of the above 12. For a repeated-measures ANOVA, which of the following is computed differently from that in an independent-measures ANOVA? a. total SS b. between treatment SS c. within treatment SS d. the denominator of the F-ratio

Answer d

page 349

Answer b page 349

13. In Stage 1 of the repeated-measures ANOVA, total degrees of freedom is partitioned into _ a. error df and between subjects df b. within treatments df and between subjects df c. between treatments df and between subjects df d. between treatments df and within treatments df 14. In Stage 2 of arepeated-measures ANOVA, the within treatments degrees of freedom is partitioned into a. between treatments df and error df b. between subjects df and error df c. between treatments df and between subjects df d. total df and error df

Chapter 14: Exam Items - page 123

Answer a

15.

A two-factor experiment means that the experimental design includes a. two independent variables b. two dependent variables c. an interaction between the independent and the dependent variable d. exactly two separate groups of subjects

16.

: Two variables are said to interact when a. the effect of one independent variable depends on the levels of the second variable b. both variables are equally influenced by a third factor c. the two variables are differentially affected by a third variable d. both variables produce a change in the subjects’ scores

page 355

Answer a page 360 WWW

17. Ina two-factor analysis of variance a main effect is defined as a. the mean differences among the levels of one factor b. the mean differences among all treatment conditions c. the mean difference between the two factors d. the difference between the largest treatment mean and the smallest treatment mean

18. A significant interaction between factors A and B indicates a. either the main effect for factor A or the main effect for factor B is also significant, but not both b. both the main effect for factor A and for factor B must also be significant fo) . neither of the two main effects is significant d. The significance of the interaction has no implications for the main effects.

Answer c

page 363

19. The results from a two-factor analysis of variance show that both main effects are significant. From this information you can conclude that a. the interaction also must be significant b. the interaction cannot be significant c. You can make no conclusions about the significance of the interaction.

20. How many separate samples would be needed for a two-factor, independentmeasures research study with two levels of factor A and three levels of factor B? a

2

b.3 C5 d.6

Chapter 14: Exam Items - page 124

Answer c

21.

page 364

n= 5 in each treatment condition, the degrees of freedom within treatments would be equal to to)pe 2 . 24 aos 8

Inan experiment with two levels of factor A, three levels of factor B, and

22. A two-factor research study has two levels of factor A and three levels of factor B with a separate sample of n = 5 subjects in each treatment condition.

Based on this information, the F-ratio evaluating the main effect of factor A would have df values a. df= 1, 24 be df= 1, 29 oe, df=2, 24 d. df= 2, 29 23. The analysis of variance for a two-factor experiment produces a. one overall F-ratio followed by a series of required post hoc tests b. two separate F-ratios c. three separate F-ratios d. four separate F-ratios Answer d

page 367

Answer d

page 361

24. Ifthe mean and variance are computed for each sample in an independentmeasures two-factor experiment, then which of the following types of sample data will tend to produce large F-ratios for the two-factor ANOVA? a. big values for the sample means b. small values for the sample means c. big values for the sample variances d. small values for the sample variances 25. Ifthe results of a two-factor experiment are presented in a line graph, then an interaction can be seen whenever a. the lines in the graph are not straight (bent) b. there is a space separating the lines c. the lines are parallel d. the lines are not parallel 26. In analysis of variance, each independent variable is referred to as a a. factor b. level c. comparison d. stage

Chapter 14: Exam Items - page 125

True/False Questions True

page 346

False page 347

True page 349

1. For a repeated-measures design, the differences between treatments may be caused by a treatment effect or by chance/error, but they cannot be caused by individual differences.

2. One major difference between the independent-measures ANOVA and the repeated-measures analysis is that the total SS is calculated differently for each type of ANOVA.

3. In the second stage of analysis for the repeated-measures ANOVA, individual differences are removed from the denominator of the F-ratio.

True page 350

4. Large amounts of individual differences will result in a large value for

False page 346

5. For the repeated-measures analysis, when Hy, is true the F-ratio on average is expected to have a value of zero.

True page 353

6. Based on the results of a repeated-measures ANOVA, a researcher reports a significant treatment effect with F(4,18)=11.57, p< .05. For this ANOVA, the null hypothesis was rejected.

False page 353

7. The result of a repeated-measures ANOVA stated that F(4,18)=11.57, p< .05.

The study compared four treatment conditions.

True page 347

8. For the repeated-measures ANOVA, between-treatment SS and within-

SSpetween subjects °

treatment SS are computed the same way as they are for the independent-measures ANOVA.

True page 346

9. A major advantage of the repeated-measures design is the removal of individual differences from the analysis.

True page 347

10. The repeated-measures ANOVA begins the same way as the independentmeasures ANOVA with the total variability partitioned into between-treatment and within-treatment components,

False

11.

A factorial experiment contains at least two dependent variables.

page 355

Chapter 14: Exam Items - page 126

True page 367

12.

For a two-factor, independent-measures experiment, the smaller the values for

the sample variances, the more likely it is that at least one of the F-ratios will be significant.

True page 356

13. A two-factor, independent-measures research study with two levels of factor A and two levels of factor B would require four separate samples.

True page 359

14. For an independent-measures two-factor experiment, the bigger the differences between the sample means, the more likely it is that at least one of the F-ratios will be significant.

True page 363

15.

True page 367

16. The variance within treatments, MSnin, IS used as the denominator for all three F-ratios in a two-factor analysis of variance.

False page 363

17. Ifthe A x B interaction is significant, then at least one of the two main effects also must be significant.

False page 366

18. For an experiment with two levels of factor A and three levels of factor B the A x B interaction would have df,,, = 6.

False page 374

19. Ifthe F-ratio for factor A has df= 2, 24 then the F-ratio for factor B also must have df = 2, 24.

True page 366

20. The results from a two-factor analysis of variance show that the F-ratio for

Atwo-factor ANOVA consists of three separate hypothesis tests.

factor A has df= 1, 30 and the F-ratio for factor B has df= 2, 30. Based on this information, there was a total of 6 different treatment conditions in this research

study.

False page 363

21. Obtaining a significant interaction means that both factors A and B have significant main effects.

True page 356

22. An independent-measures experiment with two levels of factor A and four levels of factor B would require exactly eight separate groups of subjects.

False

page 363

23. One of the disadvantages of combining two factors in the same experiment is that it makes it impossible to determine how either factor would affect the subjects’ scores if it were examined in an experiment by itself.

False page 355

24. A two-factor experimental design means that the researcher obtains two separate scores for each subject.

Chapter 14: Exam Items - page 127

False page 360

25. A researcher predicts that a special training course will be equally effective for males and females. This researcher is predicting an interaction between training and gender.

True page 360

26. A researcher is comparing a new teaching strategy relative to the old strategy.

It is predicted that the new teaching strategy will increase test scores for highly motivated students but will decrease scores for students with low motivation. This researcher is predicting an interaction between teaching strategy and student motivation.

Other Exam Items

1. One advantage of the repeated-measures ANOVA is that it eliminates individual differences as a source of variability. Explain why there are no individual differences in the numerator and in the denominator of the F-ratio.

2. A researcher studies the effect of three practice methods on memory. A sample of n = 3 subjects memorizes a list of nonsense syllables (e.g., ZOK) using method A. During a recall test, the number of errors are recorded. This procedure is repeated for methods B and C. The data are presented below.

Subject

A

# | #2 #3

3 6 6

Method B C

0 3 3

0 0 3

Is there an effect of method of practice on recall? Set alpha at .05.

Chapter 14: Exam Items - page 128

3. A psychologist examines long term retention of a learned task in a sample of n= 6 rats. At 90 days of age, rats are trained on a discrimination task. Their initial performance is recorded to

serve as a baseline for subsequent comparisons. Then their performance is tested again at 6 months, 1 year, 1 1/2 years, and 2 years of age. A partial ANOVA summary table follows.

Provide the missing values. (Hint: Begin with values for df).

Source

SS

dé

Between treatments” Within treatments Between subjects Error Total

MSs

ti‘~wCC Pa ee ae 8, A 1526

Se

4. The following data were collected from a repeated-measures study. Determine if an effect exists at the .05 level of significance.

Treatments

Subject

A

B

C

D

1 2 3 4

6 4 4 6

3 4 2 3

3 2 0 3

0 2 2 0

TJ Ss=4

ae Ss=2

G=44 =X? =172

ke ol ae S8S5=6 55=4

5. The results from a two-factor experiment can be presented in a matrix with the levels of one factor forming the rows and the levels of the second factor forming the columns, with a separate sample in each of the matrix cells. Using this matrix structure, describe what is meant by the main effects for each factor and the interaction between factors.

6. The following table summarizes the results of a two-factor ANOVA evaluating an independent-measures experiment with two levels of factor A, three levels of factor B, and n= 5

subjects in each separate sample. Fill in all missing values in the table. (Hint: Start with the df column.)

Source Between Treatments Factor A Factor B AxB Within Treatments

df —

SS 60

Fee

oink

A

ae

Ms

ee

recs cP. Bie

ae

2

Total

Chapter 14: Exam Items - page 129

7. The following data were obtained from a two-factor independent-measures experiment.

Bl

B2

B3

B4

Nt 0 Tis] Al SS=40 SS=30 SS=50 —_——————— fPcs 8 (V0 1S) A2 SS=30 SS=20 SS=40

5t=20 SS=40

n= 5 for all cells G = 100

T=40 SS=70

&X? = 800

a. State the hypotheses for each of the three separate tests included in the two-factor ANOVA. b. Calculate degrees of freedom and locate the critical region for each of the three tests. c. Calculate the three F-ratios. d. State a conclusion for each test.

Answers for Other Exam Items 1. There are no individual differences in the numerator (between treatments) because the same set of individuals participates in all treatment conditions in a repeated-measures design. There are no individual differences in the denominator because they are subtracted out in the second stage of the repeated-measures analysis.

os

a,

Source Between Treatments Within Treatments Between Subjects Error Total

SS 26 18 |e: 4 44

df 2 6 Se 4 8

==MS 13.

F(2,4) = 13.00

Source Between Treatments Within Treatments

SS 44 108

df

MS bl

P= 530

4 25

Between Subjects

68

5

Error Total

40 152

20 29

2

Chapter 14: Exam Items - page 130

4.

Source Between Treatments Within Treatments Between Subjects Error Total

SS_df 18 1 476 12: <3 i 6 3 30 7

=MS 18 F(1,3) = 9.00

With df= 1,3 the critical value is 10.13. Fail to reject the null hypothesis. There is not a significant difference between treatments.

5. Ifthe rows of the matrix correspond to the levels of factor A, then you would begin by computing an overall mean for all the subjects (scores) in each row. The differences between these row means describe the main effect for factor A. Similarly, an overall mean is computed

for each column and the differences between the column means describe the main effect for factor B. Finally, an interaction exists if there are any mean differences within the matrix that are not consistent with the overall row (or column) mean differences. For example, if the overall mean for the top row is 10 points higher than the overall mean for the second row, then this same 10-point difference must exist within every column or else there is an interaction.

6.

Source Between Treatments Factor A Factor B AxE Within Treatments Total

df SS MS a Od bo lO 10 Fee 3 2. 20-0 10.) a=.

ZO 3001S. Sel ae = 24 48 2 29 108

7. a. The hypotheses are For factor A: Hy: 1,4, =}, (no A-effect) Fy: bar F Had For factor B:

H,: pg) = [po = Mp3 = Hp, (no B-effect) H,: At least one of the B-means is different from the others.

For Ax B: H,: There is no interaction.

H,: There is an interaction. b. For factor A, the F-ratio has df = 1, 32 and the critical value is F = 4.15. For factor B and the A x B interaction, the F-ratios have df= 3, 32 and the critical value is F = 2.90.

Chapter 14: Exam Items - page 131

fon

Source Between Treatments

df 7

SS 230

MS

Factor A Factor B AxB Within Treatments Total

1 3) 3 32 39.

10 1000-5 1/0 56-67" 50 16:0) 320 = 10.00 -350

EF = 1 Fa 3:67 Ege = 07

d. The results indicate that there are significant differences among the levels of factor B, but the overall A-effect is not significant and the interaction is not significant.

Chapter 14: Exam Items - page 132

Chapter 15 Correlation and Regression

Multiple-Choice Questions Answer d page 385

1. A negative value for a correlation indicates a. a much stronger relationship than if the correlation were positive b. a much weaker relationship than if the correlation were positive c. increases in X tend to be accompanied by increases in Y d. increases in X tend to be accompanied by decreases in Y

Answer c

2. A positive correlation indicates a. increases in X tend to be accompanied by increases in Y b. decreases in X tend to be accompanied by decreases in Y

page 385 www

c. both of the above d. neither of the above Answer c

page 387

Answer ¢

page 388 WWW

Answer b page 387 WWW

3. The numerical value for a correlation a. can never be greater than 1.00 b. can never be less than -1.00 c. both of the above d. neither of the above

4. The Pearson correlation measures a. the degree of relationship without regard to the form of the relationship b. the degree to which the relationship is consistently one directional c. the degree of linear relationship d. the degree of curvilinear relationship 5. A Pearson correlation of r = -0.85 indicates that a graph of the data would show _ a. points clustered close to a line that slopes up to the right b. points clustered close to a line that slopes down to the right c. points widely scattered around a line that slopes up to the right d. points widely scattered around a line that slopes down to the right

Chapter 15: Exam Items - page 133

Answer d page 387

6. Ascatterplot shows a set of data points that are clustered close to a line that slopes down to the right. Which of the following values would be closest to the correlation for these data? a. 0.80 b. 0.40 c. -0.40

d. -0.80 Answer b

page 387

7. A-scatterplot shows a set of data points that are widely scattered around a line that slopes up to the right. Which of the following values would be closest to the correlation for these data? a. 0.80 b. 0.40 c. -0.40

d. -0.80 8. A Pearson correlation of r = +1.00 between X and Y indicates that a. each time X increases, there is a perfectly predictable increase in Y b. every change in X causes a change in Y c. every increase in X causes an increase in Y d. all of the above Answer c

9. Aset ofn=5

page 389

For this set of scores, the value of SP is a. -20 b. -28 c. 20 d. 60

10. A set ofn=5

pairs ofX and Y values has UX = 10, LY = 20, and ZXY = 60.

pairs ofX and Y values has SS, = 10, SS, = 40 and SP = 10.

For these data, the Pearson correlation is

a. r= 10/20 = 0.50 b. r= 10/400 = 0.025 c. r= V10/20 = 0.707 d. r= 40/100 = 0.40 Answer b

page 398

11. Suppose the correlation between height and weight for adults is +0.80. What proportion (or percent) of the variability in weight is accounted for by the relationship with height? a. 80% b, 64% c. 100 - 80 =20% d. 40%

Chapter 15: Exam Items - page 134

12. For a hypothesis test for the Pearson correlation, the null hypothesis states that a. there is a non-zero correlation for the general population b. the population correlation is zero c. there is a non-zero correlation for the sample d. the sample correlation is zero Answer b

page 389

Answer b

page 385

13. For the following data, SP equals

a. 6

be

b. -5 c. 43 d. none of the above

2 4 oe 82 3) 85 LPS

14. For these data, the Pearson correlation

a. is positive b. is negative

bee 6 2 4

Sis Zero d. cannot be determined

Supe 35

BOs Answer b

page 404 www

Answer Cc

page 406

15. The Spearman correlation should be used when a. the researcher's primary interest is the linearity of the relationship b. the original data are measured on an ordinal scale of measurement c. the Pearson is too difficult to compute d. all of the above

16. Which of the following characteristics of the relationship between X and Y is not measured by a Spearman correlation? a. the direction of the relationship b. the degree of consistency of the relationship c. the exact form of the relationship d. All of the above are measured by the Spearman correlation.

