A Student Solutions Manual for First Course in Statistics [10 ed.] 0136045197, 9780136045199

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SrUDENT ANUAL S’••■> TLi-.-'j„:y / h".'

Nancy S. Boudreau

H

A FIRST COURSE IN

riSTi t

TENTH EDITION

STUDENT SOLUTIONS MANUAL Nancy S. Boudreau Bowling Green Stole University

A FIRST COURSE IN

STATISTICS McClave I Sincich PEARSON Prentice HaU Upper Saddle River, NJ 07458

TENTH EDITION

Editor-in-Chief, Mathematics & Statistics: Deirdre Lynch Print Supplement Editor: Joanne Wendelken Senior Managing Editor: Linda Mihatov Behrens Project Manager, Production: Kristy S. Mosch Art Director: Heather Scott Supplement Cover Manager: Paul Gourhan Supplement Cover Designer: Victoria Colotta Senior Operations Supervisor: Diane Peirano

PEARSON Prentice Hall

© 2009 Pearson Education, Inc. Pearson Prentice Hall Pearson Education, Inc. Upper Saddle River, NJ 07458

All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher. Pearson Prentice HalH“ is a trademark of Pearson Education, Inc. The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. This work is protected by United States copyright laws and is provided solely for teaching courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. Printed in the United States of America 10

98765432

1

ISBN-13:

978-0-13-604519-9

ISBN-10:

0-13-604519-7

Pearson Education Ltd., London Pearson Education Singapore, Pte. Ltd. Pearson Education Canada, Inc. Pearson Education—Japan Pearson Education Australia PTY, Limited Pearson Education North Asia, Ltd., Hong Kong Pearson Educacion de Mexico, S.A. de C.V. Pearson Education Malaysia, Pte. Ltd. Pearson Education Upper Saddle River, New Jersey

Contents

Preface

v

Chapter 1

Statistics, Data, and Statistical Thinking

1

Chapter 2

Methods for Describing Sets of Data

6

Chapters

Probability

59

Chapter 4

Random Variables and Probability Distributions

85

Chapter 5

Inferences Based on a Single Sample: Estimation with Confidence Intervals

132

Chapter 6

Inferences Based on a Single Sample: Tests of Hypothesis

157

Chapter?

Comparing Population Means

185

Chapter 8

Comparing Population Proportions

231

Chapter 9

Simple Linear Regression

279

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Preface

This solutions manual is designed to accompany the text, A First Course in Statistics, Tenth Edition, by James T. McClave and Terry Sincich. It provides answers to most odd-numbered exercises for each chapter in the text. Other methods of solution may also be appropriate; however, the author has presented one that she believes to be most instructive to the beginning Statistics student. The student should first attempt to solve the assigned exercises without help from this manual. Then, if unsuccesslul, the solution in the manual will clarify points necessary to the solution. The student who successfully solves an exercise should still refer to the manual's solution. Many points are clarified and expanded upon to provide maximum insight into and benefit from each exercise. Instructors will also benefit from the use of this manual. It will save time in preparing presentations of the solutions and possibly provide another point of view regarding their meaning, Some of the exercises are subjective in nature and thus omitted from the Answer Key at the end of A First Course in Statistics, Tenth Edition. The subjective decisions regarding these exercises have been made and are explained by the author. Solutions based on these decisions are presented; the solution to this type of exercise is often most instmctive. When an alternative interpretation of an exercise may occur, the author has often addressed it and given justification for the approach taken. I would like to thank Kelly Barber for creating the art work and her assistance typing this work.

Nancy S. Boudreau Bowling Green State University Bowling Green, Ohio

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(r)P(7)P(DP(7)P(7)=.5(.5)(.5)(.5)(.5)(.5)(.5)(.5)(.5)(.5) = .00098 Thus, any of these 3 sequences are equally likely. In fact, any specific sequence has the same probability. Now, suppose we compute the probability of getting all heads or all tails: P(A11 heads or all tails) = /’(All heads u All tails) = P(A11 heads) + P(A11 tails) - P(A11 heads n All tails) = .00098 + .00098 - 0 = .00196 Consequently, the probability of getting a mix of heads and tails is: /•(lO coin tosses result in a mix of heads and tails) = 1 - P(A11 heads or all tails) = 1 - .00196 = .99804 So, even though any particular sequence has a probability of .00098, we just found out the probability of getting all heads or all tails is only .00196 and the probability of getting a mix of heads and tails is .99804. If we know that one of these sequences actually occurred, then we would conclude that it was probably the one with the mix of heads and tails because the probability of a mix is close to 1.

84

Chapter 3

Random Variables and Probability Distributions

Chapter

4.1

A random variable is a rule that assigns one and only one value to each simple event of an experiment.

4.3

a.

Since we can count the number of words spelled correctly, the random variable is discrete.

b.

Since we can assume values in an interval, the amount of water flowing through the Hoover Dam in a day is continuous.

c.

Since time is measured on an interval, the random variable is continuous.

d.

Since we can count the number of bacteria per cubic centimeter in drinking water, this random variable is discrete.

e.

Since we cannot count the amount of carbon monoxide produced per gallon of unleaded gas, this random variable is continuous.

f

Since weight is measured on an interval, weight is a continuous random variable.

a.

The reaction time difference is continuous because it lies within an interval.

b.

Since we can count the number of violent crimes, this random variable is discrete.

c.

Since we can count the number of near misses in a month, this variable is discrete.

d.

Since we can count the number of winners each week, this variable is discrete.

e.

Since we can count the number of free throws made per game by a basketball team, this variable is discrete.

f

Since distance traveled by a school bus lies in some interval, this is a continuous random variable.

4.5

4.7

The values x can assume are 0, 1, 2, 3,4,.... Thus, x is a discrete random variable.

4.13

The probability distribution of a discrete random variable can be represented by a table, graph, or formula that specifies the probability associated with each possible value the random variable can assume.

Random Variables and Probability Distributions

85

4.15

The expected value of a random variable, E(x), does not always equal a specific value of the random variable, x. The E(x) is the mean of the probability distribution and does not have to equal a specific value of x.

4.17

a.

We know ^ p(x) = 1. Thus, p{2) +p(3)+p(5)+ p(8) + p( 10) = 1 i

^p(5) = 1 -pi2) -pi3) -p(S) -p(\0) = 1 - .15 - .10 -.25 - .25 = .25

4.19

b.

P(A: = 2orx= 10) = /’(JC = 2) + P(x= 10) = .15 + .25 = .40

c.

P(x < 8) = P(jc = 2) + P(x = 3) + P(jc = 5) +

a.

P(x 12)= 1-P(;c< 12)= 1-.7 = .3

c.

P(x < 14) = P(jc = 10) + P(x = 11) + P(x = 12) + P(;c = 13) + Pix = 14)

= 8) = .15 + .10 + .25 + .25 = .75

= .2 + .3 + .2 + .l + .2= 1

4.21

d.

P(x=14) = .2

e.

P(jc< 11 orx> 12) = P(x< ll) + P(jc> 12) = P{x = 10) + P(x = 11) + P(jc = 13) + P(x = 14) = .2 + .3 + .l + .2 = .8

a.

p = E(x) = y^xpjx) = 1(.2) + 2(.4) + 4(.2) + 10(.2) = .2 + .8 + .8 + 2 = 3.8

b.

o^ = E[{x - //)^] = ^ (x - pf p{x) = (1 - 3.8)^(.2) + (2 - 3.8)2(.4) + (4 - 3.8)^(.2) + (10-3.8)^(.2) = 1.568 + 1.296 + .008 + 7.688 = 10.56

4.23

c.

(7= VlO.59 =3.2496

d.

The average value of x over many trials is 3.8.

e.

No. The random variable can only take on values 1, 2, 4, or 10.

f.

Yes. It is possible that p can be equal to an actual value of x.

a.

It would seem that the mean of both would be 1 since they both are symmetric distributions centered at 1.

b.

P{x) seems more variable since there appears to be greater probability for the two extreme values of 0 and 2 than there is in the distribution of;^.

86

Chapter 4

c.

Forx:

/i = Eix) = J]xp(x) = 0(.3) + 1(.4) + 2(3) = 0 + .4 + .6 = 1

(/ = £[(x - /.if] = ^ (x - /if p(x) = (0 - 1)^.3) + (1 - \f(A) + (2 - 1)'(.3) = .3 + 0 + .3 = .6 For>^:

/i = E(y)^

= 0(.l) + 1(.8) + 2(.l) = 0 + .8 + .2 = 1

o^ = E[(y-/if]=^Y. 10) = P(jc= 11)=.274.

Suppose we define the following events:

P\ {Student scores perfect 1600} N'. (Student does not score perfect 1600} From the text, P{P) = 5/10,000 = .0005. Thus, P(7V) = 1 - P{P) = 1 -.0005 = .9995 a.

If three students are randomly selected, the possible outcomes for the 3 students are as follows:

PPP PPN PNP NPP PNN NPN NNP NNN Each of the sample points listed is really the intersection of independent events. Thus, P(PPP) = P(PnPnP) = P(P)P(P)P(P) = .0005(.0005)(.0005) = .0000

P(PPN) = P(PnPnN)= P(P)P(P)P(N) = .0005(.0005)(.9995) = .0000 = P(PNP) =P(NPP) P{PNN) = P(P nNnN) = P(P)P(N)P(N) = .0005(.9995)(.9995) = .0005 = P(NPN) = P(NNP) P(NNN) = P(NnNnN) = P(N)P(N)P(N) = .9995(.9995)(.9995) = .9985 Thus, P(x = 0) = P(NNN) = .9985 P(x = 1) = P(PNN) + P(NPN) + P(NNP) = .0005(3) = .0015 P(x = 2) = P(PPN) + P(PNP) + P(NPP) = .0000(3) = .0000 P(x = 3) = P(PPP) = .0000

88

Chapter 4

4.31

b.

The graph ofp(x) is:

c.

P(x< 1) = P(x = 0) + />(a:= 1) = .9985 + .0015 = 1.0000

a.

Define the following events: M: {Miami Beach, FL} C: (Coney Island, NY} S: (Surfside, CA} B: (Monmouth Beach, NJ} O: (Ocean City, NJ) L: (Spring Lake, NJ} The possible pairs of beach hotspots that can be selected are: MC, MS, MB, MO, ML, CS, CB, CO, CL, SB, SO, SL, BO, BL, OL

b.

If we are just selecting 2 hotspots from 6, each of the 15 combinations are equally likely. Thus, each would have a probability of 1/15.

c.

The value ofx for each of the pairs is: Pair MC MS MB MO ML CS CB CO

X

0 0 1 0 1 0 1 0

Random Variables and Probability Distribntions

Pair CL SB SO SL BO BL OL

X

1 1 0 1 1 2 1

89

d.

The probability distribution for x is:

.X 0 1 2 e.

=

P(x) 6/15 8/15 1/15

= 0(6/15) +1(8/15)+2(1/15)-10/15 = .667

The average total number of hotspots with a planar nearshore bar condition in each pair is .667. 4.33

a.

p(l) = (.23)(.77)'"'= .23(.77)°-23.

The probability that a contaminated cartridge is

selected on the first sample is .23.

4.35

b.

p(5) = (.23)(.77)^"' = = .081. The probability that the first contaminated cartridge will be selected on the 5*’’ sample is .081.

c.

f’(jc > 2) = 1-/'(;c < 1) = 1-/?(!) = 1 - .23 = .77. The probability that the first contaminated cartridge will be selected on the second trial or later is .77.

Assigning points according to the directions is: OUTCOME OF APPEAL Plaintiff trial win - reversed

Number of cases 71

Points awarded, x

-1

Plaintiff trial win affirmed/dismissed Defendant trial win - reversed

240

5

68

Defendant trial win affirmed/dismissed TOTAL

299

-3 5

678

To find the probabilities for.x, we divide the frequencies by the total sample size. The probability distribution for jc is: X

-3

-1 5 TOTAL

90

p{x)

68/678 = .100 71 / 678 = .105 (240+299)/678 = .795

1.000

Chapter 4

Using MTNITAB, the graph of the probability distribution is:

4.37

For a $5 bet, you will either win $5 or lose $5 (-$5). The probability distribution for the net winnings is: X

p{x) 20738

38,

.263

1

II

20"

18/38 oo 1 oo

fj. = E{x) = Y.xp(x) = -5

+

-5 5

Over a large number of trials, the average winning for a $5 bet on red is -$0,263. 4.3’9

a.

Each point in the system can have one of 2 status levels, “free” or “obstacle”. Define the following events: Ap: {Point A is free) Bp: (Point B is free} Cp: (Point C is free}

Aq: (Point A is obstacle} Bq: (Point B is obstacle} Cq: (Point C is obstacle}

Thus, the sample points for the space are:

ApBpCp, ApBpCo, ApBoCp, ApBoCo, AoBpCp, AoBpCo, AoBqCp, AoBqCo

Random Variables and Probability Distributions

91

b. Since it is stated that the probability of any point in the system having a “free” status is .5, the probability of any point having an “obstacle” status is also .5, Thus, the probability of each of the sample points above is P(AiBiCi) = .5(.5)(.5) = . 125. The values of x, the number of free links in the system, for each sample point are listed below. A link is free if both the points are free. Thus, a link from A to B is free if A is free and B is free. A link from B to C is free if B is free and C is free. Sample point

X

Probability

ApBpCp

2

.125

ApBpCo

1

.125

ApBoCp

0

.125

ApBoCo

0

.125

AoBpCp

1

.125

AoBpCo

0

.125

AoBqCp

0

.125

AqBoCo

0

.125

The probability distribution for jc is:

4.41

0

.625

1

.250

2

.125

d. e.

The experiment consists of n identical trials. There are only two possible outcomes on each trial. We will denote one outcome by S (for Success) and the other by F (for Failure). The probability of .S'remains the same from trial to trial. This probability is denoted by p, and the probability of F is denoted by q. Note that q = 1 -p. The trials are independent. The binomial random variable x is the number of 5^s in n trials.

a.

There are « = 5 trials.

b.

The value oip ’\sp = .7.

c.

92

Probability

The five characteristics of a binomial random variable are: a. b.

4.43

JC

Chapter 4

4.45

a.

/’(jc=l)=—(.2y(.8)'’= 5-4-3-2-1 (.2)’(.8)" = 5(.2y(.8/= .4096 1!4! (l)(4-3-2-l)

I

b. P{x = 2) = —(.6)"(.4)^ = (.6)^(.4)^ = 6(.6)\.4)^ = .3456 2!2! (2-l)(2-l)

c.

P{x = 0)

' -)°(.3)^=:. -(.7)°(.3)'=:—^—( OB! (l)(3-2-l)

d.

P{x = 3)

—(.1)^(.9)^ =-(-1) (.9)^ = 10(.1)\.9)^ = .00; 312! (3-2-l)(2-l)

e.

P{x = 2) - -il_(.4)2(.6)2 =

212!

4.47

f.

P(x = 1) = —(.9)'(.1)^ =

a.

M=np = 25(.5)=\2.5

112!

= 1(.7)°(.3)^ = .027

— (A)\.6f = 6(A)\.6f = .3456

(2-1)(2-1)

0)(2-l)

(•9y(.l)^ = 3(.9y(.l)^ = .027

= np(l -p) = 25(.5)(.5) = 6.25 (T=

b.

=

V6.25 = 2.5

// = «/? = 80(.2) = 16 (p- = np{\ -p) = 80(.2)(.8) = 12.8 cr=

c.

/i =

= Vi2^ =3.578 = 100(.6) = 60

= np{\ -p)=^ 100(.6)(.4) = 24 0-= V^ = V^ = 4.899 d.

// = «/? = 70(.9) = 63 o^ = np{\ -/?) = 70(.9)(.1) = 6.3 (T= V? = V^ =2.510

Random Variables and Probability Distributions

93

e.

f^ = np = 60(.8) = 48 c? = npi\ -p) = 60(.8)(.2) = 9.6 £7= 4^ ^4^ =3.098

f.

/i =

= 1,000(.04) = 40

c? = np{\ -p) = 1,000(.04)(.96) = 38.4 cr= 44^ = S^ =6.197 4.49

4.51

a.

?(jc< 10) = P(jc 10)= 1-/’(;c (47.5, 1,159.9) 97 of the 98 values fall in this interval. The proportion is .99. This is fairly close to the 1.00 we would expect if the data were normal. From this method, it appears that the data may be normal. Next, we look at the ratio of the IQR to 5. IQR = Qu-Ql = 724.3 - 475 = 249.3. IQR

249.3

= 1.3 This is equal to the 1.3 we would expect if the data were normal. This 185.4 method indicates the data may be normal 5

112

Chapter 4

Finally, using MINITAB, the normal probability plot is:

ProbabMity Plot of DrSvHead

ma n hti

n

Since the data form a fairly straight line, the data may be normal. From the 4 different methods, all indications are that the driver’s head injury rating data are normal. 4.111

The lowest score possible is 0. The sample mean is 7.62 and the standard deviation is 8.91. Since the lower limit of 0 is less than one standard deviation below the mean, this implies that the data are not normal, but skewed to the right.

4.113

The binomial probability distribution is a discrete distribution. The random variable can take on only a limited number of values. The normal distribution is a continuous distribution. The random variable can take on an infinite number of values. To get a better estimate of probabilities for the binomial probability distribution using the normal distribution, we use the continuity correction factor.

4.115

a.

In order to approximate the binomial distribution with the normal distribution, the interval

np ± 3^fnpq should lie in the range 0 to n.

When n = 25 and p = .4, np ± 3^^ ^ 25(.4) ± 3V25(.4)(l-.4) =>10±3V6 => 10 ± 7.3485 ^ (2.6515,17.3485) Since the interval calculated does lie in the range 0 to 25, we can use the normal approximation. b.

/,/=«/? = 25(.4) = 10 = npq = 25(.4)(.6) = 6

c.

P{x > 9) = 1 -- P{x < 8) = 1 - .274 = .726

Random Variables and Probability Distributions

(Table II, Appendix A)

113

d.

P{x > 9) w P z >

(9-.5)-10

S

.

= P(z>-.61) = .5000+ .2291 = .7291 (Using Table III in Appendix A.) 4.117

jc is a binomial random variable with a7 = 100 and p = .4. P±2)(5 ^ np ± 3-yjnpq ^ 100(.4) ± 3-^100(.4)(l - .4) 40 ± 3(4.8990) r:5> (25.303, 54.697) Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the probabilities.

a.

( (35+ .5)-40 P(;c(-1.985, 3.985) Since the interval does not lie in the range 0 to 100, we cannot use the normal approximation to approximate the probabilities. i b.

For

100 and/? = .5: _

_

\

//±3cr=>n/7± 3^jnpq => 100(.5)± 3.^100(.5)(.5) => 50 ± 3(5) => 50 ± 15 => (35, 65) Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the probabilities. c.

For n = 100 and/? = .9: ±30-=>«/?+ 37w => 100(.9)± 3Vl00(.9)(.l) => 90 ± 3(3) ^ 90 ± 9 => (81, 99) Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the probabilities.

4.127

Let jc = number of beech trees damaged by fungi in 200 trees. Then x is a binomial random variable with n = 200 and /? = .25. /i = n/? = 200(.25) = 50 or =

= 7200(.25)(.75) =

P(x>100)«/’ z >

(100 + .5)-50 6.124

= 6.124

= /’(z>8.25)«0

Since this probability is so small, it is almost impossible for more than half of the trees to have fungi damage. 4.129

a.

If 80% of the passengers pass through without their luggage being inspected, then 20% will be detained for luggage inspection. The expected number of passengers detained will be; £(x) = «/? = U500(.2) = 300

b.

For n = 4,000, E{x) = np = 4,000(.2) = 800

c.

P(x > 600)« P z >

(600+ .5)-800 = P(z>-7.89) = .5 + .5 = 1.0 V4000(.2)(.8)

116

Chapter 4

4.131

A population parameter is a numerical descriptive measure of a population. Because it is based on the observations in a population, its value is almost always unknown. A sample statistic is a numerical descriptive measure of a sample. It is calculated from the observations in the sample.

4.133

a-b. The different samples of n = 2 witli replacement and their means are:

c.

Possible Samples

x

Possible Samples

jc

0,0 0,2 0,4 0,6 2,0 2,2 2,4 2,6

0 1 2 3 1 2 3 4

4,0 4,2 4,4 4,6 6,0 6,2 6,4 6, 6

2 3 4 5 3 4 5 6

Since each sample is equally likely, the probability of any 1 being selected is

1

1 16

d.

P(x =0) =

JC

p(x)

0

1/16 2/16 3/16 4/16 3/16 2/16 1/16

16

1

P{x =2) =

16 1

16 1

16 1

4

16 3

16

6

3

-j-)- = -

P(x =3) =

16 16 16 1 1 1 -j--j-

P{x =4) =

16 16 16 1-1 1 -j-j-

o

1

1 2 3 4 5

16

16

16

1

1

2

16 -I-=

=16

P(x = 5)= — + — = — ^ 16 16 16 P(x =6)= — 16

Random Variables and Probability Distributions

117

4.135

If the observations are independent of each other, then P(1,1)=/.(1M1) = .2(.2) = .04 P(1,2)=/7(1M2) = .2(.3)=.06 P(1,3)=;?(1M3) = .2(.2)=.04 etc. Possible Samples 1,1 1,2 1,3 1,4 1,5 2,1 2,2 2,3 2,4 2,5 3,1 3,2 3,3

JC

1 1.5 2 2.5 3 1.5 2 2.5 3 3.5 2 2.5 3

pix)

Possible Samples

.04 .06 .04 .04 .02 .06 .09 .06 .06 .03 .04 .06 .04

3,4 3,5 4, 1 4,2 4,3 4,4 4,5 5,1 5,2 5,3 5,4 5,5

X

3.5 4 2.5 3 3.5 4 4.5 3 3.5 4 4.5 5

P(x) .04 .02 .04 .06 .04 .04 .02 .02 .03 .02 .02 .01

Summing the probabilities, the probability distribution of a: is: X

1 1.5 2 2.5 3 3.5 4 4.5 5

118

Pix)

.04 .12 .17 .20 .20 .14 .08 .04 .01

Chapter 4

4.137

4.141

c.

i’C 3c >4.5) = .04+ .01 =.05

d.

No. The probability of observing x = 4.5 or larger is small (.05).

a.

For a sample of size n = 2, the sample mean and sample median are exactly the same. Thus, the sampling distribution of the sample median is the same as that for the sample mean (see Exercise 4.135a).

b.

The probability histogram for the sample median is identical to that for the sample mean (see Exercise 4.135b).

The mean of the sampling distribution of 3c, ju^ , '\s the same as the mean of the population from which the sample is selected.

4.143

The Central Limit Theorem states: Consider a random sample of n observations selected from a population (any population) with mean fi and standard deviation cr. Then, when n is sufficiently large, the sampling distribution of x will be approximately a normal distribution with mean

= fj. and standard deviation dj =

• The larger the sample size, the

better will be the normal approximation to the sampling distribution of 3c .

4.145

a.

/^i = A = 100,cr-

b.

=/« = 100, CTj

c.

(j

4n

V?

4n

4^

a

VlOO

4n

VlOO

5

2

1

Random Variables and Probability Distributions

119

d.

or _VI^

iu- = iu = \m,(Tj

= 1.414

4n~ 4^

e.

IJl =

4n

f.

4.147

a.

=/i = 100, o-j

4m ~ Vsoo

o- _

100, (Tj

= .447

or _ VT^

V^i" ~ ViM

i

= .316

//= ^jc;7(jc) = l(.l) + 2(.4) + 3(.4) + 8(.l) = 2.9 0^= Y^{x-^ifp{x) =(1 -2.9f(.l) + (2-2.9)'(.4) + (3-2.9)'(.4) + (8-2.9rt.l) = .361 + .324 + .004 + 2.601 = 3.29 £7=7129=1.814

b.

The possible samples, values of x, and associated probabilities are listed: Possible Samples

X

P(^)

Possible Samples

X

p{x)

1,1 1,2 1,3 1,8 2,1 2,2 2,3 2,8

1 1.5 2 4.5 1.5 2 2.5 5

.01 .04 .04 .01 .04 .16 .16 .04

3,1 3,2 3,3 3, 8

2 2.5 3 5.5 4.5 5 5.5 8

.04 .16 .16 .04 .01 .04 .04 .01

8,1 8,2 8,3 8,8

P(l,l)=p(lMl) = . 1(.1) = .01 P{\,2)=p{\)p{2) = . 1(.4) = .04 P(l,3)=/.(l)p(3) = . 1(.4) = .04 etc.

120

Chapter 4

The sampling distribution of x is:

c.

X

p{x)

1 1.5 2 2.5 3 4.5 5 5.5 8

.01 .08 .24 .32 .16 .02 .08 .08 .01 1.00

lii-,=E{x) = YjMx) = 1(.01) + 1.5(.08) + 2(.24) + 2.5(.32) + 3(.16) + 4.5(.02) + 5(.08) + 5.5(.08) + 8(.01) = 2.9 = ju 0-1=

p{x) = (1 - 2.9)^(.01) + (1.5 - 2.9)2(.08) + (2 - 2.9)^(.24) + (2.5 - 2.9)^(.32) + (3 - 2.9)2(.16) + (4.5 - 2.9)\.02) + (5 - 2.9/(.08) + (5.5 - 2.9)^(.08) + (8 - 2.9rt.01) = .0361 + .1568 + .1944 + .0512 + .0016 + .0512 + .3528 + .5408 + .2601 = 1.645

CT- =Vl.645 = 1.283 (j-=al4n ^\.Z\AI42 = 1.283

4.149

a.

/4=/i = 30,

CT-=-1=^ =-7= = 1.6

b.

By the Central Limit Theorem, the sampling distribution of x is approximately normal.

c.

P{x >28)= P z >

d.

^(22.1 < T -1.25) = .5 + .3944 = .8944

1.6

22.1-30

I

-1.88) = .5 + .4699 = .9699

1.6

Random Variables and Probability Distributions

121

4.151

a.

By the Central Limit Theorem, the sampling distribution of x is approximately normal. The mean of the x distribution is X

b.

fj.— 320 and the standard deviation of the

cr 100 100 = —j= = --- , ^ = 10.

distribution is

4n

4m

10

„r300-320 310-320 P(300 80) = P(z > 0.43) = .5 -. 1664 = .3336

a.

P(x) = //j =// = .10 and F(x) = cr_ =

.10^

2

n b.

c.

(Using Table III, Appendix A.)

=

.0002

50

By the Central Limit Theorem, the sampling distribution of x is approximately normal if the sample size is sufficiently large. In this case, n = 50 is considered large. ^ .13-.lO'' P(3c>.13) = P z > = P(z > 2.12) = .5 - .4830 = .0170

4.0002 4.157

By the Central Limit Theorem, the sampling distribution of x is approximately normal if the sample size is sufficiently large. In this case, n = 50 is considered large.

/ij=A = .53,

b.

o-j=-^ = ^ = .0273 V« v50

.58-.53^ P(T>.58) = P z >■ = P(z > 1.83) = .5 - .4664 = .0336 V .0273 y (Using Table III, Appendix A)

122

Chapter 4

c.

( .59-.58 = P(z>.37) = .5-.1443 = .3557 P{x >.59\jLi =-5^) = P z >■ V .0273 j (Using Table III, Appendix A) The above is the probability after tensioning and loading. f ■59-.53^ P(T>.59|// = .53) = P z > = P{z > 2.20) = .5 - .Ain = .0228 .0273 (Using Table III, Appendix A) The above is the probability before tensioning. Since the probability after tensioning is not small, this would not be an unusual event. Since the probability before tensioning is very small, this would be an unusual event. The sample measurements were probably obtained after tensioning and loading.

4.159

By the Central Limit Theorem, the sampling distribution of jc is approximately normal if the sample size is sufficiently large. In this case, n = 326 is considered large.

Pj

=// = 6,

10 O’? =

Pix > 7.5) = P\z>

7.5-6

= .5538

P(z> 2.71) = .5-.4966 = .0034

.5538

b.

f

P(3c>300) = P z >•

300-6

= P(z> 530.88) «.5-.5 = 0

.5538 Since it is essentially impossible to get a sample mean of 300 if the population mean was 6, we would conclude that the mean PFOA in people who live near DuPont’s Teflon-making facility is not 6 ppb but something much larger. 4.161

Handrubbing

^

^

8.344

( 30-35 = P{z < -.60) = .5 - .2257 = .2743 P(T (101.35, 102.65)

.3 => 15 ± 1.96-p= => 15 ± .0588 => (14.9412, 15.0588)

4.05 ±.163 =>(3.887, 4.213)

e. No. Since the sample size in each part was large (« ranged from 75 to 200), the Central Limit Theorem indicates that the sampling distribution of 3c is approximately normal.

132

Chapter 5

5.11

a.

For confidence coefficient .95, a= .05 and all = .05/2 = .025. From Table III, Appendix A, Z025 = 1-96. The confidence interval is: x±z all

83.2 + 1.96

4n

6.4

Vl^

83.2+ 1.25 =>(81.95, 84.45)

b.

The confidence coefficient of .95 means that in repeated sampling, 95% of all confidence intervals constructed will include ju.

c.

For confidence coefficient .99, q:= .01 and all = .01/2 = .005. From Table III, Appendix A, zqos = 2.58. The confidence interval is: x + z„/2^=^ 83.2 +2.58-^ => 83.2+ 1.65 =^(81.55, 84.85) 4n VlOO

5.13

d.

As the confidence coefficient increases, the width of the confidence interval also increases.

e.

Yes. Since the sample size is 100, the Central Limit Theorem applies. This ensures the distribution of x is normal, regardless of the original distribution.

a.

The point estimate for the average number of latex gloves used per week by all healthcare workers with latex allergy is T = 19.3 .

b.

For confidence coefficient .95, a = .05 and all = .05/2 = .025. From Table III, Appendix A, zo25 = 1.96. The 95% confidence interval is:

^ — ^025^?

5.15

x + 1.96

a

.11.9 19.3 + 1.96 V46

7^

19.3 + 3.44=^(15.86, 22.74)

c.

We are 95% confident that the average number of latex gloves used pr week by all healthcare workers with a latex allergy is between 15.86 and 22.74.

d.

We must assume that we have a random sample from the target population and that the sample size is sufficiently large.

a.

Some preliminary calculations are: 181.56

= .3602

504

74.6546 -

181.56^ 504

n -1

504-1

9.24977143

= .018389

503

5 = V.018389 =.13561

Inferences Based on a Single Sample: Estimation with Confidence Intervals

133

For confidence coefficient .90, a = .10 and a/2 = .10/2 = .05. From Table III, Appendix A, zo5 = 1.645. The 90% confidence interval is;

X±

=> jc ± 1.645

a

.3602 + 1.645

n 3602±.00994

,13561 V504

(.35026, .37014)

We are 90% confident that the true mean visible albedo value of all Canadian Arctic ice ponds is between .35026 and .37014. In repeated sampling 90% of all intervals constructed in the same manner will contain the true mean and 10% will not. First-year Ice: Some preliminary calculations are; 26.63 X

= .3026

=

n

88

10.6923-

26.63^

88

n

2.633698864

88-1

n -1

= .03027

87

s = ylmQ21 = .17399 The 90% confidence interval is:

jc ± Zo5f7j. => I+ 1.645-pr=>.3026+ 1.645

.17399 ^ v88

=>.3026+.03051 =>(.27209, .33311) We are 90% confident that the true mean visible albedo value of all First-year Canadian Arctic ice ponds is between .35026 and .37014. Landfast Ice: Some preliminary calculations are:

X

71.04

= .3624

=

n

196

30.161n

=

n-\

71.04^ 196

196-1

4.41262449 195

= .02263

5 = V.02263 = .15043

134

Chapter 5

The 90% confidence interval is:

±z^^or- z^x±\.645—:=^ .3624± 1.645

4n

Vl96

=>.3624+ .01768 =>(.34472, .38008) We are 90% confident that the true mean visible albedo value of all Landfast Canadian Arctic ice ponds is between .34472 and .38008. Multi-year Ice: Some preliminary calculations are: 83.89

220

5^

=

s

= .3813

(ST

33.8013

83.89^

220

n n -1

1.81251773

220-1

= .00827

219

5 = V.00827 = .09097 The 90% confidence interval is:

T ± ZojCT-jc ± 1.645-^ yjn

.3813± 1 .645^5?£EZ v220

3:^ .3813+ .01009::^ (.37121, .39139) We are 90% confident that the true mean visible albedo value of all Multi-year Canadian Arctic'ice ponds is between .37121 and .39139. 5.17

a.