Answer d page 409

17. Before you use the special formula for the Spearman correlation you should be sure that a. there are at least n = 20 individuals in the sample b. both X and Y variables form normal-shaped distributions c. there are no undetermined or infinite scores in the original data (before ranking) d. there are only a few, if any, tied scores

Chapter 15: Exam Items - page 135

Answer b

page 411

Answer c

page 411

18. In the general linear equation Y = bX + a, the value of a is called a. the slope constant b. the Y-intercept c. the X-intercept d. the beta factor 19. In the linear equation Y = 3X + 1, when X increases by 4 points, Y will increase by a. 4 points b. 7 points c. 12 points d. 13 points 20. In the general linear equation Y = bX + a, the value of b is called the a. X intercept b. Y intercept c. correlation between X and Y d. slope

Answer a

page 411

Answer d page 414

21. What information cannot be determined from the general linear equation Y=bX +a? a. the correlation between X and Y b. the slope c. the Y intercept d. aandc 22. The regression equation is determined by minimizing a. the total error between the X and Y values b. the total error between the predicted Y values and the actual Y values c. the total squared error between the X and Y values d. the total squared error between the predicted Y values and the actual Y values

Answer b page 411 www

23. Which of the following points is on the line defined by the equation ¥ =4X4 2? a.

X=land Y=4 b. X = 0.5 and Y = 4 c.

X=2 and Y=8 d.

X=3 and Y=9

Chapter 15: Exam Items - page 136

:

Answer c

page 411

24. The entrance fee for a theme park is $20. Tickets for each ride and attraction are $3 apiece. Which of the following equations describes the relation between the total cost (Y) and the number of tickets purchased (X) in a single visit to the park?

a. Y=20X+3 b. Y = 60X ce Y=3X+20 dA Soy + 20

Answer d page 414 WWw

25. A set ofX and Y scores has X = 4, SS, = 15, Y = 5, SP = 30. What is the regression equation for predicting Y from X?

a. Y¥=3X+2 6.2V S249

e. ¥=10X413 do Y¥S2xX-35 Answer b page 414

26. If there is a negative correlation between X and Y then the regression equation, Y = bX + a, will have a. b>0 b. b<0 ea>Q d.a<0

True/False Questions False page 385

1. A negative correlation indicates that decreases in the X variable tend to be accompanied by decreases in Y.

True page 385

2. Ina fair world, there should be a positive correlation between exam grades and the number of hours spent studying.

False

3. If the value of the Pearson correlation is r = 0, then all data points on a scatterplot would fall on a straight line with a Y intercept of zero.

page 387 True

page 387

4. A Pearson correlation of r= -1.00 means that all the data points fit perfectly on a straight line.

Chapter 15: Exam Items - page 137

True page 391

5. Anytime the sum of products, SP, has a value less than zero, the correlation

False page 387

6. A correlation of r = -.90 means that there is essentially no consistent relationship between X and Y.

False page 385

7. Suppose it was observed that there is a correlation of r = -0.81 between a driver's age and the cost of the car insurance. This correlation would mean that, in general, older people pay more for car insurance.

True

8. A researcher obtained a correlation of r = +0.62 between the amount of time spent watching television and level of blood cholesterol. This means that there is a general tendency for people who watch less television also to have lower blood cholesterol.

page 385

False page 395

True page 400

also must be negative.

9. Suppose there is a correlation of +0.87 between the length of time a person is in prison and the amount of aggression the person displays on a psychological inventory. This means that spending a longer amount of time in prison causes people to become more aggressive.

10. Ina test for significance of a correlation, the null hypothesis states that the population correlation is zero.

True

11. Ina

page 401

value by using df= n - 2.

True page 400

12. It is possible for a sample to have a real, non-zero correlation, even though the correlation for the general population is zero.

False

13. Using a two-tailed test with a = .05, a sample correlation of r = .400 based on a sample of n= 18 individuals is large enough to conclude that there is a real correlation in the general population.

page 401

test for significance for the Pearson correlation, you determine the critical

page 416

14. It is possible for two sets of data to have the same regression equation but different correlations.

True page 411

15. The point defined by X = 2 and Y = 5 is located on the line defined by the equation Y = 4X - 3.

False page 416

16. A correlation of r = +0.68 will result in more accurate predictions than a situation where r = -0.95.

True page 411

17. In the linear equation Y = 2X = 3, the slope is equal to 2.

True

Chapter 15: Exam Items - page 138

False page 411

18. Ifthe Y intercept is positive, then the regression equation will have a positive slope.

False page 400

19. When testing a hypothesis about a correlation, the null hypothesis states that the population correlation is one.

True page 419

20. The regression equation is structured so that when X = X, the predicted value

True

page 406

21. The Spearman correlation measures the degree to which X and Y are monotonically related.

True page 406

22. The range of possible values for the Spearman correlation is from +1.00 to -1.00.

False page 406

23. A Spearman correlation of zero indicates the relationship between X and Y is perfectly monotonic.

False page 406

24. The Spearman correlation can never be less than zero.

True page 405

25. Fora set of data, the value of Y increases every time the value of X increases.

For these data the Spearman correlation would be +1.00.

ofY is equal to Y.

Other Exam Items

1. Describe what is meant by a negative relationship between two variables.

2.

For the data below, compute the Pearson correlation. ee 3 4 2 3 9 64.2

Chapter 15: Exam Items - page 139

3. For the data below, compute the Pearson correlation and find the regression equation for predicting Y from X. ee 1 2 Des 4 2 6 oS 4. Find the regression equation for predicting Y from X for the following set of scores.

Re 4 Oro i 1 PG 5. A sample of n= 20 pairs of scores (X and Y values) produces a correlation of r = -0.46. Are these sample data sufficient to conclude that there is a significant non-zero correlation between X and Y in the population? Test at the .05 level of significance.

6. A psychologist would like to determine whether there is a consistent relationship between verbal skills as measured by the Scholastic Achievement Test (SAT) and performance on an anagram task. (Anagrams are words with the letters scrambled. The task is to unscramble the letters and identify the word.) Each subject is given a set of 10 anagrams, and the psychologist records how much time is needed to solve all 10. The anagram times and each subject's SAT score are as follows. (Note that two subjects failed to complete all 10 anagrams.) Use a Spearman correlation to measure and describe the relationship between SAT and anagram performance.

Subject 1 Zz 3 4 5 6 iy 8 9 10

Anagram

SAT

Time 104 195 50 93 failed 170 failed 67 45 58

Score 480 410 590 510 420 460 380 570 640 600

Chapter 15: Exam Items - page 140

Answers for Other Exam Items 1. Ina negative relationship there is a tendency for one variable to decrease as the other variable increases. The two variables tend to move in opposite directions.

2. SS, = 14, SS, = 26, SP =-17, r=-0.89.

3. SS, =30, SS, = 22, SP = 22, r=+0.86,

Y = 0.73X + 1.08.

4. SS, =2, SP=2,X=1,and Y=9. Y=(1)X+8.

5. Ho: p= 0. For these data, df= 18 and the critical value for r from Table B.6 is 0.444. Hy is rejected, p < .05.

6. After ranking the data for SAT and anagram times, the value of the Spearman correlation is Tr, = -0.967.

Chapter 15: Exam Items - page 141

Chapter 16 Hypothesis Tests with Chi-Square

Multiple-Choice Questions Answer d

page 427 www

Answer a

page 430

1. Acharacteristic of nonparametric tests is that a. they require a numerical score for each individual b. they require assumptions about the population distribution(s) c. they evaluate hypotheses about population means or variances d. none of the above 2. The term observed frequency refers to a. the frequencies found in the sample data b. the frequencies found in the population being examined c. the frequencies computed from the null hypothesis d. the frequencies that are hypothesized for the population being examined 3. The observed frequencies a. are always whole numbers b. can contain fractions or decimal values c. can contain both positive and negative values d. both b and c

Answer b page 431

4. The expected frequencies a. are always whole numbers b. can contain fractions or decimal values c. can contain both positive and negative values d. both b and c

Answer b page 432

5. The chi-square distribution 1s a. symmetrical with a mean of zero b. positively skewed with all values greater than or equal to zero c. negatively skewed with all values greater than or equal to zero d. symmetrical with a mean equal to n - 1

Chapter 16: Exam Items - page 142

Answer a

page 433

6. A researcher is using a chi-square test to determine whether there are any preferences among four brands of cola. With a = .05 and a sample of n = 30, the critical region for the hypothesis test would have a boundary of Ae . 9.49 . 42.56 fp aowg 43.77 7. As the differences between f, and f, increase a. the likelihood of rejecting Hy also increases b. the likelihood of rejecting H) decreases c. the critical value for chi-square increases d. the critical value for chi-square decreases 8. A researcher is using a chi-square test to determine whether people have any preferences among three brands of televisions. The null hypothesis for this test would state that a. there are real preferences in the population

b. one-third of the sample will prefer each brand c. one-third of the population prefers each brand d. in the population, one brand is preferred over the other two Answer ¢

page 431

9. A chi-square test for goodness of fit is used to test whether or not there are any preferences among three brands of cola. If the study uses a sample of n = 60 subjects, then the expected frequency for each category would be Avot

Biolls c..20

d. 60 Answer c

page 428 Www

Answer b

page 432

10.

The chi-square test for goodness of fit evaluates a. the relationship between two variables b. the mean differences between two or more treatments c. the shape or proportions for a population distribution d. none of the above

11. The chi-square test for goodness of fit will produce a large value for chisquare when a. the sample proportions match the hypothesized population proportions b. the sample proportions are much different than the hypothesized population proportions Q . the sample mean is close to the population mean d. there is a large difference between the sample mean and the population mean

Chapter 16: Exam Items - page 143

12.

Inachi-square test for independence or goodness of fit a. SL =n b) 2h = 2 c. both of the above d. neither of the above

Answer b page 428

13.

The null hypothesis for the chi-square test for goodness of fit specifies___ a. proportions for a sample distribution b. proportions for a population distribution c. frequencies for a sample distribution d. frequencies for a population distribution

Answer d page 433

14. Inthe test for goodness of fit, the chi-square statistic has degrees of freedom equal to

Answer c

page 450

a. n- | b. n-2

c. n-C (where C is the number of categories) d. none of the above

15. A chi-square test for goodness of fit is used in a situation with n = 60 subjects distributed over three categories. The df value for this test is ane bes c. 58 d. 59 Answer b

page 439

Answer c

page 432

16.

The null hypothesis for the chi-square test for independence states that a. there is a relationship between the two variables b. there is no relationship between the two variables c. both variables have the same frequency distribution d. the two variables have different frequency distributions

17. For a fixed level of significance, the critical value for chi-square will a. increase as n increases b. decrease as n increases c. increase as df increases d. decrease as df increases

18. The chi-square test for independence is used to test for a. a mean difference between two populations b. a difference between a sample distribution and a population distribution c. a difference in variance between two populations d. a relationship between two variables

Chapter 16: Exam Items - page 144

Answer c

page 439

19. The chi-square test for independence can be used to evaluate a. the relationship between two variables b. differences between two or more population frequency distributions c. both of the above d. neither of the above

Answer d page 443

20. The chi-square test for independence has degrees of freedom given by

Answer c

21. A basic assumption for a chi-square hypothesis test is

Ye RC b. Rx C(n-1) con Re ad.(Red}(C= 1)

the population distribution(s) must be normal . the scores must come from an interval or ratio scale the observations must be independent Pp Bor . all of the above

page 448

Answer b page 448

Answer c

page 442 Www

. The chi-square statistic should not be used if a. f.> 5 for any cell b. f, < 5 for any cell c. f. =f, for any cell d. none of the above

tOtO

23. A sample of 100 people is classified by gender (male/female) and by whether or not they are registered voters. The sample consist of 60 females, of whom 50 are registered voters, and 40 males, of whom 25 are registered voters. If these data were used for a chi-square test for independence, the expected frequency for registered males would be De Bs)

be 25 c. 30

d. 45 24. A sample of 100 people is classified by gender (male/female) and by whether or not they are registered voters. The sample consist of 60 females, of whom 50 are registered voters, and 40 males, of whom 25 are registered voters. If these data were used for a chi-square test for independence, the chi-square statistic would have degrees of freedom equal to a.

|

b. 3 c. %

d. 99

Chapter 16: Exam Items - page 145

Answer a page 443

25. A chi-square test for independence is being used to evaluate the relationship between two variables, one of which is classified into 3 categories and the second of which is classified into 4 categories. The chi-square statistic for this test would ; have df equal to a. 6

be?

G0) d.11 Answer a

page 443

26. A researcher is using a chi-square test for independence to examine the relationship between TV preferences and gender for a sample of n= 100 children.

Each child is asked to select his/her favorite from a fixed set of three TV shows and each child is classified as male or female. The chi-square statistic for this study would have df equal to

True/False Questions False page 430

1. The observed frequencies for a chi-square test represent an ideal, hypothetical distribution that would be obtained if the sample were in perfect agreement with the null hypothesis.

False page 430

2. The data for a chi-square test are called expected frequencies.

True page 431

3. Ina

True page 448

4. For the chi-square test for independence, each individual is counted in one and

True page 450

5. For either chi-square test, the sum of the expected frequencies must be equal to the sum of the observed frequencies.

chi-square test, the observed frequencies are always whole numbers.

only one category.

Chapter 16: Exam Items - page 146

True page 432

6. The value obtained for the chi-square statistic will always be greater than or

True page 442

7. The same formula is used to compute chi-square for the goodness-of-fit test and for the test for independence.

False

8. In general, a large value for the chi-square statistic indicates that the null hypothesis is correct.

page 432

equal to zero.

True page 450

9. The value of df for a chi-square tests does not depend on the sample size (n).

True page 432

10. For a fixed level of significance, the critical value for chi-square increases as the degrees of freedom increase.

True page 432

11. It is impossible to obtain a value less than zero for the chi-square statistic, unless a mistake is made.

False page 430

12. Ifthe null hypothesis for a goodness-of-fit test predicts no preference, then the researcher should select a sample with an equal number of individuals in each category.

True page 433

13. A researcher uses a chi-square test for goodness of fit to evaluate preferences among four leading brands of ice cream. The chi-square statistic for this test will have di = 3.

True page 431

14. Ifthe null hypothesis for a goodness-of-fit test predicts no preference, then the expected frequency will be the same for every category.

False page 431

15. Inthe chi-square test for goodness of fit the null hypothesis always predicts equal frequencies for all categories.

False page 433

16. A researcher is using a chi-square test for goodness of fit to determine whether there is any preference between two brands of chocolate-chip cookies.

With a = .05 and a sample of n = 25, the critical value for the chi-square statistic would be 36.42.

True

page 438

17. The chi-square test for independence requires that each individual be categorized on two separate variables.

False page 432

18. In the chi-square test for independence, a positive value for the chi-square statistic indicates a positive correlation between the two variables.

Chapter 16: Exam Items - page 147

True page 439

19. The null hypothesis for the chi-square test for independence specifies that the

proportions in one row of the matrix are identical to the proportions in any other

row.

True page 439

20. The chi-square test for independence is similar to a correlation in that it evaluates the relationship between two variables.

page 439

21. The chi-square test for independence is similar to an independent-measures t test in that it can be used to evaluate the difference between two populations or two treatment conditions.

True page 441

22. Ina chi-square test for independence the row totals and column totals for the expected frequencies must equal the corresponding row totals and column totals for the observed frequencies.

False page 442

23. Ina chi-square test for independence, the expected frequencies in one row of the matrix are identical to the expected frequencies in any other row.

False

24. It is impossible for a chi-square test for independence to have df= 1.

True

page 443 False

page 448

25. For a chi-square test for independence it is possible for one individual to be counted in two different cells provided the two cells are not in the same row or in the same column.

True

page 448

26. A chi-square test should not be used if any expected frequency is less than five.

Other Exam Items 1. Explain how the chi-square tests differ from parametric tests (such as t tests or ANOVA) with respect to a. the hypotheses, b. the data, and c. the assumptions underlying the test.

Chapter 16: Exam Items - page 148

2. A sample of thirty people is obtained and each person tastes three brands of instant coffee and identifies his/her preference. The resulting frequency data are as follows: Brand A BrandB BrandC

a. A researcher wants to use these data to test for preferences among the three brands. State the hypotheses and find the expected frequencies for the chi-square test. b. Consumer data show that ten years ago Brand A was preferred by 20% of the population and Brands B and C were each preferred by 40% of the population. A researcher wants to use the data to test whether there has been a significant change in preferences during the past ten years. State the hypotheses and find the expected frequencies for the chi-square test.