The parameter of interest is the mean effect size for all psychological studies of personality and aggressive behavior.

b.

No, the distribution does not look normal. The data appear to be skewed to the right. The shape of this distribution is not of interest because the sample size is large, n = 109. By the Central Limit Theorem, the sampling distribution of x is normal regardless of the original distribution.

c.

From the printout, the 95% confidence interval for p is (0.4786, 0.8167). We are 95% confident that the true mean effect size is between 0.4786 and 0.8167.

d.

Yes. Since 0 is not contained in the 95% confidence interval, it is not a likely value for the true mean effect size. Since all the values in the 95% confident interval are above 0, the, researchers are justified in concluding the true effect size mean is greater than 0 or tliat those who score high on a personality test are more aggressive than those who score low.

Inferences Based on a Single Sample: Estimation with Confidence Intervals

135

5.19

a.

For confidence coefficient .99, a~ .01 and a/2 = .01/2 = .005. From Table III, Appendix A, z.oos = 2.58. The confidence interval is: jc ± z all

rn

1.13±2.58

2.21

1.13+ .672 =>(.458, 1.802)

We are 99% confident that the true mean number of pecks made by chickens pecking at blue string is between .458 and 1.802.

5.21

b.

Yes, there is evidence that chickens are more apt to peck at white string. The mean number of pecks for white string is 7.5. Since 7.5 is not in the 99% confidence interval for the mean number of pecks at blue string, it is not a likely value for the true mean for blue string.

a.

For confidence coefficient .95, a = ,05 and a/2 = .05/2 = .025. From Table III, Appendix A, z.025 - 1-96. The confidence interval is: s

b.

65

^19±7.826 =>(11.174, 26.826)

For confidence coefficient .95, ar= .05 and cdl = .05/2 = .025. From Table III, Appendix A, Z025 = 1.96. The confidence interval is:

X

5.23

z:;>19±1.96

49 ± z„/, —f= ^ 7 ± 1.96- --

4n

7 ±5.90 ^(1.10, 12.90)

c.

The SAT-Mathematics test would be more likely to have a mean change of 15 because 15 is in the 95% confidence interval for the mean change in SAT-Mathematics. Since 15 is in the confidence interval, it is a likely value. The value 15 is not in the 95% confidence interval for the SAT-Verbal. Thus, it is not a likely value.

a.

For confidence coefficient .95, a = .05 and a/2 = .05/2 = .025. From Table III, Appendix A, Z025 = 1.96. For Males, the 95% confidence interval is:

X ± 2o25^x

^ + 1.96-^

16.79 ± 1.96=>16.79 ± 2.35 => (14.44, 19.14)

Vl28 For Females, the 95% confidence interval is: a 11.53 ^ ± 2.025^^^ => X +1.96^ => 10.79 +1.96-^ => 10.79 ± 1.67 => (9.12, 12.46)

^/m

136

Chapter 5

b.

Since the two intervals are independent, the probability that at least one of the 2 confidence intervals will not contain the population mean is equal to 1 minus the probability that neither of the 2 confidence intervals will not contain the population mean. Thus the probability is: /'(at least one interval will not contain mean) = 1 - /’(neither interval will not contain the mean) = 1 - /(both will contain the mean) = 1 - .95(.95) = 1 - .9025 = .0975.

c.

5.25

The 95% confidence interval for the males is (14.44, 19.14). The 95% confidence interval for the females is (9.12, 12.46). Since all of the values in the male interval are larger than all the values in the female interval, we can infer that the males consume the most alcohol, on average, per week.

The two problems (and corresponding solutions) with using a small sample to estimate p, are: 1.

The shape of the sampling distribution of the sample mean x now depends on the shape of the population that is sampled. The Central Limit Theorem no longer applies since the sample size is small. The population standard deviation a is almost always unknown. Although it is still true that (Ty =

»the sample standard deviation s may provide a poor approximation for a

when the sample size is small. 5.27

5.29

a.

If X is normally distributed, the sampling distribution of is normal, regardless of the sample size.

b.

If nothing is known about the distribution ofx, the sampling distribution of is approximately normal if n is sufficiently large. If n is not large, the distribution of is unknown if the distribution of x is not known.

a.

/(/>/o) = .025 where df= 10 to ~ 2.228

b.

/(/>/o) = .01 where df = 17 to^ 2.561

c.

/(/ -to) = .005 where df = 6 to = -3.707

d. /(t < to) = .05 where df = 13 to = -1.771

Inferences Based on a Single Sample: Estimation with Confidence Intervals

137

5.31

First, we must compute x and s.

^2 ^

,76_(25)!

^

6_

n

_

^_

n—l

6-1

5 = ^^52 = 2.2804 a.

For confidence coefficient .90, a = 1 — .90 = .10 and all = .10/2 = .05. From Table IV, Appendix A, with df = « - 1 = 6 — 1 = 5, t^s — 2.015. The 90% confidence interval is: 0 0 Q f\A

T + fo3-^=>5+2.015 v« b.

^

' ^ 5 + 1.876 =^>(3.124, 6.876) 76

For confidence coefficient .95, a = 1 - .95 = .05 and all = .05/2 = .025. From Table IV, Appendix A, with df = «-l=6-l=5, tojs - 2.571. The 95% confidence interval is: o o or\/i

J + /o254=^ 5 ± 2.571 V«

•

^ 5 ± 2.394 => (2.606, 7.394) 76

c. For confidence coefficient .99, «= 1 - .99 = .01 and a/2 = .01/2 = .005. From Table IV, Appendix A, with df = « - 1 = 6 - 1 = 5, foos = 4.032. The 99% confidence interval is: ^

x±toos^^ 5 + 4.032 ^|n d. a)

OO

/I

5 ±3.754 ^ (1.246, 8.754)

76

For confidence coefficient .90,

1 - .90 = .10 and all = .10/2 = .05. From

Table IV, Appendix A, with df = « - 1 = 25 - 1 = 24, /qs = 1.711. The 90% confidence interval is: JC ±/.05

5 ± 1.711

5 + .780 =>(4.220, 5.780)

425

7^ b)

2.2804

For confidence coefficient .95, « = 1 - .95 = .05 and a/2 = .05/2 = .025. From Table IV, Appendix A, with df = « - 1 = 25 - 1 = 24, ro25 = 2.064. The 95% confidence interval is: s 2 2 S 04 x±ro25^=^5±2.064' =>5±.941 =>(4.059,5.941) ^|n v25

c)

For confidence coefficient .99, a = 1 - .99 = .01 and a/2 = .01/2 = .005. From Table IV, Appendix A, with df = « - 1 = 25 - 1 = 24, /qos = 2.797. The 99% confidence interval is: x±t.005

5 + 2.797 4^

2.2804 7^

5 ± 1.276 =>(3.724, 6.276)

Increasing the sample size decreases the width of the confidence interval.

138

Chapter 5

5.33

a.

^

39

X = --^ =— =

n

1.857

21

28.57143

20

1.4286

5 = Vr4286 =1.1952 b. Using MINITAB, a histogram of tlie data is:

Altliough this is a histogram of the sample, it should refleet the distribution of the population fairly,well. The population distribution appears to be skewed to the right. However, the distribution is fairly mound-shaped. Since the sample size of n = 21 is somewhat close to 30, the sampling distribution of x will be approximately normal. c. For confidence coefficient .90, a= .10 and all = .10/2 = .05. From Table IV, Appendix A, with df = « - 1 = 21 - 1 = 20, fos = 1.725. The 90% confidence interval is: ‘S: 1 or-7 1 1-1952 ^ — ^.05,20 ^ 1 -857 ± 1.725—y==—

1.857±.450^(1.407, 2.307)

d.

We are 90% confident that the true mean number of individual suicide bombings and/or attacks per incident is between 1.407 and 2.307.

e.

With repeated sampling, 90% of all confidence intervals constructed in this manner will contain the tme mean and 10% will not.

Inferences Based on a Single Sample; Estimation with Confidence Intervals

139

5.35

a.

The point estimate for the mean amount of cesium in lichen specimens collected in Alaska is jc =.009027.

b.

For confidence coefficient .95, a = .05 and od'l — .05/2 = .025. From Table IV, Appendix A, with df = « - 1 = 9 - 1 = 8, /025 = 2.306.

c.

The 95% confidence interval is:

^

5.37

^ .009027 ± 2.306^^5^ v9

.009027 ± .003731

(.005296, .012758)

d.

This interval is the same as that found on the printout.

e.

We are 95% confident that the mean amount of cesium in lichen specimens collected in Alaska is between .005296 and .012758 microcuries per milliliter.

Some preliminary calculations: 6.44 - = — = 1.073 n

6 0

(yjcf ’

6 44' 7.1804-^^

==-s—=-^ = .0536 n—\ 6—1 5 = V.0536 = .2316

a.

For confidence coefficient .95, a= .05 and a/2 = .05/2 = .025. From Table IV, Appendix A, with df = «-l=6-l=5, /'025 = 2.571. The confidence interval is:

X+

1.073 ± 2.571^=^ ^ 1.073 + .243 =:> (.830, 1.316)

V«

v6

We are 95% confident that the true average decay rate of fine particles produced from oven cooking or toasting is between .830 and 1.316.

140

b.

The phrase “95% confident” means that in repeated sampling, 95% of all confidence intervals constructed will contain the true mean.

c.

In order for the inference above to be valid, the distribution of decay rates must be normally distributed.

Chapter 5

5.39

a.

Both Untreated: For confidence coefficient .90, a = ,10 and all = .10/2 ^ .05. From Table IV, Appendix A, with df = « - 1 = 29 - 1 = 28, fos = 1.701. The 90% confidence interval is:

^ ± ^05 28

s V«

3 34 20.9 ± 1.701^ V29

20.9 ± 1.055 => (19.845, 21.955)

Male Treated: For confidence coefficient .90, a= .10 and c(/2 = .10/2 = .05. From Table IV, Appendix A, with df = « - 1 = 23 - 1 = 22, (os = 1.717. The 90% confidence interval is: X ±t .05,22 -4=

20.3 +1.717^ V23

20.3 ± 1.253

(19.047, 21.553)

Female Treated: For confidence coefficient .90, a= .10 and a/2 = .10/2 = .05. From Table IV, Appendix A, with df=«-l = 18-l = 17, fos = 1.740. The 90% confidence interval is:

X ±t .05,17

22.9 ± 1.740^ => 22.9 + 1.792 => (21.108, 24.692)

^/n

vl 8

Both Treated: For confidence coefficient .90, a= .10 and a/2 = .10/2 = .05. From Table IV, Appendix A, with df = « - 1 = 21 - 1 = 20, /o5 = 1.725. The 90% confidence interval is: x±t.05,20 -^=>18.6 + 1.725 yjn

5.41

2.11

=>18.6+ 0.794 =>(17.806, 19.394)

b.

The female/male pair that appears to produce the highest mean number of eggs is the Female treated pair. The 90% confidence interval for this pair is the highest. This interval does overlap with the Both treated pair and the Males treated pair, but just barely.

a.

Some preliminary calculations are: -

yX

160 9

n

22

;c=^=-^^ = 7.314

1,389.1-1^ ^2

^-n— ^-22_ 3,10.1112 n-l 22-1

5 = V10.1112 -3.180 For confidence coefficient .95, a= .05 and a/2 = .05/2 = .025. From Table IV, Appendix A with df = « - 1 = 22 - 1 = 21, /025 2.080. The 95% confidence interval is:

+ ^^/2^ => 7.314 + 2.080^^ => 7.314 + 1.410 => (5.904, 8.724) v22

Inferences Based on a Single Sample: Estimation with Confidence Intervals

141

b.

We are 95% confident that the mean PMI for all human brain specimens obtained at autopsy is between 5.904 and 8.724.

c.

We must assume that the population of all PMI's is normally distributed. From the dot plot in Exercise 2.25, the distribution does not appear to be normal.

d.

"95% confidence" means that if repeated samples of size n were selected from the population and 95% confidence intervals formed for //, 95% of the intervals formed will contain the true mean and 5% will not.

5.43

a.

For confidence coefficient .95, a = .05, and all = .05/2 = .025. From Table IV, Appendix A, with df=w-l = 15-l = 14, to25 = 2.145. The confidence interval is:

3^ ±?025 v«

=:> 37.3 ±2.145^ => 37.3 ± 7.70 => (29.60, 45.00) vl5

b.

We are 95% confident that the true mean adrenocorticotropin level in sleepers one-hour prior to anticipated waking is between 29.60 and 45.00

c.

Suppose we assume that the true mean adrenocorticotropin level of sleepers three hours before anticipated wake-up time is 25.5. Since the 95% confidence interval for the true mean adrenocorticotropin level in sleepers one-hour prior to anticipated waking (29.60, 45.00) does not contain 25.5, there is evidence to indicate that the true mean adrenocorticotropin level of sleepers three hours before anticipated wake-up time is different from that of sleepers one-hour prior to anticipated waking.

5.45

An unbiased estimator is one in which the mean of the sampling distribution is the parameter of interest, i.e., E{^p) =p.

5.47

a.

The sample size is large enough if both

> 15 and nq >-15.

np = 196(.64) = 125.44 and nq = 196(.36) = 70.56 . Since both of these numbers are greater than or equal to 15, the sample size is sufficiently large to conclude the normal approximation is reasonable. b.

For confidence coefficient .95, a= .05 and all = .025. From Table III, Appendix A, 2.025 = 1-96. The confidence interval is:

P ± ^.025 J— => .64 ± 1.96 V «

142

^ .64 ± .067 V

(.573, .707)

196

c.

We are 95% confident the true value ofp is between 311 and .468.

d.

"95% confidence" means that if repeated samples of size 196 were selected from the population and 95% confidence intervals formed, 95% of all confidence intervals will contain the true value of p.

Chapter 5

5.49

The sample size is large enough if both np>\5 and nq>\5 . a.

np = 500(.05) = 25 and

= 500(.95) = 475. Since both of these numbers are greater

than or equal to 15, the sample size is sufficiently large to conclude the normal approximation is reasonable. b.

np = 100(.05) = 5 and nq = 100(.95) = 95 . Since the first number is not greater than 15, the sample size is not sufficiently large to conclude the normal approximation is reasonable.

c.

np = 10(.5)=:5 and

= 10(.5) = 5 . Since neither of these numbers is greater

than or equal to 15, the sample size is not sufficiently large to conclude the normal approximation is reasonable. d.

np -10(.3) = 3 and nq = 10(.7) = 7 . Since neither of these numbers is greater than or equal to 15, the sample size is not sufficiently large to conclude the normal approximation is reasonable.

5.51

a.

The population of interest is the set of gun ownerships (Yes or No) of all adults in the U.S.

b.

The parameter of interest is the true percentage of all adults in the U.S. who own at least one gun.

c.

The estimate of the population proportion is ^ = .26 . The estimate of the population percentage is .26(100%) = 26%.

d.

The sample size is large enough if both np>\S and nq>\5. np - 2,770(.26)

720.2 and nq = 2,770(.74) = 2,049.8. Since both of these numbers

are greater than or equal to 15, the sample size is sufficiently large to conclude the normal approximation is reasonable. For confidence coefficient .99, a= .01 and a/2 = .01/2 = .005. From Table III, Appendix A, z.oos == 2.575. The 99% confidence interval is:

p ± z.oos-^

2.575=>.26 ±2.575^-^^^^-^^^

.26 ± .021 => (.239, .281)

The 99% confidence interval for the true percentage is (23.9%, 28.1%). e. We are 99% confident that the true percentage of adults in the U.S who own at least one gun is between 23.9% and 28.1%. tf

f

“99% confidence” means that in repeated sampling 99% of all confidence intervals censtructed in this manner will contain the true percentage.