3. A statistics instructor would like to know whether it is worthwhile to require students to do weekly homework assignments. For one section of the statistics course, homework is assigned, collected, and graded each week. For another section, the same problems are suggested each week, but the students are not required to turn in their homework. At the end of the semester, all students take the same final exam. The grade distributions for the two sections are as follows: FINAL EXAM GRADE A B C D bid

Homework

ja] i5 | | s | |

No Homework

Do these data indicate a significant difference between the grade distributions for students with homework versus students with no homework?

Test with a = .05.

4. A researcher would like to know whether there is a consistent, predictable relationship between verbal skills and math skills for high school students. A sample of 300 students is obtained and each student is given a standardized English test and a standardized math test.

Based on the test results, students are classified as high, medium, or low for verbal skills and for math skills. The results are summarized in the following frequency distribution: Verbal Skills High Math Medium Math Low Math

Based on these results, can the researcher conclude that there is a significant relationship between verbal skills and math skills? Test at the .05 level of significance.

Chapter 16: Exam Items - page 149

Answers for Other Exam Items 1. As nonparametric tests, the chi-square tests state general hypotheses about the entire population without any reference to a specific population parameter. The data for a chi-square test consist of frequencies, but the data for a t test or ANOVA consist of scores that can be added, multiplied, squared, etc. Finally, the chi-square tests do not require any assumptions about population parameters. The t tests and ANOVA require normal populations and homogeneity of variance for tests with independent-measures designs.

2. a. The null hypothesis states that the three brands are equally preferred in the population (p = 1/3 for each brand). The expected frequency is 10 for each of the three brands. b. The null hypothesis states that there has been no change in the population proportions; that is, 20% still prefer Brand A, 40% prefer Brand B, and 40% prefer Brand C. The expected frequencies are 6 for Brand A, 12 for Brand B, and 12 for Brand C.

3. The null hypothesis states that there is no difference between the grade distributions for the two sections. With a = .05 and df= 4, the critical value is 9.49. The expected frequencies are as follows: FINAL EXAM GRADE

Kee

Homework

Bros,

Ces)

Deak

|e |12 | 15 [io[s |

No Homework

For these data, chi-squared = 13.08. difference between the two sections,

Reject H) and conclude that there is a significant

4. The null hypothesis states that there is no relationship between the two variables. With a = .05 and df = 4, the critical value is 9.49. The expected frequencies are as follows: Verbal Skills

High

Medium

Low

High Math Medium Math Low Math

For these data, chi-squared = 3.43. Fail to reject Hj. The data do not indicate a significant relationship between math skills and verbal skills.

Chapter 16: Exam Items - page 150

Section II Solutions for

End-of-Chapter Problems

Note: Many of the problems in the text require several stages of computation. At each stage there is an opportunity for rounding answers. Depending on the exact sequence of operations used to solve a problem, different individuals will round their answers at different times and in different ways. As a result, you or your students may obtain answers that are slightly different from those presented here.

To help minimize this problem, we have tried to include the numerical values obtained at different stages of complex problems rather than presenting a single, final answer.

Section II: Solutions - page 151

Chapter | Introduction to Statistics 1. Descriptive statistics are used to simplify and summarize data. Inferential statistics use sample data to make general conclusions about populations.

2. A population is the entire set of individuals (or their scores) of interest in a particular study.

A sample is a set of individuals (or their scores) selected from a population, usually intended to represent the population in a research study. A parameter is a characteristic, usually a numerical value, that describes a population. A statistic is a characteristic, usually numerical, that describes a sample.

3. In experimental research the researcher manipulates an independent variable to create the different treatment conditions or groups that will be compared in the study. Then, subjects are randomly assigned to the treatment conditions. In quasi-experimental research, the researcher simply uses a pre-existing (non-manipulated) variable to define the conditions or groups that will be compared.

4. The distinguishing characteristics of the experimental method are manipulation and control.

In the experimental method, the researcher manipulates one variable and observes a second variable. All other variables are controlled to prevent them from affecting the outcome.

5. Because a sample is only a part of the whole population, the characteristics of a sample tend to be different from the characteristics of the population. The discrepancy between a statistic for a sample and the corresponding parameter for the population is called sampling error.

6. Inan experiment, the researcher manipulates one variable and observes a second variable for changes. The manipulated variable is called the independent variable and the observed variable is the dependent variable.

7. The independent variable is whether or not the individuals were given aspirin and the dependent variable is whether or not they suffered a second heart attack.

8.

This is a quasi-experimental study. It is not an experiment because the researcher is not

manipulating a variable to create the groups. Instead, the study is comparing pre-existing groups (boys versus girls).

9. The independent variable is the loudness of the buzzer. The dependent variable is reaction time (how quickly subjects respond).

Chapter 1: Solutions - page 152

. The dependent variable is the number of colds each subject has during the winter. . discrete . Tatio scale (zero means none) Pp ange . experimental (amount of vitamin C is manipulated) 1

Oe The independent variable is the number of syllables (one/two/three) which is measured on

a ratio scale.

. The dependent variable is the number of words recalled by each subject, which is also measured on a ratio scale.

12. The independent variable is the intensity of the lighting and the dependent variable is the mood score measured for each subject.

13. a. The independent variable is the brand of medication, brand X versus brand Y, which is

measured on a nominal scale. b. The dependent variable is the degree of indigestion, which is measured on a ordinal scale; a series of ordered categories.

The independent variable is the color of the room, and the dependent variable is the score on the mood questionnaire. b. Room color is measured on a nominal scale.

14.2:

15. A discrete variable consists of separate, indivisible categories. A continuous variable is divisible into an infinite number of fractional parts.

16. a. Age and height are continuous; sex and eye color are discrete. > . Age and height would be measured on ratio scales; sex and eye color would be measured on nominal scales. nominal scale . interval scale ordinal scale . ratio scale

yy Ae Be 6 en | . (2X) = (13) = 169 ay # © es Neale aosp 19. a. LmA= 1)

by Gace |, c. YXY = 40

Chapter 1: Solutions - page 153

20.

ors

he EC

ey

. oX’- 10

. Sum the scores, then subtract 4 from the total.

eke

. Square each score, then sum the squared values.

OQ

. Add 4 points to each score, then square the resulting values. Finally, sum the squared values.

22.

pn | IX? = 49

oe x(X + 1 = 75 pay

DTGt)

pk 37 _ >(K +3) =12

24.

e e ae , LY =14 el

20;

. First, subtract 4 points from each score.

Third, sum the squared values. . (X - 4) = 34 . UX -4)=0

Second, square each of the resulting values.

Chapter 1: Solutions - page 154

Chapter 2 Frequency Distributions

Lo

Ee pe ‘La .20

13% 20% ms o 40% AS 15% RN WwW KNW NW op10 10% 2. A regular table reports the exact frequency for each category or score on the scale of measurement. After the scores have been grouped into class intervals, the table reports only the overall frequency for the interval but does not indicate how many scores are in each of the individual categories.

3. A bar graph leaves a space between adjacent bars and is used with data from nominal or ordinal scales. In a histogram, adjacent bars touch at the real limits. Histograms are used to display data from interval or ratio scales.

4. A grouped frequency distribution is used when the original scale of measurement consists of more scores or categories than can be listed simply in a regular table. Around 20 or more categories is generally considered too many for a regular table.

=F 6 5 4

6 = 4

aa

£3 2 1

2 1 X ES

Z

3

4

2

6

Xx 1

Chapter 2: Solutions - page 155

2

3

a

2

6

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Relaxed

Not sure

PNHWBODH~> DWC

X

il

Z

3

4

S

6

c. 1) The distribution is negatively skewed.

2) If X = 3 is normal, the majority of the students scores above normal.

== OO © wy BNA NWOK NON ONON NNN

Chapter 2: Solutions - page 156

160 oy On eS B95GS F

1

1) 2)

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10.

2

:

4

2)

6

f

8

wa)

The distribution is symmetrical, with half the class scoring high and half scoring low.

The class is sharply divided into two distinct groups. For some the quiz was easy and for some the quiz was hard.

12

3

, N= 14

b me >.4 = 53 BL:

2

N=Xf=13,

2X =38, LX? = 128

oa. Bae

ime Mage), Ga c . UX? = 139 rae oar — —

10

COeK POD WWN Or

Chapter 2: Solutions - page 157

4 OO Fe RPHMOW

PHOW ~4 POD

1

2

3

4

5

6

ou

8

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d. Positively skewed.

. Set I: grouped table with an interval width of 5 Set II: regular table Set III: grouped table with an interval width of 100 (7 intervals) or 50 (14 intervals)

jee a.

mee 30-31 1 28-29 0 26-27 2 24-25 1 22-23 4 20-21 5 18-19 2 16-17 3 14-15 1 12-13 1

b.

_X 30-34 25-29 20-24 15-19" 10-14

f =] 10 35 2

Chapter 2: Solutions - page 158

LOicds 85-89 80-84 75-79 70-74 65-69 60-64 55-59 50-54 45-49 40-44 1. oe

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b.

a 2 0 3 4 6 6 2 3 3 |

bax 80-89 70-79 60-69 50-59 40-49

2X 2 4 3 De i

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ia 2 7 12 5 4

Set II

Ht

x 14-15 12-13 LOstT* 8-9 6-7 4-5 2-3 0-1

1a 1 20 4 Z 1 2 1 1

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Set II 3 2 be

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c. Set I forms a symmetrical distribution, centered at X = 3, with most of the scores clustered close to the center. Set II forms a flat distribution with a center near X = 7 or X = 8 but the scores are spread evenly across the entire scale.

Chapter 2: Solutions - page 159

fP VW ON Hae RhOD NOW ee

S&S PNW

i

Z

3

=

2)

6

i

1) The distribution is negatively skewed.

2) The scores are centered around X = 4 or X = 5.

3) The scores are piled up around X = 5 or X = 6, but they are spread across the entire scale.

— \O

=

me —

ool

KS WH NY RK = DOWN —

Chapter 2: Solutions - page 160

c. 1) The distribution is roughly symmetrical with scores piled at each end of the scale.

2) The center is near X = 6.

3) The scores are spread across the entire scale.

ro

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800-899 700-799 600-699 500-599 400-499 300-399 200-299 100-199

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Chapter 2: Solutions - page 161

it

22. a. The independent variable is diet. The dependent variable is the number of errors on the discrimination task. b.

Special Reuular

Diet Dey

===

—=s

4 3 9

Z 1

c. Yes, the diet appears to have an effect. Error scores with the regular diet tend to be higher than scores with the special diet.

23.8; Imagery

5 4 cae 2 a os

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14

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16

ay

18

ug

20

10

as

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13

14

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5 d £3 2 i}

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b. For the imagery group, the graph is negatively skewed and is centered about X = 18. The graph for the no-imagery group is more symmetrical and is centered around X = 12. c. The imagery instructions seemed to produce higher memory scores.

Chapter 2: Solutions - page 162

24.

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Section

I

al

3

4 <a z i.

2

Section.

ii

1

3

4 ‘a Z 1

2

b. The scores in Section I are centered around X = 8 and form a negatively skewed distribution.

In Section II, the scores are lower (centered around X = 6) and the

distribution is more symmetrical with some tendency toward a positively skewed distribution.

Chapter 2: Solutions - page 163

Chapter 3 Central Tendency

1. The purpose of central tendency is to identify the center of a distribution, usually by determining a single score that serves as a representative for the entire distribution.

2. No single method always produces a good, representative value. Therefore, there are several methods for measuring central tendency, each with advantages and disadvantages under different circumstances.

3.

a. The mean is the balance point of a distribution because the sum of the distances above the mean is exactly equal to the sum of the distances below the mean. b. The median is the midpoint of a distribution because 50% of the scores are greater than or equal to the median and 50% are less than or equal to the median.

4.

The median is used instead of the mean when there is a skewed distribution (few extreme

scores), an open-ended distribution, or undetermined scores.

5.

The mean, median, and mode are identical for a symmetrical distribution with one mode.

6.

The mode is preferred when the scores are measured on a nominal scale.

7. With a skewed distribution, the extreme scores in the tail can displace the mean.

Specifically, the mean is displaced away from the main pile of scores and moved out toward the tail. The result is that the mean is often not a very representative value.

8.

The mean is 33/10 = 3.3, the median is 3.5, and the mode is 4.

9.

The mean is 50/10 = 5, the median is 5 and the mode is 5.

10. The mean is 45/14 = 3.21, the median is 3.5, and the mode is 4.

11. The mean is 116/20 = 5.8, the median is 5.5, and the mode is 5.

lay 2k = OG

Chapter 3: Solutions - page 164

13, ZX=— 105 14. The original 2X = 54 and the new 2X = 66. The new mean is 66/6 = 11.

15. The original 2X = 160 and the new ZX = 120. The new mean is 120/8 = 15.

16. The original 2X = 70 and the new ZX = 88. The new mean is 88/11 = 8.

17. The original

X = 30 and the new ZX = 28. The new mean is 28/4 = 7.

18. The original 2X = 80 and the new EX = 42. The score that was removed is X = 38.

19. The original 2X = 55 and the new ZX = 72. The score that was added is X = 17.

20. For the combined sample n = 10 and 2X = 32 + 72 = 104. The mean for the combined sample is 104/10 = 10.4.

21. For the combined sample n = 20 and 2X = 450 + 120 = 570. The mean for the combined sample is 570/20 = 28.5.

22. a. mode b. median (the distribution is skewed) c.

mean

23. Statement c cannot be correct. The mode is the score with the greatest frequency so there must be individuals with scores equal to the mode.

24. Statement b cannot be correct. By definition, the median always has 50% of the scores above and 50% of the scores below.

For set 1, the mean is 7, the median is 6.5, and the mode is 6.

Chapter 3: Solutions - page 165

Set

2

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6

7

8

OTTO

1

ee

Ss

ee

Ts

For set 2, the mean is 8.4, the median is 7.5 and the mode is 6.

c. The distribution for set 2 is more skewed. The greater the amount of skew, the bigger the differences between the three measures of central tendency.

26.

13 12 Average Hours of Sleep

11 10 9

8 7

i eeeSeals ee |

a

5

10 Age (in years)

15

20

a. Independent variable is the brand of coffee. Dependent variable is the flavor rating. b. nominal c. bar graph (coffee brand is a nominal scale) d.

Mean Ralang

oro Ori Gh (OTS) ss His Oo Ol OWE

A

B Brand

of

C Coffee

D

Chapter 3: Solutions - page 166

Chapter 4 Variability 1.

a. SS is the sum of squared deviation scores. b. Variance is the mean squared deviation. c. Standard deviation is the square root of the variance. standard distance from the mean.

It provides a measure of the

2. The standard deviation, o = 20 points, measures the standard distance between a score and the mean.

3. SS cannot be less than zero because it is computed by adding squared deviations. Squared deviations are always greater than or equal to zero.

4. A standard deviation of zero means that there is no variability. All the scores have exactly the same value.

5. An unbiased estimate means that on average the sample statistic will provide an accurate representation of the corresponding population parameter. More specifically, if you take all the possible samples and compute the variance for each sample, then the average of all the sample variances will be exactly equal to the population variance.

6. The variance, o”, can be found by squaring the standard deviation, o. For this example, o=4= 16.

7.

The standard deviation, s, can be found by taking the square root of the variance, s’. For

this example, s = Vs* = V100 = 10.

8. Fora 5-point quiz the scores range from X = 0 to X = 5. If the mean is 4.2, then X = 0 would be the score that is farthest from the mean and it is only 4.2 points away. If the greatest distance from the mean is only 4.2 points, it is impossible for the standard distance to be 11.4.

0:

. The sample variance is 60/5 = 12. b. The population variance is 60/6 = 10.

pak)

10. a. £X = 20 and the mean is S. b. X_ Deviation

8 43 4 «1 7 +2 —

c. The sum ofthe squared deviation is SS = 30.

Solutions: Chapter4

-

page

167

iM ie a. The definitional formula is easy to use when the mean is a whole number and there are relatively few scores. b. The computational formula is preferred when the mean is not a whole number.

SS = 9, the sample variance is 3 and the standard deviation is V3 = 1.73.

SS = 8, the sample variance is 1 and the standard deviation is 1.

SS = 54, the population variance is 9 and the standard deviation is 3.

SS = 36, the population variance is 9 and the standard deviation is 3. . a. No.

The score X = 48 is near the center of the distribution, within one standard deviation

of the mean. b. Yes. The score X = 48 is far out in the right-hand tail of the distribution.

Ly, a. Your score is 5 points above the mean. Ifthe standard deviation is 2, then your score is an extremely high value (out in the tail of the distribution). However, if the standard deviation is 10, then your score is only slightly above average. Thus, you would prefer OZ.

b. If you are located 5 points below the mean then the situation is reversed. A standard deviation of 2 gives you an extremely low score, but with a standard deviation of 10 you are only slightly below average. Here, you would prefer o = 10.

18. a. With o = 20, your sketch should show all four scores located near the center of the distribution (within one standard deviation of the mean). b. With o = 2, only one of the scores (X = 47) is located near the center of the distribution.

All of the other scores are extreme values.

19: a.

6 2) f 4 3 2 1 Xx

1

ve

3

a

o

6

q

b: tance = 7 c. interquartile range = (4.5 - 2.5) = 2 and the semi-interquartile range = 1

Solutions: Chapter4

-

page

168

20.

SS = 36, variance = 9, standard deviation = 3 SS = 36, variance = 9, standard deviation = 3 SS = 144, variance = 36, standard deviation = 6

oe . Adding a constant to each score does not change any of the deviations and does not change the standard deviation.

When each score is multiplied by a constant, the deviations and the standard deviation are also multiplied by the constant. al.

Standard deviation is 10.

. SS = 2400

. The range is 9 points, the semi-interquartile range is 3/2 = 1.5 points, SS = 86 and the sample standard deviation is 2.13. . Adding a constant to each score does not change the variability. After adding 2 points to each score, the range, the semi-interquartile range, and the standard deviation are not changed. . For the original scores, the range is 8, the interquartile range is 4, and the sample standard deviation is 2.45.

. After moving the extreme scores closer to the center, the range is reduced to 6 points, the interquartile range stays the same at 4 points, and the standard deviation is reduced to 2.07. . When the central scores are moved to the extremes, the range is still 8, the interquartile range increases to 6, and the standard deviation increases to 3.07.

Solutions: Chapter4

-

page

169

Chapter 5 z-Scores 1. A-z-score describes a precise location within a distribution. The sign of the z-score tells whether the location is above (+) or below (-) the mean, and the magnitude tells the distance from the mean in terms of the number of standard deviations.

2. The distribution of z-scores will have the same shape (positively skewed) with a mean of 0 and a standard deviation of 1.

3.

a. X= 80 is in the extreme right-hand tail at z = 2.00 b. X = 80 is slightly above the mean at z = 0.50.

4. Because the score X = 55 is above the mean, p = 45, it must have a positive z-score.

Therefore, z = -2.00 cannot be correct. a6

a

x Z 110 0.50 60 -2.00

x 140 105

b.

xX 150 105

id 2.50 0.25

xX Z 90 -0.50 80 -1.00

xX z 70 -1.50 Lise FOS

xX

Z

x

Zz

x

Z

34 1.00 28 -0.50

36 22

«61.50 -2.00

38 24

2.00 -1.50

xX Zz 33 DS 29 -0.25

x 36 26

Z =©1.50 -1.00

xX z 40 2.50 25. = b5

6a

b.

a

xX Z. 66 0.50 55 --0.42

So) ca

Ito. 97

Z 2.00 0.25

x 80 130.

xX Z 48 -1.00 70 0.83

x 84 To.

Z 2.00 4:25

x

Z

xX

z

61,2539 9=2.30 0.60 77 -040

55.

127

3-150 2.10

Z

Z -1.00 ~=—1.50

Chapter 5: Solutions - page 170

O° a:

x z tS 025 95°35 O75

x 90 60

0.50 -1.00

ae

Zz

xX

Z

L30n-2 $5

50 0.25

70 65

-0.50 -0.75

50 -1.50 100 1.00

xX Af 60

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x 35 55,

2 -1.43 1,43

xX Z 40 -0.71 42 -0.43

b.

10:

ll.

a. z=2.00 b. z=-1.00 C150 d. z=-3.00

12.

o=6

13.

o=6

14.

c=4

xX TOs 40

Z 1.50 -2.00

xX

Z

15. p=90 16. p=45

17. 1 =54 and o=8. standard deviations.

18.

The distance between the two scores is 12 points which is equal to 1.50

»=37 ando =6

19. a. With o = 2, the score X = 86 corresponds to z = 3.00 which is an extreme value. b. With o = 12, the score X = 86 corresponds to z = 0.50 which is a central value.

20. 7 = 5 fh,

= 1D

22. John’s z-score is 2.00 and Bill’s z-score is 1.00. John should expect the higher grade.

23. Mary’s statistics score corresponds to z = -0.25 and her biology score corresponds to z= -2.00. She should expect the higher grade in statistics.

Chapter 5: Solutions - page 171

24.

23.

26.

Original Score

Transformed Score

X = 68 eee) 2 aw er x= 50

Ges) X = 40 x= 70 De eS

Original Score

Transformed Score

X = 34 30 YC ave Wh X = 41 X = 46

X= 30 X= 90 Db) pS bs) X = 140

u=6ando=4 xX

Zi

I2 1S 10

50 22125 1.00

Zz x 3 -0.75 if 0.25 b -0.75

c. Original Score __Transformed Score

Gaeht ay x= 120 X= 85 a 105 x= 8)

Zi:

w=3ando=2 2 s.&c.

Original

X

z-score __Transformed X

-1.50 0.50 1.50 -1.00 0 0.50

35 55 65 40 50 ES,

z-score

Transformed X

=e 0.25 -0.25 -0.50 ber

50 62 58 56 74

u=Sando=4 om.& c. Original X 0 6

4 3 12

Chapter 5: Solutions - page 172

Chapter 6 Probability . p= 60/90 = 0.667 b. p= 15/90=0.167 . p= 5/90 = 0.056 tw

a. p = 40/50 = 0.80 b. p = 10/50 (Remember, a random sample requires replacement. )

Me

The two requirements for a random sample are (1) each individual has an equal chance of

being selected, and (2) if more than one individual is selected, the probabilities must stay constant for all selections.

4.

Sampling with replacement means that each individual selected for a sample is returned to the population before the next individual is selected. This sampling method is used to ensure that the probability of selecting any specific individual remains constant.

a:

. tail to the right, p = 0.1587 . tail to the left, p = 0.0228 . tail to the right, p = 0.0668 . tail to the left, p = .2266

. body to the left, p = 0.5987 . body to the right, p = 0.9332 . body to the left, p = 0.9599 . body to the right, p = 0.6554

. p(z> 1.50) = 0.0668 _ p(z> -2.00) = 0.9772 . p(z < 0.50) = 0.6915 . p(z <-0.75) = 0.2266 ast aed Pf pf aoe & aoe _ p= 0.5987 _ p= 0.0401 . p= 0.8159 . p= 0.1056 ® et ao z= 1.28 . z=0.25 . z= -0.84

Chapter 6: Solutions - page 173

10.

z= 0.84 z= 0.67 z=-1.04

2052 aos

11. a. p=0.0440 . p= 0.1498 p = 0.4772 . p= 0.5319 aes.

12. a. p=0.1974 . p= 0.3830 p = 0.6826 . p= 0:9500 aon L3,

Z=+1.28 Z=+1.44 z=+1.64 or +1.65 Z=+1.96 aorPp

14.

z= +0.39 z=+0.67 z=+1.04 oP a° .Z=+1.96

15. You cannot use the unit normal table to find proportions or z-scores because the distribution is not normal.

16. a. tail to the right, p = 0.4013 b. tail to the right, p = 0.2743

c. tail to the left, p = 0.3085 d. tail to the left, p = 0.2119

17. a. body to the left, p = 0.8944 b. body to the left, p = 0.7734 c. body to the right, p = 0.5987 d. body to the right, p = 0.8413

18. a.

z=2.00, p = .0228

b. z= -0.67, p = 0.2514 c.

Z= 0.67, p= 0.2514

Chapter 6: Solutions - page 174

19

Z= 1.50, p= 0.0668 . Z= 0.40, p = 0.6554 .

Z2= 0.84, X = 584 aoe . Z=-0.25, X = 475

20.

. X = 628 . X= 552 . X= 525

/2=-1.33.p=0.0918 = -1.67, p = 0.9525 . z= 1.00, p= 0.1587 2=(0.25,p=04013 Z = -0.25, p = 0.5987

z= 050, p = 0.6915 z= -0.50, p = 0.3085

z= 0.13, p= 0.4483 z = -0.87, p = 0.8078 z= 0.33, p = 0.6293 af oP z= -0.33, p = 0.3707 24.

z= -0.50, p = 0.6915

. Z=-1.50, p = 0.0668 22 OO p= 0.9772 . Z= 1.50 and z = -1.50, p = 0.8664 eonogp z=040 and z= 1.00, p= 0.1859

25.

p(z < 1.22) = 0.8888 A670 525 ~ e050) =02877 p(z > 2.11) = 0.0174 p(-1.33 < z< 1.44) = 0.8333 p(-0.33 <z< 0.33) = 0.2586

26.

X = 416 and X = 584 (z = 40.84) b. X= 3/200 X= 628 2 = 2128)

_ X = 304 and X = 696 (z = +1.96)

Chapter 6: Solutions - page 175

Chapter 7 The Distribution of Sample Means

1.

a. The distribution of sample means is the set of sample means for all the possible random samples of a specific size (n) from a specific population. b. The expected value of X is the mean of the distribution of sample means (11).

QO . The standard error of X is the standard deviation of the distribution of sample means

(og = o/Vn). . The expected value is 75. . The standard error is 10. . The expected value is 75. . The standard error is 4. pf aow

3. With n = 36 the distribution of sample means will be approximately normal even if the population is not. The expected value is p, = 65 and the standard error is 12/6 = 2.

4.

a. Standard error = 100/V4 = 50 b. Standard error = 100/V25 = 20 c. Standard error = 100/V100 = 10

5. Asample of n= 25 would be more accurate with a standard error of 3 points. A sample of n= 9 would have a standard error of 5 points.

6.

a. Whenn=1, the standard deviation, o = 6, measures the standard distance between X and UL.

b. With n= 9, og = 2 points. c. With n= 36, ox = 1 pomt.

7.

8.

a. z= b. z=

-1.00 +2.00

c. z=

+2.00

a a. standard error is 4, z = 0.50 b. standard error is 2, z= 1.00 c. standard error is 1, z = 2.00

Chapter 7: Solutions - page 176

. With a standard error of 5, X = 70 corresponds to z = 1.00 which is not extreme. . With a standard error of 2, X = 70 corresponds to z = 2.50 which is extreme.

10.

ET.

. Standard error = 8, z = 0.75, p = 0.2266 (more likely) . Standard error = 2.67, z= 1.12, p=0.1314

The distribution is normal with 1: = 100 and o = 4. . The z-score boundaries are +1.96.

The X boundaries are 92.16 and 107.84.

. X = 106 corresponds to z= 1.50. This is not in the extreme 5%.

. The distribution is normal with 1 = 80 and ox = 6. . == +1.00, p= 0.1587 . The distribution is normal with 1 = 80 and ox = 3. . Z=+2.00, p = 0.0228

. The distribution is normal with p = 55 and oz = 2. . Z=+2.00, p = 0.0228

z= +0.50,

p= 0.6915

oF mo . p(-1.00 < z< +1.00) = 0.6826 14.

. The distribution is normal with an expected value of 35 and a standard error of 4.

X = 32 corresponds to z = -0.75, p = 0.7734.

The distribution is normal with an expected value of 35 and a standard error of 2. pBos X = 32 corresponds to z = -1.50, p = 0.9332.

15.

The distribution is normal with an expected value of 80 and a standard error of 4.

X = 85 corresponds to z = 1.25, p = 0.1056 . The distribution is normal with an expected value of 80 and a standard error of 2. . X = 85 corresponds to z = 2.50, p = 0.0062 . The distribution is normal with p = 70 and o = 12. z= 0.25, p = 0.4013. . The distribution is normal with p = 70 and og = 3. z= 1.00, p = 0.1587.

Ly.

» Oe = 10,

z= 0.50, p= 0.3085

, C= 2,

Z= 2.50, p= 0.0062

b. og =5, z= 1.00, p= 0.1587 n>9 | n> 36

. No. When n= 1, the standard error is equal to the standard deviation. For any other sample size, the standard error will be smaller than the standard deviation.

19,

larger than n = 4 . larger than n = 25 larger than n = 100 Chapter 7: Solutions - page 177

20. a. Forn= 25, a; = 16 points. b. For n= 100, og = 8 points.

c. For n= 400, ox = 4 points. d. Increasing the sample size by a factor of 4 decreases the standard error by a factor of 2.

The standard error changes with the square root of n.

21.

a. The distribution of sample means is normal with p = 80 and ox = 3. b. z= 1.67, p = 0.0475 a

c. z= 1.00, p= 0.8413 d. z=-2.00, p = 0.0228 e. p(76 < X < 84) = p(-1.33 <z< 1.33) =.8164 22. a. With a standard error of 3.41 this sample mean corresponds to a z-score of z= 3.14. A z-score this extreme has a probability of only p = 0.0008. b. It would be almost impossible to obtain a sample mean this large by random sampling.

23. With n= 16, og = 0.50, p(X < 31) = p(z < -2.00) = .0228. This is a very unlikely outcome by chance alone. The inspector should suspect some problem with the machinery.

24. a.

~ 40 Mean

30

Anxiety Score

20

70

Control

Training

b. Even considering the standard error for each mean, there is no overlap between the two groups. The relaxation training does seem to have lowered scores even allowing for the standard error.

Chapter 7: Solutions - page 178

25.

Mean hours of sleep per night

AK 12 it 10

Gy {S =I7CO

ea aes4xSlaera ndvLA Ee il

5 Age

(in

Hi0) years)

ia

Chapter 7: Solutions - page 179

20

Chapter 8 Introduction to Hypothesis Testing 1.

measures the difference between the sample data and the hypothesized population a. Xmean. b. A sample mean is not expected to be identical to the population mean. The standard error indicates how much difference between X and i is expected by chance. %

to

:

a. A larger sample will produce a larger z-score (less error). b. A larger standard deviation will produce a smaller z-score (more error).

A larger difference will produce a larger z-score.

Ga

3. Lowering the alpha level causes the boundaries of the critical region to move farther out into the tails of the distribution. Lowering a reduces the probability of a Type I error. +

4. The advantage of using a smaller alpha is that it reduces the risk of aType L error. The disadvantage is that a smaller alpha requires more evidence to reject the null hypothesis. An extremely small alpha level may make the test so demanding that it is essentially impossible to obtain a treatment effect that is large enough to reject Hp. Thus, a very small alpha level can increase the risk of a Type II error. tan

a. A Type | error is rejecting a true Hy. This can occur if you obtain a very unusual sample with scores that are much different from the general population. b. A Type II error is failing to reject a false Hy. This can happen when the treatment effect is very small. In this case the treated sample is not noticeably different from the original population.

6.

a. A researcher can reduce the risk of a Type I error by lowering the alpha level for the hypothesis test. b. A researcher can reduce the standard error by taking a larger sample (increasing n).

7.

The analyses are contradictory.

The critical region for the two-tailed tests consists of the

extreme 2.5% in each tail of the distribution.

The two-tailed conclusion indicates that the data

were not in this critical region. However, the one-tailed test indicates that the data were in the extreme 1% of one tail. Data cannot be in the extreme 1% and at the same time fail to be in the ; extreme 2.5%.