Inferences Based on a Single Sample: Estimation with Confidence Intervals

143

5.53

a.

The sample size is large enough if both np>\5 and

> 15.

np = 1,000(.63) = 630 and nq = 1,000(.37) = 370. Since both of tlie numbers are greater than or equal to 15, the sample size is sufficiently large to conclude the normal approximation is reasonable. For confidence coefficient .95, a = .05 and a/1 = .05/2 = .025. From Table III, Appendix A, 2025 == 1-96. The 95% confidence interval is:

^±1.96^^ ^.63±1.96.

^±^.025

■63(.37)

1,000

.63±.030 =>(.600, .660) b.

5.55

Yes, we would be surprised. Since .70 is not in the 95% confidence interval, it is not a likely value for the true value of the population proportion of adults who would choose to sleep when home sick.

X 183 First, we compute p: p = — =-=.219 n 837 The sample size is large enough if both np>\5 and nq>\S. = 837(.219) = 183.303 and

= 837(.781) = 653.697. Since both of the numbers are

greater than or equal to 15, the sample size is sufficiently large to conclude the normal approximation is reasonable. For confidence coefficient .90, a = .10 and q/2 = .10/2 = .05. From Table III, Appendix A, zo5 = 1.645. The confidence interval is:

p±z05

~p±\.645=^> .219± 1.645^

.219(.781) 837

5.57

a.

.219±.024=>(.195, .243)

First, we compute p. There were 1333 useable responses. Of these, 450 chose “Bible is the actual word of God and is to be taken literally”. 450

= .338

1333 The sample size is large enough if both = 1,333(.338) = 450.6 and

> 15 and

> 15.

= 1,333(.662) = 882.4 . Since both of the numbers are

greater than or equal to 15, the sample size is sufficiently large to conclude the normal approximation is reasonable.

144

Chapter 5

For confidence coefficient .95, a= .05 and all = .05/2 = .025. From Table III, Appendix A, Z 025 = 1.96. The 95% confidence interval is:

.96M \ n

.338± 1.96

n

^

1,333

=>.338±.025 =>(.313, .363) b.

5.59

a.

We are 95% confident that the true proportion of all Americans who believe that the “Bible is the actual word of God and is to be taken literally” is between .313 and .363. X 97 First we must compute p: p = — =-= .524 n 185 The sample size is large enough if both np>\S and

> 15.

np = 185(.524) = 96.9 and nq -185(.476) = 88.1. Since both of the numbers are

greater than or equal to 15, the sample size is sufficiently large to conclude the normal approximation is reasonable. For confidence coefficient .99, a = .01 and all = .01/2 = .005. From Table III, Appendix A, z qos = 2.58. The confidence interval is:

P±z,„J^«p±2.5sM \ n

.514 ± 2.58,

(.524(.476)

\ fj

.524 ± .095

185 (.429, .619)

b.

5.61

We are 99% confident that the true proportion of bottlenose dolphin signature whistles that are Type "a" whistles is between .429 and .619.

The point estimate for the proportion of health care workers with latex allergy who suspect X

36

that he/she has the allergy is p = — = — = .434 . ft 83 The sample size is large enough if both np> 15 and ftq >15 . np = 83(.434) = 36 and nq = 83(.566) = 47 . Since both of the numbers are greater than or

equal to 15, the sample size is sufficiently large to conclude the normal approximation is reasonable. Since no confidence limit is given, we will use 95%. For confidence coefficient .95, a - .05 and all = .05/2 = .025. From Table III, Appendix A, z.025 = 1.96. The 95% confidence interval is:

p±

=> p ± 1 -96^ ^ .434 ± 1^ .434 ±. 107 => (.327, .541)

We ar€*95% confident that the proportion of health care workers with latex allergy who suspect that he/she has the allergy is between .327 and .541.

Inferences Based on a Single Sample; Estimation with Confidence Intervals

145

5.63

The statement “For a specified sampling error SE, increasing the confidence level (1 - a) will lead to a larger n when determining sample size” is True.

5.65

To compute the necessary sample size, use 2 _2

n

^ (^a/2) O'

where a = 1 - .95 = .05 and odl = .05/2 = .025

{SEf From Table III, Appendix A, z 025 = 1 96. Thus,

„=(.5189, .6161) We are 95% confident that the proportion of measurements in the population with characteristic A is between .5189 and .6161. b.

We will use ^ = ,5675 to estimate/p. For confidence coefficient .95, a= .05 and a/2 = .05/2 = .025. From Table III, Appendix A, z 025 = 1 -96. 1.96^(.5675X.4325)

TTj/*-sample size is « = iSEf

.02'

= 2,357.2 « 2,358.

Inferences Based on a Single Sample; Estimation with Confidence Intervals

149

5.89

5.91

The parameters of interest for tlie problems are: (1)

The question requires a categorical response. One parameter of interest might be the proportion, p, of all Americans over 18 years of age who think their health is generally very good or excellent.

(2)

A parameter of interest might be the mean number of days, p, in the previous 30 days that all Americans over 18 years of age felt that their physical health was not good because of injury or illness.

(3)

A parameter of interest might be the mean number of days, p, in the previous 30 days that all Americans over 18 years of age felt that their mental health was not good because of stress, depression, or problems with emotions.

(4)

A parameter of interest might be the mean number of days, p, in the previous 30 days that all Americans over 18 years of age felt that their physical or mental health prevented them from performing their usual activities.

a.

X 35 The point estimate ofp, the true driver phone cell use rate, is p = — =-= .030 n 1,165

b.

The sample size is large enough if both np>\5 and

> 15.

«p = l,165(.030) = 35 and «^ = 1,165(.970) = 1130 . Since both of the numbers are greater than or equal to 15, the sample size is sufficiently large to conclude the normal approximation is reasonable. For confidence coefficient .95, a= .05 and a/2 = .05/2 = .025. From Table III, Appendix A, 2.025 = 1.96. The 95% confidence interval is:

^±2o25C7.

5.93

p + 1.96^|S=:>.030±1.96^P^|^^=>.030±.010=>(.020, .040)

c.

We are 95% confident that the true driver phone cell use rate is between .020 and .040.

a.

The confidence level desired by the researchers is .90.

b.

The sampling error desired by the researchers is SE = .05.

c.

Since no value was given for p, we will use p = — = = .604 n m For confidence coefficient .90, a = .10 and a/2 = .10/2 = .05. From Table III, Appendix A, 2 05 = 1.645.

The sample size is « =

(z,o,ypq _ 1.645^(.604)(.396) (SEf

150

.05"

= 258.9 » 259.

Chapter 5

5.95

Using MINITAB, the descriptive statistics are: f

Descriptive Statistics: SeedWt Variable SeedWt

N 16

N* 0

Mean 1.373

SE Mean 0.258

StDev 1.034

Minimum 0.201

Q1 0.475

Median 1.347

Q3 2.009

Maximum 3.751

For confidence coefficient .99, a= .01 and «/2 = .01/2 = .005. From Table IV, Appendix A, with df = « - 1 = 16 - 1 = 15,7,005.= 2.947. The 99% confidence interval is:

x±t.0.05

1.373 + 2.947

1.034

=5l.373±.762=i>(.611. 2.135)

Vl6

We are 99% confident that the average weight of dry seeds in the crops of spinifex pigeons inhabiting the Western Australian desert is between .611 and 2.135 grams. 5.97

The 95% confidence interval is (.045, .091). We are 95% confident that the true proportion of people who were/are homeless is between .045 and .091.

5.99

a.

665

First, we compute p :

= .660

1007 The sample size is large enough if both np>\5 and nq > 15. np = 1,001(.660) = 665 and wg=:l,007(.340) = 342. Since both of the numbers are greater than or equal to 15, the sample size is sufficiently large to conclude the normal approximation is reasonable. For confidence coefficient .95, a= .05 and a/2 = .05/2 = .025. From Table III, Appendix A, zo^5 = 1-96. The confidence interval is:

\pg

P9

p ± z025 J^~p±l-96^^ => .660±1.96^

f.660(.340) 1007

.660±.029=^(.631, .689)

We are 95% confident that the proportion of U.S. workers who take their lunch to work is between .631 and .689.

.

b.

,

.

.

X

200

First, we compute p : p-~ =-= .301 n 665 The sample size is large enough if both np>\5 and

> 15

np = 665(.301) = 200 and nq = 665(.699) = 465 . Since both of the numbers are greater than or equal to 15, the sample size is sufficiently large to conclude the normal afJproximation is reasonable.

Inferences Based on a Single Sample; Estimation with Confidence Intervals

151

For confidence coefficient .95, a~ .05 and odl = .05/2 = .025. From Table III, Appendix A, Z025 = 1-96. The confidence interval is;

/7 + Z, i

^±1.96

.301 ±1.96,

.301(.699)

.301 ±.035 ^(.266, .336)

665

We are 95% confident that the proportion of U.S. workers who take their lunch to work who brown-bag their lunch is between .266 and .336. 5.101

a.

For confidence coefficient .99, a= .01 and all ~ .01/2 = .005. From Table III, Appendix A, Z005 = 2.58. The confidence interval is:

jc± Z005

V 884 -4- ^.044 ±2.58^^ 4n 4\91

zz>

.044 ± .162

(-.118, .206)

We are 99% confident that the mean inbreeding coefficient for this species of wasps is between -.118 and .206.

5.103

b.

A coefficient of 0 indicates that the wasp has no tendency to inbreed. Since 0 is in the 99% confidence interval, it is a likely value for the true mean inbreeding coefficient. Thus, there is no evidence to indicate that this species of wasps has a tendency to inbreed.

a.

Using MINITAB, the descriptive statistics are:

Descriptive Statistics: CPU Time Variable CPU Time

N 52

N* 0

Mean 0.812

SE Mean 0.209

StDev 1.505

Minimum 0.0360

Q1 0.136

Median 0.275

Q3 0.595

Maximum 8.788

For confidence coefficient .95, a= .05 and all = .05/2 = .025. From Table III, Appendix A, Z025 = 1.96. The 95% confidence interval is:

± ^

=> X ± 1.96-^ => .812 ± 1.96^^ V« V52

.812 ± .409

(.403, 1.221)

We are 95% confident that the mean solution time for the hybrid algorithm is between .403 and 1.221.

b.

1.96^(1.505)^

The sample size needed would be « = {SEf

152

.25^

= 139.2 «140.

Chapter 5

5.105

For all parts to this problem, we will use a 95% confidence interval. For confidence coefficient, .95, \S and nq>\5. np - 37(.378) = 14 and nq = 37(.622) = 23 . Since the first number is not greater than or equal to 15, the sample size is not sufficiently large to conclude the normal approximation is reasonable. Since the normal approximation is not reasonable, we will use the Wilson adjusted confidence interval.

n+A

37 + 4

41

Tlie Wilson adjusted 95% confidence interval is:

p + z 025

pQ - P) V n+A

.390 ±1.96,

.390(.610)

.390+ .149 =>(.241, .539)

37 + 4

We are 95% confident that the true proportion of suicides at the jail that are committed by inmates charged with murder/manslaughter is between .241 and .539. b.

Some preliminary calculations: p =jc/« = 26/37 = .703 The sample size is large enough if both

> 15 and nq>\5.

np = 37(.703) = 26 and nq = 37(.297) = 11. Since the second number is not greater than or equal to 15, the sample size is not sufficiently large to conclude the normal approximation is reasonable. Since the normal approximation is not reasonable, we will use the Wilson adjusted confidence interval. p

x + 2 26 + 2 28 , „ =-=-= — = .683 n + A 37 + 4 41

The Wilson adjusted 95% confidence interval is:

p ±ZQ2fJ——~ ^ -683 + 1.96 n+A V

.683(.317)

.683+ .142 =^(.541, .825)

37 + 4

We are 95% confident that the true proportion of suicides at the jail that are committed at night is between .541 and .825.

Inferences Based on a Single Sample: Estimation with Confidence Intervals

153

c.

Some preliminaiy calculations are: 1532

41.405

- = -^=

n

37

223.606 n

=

= 4,449.2477 37-1

n-\

= ^4,449.2477 = 66.703 The confidence interval is:

3^ + ^

41.405 i 1.96 4n

r4> 41.405 ±21.493 =>(19.912,62.898) W

We are 95% confident that the true average length of time an inmate is in jail before committing suicide is between 19.912 and 62.898 days. d.

Some preliminary calculations: p =jc/«= 14/37 = .378 The sample size is large enough if both np>\5 and

> 15.

np = 37(.378) = 14 and nq = 37(.622) = 23 . Since the first number is not greater than or equal to 15, the sample size is not sufficiently large to conclude the normal approximation is reasonable. Since the normal approximation is not reasonable, we will use the Wilson adjusted confidence interval.

«+4

37 + 4

41

The Wilson adjusted 95% confidence interval is:

p ± z 025

«+4

.390 ± 1.96

=> .390 ±. 149 => (.241, .539) 37 + 4

We are 95% confident that the true percentage of suicides at the jail that are committed by white inmates is between 24.1% and 53.9%.

154

Chapter 5

5.107

a.

The point estimate for the fraction of the entire market who refuse to purchase bars is: p = — =-= .094 n 244

b. The sample size is large enough if both np>\S and

> 15.

np = 244(.094) = 23 and nq = 244(.906) -221. Since both of the numbers are greater than or equal to 15, the sample size is sufficiently large to conclude the normal approximation is reasonable. c.

For confidence coefficient .95, a = 1 - .95 = .05 and all = .05/2 = .025. From Table III, Appendix A, Z025 = 1.96. The confidence interval is: p±z.025

5.109

.094+ 1.96,

.094(.906)

z:>.094+ .037 =>(.057, .131)

244

d.

The best estimate of the true fraction of the entire market who refuse to purchase bars six months after the poisoning is .094. We are 95% confident the true fraction of the entire market who refuse to purchase bars six months after the poisoning is between .057 and. 131.

a.

For confidence coefficient .99, «= .01 and «/2 = .01/2 = .005. From Table IV, Appendix A, with df=n-l=3-l=2, foos = 9.925. The confidence interval is: s X ± foo5—=> 49.3 + 9.925 ill

s

Tn

49.3

± 8.60 => (40.70, 57.90)

b.

We are 99% confident that the mean percentage of B(a)p removed from all soil specimens using the poison is between 40.70% and 57.90%.

c.

We must assume that the distribution of the percentages of B(a)p removed from all soil specimens using the poison is normal.

d. For confidence coefficient .99, Appendix A, zqos = 2.58.

^^1,9_6-_^5)1^ ^7.

^2

{SEY

1

o/r2/i

.01 and all = .01/2 = .005. From Table III,

c\2

34^57 ^35

s

Inferences Based on a Single Sample; Estimation with Confidence Intervals

155

5.111

a.

From Chebyshev’s Rule, we know that at least 1

the observations will fall within

k standard deviations of the mean. We need to find k so that 1 — — =.60. k 1—L

e

= .60=e>.40=-V=>A:"= —= 2.5i=>A: = V^ = 1.58 e .40

80'* percentile -60'* percentile

73,000-35,100

k

1.58

For confidence coefficient .98, a= .02 and a/2 = .02/2 = .01. From Table III, Appendix A, z.oi = 2.33. The sample size required is: ^2.33^(23,987.34)^ 2,000^

5.113

b.

See part a.

c.

We must assume that any sample selected will be random.

a.

As long as the sample is random (and thus representative), a reliable estimate of the mean weight of all the scallops can be obtained.

b.

The government is using only the sample mean to make a decision. Rather than using a point estimate, they should probably use a confidence interval to estimate the true mean weight of the scallops.

c.