Chapter 8: Solutions - page 180

. Independent variable is presence or absence of the hormone. Dependent variable is weight at 10 weeks of age. . The null hypothesis states that the growth hormone will have no effect on the weight of rats at 10 weeks of age. . Hy p= 950 Hy: p #950 . The critical region consists of z-score values beyond 1.96 or -1.96 in a normal distribution.

. For this sample oy = 6 and z = 4.00.

Reject the null hypothesis and conclude that the hormone has a significant effect on weight.

9,

. Hy p= 40. The sample mean X = 42 corresponds to z = 1.00. This is not sufficient to reject the null hypothesis. You cannot conclude that the program has a significant effect.

- Hy = 40. The sample mean X = 44 corresponds to z = 2.00. This is sufficient to reject the null hypothesis and conclude that the program does have a significant effect. . The 2-point difference between the sample and the hypothesis in part a was not sufficient to reject H, however, the 4-point difference in part b is significantly more than chance.

10,

. Hy = 70. With o = 20, the sample mean corresponds to z = 1.50. This is not sufficient to reject the null hypothesis. You cannot conclude that the course has a significant effect. . H, p= 70, With o = 10, the sample mean corresponds to z = 3.00. This is sufficient to reject the null hypothesis and conclude that the course does have a significant effect. . There is a 6-point difference between the sample and the hypothesis. In part a, the standard error is 4 points and the 6-point difference is not significant. However, in part b, the standard error is only 3 points and the 6-point difference is now significantly more than is expected by chance.

. The dependent variable is SAT score and the independent variable is whether they take the special course or not. . Hy, p= 500, Hy: p #500. The critical boundaries are z = +1.96. For these data, z = 2.16, # our decision is to reject H, and conclude that the course did affect SAT scores. . With

a = .01, the critical region consists of z-score values greater than 2.58 or less than

-2.58. . With a = .01, the decision is to fail to reject H, With a =.01 you have less risk of a Type I error. Therefore, the test requires a larger treatment effect in order to reject H,.

H,: p = 50. With a = .05 the critical boundaries are z = +1.96. For these data, oy = 3 and z = -2.33. Reject Hy. b. With « = .0] the critical boundaries are z = +2.58. The z-score of z = -2.33 is not in the critical region. Fail to reject Hy. As the alpha level decreases, the critical boundaries are moved farther out and it becomes more difficult to reject H,,.

Chapter 8: Solutions - page 181

is . Hy: p = 100. The critical boundaries are z = +1.96. With o = 12, the standard error 6x = 2 and z= 3.00. Reject Hp. b. With o = 30, oz = 5 and z = 1.20. Fail to reject Hp.

_ There is a 6-point difference between the sample mean and the hypothesized population value. In part a, the standard error is 2 points and the 6-point difference is significantly more than is expected by chance. However, in part b, the standard error is 5 points and the 6-point difference is not sufficient to reject the null hypothesis.

14.

. Hy: p= 70. The critical boundaries are z = +1.96. With n= 4 the standard error is ox = 5 and z= 1.00. Fail to reject Hp. _ With n = 25 the standard error is oz = 2 and z = 2.50. Reject Hp. . A larger sample produces a smaller standard error and increases the chances of rejecting the null hypothesis.

15. The null hypothesis states that the mean completion time for the MS population is not different from the mean time for the regular population; Hy: p = 185 seconds. With a = .01 the critical value is z = +2.58. For this sample mean, the standard error is 3.77 and z = 7.16.

Reject the null hypothesis and conclude that the MS population is significantly different.

16. The null hypothesis states that the mean completion time the brain-damaged population is not different from the mean time for the regular population; Hy: p = 92 seconds. With a = .01 the critical value is z = +2.58.

For this sample mean, the standard error is 4.92 and z = 4.67.

Reject the null hypothesis and conclude that the brain-damaged population is significantly different.

17, a. For n=25, oz = 2 and z= 1.50. Fail to reject the null hypothesis. There is no significant difference in vocabulary skills between only-children and the general population. b. For n= 100, og = 1 and z=3.00. Reject the null hypothesis. There is a significant difference in vocabulary skills between only-children and the general population. c. The larger sample results in a smaller standard error. With n= 100 the difference between the data and the null hypothesis is significant.

18. For this sample og = 0.32 and z = -4.06. Reject H) and conclude there has been a significant change in homework hours.

19. For the one-tailed test, the null hypothesis states that there is no memory impairment, H,: » > 50. The critical region consists of z-scores less than -2.33. For these data og = 1.28 and z = -2.34. Reject H, and conclude the alcoholics have significantly lower memory scores.

20. a. Hy: » = 50. The standard error is 2 and the sample mean corresponds to z = 2.25. Reject H, and conclude that people who buy red cars are significantly different from the general population. b. With a = .01 the critical value is

z= +2.58.

The calculated z-score of 2.25 is not in the

critical region. Fail to reject Hp.

Chapter 8: Solutions - page 182

. For a one-tailed test with a = .01 the critical value is z = 2.33. The calculated z-score

does not reach this standard. Fail to reject Hp.

ale

- Ho: p= 2.95. og = 0.11 and z= 2.73. The calculated z-score is beyond the critical boundary of 2.58, so we reject H, and conclude that students who sit in the front have significantly different grade point averages than the general population. . Ho: # < 2.95 (not higher). For the one-tailed test, the critical boundary is 2.33, so we reject H, and conclude that students who sit in the front have grade point averages that are significantly higher than the general population.

22.

- Ho: H= 65. og =4 and z= -1.25. Fail to reject Hy. b. Hy: p= 55. op =4 and 2= 1.25.0 Fail tolreject H,. . With a sample mean of X = 60 it is reasonable to expect that the population mean has a value near 60. Any value that is reasonably near to 60 will be an acceptable hypothesis for p. The standard error and the alpha level determine the range of acceptable values.

me

. The null hypothesis states that the training program has no effect on the problem solving scores (Ho: = 80). The critical region consists of z-score values beyond £1.96. For these data, the standard error is 2.36 and z = 1.88.

Fail to reject H,. There is not

sufficient evidence to conclude that the training program has any effect.

- For a one-tailed test the null hypothesis would state that training does not improve problem solving ability; Hj: p < 80. The critical region consists of z-score values greater than +1.65. For the one-tailed test, the z-score is in the critical region and the decision is to reject Hy. For a = .025, the null hypothesis would not be rejected.

24.

. Hy: pb = 26 (no decrease). The critical region consists of z-scores less than -2.33. For these data, the standard error is 1.07 and z = -2.47. Reject the null hypothesis and conclude that anxiety significantly reduces the number of anagrams solved. . Hy: = 26 (no difference). For a two-tailed test, the critical region consists of z-scores

greater than 2.58 or less than -2.58. The data produce z = -2.47, which is not in the critical region. Fail to reject the null hypothesis.

25. Ho: p< 12 (no increase during hot weather). H,: t1> 12 (there is an increase). The critical region consists of z-score values greater than +1.65. For these data, the standard error is 1.50, and z = 2.33 which is in the critical region, so we reject the null hypothesis and conclude that there is a significant increase in the mean number of hit players during hot weather.

26. Hy: = 55 (patients' scores are not different from the normal population). The critical region consists of z-scores greater than +2.58 or less than -2.58. For these data, X = 76.62, the standard error is 2.62, and z = 8.25. Reject the null hypothesis. Scores for depressed patients are significantly different from scores for normal individuals on this test.

Chapter 8: Solutions - page 183

Chapter 9 Introduction to the t Statistic

1. A z-score is used when the population standard deviation (or variance) is known. The t statistic is used when the population variance or standard deviation is unknown. You use the sample data to estimate the variance and the standard error.

2.

a s=10ands,=2. b. The standard deviation describes how the scores are spread around the mean and the standard error describes how much error, on average, is expected between the sample mean and the population mean.

3.

a. As variability increases, t becomes smaller (closer to zero.)

b. As sample size increases, the standard error decreases and the t value increases. c. The larger the difference between X and 1, the larger the t value.

4. The sample variance (s’) in the t formula is an estimated value and contributes to the variability.

5.

As the value of df increases, the t distribution becomes less variable (less spread out) and

more like a normal distribution. increases.

6.

a s=6ands;=3 b. s=6 and sg = 1.50

7.

a. X=6ands=4 b. sx = 2 points

8

a

For a = .05, the critical t values move toward +1.96 as df

X=6ands=6

b. sx =3 points

9.

a, X=Sands=4 b. sx = 2 points

10. a. For sample I, s> = 16 and

s=4. For sample II, s’ = 4 and s = 2.

b. For sample I, sg = 2. For sample II, sg = 1. c. Sample variance is one component ofthe estimated standard error. As variance increases, the standard error also increases.

Solutions: Chapter9

-

page 184

— —.

iY

. For sample I, s* = 9 and s = 3. For sample Il, s* = 9 and s =3.

a

. For sample J, sy = 1.50. For sample II, sy = 1.

c. Sample size is one component of the estimated standard error. As sample size increases, the estimated standard error decreases.

12. a. 6 points b. Sy = 2

c. For these data, t= 3. With df= 8 the critical value is t = +2.306. Reject H, and conclude that there is a significant effect.

13. a. With a = .05 and df= 24, the critical boundaries are t = +2.064.

For these data, the

estimated standard error is 2 and t = 2.50. Reject H,. b. With a = .01 and df = 24, the critical boundaries are t = +2.797.

For these data, the

estimated standard error is 2 and t = 2.50. Fail to reject Hy. c. The test with a = .01 involves less risk of a Type I error but also requires more evidence from the data. In this problem, the data are sufficient to reject H, at the .05 level, but are not sufficient to satisfy the criteria for a = .01.

14.

The null hypothesis states that the population mean is still $81, the same as last year. With

a = .05 and df= 15, the critical boundaries are t = +2.131.

For these data, the estimated standard

error is 5 and t = 0.80. Fail to reject Hy.

15. a. With df = 15 the critical boundaries are +2.947. For these data, the sample variance is 36, the estimated standard error is 1.50 and t = 4.00. Reject H,. b. For the one-tailed test, the critical boundary is 2.602. For the data, t = 4. Reject H, and concluded that hand-eye coordination is significantly better for those who are above average in mathematics performance.

16. Because df = 99 is not in the table, use df = 60 to obtain a critical boundary of 1.671.

The

sample variance is 4, the estimated standard error is 0.20, and t = -11.00. Reject H, 17. Because df= 35 is not in the table, use df = 30 to obtain critical boundaries of +2.042.

The

sample variance is 0.36, the estimated standard error is 0.10, and t = -2.00. Fail to reject Hy. 18. a. With n= 25 the critical boundaries are +2.064. For these data, the standard error is 2.00 and the t statistic is (24) = -2.00. Fail to reject Hp. bp. Because df = 399 is not in the table, use df = 120 to obtain critical boundaries of +1.98.

With n = 400 the standard error is 0.50 and the t statistic is t(399) = -8.00. Reject H,. c. The larger the sample, the smaller the standard error (denominator of the t-statistic). If other factors are held constant, a smaller standard error will produce a larger t statistic which is more likely to be significant.

Solutions: Chapter9

-

page 185

19. a. The critical boundaries are +2.131.

For these data, the estimated standard error is 0.50

and t = 6.00. Reject Hp. 5.00 b. The critical boundaries are +2.131. For these data, the estimated standard error is and t = 0.60. Fail to reject Ho. t c. The larger variance produces a larger estimated error which produces a smaller value for (closer to zero) and reduces the probability of a significant result.

20. Hy: 1 =21. Because df= 99 is not in the table, use df= 60 to obtain critical boundaries of 42.00. With a standard error of 0.50, t(99) = -4.60. Reject Hy and conclude that humidity has a significant effect on the rats’ eating behavior.

21. Hy: p = 40. Because df= 35 is not in the table, use df= 30 to obtain critical values of t = +2.042.

For these data, s? = 144, the standard error is 2.00, and t(35) =2.25.

Reject Hy and

conclude that depression for the elderly is significantly different from depression for the general population.

22. Hy: p < 20 (not more than chance). The critical region consists of t values greater than 1.860. For these data, X = 22.22, s’ = 17.44, the standard error is 1.39, and t(8) = 1.60. Fail to reject H). The performance for this sample is not significantly better than chance.

23. Hy: p = 14 (no change). With df= 4 the critical boundaries are +2.776. For these data, X = 12, s? = 2.5, the estimated standard error is 0.71, and t(4) = -2.82. Reject Hy and conclude that there has been a significant change in the mean age at which drinking begins.

24. Hy: p> 15. The critical region consists of tvalues less than -1.943. For these data, X = 11.71. s? = 12.24, the standard error is 1.32, and t(6) = -2.49. Reject H, and conclude righthemisphere brain damage does significantly reduce spatial skills.

Solutions: Chapter9

-

page 186

Chapter 10 The t Test for Two Independent Samples

1. An independent-measures study requires a separate sample for each of the treatments or populations being compared. An independent-measures t statistic is appropriate when a researcher has two samples and wants to use the sample mean difference to test hypotheses about the population mean difference.

2. The standard error for the independent measures t provides an estimate of the standard distance between a sample mean difference (X, - X,) and the population mean difference (1; - H). When the two samples come from the same population (when H, is true), the standard error indicates the standard amount of error (distance) between the two sample means.

3. The homogeneity of variance assumption specifies that 0? = 03 for the two populations from which the samples are obtained. If this assumption is violated, the t statistic can cause misleading conclusions for a hypothesis test.

4. As the difference between the two sample means increases, the value of the t statistic increases. As the variability of the scores increases, the value of t decreases (becomes closer to ZeTOo).

5.

a. The first sample has s* = 8 and the second has s* = 12. b. The pooled variance is 164/16 = 10.25.

6.

a. The first sample has s* = 7 and the second has s’ = 5. The pooled variance is 108/18 = 6 (halfway between). b. The first sample has s* = 7 and the second has s’ = 3. The pooled variance is 108/24 = 4.5.

7.

a. The first sample has a variance of 8, the second sample variance is 12, and the pooled variance is 10 (halfway between). b. The first sample has a variance of 4, the second sample variance is 16, and the pooled variance is 7.27 (closer to the variance for the larger sample).

8.

a. estimated standard error = 4 b. estimated standard error = 3

Chapter 10: Solutions - page 187

. The pooled variance is 120. b. The standard error is 4.00.

_ A mean difference of 8 would produce t = 8/4 = 2. With df= 28 the critical values are +2.048. Fail to reject Hp. _ A mean difference of 12 would produce t = 12/4 = 3. With df= 28 the critical values are +2.048. Reject Hy.

1a: The pooled variance is 90. b. The estimated standard error is 5.

A mean difference of 10 produces t = 2.00. With critical boundaries of +2.160, fail to reject Hy d. A mean difference of 15 produces t = 3.00. With critical boundaries of +2.160, reject Hy

Cc.

11. The null hypothesis states that there is no difference in the mother/child relationship between easy delivery and difficult delivery daughters, Hp: fy - 4 = 0. With df= 22 and a = .05, the critical region consists of t values beyond +2.074. For these data, the pooled variance is 96, the estimated standard error is 4, and t(22) = -1.25. Fail to reject Hp. The data are not sufficient to indicate a significant difference between the two birth conditions. } eae

For treatment 1, the variance is 84/3 = 28 and for treatment 2 the variance is 108/3 = 36.

The pooled variance is 192/6 = 32. b. The null hypothesis states that there is no difference between the two treatments, Ho: ft - Hy = 0. With df= 6 and a = .05, the critical region consists of tvalues beyond +2.447.

For these data, the estimated standard error is 4, and t(6) = 1.50.

Fail to reject

H,. The data are not sufficient to conclude that there is a significant difference between the two treatments. . For treatment 1, the variance is 420/15 = 28 and for treatment 2 the variance is 540/15 =

36. The pooled variance is 960/30 = 32. . The null hypothesis states that there is no difference between the two treatments, H): ) - Wy = 0. With df= 30 and a =.05 the critical region consists of tvalues beyond +2.042. For these data, the estimated standard error is 2, and t(30) = 3.00. Reject Hp. The data are sufficient to conclude that there is a significant difference between the two treatments. . The samples in this problem are much larger than the samples in problem 12. The larger samples produce a smaller standard error so that the sample mean difference is now significant. t4. 2. Cohen’s d = 1.06 5for both problem 12 and problem 13.

b. For problem 12, r = 0.27 (27%) and for problem 13, r° = 0.23 (23%). . The sample size is not a part of Cohen’s, but appears in the r° formula as part of the df value.