We will form a 95% confidence interval for the mean weight of the scallops. Using MINITAB, the descriptive statistics are: Descriptive Statistics: Weight Variable Weight

N 18

N* 0

Variable Weight

Maximum 1.1400

Mean 0.9317

SE Mean 0.0178

StDev 0.0753

Minimum 0.8400

Q1 0.8800

Median 0.9100

Q3 9800

For confidence coefficient .95, a= .05 and a/l = .05/2 = .025. From Table IV, Appendix A, withdf=«- 1 = 18- 1 = 17, /o25 = 2.110. t The 95% confidence interval is;

^ ± ^25 -7- ^ .9317 ± 2.110^^^ => .9317 ± .0374 =:> (.8943, .9691) y/n V18 We are 95% confident that the true mean weight of the scallops is between .8943 and .9691. Recall that the weights have been scaled so that a mean weight of 1 corresponds to 1/36 of a pound. Since the above confidence interval does not include 1, we have sufficient evidence to indicate that the minimum weight restriction was violated.

156

Chapter 5

Inferences Based on a Single Sample: Tests of Hypothesis 6.1

The null hypothesis is the "status quo" hypothesis, while the alternative hypothesis is the research hypothesis.

6.3

The "level of significance" of a test is a. This is the probability that the test statistic will fall in the rejection region when the null hypothesis is true.

6.5

The four possible results are: 1. 2. 3. 4.

6.7

Rejecting the null hypothesis when it is true. This would be a Type I error. Accepting the null hypothesis when it is true. This would be a correct decision. Rejecting the null hypothesis when it is false. This would be a correct decision. Accepting the null hypothesis when it is false. This would be a Type II error,

When you reject the null hypothesis in favor of the alternative hypothesis, this does not prove the alternative hypothesis is correct. We are 100(1 - a)% confident that there is sufficient evidence to conclude that the alternative hypothesis is correct. If we were to repeatedly draw samples from the population and perform the test each time, approximately 100(1 - a)% of the tests performed would yield the correct decision.

6.9

Let p = proportion of liars correctly detected by the new thermal imaging camera. To determine if the camera can correctly detect liars 75% of the time, we test: //o: p=.75

//a: ;?^.75 6.11

a.

To determine if the percentage of senior women who use herbal therapies to prevent or treat health problems is different from 45%, we test: //o: p = .45 //a: ^9^.45

b.

To determine if the average number of herbal products used in a year by senior women who use herbal therapies is different from 2.5, we test: //o: //=2.5

//a: /^^2.5 6.13

Let p = error rate of the DNA-reading device. To determine if the error rate of the DNAreading device is less than 5%, we test: //o: ;7 = .05 H,: p < .05

Inferences Based on a Single Sample: Tests of Hypothesis

157

6.15

a.

To determine if the average level of mercury uptake in wading birds in the Everglades today is less than 15 parts per million, we test: Ho.^= 15

6.17

b.

A Type I error is rejecting Hq when Ho is true. In terms of this problem, we would be concluding that the average level of mercury uptake in wading birds in the Everglades today is less than 15 parts per million, when in fact, the average level of mercury uptake in wading birds in the Everglades today is equal to 15 parts per million.

c.

A Type II error is accepting Ho when Ho is false. In terms of this problem, we would be concluding that the average level of mercury uptake in wading birds in the Everglades today is equal to 15 parts per million, when in fact, the average level of mercury uptake in wading birds in the Everglades today is less than 15 parts per million.

a.

The null hypothesis is: Ho'. No intrusion occurs

b.

The alternative hypothesis is: Ha'. Intrusion occurs

c.

a = Probability of a Type I error = Probability of rejecting Ho when it is true = P(system provides warning when no intrusion occurs) = 1 / 1000 = .001. Probability of a Type II error = Probability of accepting Ho when it is false = P(system does not provide warning when an intrusion occurs) = 500 / 1000 = .5.

6.19

6.21

There are 2 conditions required for a valid large-sample hypothesis test for /^. They are: 1.

A random sample is selected from a target population.

2.

The sample size n is large, i.e., n > 30. (Due to the Central Limit Theorem, this condition guarantees that the test statistic will be approximately normal regardless of the shape of the underlying probability distribution of the population.)

a.

Probability of Type I error = P(z > 1.96) = .5 - .4750 = .0250 (From Table III, Appendix A)

b. Probability of Type I error = P(z > 1.645) = .5 - .4500 = .05 (From Table III, Appendix A)

c. Probability of Type I error = P(z>2.58) = .5 -.4951 = .0049 (From Table III, Appendix A)

Rajectton region

158

Chapter 6

d. Probability of Type I error = P{z < -1.28) = .5 - .3997 = .1003 (From Table III, Appendix A) Rejection region

e. Probability of Type I error = P(z 1.645) = .5 - ,4500 + .5 - .4500 = .05 + .05 = .10 (From Table III, Appendix A) Rejection region

Rejection region

f. Probability of Type I error = F(z 2.58) = .5 - .4951 + .5 - .4951 = .0049 + .0049 = .0098 (From Table III, Appendix A)

6.23

a.

Hq. //= 100 H^\

//>

100

The test statistic is z =

X- jLl^ _ X-/Jq

(Tj

a/^fn

110-100

1.67

60/VlOO

The rejection region requires a= .05 in the upper tail of the z distribution. From Table III, Appendix A, zqs = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic falls in the rejection region, (z = 1.67 > 1.645), Hq is rejected. There is sufficient evidence to indicate the true population mean is greater than 100 at a = .05. b.

Ho:

100

... x-Uq 110-100 , The test statistic is z =-^ =-7= = 1.67 (T60/vlOO The rejection region requires a/2 = .05/2 = .025 in each tail of the z distribution. From Table III, Appendix A, Z025 = 1.96. The rejection region is z < -1.96 or z > 1.96. Since the observed value of the test statistic does not fall in the rejection region, (z = \.61 i> \ ,96), Ho is not rejected. There is insufficient evidence to indicate n does not equal 100 at a = .05.

Inferences Based on a Single Sample: Tests of Hypothesis

159

c.

In part a, we rejected Hq and concluded the mean was greater tlian 100. In part b, we did not reject Hq. There was insufficient evidence to conclude the mean was different from 100. Because the alternative hypothesis in part a is more specific than the one in

b, it is easier to reject Hq. 6.25

a.

A Type I error would be to conclude that the mean response for the population of all New York City public school children is not 3 when, in fact, the mean response is equal to 3. A Type II error would be to conclude that the mean response for the population of all New York City public school children is equal to 3 when, in fact, the mean response is not equal to 3.

b.

To determine if the mean response for the population of all New York City public school children is different from 3, we test; //o:

^=3

//al

The test statistic is z = -—— = (jy

I 05

^ = -85.52

Vl 1,160 The rejection region requires a/2 = .05/2 = .025 in each tail of the z distribution. From Table III, Appendix A, Z025 = 1.96. The rejection region is z < -1.96 or z > 1.96. Since the observed value of the test statistic falls in the rejection region (z = -85.52 < -1.96), Ho is rejected. There is sufficient evidence to indicate the mean response for the population of all New York City public school children different from 3 at cr= .05. c.

To determine if the mean response for the population of all New York City public school children is different from 3, we test: //o:

M=3

The test statistic is z = -—— =

^ = -85.52 Vl 1,160

The rejection region requires a/2 = .10/2 = .05 in each tail of the z distribution. From Table III, Appendix A, z.os = 1.645. The rejection region is z < -1.645 or z > 1.645. Since the observed value of the test statistic falls in the rejection region (z = -85.52 < -1.645), Hq is rejected. There is sufficient evidence to indicate the mean response for the population of all New York City public school children different from 3 at 0?= .10.

160

Chapter 6

6.27

To determine if the true mean heart rate during laughter exceeds 71 beats/minute, we test: //o:

/4: /W>71 X

—

u

The test statistic is z =-—

The rejection region requires a = .05 in the upper tail of the z distribution. From Table III, Appendix A, zqs = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic falls in the rejection region (z = 3.95 > 1.645). Hq is rejected. There is sufficient evidence to indicate the true mean heart rate during laughter exceeds 71 beats/minute at «= .05. 6.29

a.

To determine if the true mean PTSD score of all World War II avaiator POWs is less than 16, we test: //o: //=16

//
2.58. Since the observed value of the test statistic falls in the rejection region (z = 4.03 > 2.58), Hq is rejected. There is sufficient evidence to indicate that the true population mean ratio of all bones of this particular species differs from 8.5 at or = .01. b. The practical implications of the test in part a is that the species from which these particular bones came from is probably not species A. 6.33

a.

Since the hypothesized value of /41 ($60,000) falls in the 95% confidence interval, it is not an unusual value. There is no evidence to reject it. There is no evidence to indicate that the mean salary of all males with post-graduate degrees differs from $60,000.

b.

To determine if the mean salary of all males with post-graduate degrees differs from $60,000, we test: Hq\ fj^ — 60,000 F/a- //m

60,000 61,340-60,000

The test statistic is

0.61

2,185 The rejection region requires dl = .05/2 = .025 in each tail of the z distribution. From Table III, Appendix A, Z025 = 1.96. The rejection region is z < -1.96 or z > 1.96. Since the observed value of the test statistic does not fall in the rejection region (z = 0.61 -i> 1.96), 7/0 is not rejected. There is insufficient evidence to indicate the mean salary of all males with post-graduate degrees differs from $60,000 at a= .05. c. The inferences in parts a and b agree because the same values are used for both. The z value used in the 95% confidence interval is the same z used for the rejection region. The value of x and 5- are the same in both the confidence interval and the test statistic. d.

Since the hypothesized value of //p ($33,000) falls in the 95% confidence interval, it is not an unusual value. There is no evidence to reject it. There is no evidence to indicate that the mean salary of all females with post-graduate degrees differs from $33,000.

e. To determine if the mean salary of all females with post-graduate degrees differs from $33,000, we test: ^0' /4i ~ 33,000 74/1^33,000

The test statistic is

X-/4

32,227-33,000 932

162

-0.83

Chapter 6

The rejection region requires dl - .05/2 = .025 in each tail of the z distribution. From Table III, Appendix A, Z025 - 1.96. The rejection region is z < -1.96 or z > 1.96. Since the observed value of the test statistic does not fall in the rejection region (z = -0.83 ft -1.96), //o is not rejected. There is insufficient evidence to indicate the mean salary of all females with post-graduate degrees differs from $33,000 at a= .05. f.

6.35

The inferences in parts d and e agree because the same values are used for both. The z value used in the 95% confidence interval is the same z used for the rejection region. The value of x and s- are the same in both the confidence interval and the test statistic. To determine if the mean social interaction score of all Connecticut mental health patients differs from 3, we test:

H,:

/i = 3

Ha. The test statistic is z _

2.95-3

1.10/V6,681

= -3.72

The rejection region requires dl = .01/2 = .005 in each tail of the z distribution. From Table III, Appendix A, zoo5 = 2.58. The rejection region isz < -2.58 orz > 2.58. Since the observed value of the test statistic falls in the rejection region (z = -3.72 < -2.58), T/o is rejected. There is sufficient evidence to indicate that the mean social interaction score of all Connecticut mental health patients differs from 3 at or = .01.

6.37

b.

From the test in part a, we found that the mean social interaction score was statistically different from 3. However, the sample mean score was 2.95. Practically speaking, 2.95 is very similar to 3.0. The very large sample size, n = 6,681, makes it very easy to find statistical significance, even when no practical significance exists.

c.

Because the variable of interest is measured on a 5-point scale, it is very unlikely that the population of the ratings will be normal. However, because the sample size was extremely large, {n = 6,681), the Central Limit Theorem will apply. Thus, the distribution of 3c will be normal, regardless of the distribution of jc. Thus, the analysis used above is appropriate.

In general, small /7-values support the alternative hypothesis, The /7-value is the probability of observing your test statistic or anything more unusual, given Hq is true. If the /?-value is small, it would be very unusual to observe your test statistic, given Ha is true. This would indicate that Ha is probably false and H^ is probably true.

Inferences Based on a Single Sample: Tests of Hypothesis

163

6.39

6.41

a.

Since the /7-value = .10 is greater than a = .05, //o is not rejected.

b.

Since the/?-value = .05 is less than a= .10, i/o is rejected.

c.

Since the/?-value = .001 is less than a = .01, //q is rejected.

d.

Since the /7-value = .05 is greater than a = .025, //o is not rejected.

e.

Since the /7-value = .45 is greater than « = .10, //o is not rejected.

/7-value = P{z > 2.17) = .5 - P(0 < z < 2.17) = .5 - .4850 = .0150 (using Table III, Appendix A) X-

6.43

jiiQ _ 49.4-50

o-j

” 4.1/VT^

/7-value = P{z > -1.46) = .5 + .4279 = .9279 There is no evidence to reject 6.45

a.

a < .10.

The/7-vaIue reported by SPSS is for a two-tailed test. Thus, P{z < -1.63) + P{z > 1.63) = .1032. For this one-tailed test, the /7-value = P{z < -1.63) = .1032/2 = .0516. Since the/7-value = .0516 > a = .05, Hq is not rejected. There is insufficient evidence to indicate //< 75 at «= .05.

b.

For this one-tailed test, the/7-value = P(z < 1.63). Since P(z cr= .10, TTo is not rejected. There is insufficient evidence to indicate /y< 75 at a= .10.

c.

For this one-tailed test, the/7-value = P(z > 1.63) = .1032/2 = .0516.

Since the /7-value = .0516 < Qr= .10, T/q is rejected. There is sufficient evidence to indicate //> 75 at cr= .10. d.

For this two-tailed test, the /7-value = .1032. Since the /7-value = .1032 > a = .01, i/o is not rejected. There is insufficient evidence to indicate 75 at a = .01.

6.47

164

a.

From Exercise 6.26, z = -.40. The/7-value is/7 = P{z < -.40) = .5 - . 1554 = .3446 (using Table III, Appendix A).

b.

The/7-value is/7 = .3446. Since this is greater than a = .01, Hq is not rejected. There is insufficient evidence to indicate the mean number of latex gloves used per week by hospital employees diagnosed with a latex allergy from exposure to the powder on latex gloves is less than 20 at a = .01.

Chapter 6

6.49

From the printout, thep-value < .0001. Since the/?-value < .0001 < a = .01, is rejected. There is sufficient evidence to indicate that the true population mean ratio of all bones of this particular species differs from 8.5 at a= .01.

6.51

a.

The test statistic is z = ——— = =2.3 cTj 31.3/V50

b.

To determine if the mean level of feminization differs from 0%, we test: //o: /^=0 //a-

^0

Since the alternative hypothesis contains this is a two-tailed test. The /?-value is p = P{z< -2.3) + P{z > 2.3) = .5 - P{-23 < z < 0) + .5 - i’(0 < z < 2.3) = .5 - .4893 + .5 - .4893 = .0214.

b.

The test statistic is z = ——— - ——%= =4.22. o-j 25.1/V50 Since the alternative hypothesis contains this is a two-tailed test. The p-va\uQ is p = P{z< -4.22) + P(z > 4.22) = .5 - P(-4.22 < z < 0) + .5 - P(0 < z < 4.22)« .5-.5+ .5-.5 = 0.

6.53

a.

z = ^-^° =52-^~^^=1.29 .a7.1/V50 /?-value = P(z < -1.29) + P(z > 1.29) = (.5 - .4015) + (.5 - .4015) = .0985+ .0985 = .1970 ^52.3-51^^ 29

b.

/j-value = P(z > 1.29) = (.5 - .4015) = .0985

c. 0-3,

10.4/V^

;?-value = P(z < -0.88) + P(z > 0.88) = (.5 - .3106) + (.5 - .3106) = .1894 + .1894 = .3788 d.

For part a, any value of a greater than .1970 would lead to the rejection of the null hypothesis. For part b, any value of a greater than .0985 would lead to the rejection of the null h^othesis. For part c, any value of or greater than .3788 would lead to the rejection of the null hypothesis.

Inferences Based on a Single Sample; Tests of Hypothesis

165

e.

For /7-value of .01 and a two-tailed test, we need to find a z-value so that .01/2 - .005 is to the right of it. From Table III, Appendix A, z qo5 2.58.