Chapter 10: Solutions - page 188

15. a. The null hypothesis states that there is no difference in the performance scores between the students with pets and those without pets, Hy: u, - u, = 0. With df= 23 and a = .05, the critical region consists of t values beyond +2.069. For these data, the pooled variance is 150, the estimated standard error is 5, and t(23) = 2.40. Reject Hy. The data are sufficient to conclude that there is a significant difference between the two groups. b. For these data, d = 0.98.

16. The null hypothesis states that personality has no effect on frustration, Hy: p, - w= 0. With a = .01, the critical region consists of t values beyond +2.977. The pooled variance is 100, the standard error is 5, and t(14) = 2.60. Fail to reject H, and conclude that there is no significant difference between the two personality groups.

17. The null hypothesis states that there is no difference in difficulty between the two problems, Ho: f) - H) = 0. With a = .01, the critical region consists of tvalues beyond +3.355. With the tacks in the box, X = 119.2 and SS = 2746.80. With the tacks and box separate, X = 43.2 and SS = 1066.80. The pooled variance is 476.7, the standard error is 13.81, and t(8) = 5.50. Reject Hy. The data are sufficient to indicate a significant difference between the two conditions.

18. a. The null hypothesis states that owning a pet has no effect on health, H,: p, - p, =0. The critical boundaries are t = +2.228. For the control group, X = 11.14 and SS = 56.86.

For the dog owners, X = 6.4 and SS = 17.2. The pooled variance is 7.41 and t(10) = 2.98. Reject Hy. The data show a significant difference between the dog owners and the control group.

b. For these data, r° = 0.47 (47%) 19. a. There is a total of 26 subjects in the two samples combined. b. With df= 24 and a = .05, the critical region consists of tvalues beyond +2.064. The t statistic is in the critical region. Reject H, and conclude that there is a significant difference. c. With df= 24 and a = .01, the critical region consists of tvalues beyond +2.797. The t statistic is not in the critical region. Fail to reject H, and conclude that there is no significant difference.

20. a. For treatment 1, the mean is X = 4 and for treatment 2, the mean is X = 11. b. For treatment 1, SS = 20 and for treatment 2, SS = 20. The pooled variance is 6.67 and

the estimated standard error is 1.83. c. The null hypothesis states that there is no difference between the two treatments, Hy: B, - po = 0. With df= 6 and a = .05, the critical region consists of tvalues beyond +2.447. For these data, t(6) = -3.83. Reject Hj. The data are sufficient to conclude that there is a significant difference between the two treatments.

Chapter 10: Solutions - page 189

ral

. For treatment 1, the mean is X = 4 and for treatment 2, the mean is X =11.

The means

are the same as in problem 20. . For treatment 1, SS = 96 and for treatment 2, SS = 100. The pooled variance is 32.67 and the estimated standard error is 4.04. The variances and the standard error are much

larger than they were in problem 20. . The null hypothesis states that there is no difference between the two treatments, Ho: 1) - b> = 0. With df= 6 and a = .05, the critical region consists of t values beyond +2.447. For these data, t(6) =-1.73. Fail to reject Hy). The data are not sufficient to conclude that there is a significant difference between the two treatments. . The difference between problem 20 and problem 21 is the magnitude of the sample variance. In problem 21, the large variance produces a large standard error with the result that the difference between sample means is not significant.

D2,

. For treatment 1, the mean is X = 4 and for treatment 2, the mean is X =6.

The means

are much closer together than they were in problem 20. . For treatment 1, SS = 20 and for treatment 2, SS = 20. The pooled variance is 6.67 and

the estimated standard error is 1.83. The variances and error are exactly the same as they were in problem 20. . The null hypothesis states that there is no difference between the two treatments, Hp: fy, - Wy = 0. With df= 6 and a = .05, the critical region consists of t values beyond +2.447. For these data, t(6) =-1.09. Fail to reject H). The data are not sufficient to conclude that there is a significant difference between the two treatments. . The difference between problem 20 and problem 22 is the magnitude of the difference between the sample means. In problem 20 there is a 7-point difference between means which is large enough to be significant. In this problem the sample mean difference is only 2 points which is not big enough to be considered statistically significant.

25%

. The null hypothesis states that feedback will have no effect on estimation of length, Hy: 1, - > = 0. With df= 6 and a = .05, the critical region consists of t values beyond +2.447. For the no-feedback group, X = 6 and SS = 6. For the feedback group, X = 3 with SS = 8. For these data, the pooled variance is 2.33, the estimated standard error is

1.08, and t(6) = 2.77. Reject Hy). The data are sufficient to conclude that there is a significant difference between the two feedback conditions. . For these data, Cohen’s d = 1.96 (a very large effect) and r° = 0.56 (56%).

Chapter 10: Solutions - page 190

Chapter 11 The t Test for Two Related Samples 1.

a. This is an independent measures experiment with two separate samples. b. This is repeated measures. The same sample is measured twice.

2. The primary advantage of a repeated-measures design over an independent-measures design is that the repeated-measures study eliminates variability due to individual differences. This usually results in smaller variability and produces a smaller standard error which increases the likelihood of a significant difference.

3. For a repeated-measures design the same subjects are used in both treatment conditions. Ina matched-subjects design, two different sets of subjects are used. However, in a matched-subjects design, each subject in one condition is matched, with respect to a specific variable, with a subject in the second condition so that the two separate samples are equivalent with respect to the matching variable. a

An independent-measures design would require two separate samples, each with 10 subjects, for a total of 20 subjects. b. A repeated-measures design would use the same sample of n = 10 subjects in both treatment conditions. . A matched-subjects design would require two separate samples with n = 10 in each, for a total of 20 subjects. . For sample 1 the scores pile up around D = 4 and the majority of the scores are spread over a range from -6 to 14. A value of zero is located near the center of the distribution.

For sample 2 the scores pile up around D = 12 and the majority of the scores are spread over a range from 2 to 22. A value of zero is an extreme value, not located near the center of the distribution. . Sample 2 is more likely to reject a hypothesis of 1, = 0. For this sample there is a clear difference between the center of the sample and the value of zero. . For sample | the scores pile up around D = 4 and the majority of the scores are spread over a range from -6 to 14. A value of zero is located near the center of the distribution.

For sample 2 the scores pile up around D = 4 but now the majority of the scores are contained between 2 and 6. A value of zero is an extreme value, clearly outside the center of the distribution. . Sample 2 is more likely to reject a hypothesis of 11, = 0. For this sample there is a clear difference between the center of the sample and the value of zero.

Chapter 11: Solutions - page 191

Ts.2 . The estimated standard error is 1.35. b . D =3 produces t = 2.22. The obtained t value is beyond the critical boundaries of

#2,131. Reject H,.

8.a _ With n = 4, the estimated standard error is 2 and t(3) = 2.50.

With critical boundaries of

+3,182, fail to reject Hy b _ With n = 16, the estimated standard error is 1 and t(15) = 5.00.

ora

13

With critical boundaries

teect a.

9. The null hypothesis says that the magnetic therapy has no effect, Hy: tp = 0. With df= 8 and a. = .01, the critical values are t = +3.355. For these data, the standard error is 4, and t(8) = 3.13.

Fail to reject H, and conclude that the magnetic therapy has no significant effect on chronic pain.

10. a. The null hypothesis says that the increased lighting has no effect, Hy: tp = 0. Because df = 35 is not in the table, use df = 30 to obtain critical values of t= +2.042. For these data,

the standard error is 1.50, and t(35) = 5.73. Reject Hy) and conclude that the extra lighting has a significant effect on mood. . With a mean difference of 8.6 and a standard deviation of 9, Cohen’s d = 0.96.

Hl. &

The null hypothesis says that the medication has no effect, Hy: 1p = 0. Because df= 35 is not in the table, use df = 30 to obtain critical values of t = +2.042.

For these data, the

standard error is 1.33, and t(35) = 18.05. Reject H, and conclude that the medication produces a significant change in reaction time.

. For these data, r° = 0.90 (90%). . The null hypothesis says that there is no increase in pain tolerance, Hy: Up < 0. For a onetailed test with a = .01 the critical region consists of t values greater than 2.602. For these data, s’ = 64, the standard error is 2, and t(15) = 5.25. Reject H) and conclude that there has been a significant increase in pain tolerance.

. For these data, r° = 0.65 (65%).

13. a. The null hypothesis says that the drug does not increase the pain threshold, Hy: tp < 9.

With a = .05 the one-tailed critical region consists of t values greater than 1.860. For these data, s’ = 0.16, the standard error is 0.13, and t(8) = 12.31. Reject H, and conclude that there has been a significant change.

b. For these data, r° = 0.95 (95%) 14. The null hypothesis says that stress has no effect on lymphocyte count, Hy: Up = 0. With a = .05 the critical region consists of t values beyond +2.093. For these data, the standard error is 0.018, and t(19) = 5.00. Reject H, and conclude that the data show a significant change in lymphocyte count.

Chapter 11: Solutions - page 192

12

tas

The difference scores are 9, 5, 1, and 13. D=7.

. SS = 80, sample variance is 26.67, and the estimated standard error is 2.58. . With df= 3 and a = .05, the critical values are t = +3.182. For these data, t = 2.71. Fail to reject Hy. There is not a significant treatment effect.

1G. a: The difference scores are 12, 6, 4, and 6.

D=7. The average difference is the same as in

problem 15. . SS = 36, sample variance is 12, and the estimated standard error is 1.73. The sample variance and the estimated standard error are smaller than in problem 15. . With df= 3 and a = .05, the critical values are t = +3.182.

For these data, t = 4.05.

Reject Hy. There is a significant treatment effect. . With smaller variance and smaller error, these data are more likely to reject H, than the data in problem 15.

Lp

na: For problem 15, Cohen’s d is 1.36 and for problem 16 d = 2.02.

The percentage of variance accounted for is r = .71 for problem 15 and r’ = .85 for problem 16. . In general, the smaller sample variance (problem 16) produces the larger effect size.

1S. 2a: The null hypothesis states that there is no difference between the two treatments,

Ho: H, - #, = 0. With a = .05, the critical region consists of t values beyond 42.306. The pooled variance is 35.5, the standard error is 3.77, and t(8) = -0.53. Fail to reject Hp. There is no significant difference between the two treatments. . For the repeated-measures design, Hy: tp = 0. With a = .05, the critical region consists of t values beyond +2.776. For these data, s = 1, the standard error is 0.45, and t(4) = 4.44. Reject H, and conclude that there is a significant difference between treatments. . The repeated-measures design removes the individual differences which produces a smaller variance and a smaller standard error.

The smaller standard error results in a

larger t statistic which is big enough to be significant.

19. The null hypothesis says that the oatmeal has no effect, Ho: pp = 0. With df= 8 and a = .05, the critical values are t = +2.306.

For these data, D = -7, SS = 288, the standard error is 2, and

t(8) = -3.50. Reject H, and conclude that the oatmeal has a significant effect on cholesterol.

20. The null hypothesis says that forming images has no effect on visual perception, H): Up = 0.

With df= 6 and a = .05, the critical values are t = +2.447.

For these data, D= -4, SS = 68, the

estimated standard error is 1.27, and t(6) = -3.15. Reject H, and conclude that the mental images do have a significant effect on visual perception.

21. The null hypothesis says that the treatment has no effect on agoraphobia, Hy: pp) = 0. With df= 6 and a = .05, the critical values are t = +2.447.

For these data, D = 8, SS = 238, the

estimated standard error is 2.38, and t(6) = 3.36. Reject Hy and conclude that the treatment does have a significant effect on agoraphobia.

Chapter 11: Solutions - page 193

22. The null hypothesis says that pets do not have an effect on mood, Hy: py = 0. With df= 5 and a = .05, the critical values are t = +2.571. For these data, D = 5, SS = 120, the estimated standard error is 2.00, and t(5) = 2.50. Fail to reject Hy and conclude that pets do not have a significant effect on mood.

23. The null hypothesis says that there is no difference between shots fired during versus between heart beats, Hp: , = 0. For these data, D = 2.83, SS = 34.83, s’ = 6.97, the standard

error is 1.08, and t(5) = 2.62. With a = .05 the critical region consists of tvalues beyond +2.571.

Reject H, and conclude that the timing of the shot has a significant effect on the marksmen's scores.

Chapter 11: Solutions - page 194

Chapter 12 Estimation

1.

The general purpose ofa hypothesis test is to determine whether or not a treatment effect exists. A hypothesis test always addresses a "yes/no" question. The purpose of estimation is to determine the size of the effect. Estimation addresses a "how much" question.

2. Ifa hypothesis test results in failure to reject Ho, the conclusion is that there is no significant treatment effect. In this case it would be pointless to attempt to determine the size of the effect. ae

ca . The larger the sample, the narrower the interval. b. The larger the sample standard deviation (s), the wider the interval. c . The higher the percentage of confidence, the wider the interval.

4.

a . Estimation (how much) and independent measures (two separate samples) b . Hypothesis test, repeated measures c . Estimation, repeated measures d . Hypothesis test, independent measures

x

2 . Use the sample mean, X = 7.4 for the point estimate. b . Using a single-sample t statistic with df = 24, the standard error is 0.60, the t score boundaries for 90% confidence are +1.711, and the interval extends from 6.373 to 8.427.

Ga . Using a single-sample t statistic with df= 15, the standard error is 5, the t score

boundaries for 80% confidence are 1.341, and the interval extends from 39.295 to 52-705. . For 90% confidence the t score boundaries are +1.753, and the interval extends from

31,23) 10 54,765. . For 95% confidence the t score boundaries are +2.131, and the interval extends from

35.345 to 56.655. . As the percentage of confidence increases, the width of the interval also increases.

Chapter 12: Solutions - page 195

7.

a. Using a single-sample t statistic with df= 3, the standard error is 3, the t score boundaries for 90% confidence are +2.353, and the interval extends from 26.941 to 41.059.

b. Using a single-sample t statistic with df= 15, the standard error is 1.5, the t score boundaries for 90% confidence are +1.753, and the interval extends from 31.370 to 36.630.

c. Using a single-sample t statistic with df = 35, the standard error is 1, the t score boundaries for 90% confidence are +1.697 (using df = 30 because 35 is not listed), and the interval extends from 32.303 to 35.697. d. The larger the sample size, the narrower the confidence interval.

8. The width of the interval depends on standard error which varies inversely with the square root of n. To decrease the interval width by a factor of 2, you must increase the sample size by a factor of 4, from n= 25 to n= 100.

. Use the sample mean difference, 3.6 seconds, for the point estimate. . Using an independent-measures t statistic with df = 18, the pooled variance is 5, the standard error is 1, the t score boundaries for 80% confidence are +1.330, and the interval extends from 2.27 to 4.93 seconds.

. Use the sample mean, 62 miles per hour, for the point estimate. . Using a single-sample t statistic with df = 99, the sample variance is 25 and the estimated standard error is 0.50. Using df = 60 because 99 is not listed, the t score boundaries for 95% confidence are +1.671, and the interval extends from 61.164 to 62.836 miles per

hour. . Use the sample mean difference, 3.5 miles per hour, for the point estimate.

. Using an independent-measures t statistic with df = 198, the pooled variance is 27.5 and the standard error is 0.74. Using df= 120 because 198 is not listed, the t score boundaries for 95% confidence are +1.980, and the interval extends from 2.035 to 4.965 miles per hour.

12.

. Use the sample mean difference, 5.7, for the point estimate. . Using a repeated-measures t statistic with df = 24, the sample variance is 36, the estimated standard error is 1.20, the t score boundaries for 95% confidence are +2.064, and the interval extends from 3.223 to 8.177.