2.58 =

Z-'

52.3-51

sjSo

2.58

= 1.3

j=

V50

1.3(V^) 5

= 3.56

2.58

For any value of s less than or equal to 3.56, the/?-value will be less than or equal to .01 for a two-tailed test. Forp-value of .01 and a one-tailed test, we need to find a z-value so that .01 is to the right of it. From Table III, Appendix A, zoi = 2.33.

z=

2.33 = CT^

52.3-51

s/JsO

= 1.3=>5 =

2.33 V50

1.3(7^) 5

= 3.95

2.33

For any value of 5 less than or equal to 3.95, the/?-value will be less than or equal to .01 for a one-tailed test. 6.55

We should use the t distribution in testing a hypothesis about a population mean if the sample size is small, the population being sampled from is normal, and the variance of the population is unknown.

6.57

a = P(Type I error) = P(Reject Hq when Hq is true)

6.59

a.

a = P(t > 1.440) where df = 6 = .10 Table IV, Appendix A

b.

c.

a = P(r 1.782) = .05 Table IV, Appendix A a = P{t < -2.060 or / > 2.060) where df = 25 = 2P(/> 2.060) = 2(.025) = .05 Table IV, Appendix A

a.

Hq. //=6

The test statistic is t

^ sl4n

^ = -2.064. 1.3/V5

The necessary assumption is that the population is normal. The rejection region requires a= .05 in the lower tail of the t distribution with df= w—1 = 5—1=4. From Table IV, Appendix A, tos ~ 2.132. The rejection region is r 2.776.

/025

= 2.776. The rejection region

Since the observed value of the test statistic does not fall in the rejection region {t = -2.064 ■/: -2.776), Hq is not rejected. There is insufficient evidence to indicate the mean is different from 6 at a = .05. c.

For part a, the /7-value = P{t < -2.064). From Table IV, with df= 4, .05 .076. There is evidence to indicate the mean is different than 1000 for a> .076.

Inferences Based on a Single Sample: Tests of Hypothesis

167

6.63

a.

To determine if the mean breaking strength of the new bonding adhesive is less than 5.70 Mpa, we test: Hq\ //=5.70 /i 2.056. Since the observed value of the test statistic does not fall in the rejection region {t = -.43 ft -2.056), T/o is not rejected. There is insufficient evidence to indicate the mean Dental Anxiety Scale score for college students differs from 11 at a = .05. 6.67

To determine if the mean amount of cesium in lichen specimens differs from .003, we test: //o: //=.003 H^\ jLi ^ .003 From the printout, the test statistic is t = 3.725 and the/?-value \sp = .0058. Since the p-value is less than a{p = .0058 < a= .10), //o is rejected. There is sufficient evidence to indicate the mean amount of cesium in lichen specimens differs from 003 at

a:=.10.

168

Chapter 6

6.69

From Exercise 5.40, x = 358.45 and s = 117.8172. To determine if the mean skidding distance is less than 425 meters, we test: Ho: //=425 Ha'. ju< 425 X — JU

358.45-425

si yin

117.8172

The test statistic is t =-p

= -2.53

The rejection region requires a = .10 in the lower tail of the /-distribution with df = « - 1 = 20 ~ 1 = 19. From Table IV, Appendix A, / lo - 1.328. The rejection region i5 / < -1.328. Since the observed value of the test statistic falls in the rejection region (/ = -2.53 < -1.328), Ho is rejected. There is sufficient evidence to indicate the mean skidding distance is less than

425 meters at a = .10. Thus, there is enough evidence to refute the claim. 6.71

To determine if the true mean PST score of all patients with mild to moderate traumatic brain injury exceeds the "normal" value of 40, we test: Ho: /i = 40 H^: /i>40

... x-LL 48.43-48 , The test statistic is / =-=-j= = 1.95 sNn 20.76/V23 The rejection region requires a = .05 in the upper tail of the t distribution with df = « - 1 = 23 - 1 = 22. From Table IV, Appendix A, t,os = 1.717. The rejection region is /> 1.717. Since the observed value of the test statistic falls in the rejection region (/= 1.95 > 1.717), Ho is rejected. There is sufficient evidence to indicate that the true mean PST score of all patients with mild to moderate traumatic brain injury exceeds the "normal" value of 40 at a=.05. 6.73

Typically, qualitative data are associated with making inferences about a population proportion.

6.75

The sample size is large enough if both npo and nqo are greater than or equal to 15. a.

n/7o = 500(.05) = 25 and = 500(1 - .05) = 500(.95) = 475. Since both of these values are greater than 15, the sample size is large enough to use the normal approximation.

b.

= 100(.99) = 99 and = 100(1 - .99) = lOO(.Ol) = 1. Since the second number is less than 15, the sample size is not large enough to use the normal approximation.

c.

npo - 50(.2) = 10

d.

= 20(.2) = 4 and wg-o = 20(1 - .2) = 20(.8) = 16. Since the first number is less than 15, the sample size is not large enough to use the nonnal approximation.

and nq^ = 50(1 - .2) = 50(.8) = 40. Since the first number is less than 15, the sample size is not large enough to use the normal approximation.

Inferences Based on a Single Sample; Tests of Hypothesis

169

e.

6.77

npo = 10(.4) = 4 and wg'o = 10(1 - .4) = 10(.6) = 6. Since both numbers are less than 15, the sample size is not large enough to use the normal approximation.

z=

P-Pa

.84-.9

IPo^o

•9(.l)

n

100

2.00

= -

.75(.25) •9(.l) The denominator in Exercise 6.76 is . --=.0433 as compared to J———-03 in part V 100 V 100 a. Since the denominator in this problem is smaller, the absolute value of z is larger. c.

The rejection region requires a = .05 in the lower tail of the z distribution. From Table

in. Appendix A, zqs = 1.645.

The rejection region is z < -1.645.

Since the observed value of the test statistic falls in the rejection region (z = -2.00 < -1.645), Bo is rejected. There is sufficient evidence to indicate the population proportion is less than .9 at a= .05. d. 6.79

Thep-value = F(z < -2.00) = .5 - .4772 = .0228 (from Table III, Appendix A).

From Exercise 5.50, « = 50 and sincep is the proportion of consumers who do not like the snack food, will be: . Number of O's in sample 29 p =-= —= .58 « 50 In order for the inference to be valid, the sample size must be large enough. The sample size is large enough if both npo and nqo are greater than or equal to 15. npo - 50(.5) - 25 and nqo - 50(1 - .5) = 50(.5) = 25. Since both of these values are greater than 15, the sample size is large enough to use the normal approximation. a.

Ho. p^.S H,: p>.5 The test statistic is z =

=1.13 Ml

in

-5(1-.5)

V

50

The rejection region requires a = .10 in the upper tail of the z distribution. From Table III, Appendix A, Z io = 1.28. The rejection region is z > 1.28. Since the observed value of the test statistic does not fall in the rejection region (z = 1.13 5^^ 1.28), Bo is not rejected. There is insufficient evidence to indicate the proportion of customers who do not like the snack food is greater than .5 at a = .10. b.

170

;7-value = P(z > 1.13) = .5 - .3708 = . 1292

Chapter 6

6.81

a.

The population parameter of interest is p, the proportion of items that had the wrong price when scanned through the electronic checkout scanner at California Wal-Mart stores.

b. To determine if the true proportion of items scanned at California Wal-Mart stores exceeds the 2% NIST standard, we test: Hq. p = .02 //a: p > .02

c. The test statistic is z = ^ &

in

14.23 •02(.98)

i

1000

The rejection region requires «= .05 in the upper tail of the z-distribution. From Table III, Appendix A, zqs = 1.645. The rejection region is z > 1.645. d.

Since the observed value of the test statistic falls in the rejection region (z = 14.23 > 1.645), No is rejected. There is sufficient evidence to indicate the true proportion of items scanned at California Wal-Mart stores exceeds the 2% NIST standard at or = .05.

e.

In order for the inference to be valid, the sample size must be large enough. The sample size is large enough if both npo and nqo are greater than or equal to 15. npo - 1000(.02) = 20 and nqo - 1000(1 - .02) = 1000(.98) = 980. Since both of these values are greater than 15, the sample size is large enough to use the normal approximation.

6.83

Some preliminary calculations: X _ 401

P

-.48

n ~ 835

In order for the inference to be valid, the sample size must be large enough. The sample size is large enough if both npo and nqo are greater than or equal to 15. npo = 835(.45) = 375.75 and nqo = 835(1 - .45) - 835(.55) = 459.25. Since both of these values are greater than 15, the sample size is large enough to use the normal approximation. To determine if more than 45% of male youths are raised in a single-parent family, we test: /7o: p = .45 //a: p> AS ... * p — Po .48-.45 , _. The test statistic is z = . ■ = , — ^ = 1.74

\Pq% in

A5(.55) i

835

Inferences Based on a Single Sample: Tests of Hypothesis

171

The rejection region requires a= .05 in the upper tail of the z distribution. From Table III, Appendix A, 2.05 = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic falls in the rejection region (z = 1.74 > 1.645), Hq is rejected. There is sufficient evidence that more than 45% of male youths are raised in a single-parent family at a= .05.

6.85

The point estimate for

,

is p = — n

71 + 68

139

678

678

= .205

In order for the inference to be valid, the sample size must be large enough. The sample size is large enough if both np^ and nqo are greater than or equal to 15. npo = 678(.25) = 169.5 and nq^ = 678(1 - .25) = 678(.75) = 508.5. Since both of these values are greater than 15, the sample size is large enough to use the normal approximation. To determine if the true percentage of appealed civil cases that are actually reversed is less than 25%, we test: H,-. p = 25 H,: p 7) where x is a binomial random variable with n = 10 and p= .5. From Table II, /7-value - lP{x > 7) = 2(1 - P(x < 6)) = 2(1 - .828) = .344 Sinee the /7-value = .344 > a = .05, Ho is not rejected. There is insufficient evidence to indicate the median is different than 9 at «= .05.

c.

Ho". 7] = 29 H,: 7] 9) where x is a binomial random variable with 77 = 10 and p = .5. From Table II, /7-value = P{x > 9) = I - P{x < 8) = 1 - .989 = .011 Since the /7-value = .011 < a = .05, Ho is rejected. There is sufficient evidence to indicate the median is less than 20 at a = .05.

d.

Ho: 7] = 29 H^: rj^ 29 S\ = (Number of observations less than 20} = 9 and iS'2 = (Number of observations greater than 20} = 1 The test statistic is iS" = larger of S\ and S2 = 9. The /7-value = 2P{x > 9) where x is a binomial random variable with « = 10 and p= .5. From Table II, /7-value = 2P(x > 9) = 2(1 - P(x < 8)) = 2(1 - .989) = .022 Since the /7-value = .022 < a = .05, Ho is rejected. There is sufficient evidence to indicate the median is different than 20 at a = .05.

e.

For all parts, p = np= 10(.5) = 5 and a= -^npg = ^J\ 0(.5)(.5) = 1.581.

( For part a, P{x >7)x P z >

a-.5)-s) _ 1.581

/’(z>.95) = .5 -.3289 = .1711

j

This is close to the probability .172 in part a. The conclusion is the same.

(

For part b, 2P{x > 7)« 2P z > ■ V

(7-.5)-s'!

2P{z > .95) = 2(.5 - .3289)

1.581 = .3422

Inferences Based on a Single Sample; Tests of Hypothesis

175

This is close to the probability .344 in part b. The conclusion is the same. For part c, P{x > 9)

z>

(9-.5)-5^ 1.581

J = /’(z>2.21) = .5-.4864 = .0136

This is close to the probability .011 in part c. The conclusion is the'same. For part d, 2P{x > 9)» 2P z >

(9-.5)-5

= 2P(z> 2.21 ) = 2(.5 - .4864)

1.581 = .0272 This is close to the probability .022 in part d. The conclusion is the same. f.

6.97

We must assume only that the sample is selected randomly from a continuous probability distribution.

To determine if the median cesium amount of lichen differs from .003, we test: Ho.

T]=.m

Ha’. T] ^ .003 The test statistic is .S' = {number of observations greater than .003} = 8 (from printout). The /?-value is/? = .0391. Since the p-value is less than a = .10, //o is rejected. There is sufficient evidence to indicate the median cesium amount of lichen differs from .003 at ar = .10. This result agrees with the t-test in Exercise 6.65. 6.99

a.

To determine if the median daily ammonia concentration for all afternoon drive-time days exceeds 1.5 ppm, we test: Ho: 7;=1.5 Ha. r]>\.5

b.

The test statistic is 5” = number of measurements greater than 1.5 = 3.

c.

The p-value = P{x > 3) where x is a binomial random variable with « = 8 and p = .5. From Table II, Appendix A, P(x>3)= 1 -P(x .05), Ho is not rejected. There is insufficient evidence to indicate that the median daily ammonia concentration for all afternoon drive-time days exceeds 1.5 ppm at a= .05.

Chapter 6

6.101

a.

To detennine if the population median chromatic contrast of spiders on flowers is less than 70, we test: ■Hq\ 7 = 70 /7a: /? < 70

b.

The test statistic is .S' = number of measurements less than 70 = 7.

c.

The /?-value = P(x > 7) where x is a binomial random variable with « = 10 and p= .5. From Table II, Appendix A, P(jc > 7) = 1 -P(x < 6) = 1 - .828 = .172

6.103

d.

Since the p-value is greater than «= .10 (/? = .172 > .10), //o is not rejected. There is insufficient evidence to indicate the population median chromatic contrast of spiders on flowers is less than 70 at cr= .10.

a.

To determine whether the median biting rate is higher in bright, sunny weather, we test: Ho: t] = 5 7>5

H,:

The test statistic is z

(S~.5)-.5n

.5V^

(95-.5)-.5(122) _ = 6.07

.5Vm

(where S = number of observations greater than 5) The /7-value is /? = P(z > 6.07). From Table III, Appendix A,p = P(z > 6.01)« 0.0000. .

6.105

c.

Since the observed /7-value is less than a(p = 0.0000 < .01), Ho is rejected. There is sufficient evidence to indicate that the median biting rate in bright, sunny weather is greater than 5 at cr= .01.

To determine if 50% of the ingots have a freckle index of 10 or higher, we test: Ho-.

7=10

H,: 7 >10 S = number of measurements greater than 10 = 10.

The test statistic is

(6'-.5)-.577 .5^77

(10-5)^5(18) ^0 ^^ .5Vl8

The /7-value = P(z > 0.24). From Table III, Appendix A,p= P{z > 0.24) = .5 - .0948 = .4052. Since the/>-value is greater than a= .0\ (p = .4052 > .01), Ho is not rejected. There is insyfificient evidence to indicate that 50% of the ingots have a freckle index of 10 or higher at cr= .01.

Inferences Based on a Single Sample; Tests of Hypothesis

177

6.107

Tlie smaller the /?-value associated with a test of hypothesis, the stronger the support for the alternative hypothesis. The p-value is the probability of observing your test statistic or anytliing more unusual, given the null hypothesis is true. If this value is small, it would be very unusual to observe this test statistic if the null hypothesis were true. Thus, it would indicate the alternative hypothesis is true.

6.109

The elements of the test of hypothesis that should be specified prior to analyzing the data are: null hypothesis, alternative hypothesis, and significance level.

6.111

The larger the /7-value associated with a test of hypothesis, the stronger the support for the null hypothesis. The /7-value is the probability of observing your test statistic or anything more unusual, given the null hypothesis is true. If this value is large, it would not be very unusual to observe this test statistic if the null hypothesis were true. Thus, it would lend support that the null hypothesis is true.

6.113

a.

I

H^: p = 35 p 1) where x is a binomial random variable with « = 10 andp = .5. From Table II, /?-value = P{x > 7) = 1 - P{x < 6) = 1 - .828 = .172

6.117

b.

Since the /j-value = . 172 > or = .05, Hq is not rejected. There is insufficient evidence to indicate the median is greater than 150 at .05.

a.

Since the company must give proof the drug is safe, the null hypothesis would be the drug is unsafe. The alternative hypothesis would be the drug is safe.

b.

A Type I error would be concluding the drug is safe when it is not safe. A Type II error would be concluding the drug is not safe when it is. a is the probability of concluding the drug is safe when it is not. y^is the probability of concluding the drug is not safe when it is.

6.119

c.

In this problem, it would be more important for a to be small. We would want the probability of concluding the drug is safe when it is not to be as small as possible.

a.

To determine if the true mean score of all sleep-deprived subjects is less than 80, we test: Ho\ /^=80 H^\ //8) = F(x =8) + F(x = 9) + F(x=lO) + F{x = 11) + P(x = 12)

v8y

'12'

'12"

Uy

JO,

ri2^

12 cl2-12

,5'^5

vlly

12!

12!

12!

5"_5i2-ii

10 C12-10

.5‘".5

-.5*.5''+.5".5'+10!(12-10)! 8!(12-8)! 9!(12-9)! + ■

12! 12! ■.5".5‘ +■ 12!(12-12)! 11!(12-11)!

12! .,2 , 12! ,,2 ,

12!

5‘"+—.5‘"+8!4! 9!3! 10!2!

. 12! ,,2 . 11!1!

12! ,,2 12!0!

=. 1208 + .0537 + .0161 + .0029 + .0002 =. 1937 Since the /7-value = .1937 > a = .05,

is not rejected. There is insufficient evidence

to indicate the median score of sleep deprived students is below 80 at a = .05. 6.121

a.

Let p = proportion of the subjects who felt that the masculinization of face shape decreased attractiveness of the male face. If there is no preference for the unaltered or morphed male face, then p = .5. For this problem, p = xin = 58/67 = .866. In order for the inference to be valid, the sample size must be large enough. The sample size is large enough if both npo and nq^ are greater than or equal to 15. npo = 67(.5) = 33.5 and nqo = 67(1 - .5) = 67(.5) = 33.5. Since both of these values are greater than 15, the-sample size is large enough to use the normal approximation. To determine if the subjects showed a preference for either the unaltered or morphed face, we test: Hq: p = .5 p^,5

180

Chapter 6

b.

The test statistic is z =

= Mo. in

c.

5 99

•50(.50) V

67

By definition, the p-value is the probability of observing our test statistic or anything more unusual, given Ho is true. For this problem,p = P(z < -5.99) + P{z > 5.99) = .5 -P(-5.99 < z < 0) + .5 -P(0 < z < 5.99) « .5 - .5 + .5 - .5 = 0. This corresponds to what is listed in the Exercise.

d.

The rejection region requires odl = .01/2 = .005 in each tail of the z distribution. From Table III, Appendix A, zqos = 2.575. The rejection region is z < -2.575 or z > 2.575. Since the observed value of the test statistic falls in the rejection region (z = 5.99 > 2.575), Ho is rejected. There is sufficient evidence to indicate that the subjects showed a preference for either the unaltered or morphed face at a = .01.

6.123

To determine if the mean alkalinity level of water in the tributary exceeds 50 mpl, we test:

//o:/i=50 H^'. //> 50 ... 3c-r/o 67.8-50 The test statistic is z =-— =-;== = 12.36 14.4/VI00 The rejection region requires a = .01 in the upper tail of the z distribution. From Table III, Appendix A, zoi = 2.33. The rejection region is z > 2.33.

.

6.125

Since the observed value of the test statistic falls in the rejection region (z = 12.36 > 2.33), Ho is rejected. There is sufficient evidence to indicate that the mean alkalinity level of water in the tributary exceeds 50 mpl at a= .01. Using MINITAB, the descriptive statistics are: Descriptive Statistics: Weight Variable Weight

N 16

N* 0

Mean 1.373

SE Mean 0.258

StDev 1.034

Minimum 0.201

Q1 0.475

Median 1.347

Q3 2.009

Maximum 3.751

To determine whether the mean weight in the crops of all spinifex pigeons differs from 1 gram, we test:

Ho: p=\ H,: 1 ...

Pn

1.373-1

, . .

The test statistic \s t- —=-7-7==- = 1.44 I.O34/VI6

s/^

The rejection region requires a/2 = .05/2 = .025 in each tail of the t distribution with df= n-l = 16-l = 15. From Table IV, Appendix A, fo25 = 2.131. The rejection region is t2.131.

Inferences Based on a Single Sample: Tests of Hypothesis

181

Since the observed value of the test statistic does not fall in the rejection region (r = 1.44 > 2.131), Hq is not rejected. There is insufficient evidence to indicate the mean weight in the crops of all spinifex pigeons differs from 1 gram at of = .05. We must assume that the population being sampled from is normal. The stem-and-leaf display of the data is;

>

Stem-and-Leaf Display: Weight stem-and-leaf of Weight Leaf Unit = 0.10

N

=16

4 0 2234 7 0 588 (4) 1 3344 5 16 4 2 1 3 2 59 1 3 13 7

,

‘

The data do not appear to be mound-shaped. Thus, the test above may not be valid. 6.127

a.

To determine if the true mean inbreeding coefficient /i for this species of wasp exceeds 0, we test: Hq. //=0 //a: /«>0 _ ... 3c - /7o The test statistic is z =-^ CTj

.044 - 0

= 0.70

.884/vl^

The rejection region requires or = .05 in the upper tail of the z distribution. From Table III, Appendix A, z.os = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic does not fall in the rejection region (z = 0.70 > 1.645), Hq is not rejected. There is insufficient evidence to indicate that the true mean inbreeding coefficient p for this species of wasp exceeds 0 at or = .05.

6.129

b.

This result agrees with that of Exercise 5.101. The confidence interval in Exercise 7.101 was (-.118, .206). Since this interval contains 0, there is no evidence to indicate that the mean is different from 0. This is the same conclusion that was reached in part a.

a.

To determine if the production process should be halted, we test: //o: yw=3 //a: //>3 where fj. = mean amount of PCB in the effluent. The test statistic is z = -—— — = 1.41 CTj .S/yJSO

182

Chapter 6

The rejection region requires a =.01 in the upper tail of the z distribution. From Table III, Appendix A, zqi = 2.33. The rejection region is z > 2.33. Since the observed value of the test statistic does not fall in the rejection region, (z = 1.41 >> 2.33), Ffo is not rejected. There is insufficient evidence to indicate the mean amount of PCB in the effluent is more than 3 parts per million at a = .01. Do not halt the manufacturing process. b.

6.131

As plant manager, I do not want to shut down the plant unnecessarily. Therefore, I want a = /*(shut down plant when // = 3) to be small. X

The point estimate for p is p- — =-= .685 n 124 In order for the inference to be valid, the sample size must be large enough. The sample size is large enough if both npo and nqo are greater than or equal to 15. npo= 124(.167) = 20.708 and = 124 (1 - .167) = 124 (.833) = 103.292. Since both of these values are greater than 15, the sample size is large enough to use the normal approximation. To determine if the students will tend to choose the three-grill display so that Grill #2 is a compromise between a more and a less desirable grill (p > .167), we test; Hq: p=.\61 //a:

p>.\61

The test statistic is z =

= ^685_T67_ _ ^^ • 1670822) in

V

124

The rejection region requires a= .05 in the upper tail of the z distribution. From Table III, Appendix A, zqs = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic falls in the rejection region (z = 15.46 > 1.645), Ho is rejected. There is sufficient evidence to indicate the students will tend to choose the three-grill display so that Grill #2 is a compromise between a more and a less desirable grill (p > .167) at a= .05.

Inferences Based on a Single Sample: Tests of Hypothesis

183

6.133

a.

A Type I error is rejecting the null hypothesis when it is true. In this problem, we would be concluding that the individual is a liar when, in fact, the individual is telling the truth. A Type II error is accepting the null hypothesis when it is false. In this problem, we would be concluding that the individual is telling the truth when, in fact, the individual is a liar. *

b.

The probability of a Type I error would be the probability of concluding the individual is a liar when he/she is telling the truth. From the problem, it stated that the polygraph would indicate that of 500 individuals who were telling the truth, 185 would be liars. Thus, an estimate of the probability of a Type 1 error would be 185/500 = .37. The probability of a Type II error would be the probability of concluding the individual is telling the truth when he/she is a liar. From the problem, it stated that the polygraph would indicate that of 500 individuals who were liars, 120 would be telling the truth. Thus, an estimate of the probability of a Type II error would be 120/500 = .24.

6.135

a.

To determine if the true mean score of all sleep-deprived subjects is less than 80, we test: Hq\ // = 80

Ha. /i1.645

=> 1.3 >1.645(5 ) If-!-+ — 159 j 1.3

=>-> s

1.645(.1564)

^

5.053 > 5p Thus, if both 5i and 52 were less than 5.053, then we would conclude that blacks, on average would have a higher pain intensity than whites. (It is possible that either 5i or 52 could be greater than 5.053 and we would still reject Hq. As long as Sp < 5.053, we would reject Hq.) c.

From part

b, we know we would reject Ho

if Sp < 5.053. Thus, we would not reject Ho if

Sp > 5.053. This could happen if both 5i and 52 are greater than 5.053. (It is possible that either 5i or 52 could be less than 5.053 and we would still reject Ho. As long as 5.053, we would reject Ho.) 7.29

In a paired difference experiment, the observations should be paired before collecting the data.

7.31

a.

The rejection region requires a = .05 in the upper tail of the t distribution with df=nd1 = 10-1=9. From Table IV, Appendix A, /05 = 1.833. The rejection region is t> 1.833.

194

b.

From Table IV, with df = t> 1.328.

— 1 = 20 - 1 = 19, ? 10 = 1.328. The rejection region is

c.

From Table IV, with df = «d - 1 = 5 - 1 = 4, tois = 2.776. The rejection region is t> 2.116.

d.

From Table IV, with df = nj - 1 = 9 - 1 = 8, toi r> 2.896.

2.896. The rejection region is

Chapter 7

7.33

a. Pair

Difference

1

3

2

2

3

2

4

4

5

0

6

1

b. IM = IU\- 1^1 c. For confidence coefficient .95, a= .05 and all = .025. From Table IV, Appendix A, with df = «() - 1 = 6 - 1 = 5, to25 - 2.571. The cpnfidence interval is;

^d±^a/2'

2 + 2.571

A

2+ 1.484 ^(.516, 3.484)

d. Ho: Md = 0 Ha. JUd^O

The test statistic is f ==

■^d-0

ijn^

2

V2/V6

3.46

The rejection region requires all = .05/2 = .025 in each tail of the t distribution with df = «d - 1 = 6 - 1 = 5. From Table IV, Appendix A, tois ~ 2.571. The rejection region is/ 2.571. Since the observed value of the test statistic falls in the rejection region (/ = 3.46 > 2.571), Ho is rejected. There is sufficient evidence to indicate that the mean difference is different from 0 at a = .05. 7.35

a.

/7o:/Ai=10

Ha. Ad

10

The test statistic is

11.7-10

-1.79

V40

Comparing Population Means

195

The rejection region requires cdl = .05/2 = .025 in each tail of the z distribution. From Table III, Appendix A, Z025 = 1.96. The rejection region is z < -1.96 or z > 1.96. Since the observed value of the test statistic does not fall in the rejection region (z =1.79 1.96), Hq is not rejected. There is insufficient evidence to indicate the difference in the population means is different from 10 at or = .05. I %

b.

Tlie /7-value is p = P{z< -1.79>fF(z > 1.79) = (.5 - .4633) + (.5 - .4633) = .0367 + .0367 = .0734

7.37

For confidence coefficient .95, a= .05 and a/2 = .05/2 = .025. From Table IV, Appendix A, with df = Wd - 1 = 26 - 1 = 25, to25 = 2.060. The 95% confidence interval is:

—

^.025

10.5 + 2.060

7.6

10.5 ±3.07 =>(7.43, 13.57)

y/26

We are 95% confident that the difference in the mean anxiety levels between the before and after the visit is between 7.43 and 13.57. Since all of these values are positive, we can conclude that mean anxiety level after the visit is significantly less than the anxiety level before the visit. 7.39

a.

Let

= mean BOLD score for the first session and jui = mean BOLD score when the

placebo was applied. The target parameter is //d = /^i - Pi, the difference in the mean BOLD scores between the first session and the session involving the placebo. b.

A paired difference design was used to collect the data. Each experimental person had two observations measured on it.

c.

To determine if there was a placebo effect, we test: Ho: Ai = 0 H^: ju0

d.

The test statistic is

0.21-0

2.19

Since no or value was given, we will use a= .05. The rejection region requires a= .05 in the upper tail of the t distribution with df = «d - I = 24 - I = 23. From Table IV, Appendix A, r025 = 1.714. The rejection region is / > 1.714. Since the observed value of the test statistic falls in the rejection region (t = 2.\9> 1.714), Ho is rejected. There is sufficient evidence to indicate that there is a placebo effect at «= .05. e.

196

Since the p-value is less than a(p = .02 3.250. Since the observed value of the test statistic does not fall in the rejection region (/= 0.57 > 3.250), Hq is not rejected. There is insufficient evidence to indicate a difference in mean standardized growth of genes in the full-dark condition and genes in the transient dark condition at or = .01.

,

Using MfNITAB, the mean difference in standardized growth of the 103 genes in the full dark condition and the transient dark condition is: Descriptive Statistics: FD-TD Variable FD-TD

N 103

Mean -0.2274

StDev 0.9239

Minimum -2.8516

Q1 -0.8544

Median -0.1704

Q3 0.2644

Maximum 2.3998

The sample mean difference is —.2274. The test above did not detect this difference. d.

= fJQ.difference in mean standardized growth between genes in transient light condition and genes in transient dark condition. Some preliminary calculations are:

X.

Gene ID SLR2067

Transient Light 1.40989

Transient Dark -1.28569

Difference 2.69558

SLR1986

1.83097

-0.68723

2.51820

SLR3383

-0.79794

-0.39719

-0.40075

SLR0928

-1.18618

SLR0335

-1.20901 1.71404

-0.02283 2.44433

SLR1459

2.14156

SLR1326 SLR1329

1.03623

-0.33174 0.30392 0.44545

SLR1327

-0.21490 1.42608

SLR1325

1.93104

13.52175 =

10

---

-0.73029

2.47330 0.73231

-0.13664

-0.66035 1.56272

-0.24820

2.17924

= 1.352175

(13.52175)' 34.02408743-'^ ’

10 =■

-1

10-1

15.74031512

= 1.748923902

5 = Vl.748923902 =1.322468866 To determine if there is a difference in mean standardized growth of genes in the transient light condition and genes in the transient dark condition, we test:

200

0

Hq.

~

//g:

^0

Chapter 7

1.352175-0

The test statistic is

1.322468866

= 3.23

VTo

From part a, the rejection region is t < -3.250 or t > 3.250. Since the observed value of the test statistic does not fall in the rejection region (t = 3.23 >■ 3.250),//o is not rejected. There is insufficient evidence to indicate a difference in mean standardized growth of genes in the transient light condition and genes in the transient dark condition at a = .01. Using MINITAB, the mean difference in standardized growth of the 103 genes in the full-dark condition and the transient dark condition is: Descriptive Statistics: TL-TD Variable TL-TD

N 103

Mean 0.192

StDev 1.499

Minimum -3.036

Q1 -1.166

Median 0.149

Q3 1.164

Maximum 2.799

The sample mean difference is .192. The test above did not detect this difference. 7.47

Let fj.\ = mean density of the wine measured with the hydrometer and //a = mean density of the wine measured with the hydrostatic balance. The target parameter /j^ = /U\- JLI2, the difference in the mean density of the wine measured with the hydrometer and the hydrostatic balance. Using MfNITAB, the descriptive statistics for the difference data are: Descriptive Statistics: Diff Variable Diff

N 40

N* 0

Mean -0.000523

Variable Diff

Median -0.000165

SE Mean 0.000204

Q3 0.000317

StDev 0.001291

Minimum -0.004480

Q1 -0.001078

Maximum 0.001580

We will use a 95% confidence inten^al to estimate the difference in the mean density of the wine measured with the hydrometer and the hydrostatic balance. For confidence coefficient .95, a= .05 and a/2 = .05/2 = .025. From Table III, appendix A, 2,025 = 1-96. The 95% confidence interval is:

^(-.000923, -.000123)

V40

We are 95% confident that the difference in the mean density of the wine measured with the hydrometer and the hydrostatic balance is between -.000923 and -.000123. Thus, we are 95% confident that the difference in the means ranges from .000123 and .000923 in absolute value. Since this entire confidence interval is less than .002, we can conclude that the differefice in the mean scores does not exceed .002. Thus, we would recommend that the winery switch to the hydrostatic balance.