. Yes, there is a significant change in errors from the first 15 minutes to the second 15 minutes.

A value of zero (no change) is not included in the 95% confidence interval and,

therefore, you can be 95% confident that a value of zero is not an acceptable value for the population mean difference.

Chapter 12: Solutions - page 196

ie ee Use the sample mean difference, 5.2, for the point estimate. . Using an independent-measures t statistic with df = 28, the pooled variance is 30, the estimated standard error is 2.00, the t score boundaries for 95% confidence are +2.048, and the interval extends from 1.104 to 9.296. . The t score boundaries for 99% confidence are +2.763, and the interval extends from

-0.326 to 10.726. . The change is significant for a = .05 because a mean difference of zero is outside the interval and therefore rejected with 95% confidence. However, a mean difference of zero is within the 99% interval and therefore is an acceptable value (fail to reject) with 99% confidence.

14. Use the sample mean difference, 14 points, for the point estimate. Using an independentmeasures t statistic with df= 28, the pooled variance is 120, the standard error is 4, the t score

boundaries for 90% confidence are +1.701, and the interval extends from 7.196 to 20.804.

15. Use the sample mean difference, 0.72, for the point estimate. Using a repeated-measures t statistic with df = 24, the standard error is 0.20, the t score boundaries for 95% confidence are +2.064, and the interval extends from 0.3072 to 1.1328.

16. -B. The sample variance, s> = 100, provides an unbiased estimate of the population variance. b. Use X = 43 to estimate up. e. Using t = 1.711, estimate pp: between 39.578 and 46.422.

ry.

X= 7.36ands =2.05 2. Use X = 7.36 as the point estimate for i. . With df= 10 and an estimated standard error of 0.43, the t score boundaries for 80% confidence are t = 1.372, and the interval extends from 6.77 and 7.950.

18

a. Use X = 95 for the boys.

With a standard error of 3 and using a range of t values from -1.341 to 1.341, the interval extends from 90.977 to 99.023. . Use X = 40 for the girls. With a standard error of 4 and using a range of t values from -1.397 to 1.397, the interval extends from 34.412 to 45.588.

. Use the sample mean difference, 55 millimeters, for the point estimate. Using an independent-measures t statistic with df= 23, the pooled variance is 144, the standard error is 5. the t score boundaries for 80% confidence are 1.319, and the interval extends from 48.405 to 61.595.

19. Use the sample mean difference, 5.5, for the point estimate.

Using a repeated-measures t

statistic with df = 15, the standard error is 2, the t score boundaries for 80% confidence are +].341, and the interval extends from 2.818 to 8.182.

ZA).

Use the sample mean difference, 12 points, for the point estimate.

D: Using an independent-measures t statistic with df = 23, the pooled variance is 150, the estimated standard error is 5, the t score boundaries for 95% confidence are +2.069, and the interval extends from 1.655 to 22.345.

Chapter 12: Solutions - page 197

cals

. Use the sample mean difference, 24, for the point estimate.

. Using a repeated-measures t statistic with df = 35, the estimated standard error is 1.33,

the t score boundaries for 95% confidence are +2.042 (using df = 30 because 35 is not listed), and the interval extends from 21.284 to 26.716.

22.

. The sample means are X = 11.14 for the control group and X = 6.4 for the dog owners.

Use the sample mean difference, 4.74, for the point estimate. . Using an independent-measures t statistic with df = 10, the pooled variance is 7.41, the estimated standard error is 1.59, the t score boundaries for 90% confidence are +1.812, and the interval extends from 1.859 to 7.621.

23s

Use the sample mean difference, 2.83, for the point estimate.

. Using a repeated-measures t statistic with df = 5, the estimated standard error is 1.08, the t score boundaries for 95% confidence are +2.571 and the interval extends from 0.053 to 5.607.

Chapter 12: Solutions - page 198

Chapter 13 Introduction to Analysis of Variance

1. When there is no treatment effect, the numerator and the denominator of the F-ratio are both measuring the same source of variability (differences due to chance/error). In this case, the F-ratio is balanced and should have a value near 1.00.

2. Both the F-ratio and the t statistic are comparing the actual mean difference between samples (numerator) with the difference that would be expected from chance factors such as experimental error (denominator). If the numerator is sufficiently bigger than the denominator, we conclude that there is a significant difference between treatments.

3.

a. As the differences between sample means increase, MS, crween also increases, and the Fratio increases. b. Increases in sample variance cause MS, jini,to increase and, thereby, decrease the F-ratio.

4. With three or more treatment conditions you need two or more t tests to evaluate all the mean differences. Each test involves a risk of a Type I error. The more tests you do the more risk there is of a Type I error. The ANOVA performs all of the tests simultaneously with a single, fixed alpha level.

5. Post tests are done after an ANOVA in which you reject the null hypothesis with three or more treatments. Post tests determine which treatments are significantly different.

6. Within each treatment all the scores are the same. There is no variability within treatments.

MS, nin Should be zero.

7 There is no difference between treatments. Both treatments have the same total, T = 9, and the same mean, X = 3. With no difference between treatments, MSyenveen = 0.

8.

SS Source 72 ts Between Treatmen 36 Within Treatments 108 Total

__df MS 2 36 4 9 11

F(2,9) = 9.00

With u = .05. the critical value is F = 4.26. Reject the null hypothesis and conclude that there are significant differences among the three treatments.

Solutions: Chapter 13

-

page 199

a

a. Reducing the differences between treatments should reduce the value Of MSyeeen ANd should lower the value of the F-ratio. la

Source Between Treatments Within Treatments Total

SS__df MS 32 2 16 36 «9S 68 11

F(2,9) = 4.00

With a = .05, the critical value is F = 4.26. Fail to reject the null hypothesis and conclude that there are no significant differences among the three treatments. Notice that the F-ratio is substantially smaller than it was in problem 8.

10.

a. Increasing the variance within treatments should also increase the value of MSiin and should lower the value of the F-ratio. b.

Source Between Treatments Within Treatments Total

SS 72 72 144

df 2 oD 11

MS 36 3

F(2,9) = 4.50

With a = .05, the critical value is F = 4.26. Reject the null hypothesis and conclude that there are significant differences among the three treatments. Although it is still significant, the F-ratio is substantially smaller than it was in problem 8.

We a. With dfremveen = 3, there must be k = 4 treatment conditions. b. Adding the two df values gives df,,,,, = 27. Therefore, the total number of subjects must be N = 28.

2,

SS

df

MS

Between Treatments

Source

66

3

22

Within Treatments Total

72 138

36 39

2

Source Between Treatments Within Treatments

SS 36 99

MS 18 3

F(2,33) = 6.00

Total

135

df 2 33 35

Source SS Between Treatments 90 Within Treatments 100 Total 190

df 3 20 23

MS 30 5

F(3,20) = 6.00

F(3,36) = 11.00

Solutions: Chapter 13

-

page 200

15:

For the German shepherds, X = 4 and s = 0.82. For the cocker spaniels, X=lands=

a.

1.15. For the pit bulls, X = 6 and s = 1.63.

Source Between Treatments Within Treatments Total

Soe 50.67 14.00 64.67

odie Mo 2 25.33 9 ~~ 1.56 11

F(2,9) = 16.24

With a = .05, the critical value is F = 4.26, Reject the null hypothesis and conclude that there are significant differences among the three breeds. . For these data, r° = 50.67/64.67 = 0.78 (78%) 16.

. For the German shepherds, X = 4 and s = 2.94. For the cocker spaniels, X = 1 and s = 2.

For the pit bulls, X = 6 and s = 3.37. The means are the same as in problem 15, but the

standard deviations are much larger.

Source

SS__df

Ms

Between Treatments

50.67

2

25.33

Within Treatments Total

72.00 122,67

9 9d

8.00

FS) = 3.17

With a = .05, the critical value is F = 4.26. Fail to reject the null hypothesis and conclude that there are no significant differences among the three breeds. . The increased sample variance produced a larger value for MS,,inin and lowered the Fratio.

_ For these data, r = 50.67/122.67 = 0.41 (41%). Although the means and mean differences are the same as in problem 15, the increase in the sample variances lowers the percentage of variance accounted for.

re,

a.

For the German shepherds, X =2 and s = 0.82. For the cocker spaniels, X = 1 and s = 1.15. For the pit bulls, X = 3 and s = 1.63. The standard deviations are the same as in problem 15, but the mean differences are much smaller.

Source Between Treatments Within Treatments Total

SS__df 8.00 2 14:00, 2.

22.00: 11

MSs 4.00 1,56

FQ,9) = 2:56

With a = .05, the critical value is F = 4.26. Fail to reject the null hypothesis and conclude that there are no significant differences among the three breeds.

_ The smaller differences between sample means produced a smaller value for MSyeeen and lowered the F-ratio. _ For these data, ° = 8/22 = .36 (36%).

Because the mean differences are smaller than in

problem 15, the percentage of variance accounted for is lower.

Solutions: Chapter 13

-

page 201

18.

SS

_df = MS

Between Treatments Within Treatments

Source

10 46

2 12

Total

56

14

5.00 3.83

F(2,12)=1.31

With a = .05, the critical value is F = 3.88. Fail to reject the null hypothesis and conclude that there are no significant differences among the three treatment environments. iB

Source Between Treatments Within Treatments Total

a.

SS 70 IG 86

df MSs 2 35.00 “2?

1/33 14

F(2,12) = 26.32

The critical value is F = 3.88. Reject H,. These data do provide evidence of differences among the three therapies. . For these data Tukey’s HSD is 1.94. According to this criterion, therapy A is significantly different from both B and C, but therapies B and C are not significantly different form each other.

20.

a.

The null hypothesis states that there are no differences among the age groups, Hp: , = Ho = Hy. The critical value for a = .05 is 5.14.

Source

SS

df MS

Between Treatments

26

2

13

Within Treatments Total

12-6: 38 8

2

F(2,6) = 6.50

Reject H, and conclude that there are significant differences among the three age groups. . Beginning with the largest difference between samples, the Scheffe’ comparisons and F-ratios are as follows: 5 versus 7-year-olds: F(2, 6) = 6.00 (significant) 6 versus 7-year-olds: F(2, 6) = 3.38 (not significant) No other differences are significant.

2s

. For treatments I and II only, the null hypothesis is, Hy: , = fy. The critical value for a = .05 is 5.99.

Source Between Treatments Within Treatments Total

SS__df 32 1 24 6 56 7

MS 32 4

F(1,6)=8.00

The obtained F-ratio is in the critical region, so we reject H) and conclude that there is a significant difference between the two treatments.

Solutions: Chapter 13

-

page 202

b. When the analysis includes all three treatments, the null hypothesis is, Hy: 11, = [) = Hs. The critical value for a = .05 is 4.26.

Source Between Treatments Within Treatments Total

Speeches 32 2 16 367 Peed 68 11

F(2,9)= 4.00

The obtained F-ratio is not in the critical region so we fail to reject Hy and conclude that there are no significant differences among the three treatments. c. When comparing only treatments J and II, there is a 4-point difference between means, X, = 1 and X, = 5. However, when the third treatment, with X, = 3, is added, the study

contains mean differences that are only 2 points (for example X, versus X,). Thus, when the third treatment is included, the average difference between means decreases and MS yerween AlSo decreases. As a result, the mean differences are no longer significant.

22. The null hypothesis states that there are no differences among the three treatments, Hp: #; = Wo = pH. The critical value for a = .05 is 3.88.

Source

SS_df

MS

Between Treatments Within Treatments

15 2 7.5 (F(2,12)= 1.50 DU ST. Total 75 14 Fail to reject Hp. No evidence of any significant differences.

23. The means and SS values are: Single Twin Triplet

X=8

X= 6

X=4

SS = 10 SS= 18 SS=14 The null hypothesis states that there are no differences in language development among the three groups, Hy: 1, = UW =. The critical value for a = .05 is 3.88. The analysis of variance produces,

Source Between Treatments Within Treatments Total

SS__df 40 2 42° 2 82 14

MS 20 3S

F(2,12)=5.71

Reject H, and conclude that there are significant differences in language development among the three groups.

Solutions: Chapter 13

-

page 203

Chapter 14 More Advanced Analysis of Variance Repeated-Measures and Two-Factor Designs

1. For an independent measures design, the variability within treatments is the appropriate error term. For repeated measures, however, you must subtract out variability due to individual differences from the variability within treatments to obtain a measure of error.

2.

df=2,

3.

df=3,27

4.

22

a k=4

b. n= 13 subjects JN

eo)

+

|

4

6. The null hypothesis states that there are no differences among the three treatments, Ho: by = Wy = uy. With df= 2, 6, the critical value is 5.14.

Source Between Treatments Within Treatments

Between Subjects Error Total

SS._df § 2 94 9

MS 4+

90

3

4+.

102.

6 = 0.67 11

F(2.6) = 6.00

Reject Hp. There are significant differences among the three treatments.

Chapter 14: Solutions - page 204

7. a. The null hypothesis states that there are no differences among the three treatments, Ho: Lb) = Wy) = py. With df= 2, 6, the critical value is 5.14.

Source SS df MS Between Treatments gS 62 4 F(2,6) = 0.27 Within Treatments 94 9 Between Subjects 4 3 Error 90 6 15 Total 102 11 Fail to reject Hy. There are no significant differences among the three treatments.

b. In problem 6 the individual differences were relatively large and consistent. When the individual differences were subtracted out, the error was greatly reduced.

8. a. First X =3, second X = 5, and third X = 1. b. The null hypothesis states that the position in the list has no effect on the number of errors, Hy: , = po = p,. The critical value is 5.14.

Source Between Treatments Within Treatments Between Subjects Error Total

SS 32 24 iz 12 56

df 2 9 3 6 11

MS 16 F(2,6) = 8.00

We

Reject H, and conclude that the number of errors differs significantly with the position in the list. Most errors are made in the middle of the list, with fewer errors at the beginning, and fewest errors on the last items in the list.

9. a.

The null hypothesis states that there are no differences between treatments, H,: u, =u) = ,. For an independent-measures design, the critical value is 3.88.

Source SS. df MS Between Treatments 40 2 20.00 F(2,12)=3.24 Within Treatments 74 12 6.17 Total 114.

14 b. The null hypothesis states that there are no differences between treatments, H,: , =~) = U,. For a repeated-measures design, the critical value is 4.46.

3 Source Between Treatments Within J reatments

SS 40 74

_df 2 12

Between Subjects Error

54 20

4 om

Total

114

14

MS 20.0

F(2,8) = 8.00

ee

c. The independent measures design includes all the individual differences in the error term (MS ,»,,). AS a result the F-ratio, F(2,12) = 3.24 is not significant. With a repeated measures design, the individual differences are removed and the result is a significant F-ratio, F(2,8) = 8.00, p < .05.

Chapter 14: Solutions - page 205

10. The null hypothsis states that the mean number of errors does not change across sessions, Ho: Wy; =H = Hy = Hy. The critical value is 4.76.

SS

df

MS

Between Treatments

Source

18

3

6.00

Within Treatments

127

as

F(3,6)=8.96

Between Subjects Ser 2 Error 4 6 0.67 Total 30) 1 Reject H, and conclude that there is a significant practice effect. bs

The means have all increased by 2 points compared to the values in problem 10, but the mean differences have not changed at all. MS, ween iS Still equal to 6.00, unchanged.

Dy MS..,.. is not changed from the value obtained in problem 10. The extra variability within treatments is all accounted for as variability between subjects which is subtracted out before MS,,,, is computed. a.

. The null hypothsis states that the mean number of errors does not change across sessions,

Ho: Hy = UW = Hy = Wy. The critical value is 4.76. The complete analysis for the new data is as follows: Source SS_df MS Between Treatments 18 3 6.00 F(3,6) = 8.96 Within Treatments 156 8 Between Subjects oe Error 4 6. 0.67 Total 174 11 Reject H, and conclude that there is a significant practice effect.

12. The null hypothesis states that there are no differences among the three food mixes, Hy: b; = bo = Hy. The critical value is 4.10.

Source Between Treatments Within Treatments

SS 28 40

df 2 15

MS 14 ~~ F(2,10) = 4.67

Between Subjects 10 5 Error a010> 23 Total 65 17 Reject H). There are significant differences among the three mixes.

3,

Source

SS

__df

MS

Between Treatments Within Treatments Between Subjects Error

200 500 140 360

4 45 9 36

50

Total

700

49

F(4,36)=5.00

10

Chapter 14: Solutions - page 206

14.

Source

SS

df

MS

Between Treatments Within Treatments Between Subjects Error Total

48 120 36 84 168

2 42 «14 28 44

24

F(2, 28) = 8.00

3

15. One hypothesis test evaluates mean differences among the rows and a second test evaluates mean differences among the columns. The test for an interaction evaluates the significance of any mean differences that are not explained by row and/or column differences.

16. a. The graph shows parallel lines with the means for A, consistently below the means for A).

There appear to be main effects for both factors, but no interaction. b. The graph shows crossing lines (the line for A, is horizontal and the line for A, slopes sharply). There is a definite interaction and there are main effects for both factors. c. The graph shows parallel, horizontal lines, with A, consistently above A,. There appears to be a main effect for factor A, but no effect for factor B and no interaction.

17. All F-ratios have the same error term, MS\ithins 2d therefore have the same df value for the denominator.

18. a. df=1,36 b. df= 1, 36 ec. df=1,36 19. a. 20

b. 0 c. 60 20. 2.

A2 10 8 6 Mean Score

Al 4

v7

Bl

B2

Chapter 14: Solutions - page 207

b. Overall the scores in A, are slightly lower than the scores in A,. There may be a small main effect for factor A. There appears to be a large main effect for factor B with scores in B, much larger than scores in B,. The lines are not parallel so there appears to be an interaction. c. The null hypotheses state that there is no mean difference between levels of factor A (H): La; = a>), no difference between levels of factor B (Hp: tp; = Hg), and no interaction.

All F-ratios have df= 1,16 and the critical value is F = 4.49.

Source Between Treatments A B AxB Within Treatments Total

SS__df 280 3 20 1 180 1 80 1 320 16 600 19

MS

20 180 80 20

F(1,16)=1.00 F(1,16) = 9.00 F(1,16) = 4.00

The A effect is not significant, but Factor B does have a significant effect at the .05 level.

The interaction does not reach significance at the .05 level.

21.

The null hypotheses state that there is no mean difference between levels of factor A (Ho: Lourer = noisy)» nO difference between levels of factor B (Ho: Upyrroverrt = Usxrrovert)» and no interaction. All F-ratios have df= 1,16 and the critical value is F = 4.49.

Source Between Treatments

A (DISTRACTION) B (PERSONALITY) AxB Within Treatments Total

SS 120

__df 3

MS

80 20

1 1

80 F(1,16)=16 20 = F(1,16)=4

20 80 200

1 16 19

20 = #F(1,16)=4 5

Distraction has a significant effect on performance. These results do not provide sufficient evidence to conclude that personality affects performance or that personality interacts with distraction.

Chapter 14: Solutions - page 208

22. a. The null hypotheses state that self-esteem has no effect (Hp: Lincy = Hrow), the audience has no effect (Ho: Uatone = Haupmnce)» and that there is no interaction. For df= 1, 20, the critical value is F = 4.35.

Source Between Treatments A (self-esteem) B (audience) Ae Within Treatments Total

SS 216 96 96 24 86 302

df MS 3 «61 96" OF(120) = 22.33 1 96 ~ F(1,20)=22.33 1 24 ~= F(1,20)= 5.58 20 43 23 The significant interaction indicates that the effect of the audience was different for high

self-esteem subjects than for low self-esteem subjects. Specifically, the audience produced an increase in errors for the low self-esteem and had little effect on high self-esteem.

23.

24.

Source Between Treatments A B AxB Within Treatments Total

So. 280 16 144 120 320 600

df 7 1 3 3 40 47

MS

Source Between Treatments A B AxB Within Treatments Total

SS 60 0 20 40 108 168

df 5 1 2 2 54 59

MS

16 48 40 8

F(1,40) = 2.00 (3,40) = 6.00 F(3,40)=5.00

O- F(1,54)= 0.00 10 = F(2,54)=5.00 20 F(2,54)= 10.00 2

Chapter 14: Solutions - page 209

Chapter 15 Correlation and Regression

1. A positive correlation indicates that X and Y change in the same direction: As X increases, Y also increases. A negative correlation indicates that X and Y tend to change in opposite directions: As X increases, Y decreases.

2. The numerical value of the Pearson correlation indicates how well the data points fit a straight line. A value of 1.00 (or -1.00) indicates a perfect linear fit and a value of zero indicates no linear trend.

3. The Pearson correlation measures the degree of linear relationship. The Spearman correlation measures the degree to which the relationship is monotonic or one directional independent of form.

4. For set I, SP = 10. For set II, SP = -11. eae

8 y values

e

6

°

4

r)

2

e

i

2

3 X

4 5 values

6

b. It appears to be a perfect positive correlation, r = 1.00 c. SS, = 10, SS, = 40, SP = 20, and r = 1.00.

Chapter 15: Solutions - page 210

b. Estimate a strong negative correlation, probably -.8 or -.9.

C SS, = 10, SS, = 40, SP =-18. r=-0.90.

values

@)

ul

Z

S 4 5 X values

6

7

8

Sa 8)

. It appears to be a strong positive correlation, about r = +.8 or +.9. . SS, = 64, SS, = 4, SP = 14, and r = +.875 . SS, = 64, SS, = 14, SP = -16, and r = -.535 . The correlation between X and Y is r = +0.60. b. The correlation between Y and Z is r = +0.60.

Based on the answers to part a and part b you might expect a fairly strong positive correlation between X and Z. . The correlation between X and Z is r = -.20. . Simply because two variables are both related to a third variable does not necessarily imply that they are related to each other.

Chapter 15: Solutions - page 211

. SS, = 52, SS, = 54, SP = -47, and r = -.887. . The correlation is still r = -.887.

Adding a constant does not change SS or SP or the correlation. . The correlation is still r = -.887. oS oma . Multiplying each score by 3 will cause all the SS and SP values to be multiplied by 9.

However, this multiplication cancels out in the correlation formula, so the correlation is unchanged.

10. a. For these data, SS,i.= 211.50, SS, and = 137.875, SP = 155.25, and the Pearson correlation is r = 0.909. b. With df= 6, the correlation is significant at the .05 level.

11. For these data, SS,,, = 64, SS,,4 = 100, SP = 76, and the Pearson correlation is r = 0.95.

12. With n= 25 the correlation must be greater than 0.396 to be significant at the .05 level.

Reject H, and conclude that these data provide evidence for a non-zero correlation in the population.

13. a. The null hypothesis states that there is no correlation in the population. With n= 18, the correlation must be greater than 0.468 to be significant at the .05 level. Fail to reject Hp. These data do not provide evidence for a significant correlation. b. With n= 102, the critical value is 0.195. Reject H, and conclude that this sample provides enough evidence to conclude that a significant, non-zero correlation exists in the population.

14. Converting the number of courses to ranks produces SS, = 42, SS, = 40, SP = -27, and tz = -0,059, Lode = 0 b. Yes, there is a strong positive relationship between the grades assigned by the two instructors.

16

a. The graph shows an accelerating increase in Y as X increases. b. The Spearman correlation is r, = +1.00. The relationship is perfectly monotonic.

Chapter 15: Solutions - page 212

17. a. SS, = 24, SS, = 252, SP = -56, andr=-.72 b. The ranks are as follows: Hours _ Errors l 2 3 a ao Pe) 4 ips ho uy pS The Spearman correlation is rs = -.58 using the Pearson formula on the ranks, or rg = -.56 using the special Spearman formula. c. After ranking, the extreme individual (0, 19) is moved back into the rest of the group producing a moderate negative correlation. Ranking the scores reduces the influence of any extreme scores.

- SSgr = 3023.88, SS ois = 89.88, SP = -402.88, and the Pearson correlation is r = -0.773. . There is a consistent, negative relation between reaction time and errors. As reaction time gets faster (shorter time), errors increase.

LY,

V=2x44

20. SS, = 20, SS, = 24, SP = -20. The regression equation is Y = (-1)X + 10

Chapter 15: Solutions - page 213

21. a, Y=3x-3 b. For each X, the predicted Y value would be,

), oe 6 ] 4 3 2 5 3 22

0 9 6 3 12 6

SS, = 10, SS, = 67.2, SP = 24,

r=+0.926.

The regression equation is, Y =2.4X + 6.8.

23. a. SSsumer = 14, SSiq = 996, SP = -97, r = -0.82. b. The regression equation is Y = -6.93X + 127.82.

24. SS, = 1385.67, SS, = 1586, SP = 920, and r= +0.621, which is smaller than the critical

value of 0.708. These data are not sufficient to conclude that there is a significant correlation between the first exam and the final grades with a = .01.

Chapter 15: Solutions - page 214

Chapter 16 The Chi-Square Statistic Tests for Goodness of Fit and Independence

1. Nonparametric tests make few if any assumptions about the populations from which the data are obtained.

For example, the populations do not need to form normal distributions, nor is it

required that different populations in the same study have equal variances (homogeneity of variance assumption). Parametric tests require data measured on an interval or ratio scale. For nonparametric tests, any scale of measurement is acceptable.

2. The null hypothesis states that there are no preferences among the three brands.

With df= 2,

the critical value is 5.99. The expected frequencies are f, = 30 for all categories, and chi-square = 3.47. Fail to reject H, and conclude that there is no significant preference.

3. Hp states that the distribution of automobile accidents is the same as the distribution of registered drivers: 16% under age 20, 28% age 20 to 29, and 56% age 30 or older. With df= 2, the critical value is 5.99. The expected frequencies for these three categories are 48, 84, and 168. Chi-square = 13.76. Reject Hy and conclude that the distribution of automobile accidents is not identical to the distribution of registered drivers.

4. H, states that the distribution of jury member ages is the same as the distribution of ages for registered voters:

20% between 18 and 29, 45% between 30 and 49, and 35% age 50 or older.

With df= 2, the critical value is 5.99. The expected frequencies for these three categories are 16, 36, and 28, Chi-square = 1.57. Fail to reject H) and conclude that the distribution of ages for jurors is not significantly different from the distribution for registered voters.

5. The null hypothesis states that there are no preferences among the four colors; p = 1/4 for each color.

With df= 3, the critical value is 7.81.

The expected frequencies are f, = 15 for all

categories, and chi-square = 5.73. Fail to reject Hy and conclude that there are no significant preferences.

6. The null hypothesis states that there are no preferences among the three factors; p = 1/3 for each. With df= 2, the critical value is 5.99. The expected frequencies are f, = 20 for all categories, and chi-square = 11.2. Reject H, and conclude that there are significant differences in the perceived importance for the three factors.

Chapter 16: Solutions - page 215

7. The null hypothesis states that there is no preference among the three photographs; p = 1/3 for all categories. The expected frequencies are f, = 50 for all categories, and chi-square = 20.28.

With df = 2, the critical value is 5.99. Reject H, and conclude that there are significant preferences.

8. H, states that the distribution of flu-related deaths is the same as the distribution of the population:

30% under 30, 40% for 30 to 60, and 30% for over 60. With df= 2, the critical

value is 5.99. The expected frequencies for these three categories are 15, 20, and 15.

Chi-square = 59.58. Reject H, and conclude that the distribution of flu-related deaths is not identical to the population distribution.

9. The null hypothesis states that the grade distribution for last semester has the same proportions as it did in 1985. For a sample of n = 200, the expected frequencies are 28, 52, 62, 38, and 20 for grades of A, B, C, D, and F, respectively. With df= 4, the critical value for chi-

square is 9.49. For these data, the chi-square statistic is 6.68. Fail to reject H, and conclude that there is no evidence that the distribution has changed.

10. The null hypothesis states that there is no preference among the three colas; one third of the population prefers each of the three colas. For a sample of n= 60, the expected frequency is 20 for each cola. For df= 2 and a = .05, the critical value for chi-square is 5.99. For these data, the

chi-square statistic is 5.20. Fail to reject H, and conclude that there is no significant preference among the colas.

11. a. The null hypothesis states that there is no preference among the three colas; one third of the population prefers each of the three colas. For a sample of n = 120, the expected frequency is 40 for each cola. For df= 2 and a = .05, the critical value for chi-square is 5.99. For these data, the chi-square statistic is 10.40. Reject H, and conclude that there are significant preferences among the colas. b. A larger sample should be more representative of the population. If the sample continues to be different from the hypothesis as the sample size increases, eventually the difference will be significant.

12. The null hypothesis states that there is no relationship between a person’s age and his/her preference for colas.

For df= 1 and a = .05, the critical value for chi-square is 3.84.

The

expected frequencies are Brand A

Brand B

Over 30

Under 30

For these data, chi-square = 10.21. Reject H) and conclude that there is a significant relationship.

Chapter 16: Solutions - page 216

13. The null hypothesis states that there are no preferences among the three books; p = 1/3 for

all categories. With df= 2, the critical value is 5.99. The expected frequencies are f, = 40 for all categories, and chi-square = 7.85. Reject H, and conclude that there are significant preferences.

14.

The null hypothesis states that the distribution of preferences is the same for both groups of

students (the two variables are independent).

With df= 2, the critical value is 5.99. The

expected frequencies are Book 1

Book2

Book3

ove Ha Chi-square = 17.32. Reject H).

15. The null hypothesis states that there is no relation between dream content and gender; the distribution of aggression content should be the same for males and females. The critical value is 9.21. The expected frequencies are Low Medium High

The chi-square statistic is 25.52. Reject H) with a = .01 and df= 2.

16. The null hypothesis states that littering is independent of the amount of litter already on the ground. With df= 2, the critical value is 5.99. The expected frequencies are Amount of Litter None Small Large

Litter

a1.25

i) ees)

eyee)

Not Litter

$7.79

$7.75

88.49

Chi-square = 22.43.

Reject Hp.

Chapter 16: Solutions - page 217

17. The null hypothesis states that the distribution of opinions is the same for those who live in the city and those who live in the suburbs.

For df= 1 and a = .05, the critical value for chi-

square is 3.84. The expected frequencies are Favor

Oppose

City

51.33 | 48.67

Suburb

| 102.67 | 97.33

For these data, chi-square = 16.68. Reject H, and conclude that opinions in the city are different from those in the suburbs.

18. The null hypothesis states that there is no relation between the level of satisfaction and attending the e-mail training. With df= 3 and a = .05, the critical value is 7.81. The expected frequencies are Very Somewhat Somewhat Very Satisfied

Satisfied

Dissatisfied

Dissatisfied

Training No Training

Chi-square = 25.33. Reject H, and conclude that there is a relationship between training and satisfaction.

19. The null hypothesis states that there is no relationship between group size and helping behavior. With df= 2, the critical value is 9.21. The expected frequencies are None

Small

Lar pe

Help

Not Help

Chi-square = 7.91. Fail to reject Hy and conclude that there is no relationship between helping and group size.

20. The null hypothesis states that there is no relation between personality and heart disease.

With df= 1, the critical value is 3.84. The expected frequencies are No Heart Disease

Chi-square = 42.29. personality.

Heart Disease

Type A

90.34

| 79.66

Type B

95.66

84.34

Reject Hy.and conclude that heart disease is significantly related to

Chapter 16: Solutions - page 218

21. The null hypothesis states that there is no relation between handedness and eye preference.

With df= 1 and a = .01 the critical value is 6.63. The expected frequencies for left handed subjects are 12 left eye and 18 right eye. For right handed subjects the expected frequencies are 48 for left eye and 72 for right eye. Chi-square = 11.11. There is a significant relation between hand and eye preference.

22. The null hypothesis states that the salaries for males and females are distributed evenly around a common median.

expected frequencies are

For df= 1 and a = .05, the critical value for chi-square is 3.84. The Female

Male

For these data, chi-square = 12.00. Reject Hy and conclude that the males and females do not share a common median for salary.

Chapter 16: Solutions - page 219

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