Comparing Population Means

201

7.49

We can obtain an estimate of

by using the sample variance

, or from an educated

guess based on the range - 5 w Range (of differences) / 4. 7.51

a.

For confidence coefficient .95, a = \ — .95 = .05 and otl2 = .05/2 = .025. From Table III, Appendix A,

z.025

= 1-96.

+ 50, from Table V, Appendix A, with n\ = 6, «2 = 6, and a = .05.

c.

The test statistic is T\, the rank sum of population A (because n\
Za/2- For a = .05 and all = .05/2 = .025, ^025 = 1.96 from Table III, Appendix A. The rejection region is z < -1.96 or z > 1.96. 7.65

Ho: The probability distributions of treatments A and B are identical 74: The probability distribution of treatment A lies to the left of that for treatment B The test statistic is

^

»2(»i+^2 + 1)

^ = J_2___2-_ I«,«2(«,+«2 +1) V

12

Comparing Population Means

/15(15)(15 + 15 + 1) ]l

= _2 47 24.1091

11

203

The rejection region requires a = .05 in the lower tail of the z distribution. From Table III, Appendix A, zqs = 1.645. The rejection region is z < -1.645. Since the observed value of the test statistic falls in the rejection region (-2.47 < -1.645), Ho is rejected. There is sufficient evidence to conclude that distribution A is, shifted to the left of distribution B for a= .05. 7.67

a.

To determine if the distribution of FNE scores for bulimic students is shifted above the corresponding distribution for female students with normal eating habits, we test: Hoi The distribution of FNE scores for bulimic students is the same as the corresponding distribution for female students with normal eating habits The distribution of FNE scores for bulimic students is shifted above the corresponding distribution for female students with normal eating habits

b. The data ranked are: Bulimic Students 21 13 10 20 25 19 16 21 24 13 14

Normal Students 13 6 16 13 8 19 23 18 11 19 7 10 15 20

Rank 21.5 8.5 4.5 19.5 25 17 13.5 21.5 24 8.5 11

T,= 174.5

Rank 8.5 1 13.5 8.5 3 17 23 15 6 17 2 4.5 12 19.5

T2 = 150.5

c.

The sum of the ranks of the 11 FNE scores for bulimic students is Ti = 174.5.

d.

The sum of the ranks of the 14 FNE scores for normal students is Tj = 150.5. j

e.

The test statistic is z =

”i(”i+”2+l)

11(11 + 14 + 1)

^ = —2 //7i/72(«i +/72+1) /11(14)(11 + 14 + 1) V

12

V

_ ^ ^2

12

The rejection region requires a =. 10 in the upper tail of the z distribution. From Table III, Appendix A, z jo = 1.28. The rejection region is z> 1.28.

204

Chapter 7

7.69

f.

Since the observed value of the test statistic falls in the rejection region (z = 1.72 > 1.28), Ho is rejected. There is sufficient evidence to indicate that the distribution of FNE scores for bulimic students is shifted above the corresponding distribution for female students with normal eating habits at a = .10.

a.

To determine if the recall of those receiving an audiovisual presentation is different from those receiving only a visual presentation, we test: Hq-. The probability distributions of those receiving an audiovisual presentation and those receiving a visual presentation are identical The probability distribution of those receiving an audiovisual presentation is shifted to the right or left of the probability distribution of those receiving a visual presentation

b.

First, we rank the data: A/V Group

Rank

Video Group

Rank

0 4 6 6 1 2 2 6 6 4 1 2 6 1 3 0 2 5 4 5

1.5 24 34.5 34.5 5 12 12 34.5 34.5 24 5 12 34.5 5 19 1.5 12 28 24 28 Ti =385.5

6 3 6 2 2 4 7 6 1 3 6 2 3 1 3 2 5 2 4 6

34.5 19 34.5 12 12 24 40 34.5 5 19 34.5 12 19 5 19 12 28 12 24 34.5 Tj = 434.5

'

The test statistic is

j,

^2(^1 + ^2

0

335 5

+

+0 -24.5

z=•

=

+«2 +1) 12 c.

20(20)(20 + 20 + l) V

-.66

36.9685

12

TJac rejection region requires a/2 = .10/2 = .05 in each tail of the z distribution. From Table III, Appendix A, z,05 = 1.645. The rejection region is z 1.645.

Comparing Population Means

205

d.

Since the observed value of the test statistic does not fall in the rejection region (z = -.66 ft -1.645), Bo is not rejected. There is insufficient evidence to indicate the recall of those receiving an audiovisual presentation is different from those receiving only a visual presentation at a = . 10.

7.71

a. Deaf Children 2.75 3.14 3.23 2.30 2.64 1.95 2.17 2.45 1.83 2.23

Hearing Children

Rank

1.15 1.65 1.43 1.83 1.75 1.23 2.03 1.64 1.96 1.37

18 19 20 15 17 10 13 16 8.5 14 Ti = 150.5

Rank 1 6 4 8.5 7 2 12 5 11 3 T2 = 59.5

Bo". The probability distributions of eye movement ra^tes are identical for deaf and hearing children Ba'. The probability distribution of eye movement rates for deaf children lies to the right of that for hearing children The test statistic can be either Ti or T2 since the sample sizes are equal. Use Ti = 150.5. The null hypothesis will be rejected if Ti > Tu where a= .05 (one-tailed), «i = 10 and «2 = 10. From Table V, Appendix A, Tb = 127. Reject//o iff) > 127. Since Ti = 150.5 > 127, we reject Bq. There is sufficient evidence to indicate that deaf children have greater visual acuity than hearing children at a= .05. b.

Bo: The probability distributions of eye movement rates are identical for deaf and hearing children f/a: The probability distribution of eye movement rates for deaf children lies to the right of that for hearing children j, The test statistic is 2 =

•

—7-

«i(«i+«2 + l) ^

10(10 + 10 + 1)

|»i»2(^i + ”2 +1) V

2

= —

12

_ ^

/10(10)(10 + 10 + 1) V

12

The rejection region requires a= .05 in the upper tail of the 2 distribution. From Table in. Appendix A, 205 = 1.645. The rejection region is 2 > 1.645. Since the observed value of the test statistic falls in the rejection region (z = 3.44 > 1.645), Bo is rejected. There is sufficient evidence to support the psychologist's claim that deaf children have greater visual acuity than hearing children at a = .05. The results agree with those found in part a.

206

Chapter 7

7.73

a.

Since the data are probably not from a normal distribution the Wilcoxon rank sum test should be used.

b. To determine if the relational intimacy scores for the participants in the CMC group are lower than the relational intimacy scores for the participants in the FTP group, we test: Hq\ The probability distributions of those in the CMC group and those in the FTF group are identical H^\ The probability distribution of the CMC group is shifted to the left of that for the FTF group c.

The rejection region requires «= .10 in the lower tail of the z distribution. From Table III, Appendix A, z.io = 1.28. The rejection region is z < -1.28.

d. First, we rank the data: CMC Group 4 3 3 4 3 3 3 3 4 4 3 4 3 3 2 4 2 4 5 4 4 4 5 3

Comparing Population Means

Rank 34.5 13.5 13.5 34.5 13.5 13.5 13.5 13.5 34.5 34.5 13.5 34.5 13.5 13.5 2 34.5 2 34.5 47 34.5 34.5 34.5 47 13.5 Ti = 578

FTF Group 5 4 4 4 3 3 3 4 3 3 3 3 4 4 4 4 4 3 3 3 4 4 2 4

Rank

34.5 34.5 34.5 13.5 13.5 13.5 34.5 13.5 13.5 13.5 13.5 34.5 34.5 34.5 34.5 34.5 13.5 13.5 13.5 34.5 34.5 2 34.5 T2 = 598

207

The test statistic is Tx

«2(«1 +«2 +1)

578-

24(24 + 24 + 1)

-10

2

«,«2(«, +«2

24(24X24 + 24 + 1)

-.21

48.4974

12

12

1

Since the observed value of the test statistic does not fall in the rejection region (z = -.21 ft -1.28), T/o is not rejected. There is insufficient evidence to indicate the relational intimacy scores for the participants in the CMC group are lower than the relational intimacy scores for the participants in the FTP group at a = .10. 7.75

a.

The Students’ t procedure requires that the populations sampled from be normal. The distributions of the zinc measurements in the 3 locations may not be normal.

b.

The ranks of the observations and the sums for the two groups are: Text-line

Rank

Intersection

Rank

.335 31A .440

4 6 8

.393 .353 .285 .295 .319

7 5 1 2 3 oo

II

Tx = 18

To determine if the distributions of zinc measurements for the text-line and the intersection have different centers of location, we test: Hq. The distributions of zinc measurements for the text -line and the intersection are identical H^: The distribution of zinc measurements for the text -line is shifted to the right or left of that for the intersection The test statistic is T] = 18. The null hypothesis will be rejected if T, < 71 or Ti > Zb where a = .05 (two-tailed), «] = 3 and «2 = 5. From Table V, Appendix A, Tl = 6 and Tu = 21. The rejection region is Ti < 6 or Ti > 21. Since the observed value of the test statistic does not fall in the rejection region {T\ i 6 and Tx ^ 21), Hq is not rejected. There is insufficient evidence to indicate the distributions of zinc measurements for the text -line and the intersection have different centers of location at a= .05.

208

Chapter 7

c.

The ranks of the observations and the sums for the two groups are; Witness-line

Rank

Intersection

Rank

.210

2

.393

9

.262

3

.353

8

.188

1

.285

4

.329

7

.295

5

.439

11

.319

6

.397

10 j

II

72 = 32

To determine if the distributions of zinc measurements for the witness-line and the intersection have different centers of location, we test; H(j‘. The distributions of zinc measurements for the witness-line and the intersection are identical H^\ The distribution of zinc measurements for the witness-line is shifted to the right or left of that for the intersection The test statistic is 7^2 = 32. The null hypothesis will be rejected if Tj < 31 or 31 > 71 where a = .05 (two-tailed), n\=6 and ^2 = 5. From Table V, Appendix A, 71 = 19 and 71 = 41. The rejection region is Tj < 19 or Tj >41. Since the observed value of the test statistic does not fall in the rejection region (71 ^ 19 and T2 ^ 41), Hq is not rejected. There is insufficient evidence to indicate the distributions of zinc measurements for the witness-line and the intersection have different centers of location at q:== .05. d.

7.77

The median zinc score for the Text-line is .374. The median zinc score for the Witness¬ line is .296. The median zinc score for the Intersection is .319. We found no difference in the distributions of zinc scores between the Text-line and the Intersection. We also found no difference in the distributions of zinc scores between the Witness-line and the Intersection. However, we did not compare the Text-line and the Witness-line zinc scores. These two distributions appear to be the furthest apart.

We assume that the probability distribution of differences is continuous so that the absolute differences will have unique ranks. Although tied (absolute) differences can be assigned average ranks, the number of ties should be small relative to the number of observations to assure validity.

Comparing Population Means

209

7.79

a.

The test statistic for this two-tailed test is T, the smaller of the positive and negative rank sums, TV and TL. The null hypothesis of identical probability distributions will be rejected \iT 25, the large sample approximation could also be used.

c.

The test statistic is 7+, the positive rank sum. Since it is necessary to reject the null hypothesis of identical probability distributions only if the distribution of A is shifted to the left of the distribution of B, small values of 7+ would imply rejection of Ho. We will reject Ho if 7+ < To where To is found in Table VI corresponding to a = .005 (one-tailed) and n = 7: There is no critical value given in the table for « = 7. Therefore, we will never reject

Ho. 7.81

a.

The hypotheses are: Ho". 77a:

b.

The two sampled populations have identical probability distributions The probability distributions for population A is shifted to the right of that for population B

Some preliminary calculations are: Treatment A

B

54 60 98 43 82 77 74 29 63 80

45 45 •87 31 71 75 63 30 59 82

Difference A-B 9 15 11 12 11 2 11 -1 4 -2

Rank of Absolute Difference 5 10 7 9 7 2.5 7 1 4

2A 7 =3.5

210

Chapter 7

The test statistic is T =3.5 The rejection region is T_ < 8, from Table VI, Appendix A, with « = 10 and a= .025. Since the observed value of the test statistic falls in the rejection region (T_ = 3.5 < 8), Ho is rejected. There is sufficient evidence to indicate the responses for A tend to be larger than those for B at a = .025. 7.83

7.85

a.

Since two measurements were obtained from each patients (before and after), the measurements are not independent of each other. Each “before” measurement is paired with an “after” measurement.

b.

The distributions for both before and after treatment are skewed to the right and the spread of the distribution before treatment is much larger than the spread after treatment.

c.

Since the /?-value is so small (p < ,0001), Ho would be rejected. There is sufficient evidence to indicate the ichthyotherapy was effective in treating psoriasis for any reasonable value of a.

a.

To determine if the distributions of the FSI scores for good and average readers differ in location, we test; Ho: The distribution of the FSI scores of good readers is identical to the distribution of the FSI scores of average readers Ha'. The distribution of the of the FSI scores of good readers is shifted to the right or left of the distribution of the FSI scores of average readers b, c.

STRATEGY Word meaning Words in context Literal comprehension Draw inference from single string Draw inference from multiple string Interpretation of metaphor Find salient or main idea Form judgment

FSI Scores Good Average Readers Readers .38 .29 .42 .60 .45 .32 .21 .73

.32 .25 .25 .26 .31 . .14 .03 .80

Difference

.06 .04 .17 .34 .14 .18 .18 -.07

Rank of Absolute Difference 2 1 5 8 4 6.5 6.5 3

d.

Positive rank sum r+= 33. Negative rank sum r_ = 3. The test statistic is the smaller of T+ and T_ and is r_ = 3.

e.

'Dj# rejection region is T+ < 4, from Table VI, Appendix A, with « = 8 and a = .05 (twotails).

Comparing Population Means

211

f.

Since the observed value of the test statistic falls in the rejection region (TV = 3 < 4), Ho is rejected. There is sufficient evidence to indicate the distributions of the FSI scores for good and average readers differ in location at a = .05.

7.87

Some preliminary calculations are: Change in Transverse Strain

_

Oct. 24 Dec. 3 Dec. 15 Feb. 2 Mar. 25 May 24

Change in Temp. -6.3 13.2 3.3 -14.8 1.7 -.2

Field Meas. -58 69 35 -32 -40 -83

3D Model -52 59 32 -24 -39 -71

Diff. -6 10 3 -8 -1 -12

Rank of Absolute Difference 3 5 2 4 1 6

n=i To determine if there is a shift in the change in transverse strain distributions between field measurements and the 3D model, we test: HqI The change in transverse strain distribution for field measurements is identical to the change in transverse strain distribution for the 3D model H^: The change in transverse strain distribution for field measurements is shifted to the right or left of the change in transverse strain distribution for the 3D model f

The test statistic is r= smaller of T_ and TV which is r+ = 7. The rejection region is 7; < 1, from Table VI, Appendix A, with « = 6 and a = .05 (two-tails). Since the observed value of the test statistic does not fall in the rejection region (r+ = 7 ft 1), Ho is not rejected. There is insufficient evidence to indicate &ere is a shift in the change in transverse strain distributions between field measurements and the 3D model at a =.05.

212

Chapter 7

7.89

Some preliminary calculations are;

Patient 1

Before Thickness 11.0

2 3 4 5 6 ' 7 8 9 10 11 12

4.0 6.3 12.0 18.2 9.2 7.5 7.1 7.2 6.7 14.2 7.3

After Thickness 11.5 6.4 6.1 10.0 14.7 7.3 6.1 6.4 5.7 6.5 13.2 7.5 7.4 7.2 6.3 6.0 7.3

13 14 15 16 17 18 19 20 21 22

8.7 6.7 10.2 6.6 11.2 8.6 6.1

23 24

10.3 7.0

7.0 5.3 9.0 6.6 6.3 7.2 7.2

25

12.0

8.0

9.7 9.5 5.6

Difference -0.5

Rank of Absolute difference 6

-2.4 0.2 2.0 3.5 1.9 1.4 0.7 1.5 0.2 1.0 -0.2 2.3 2.3 -0.7 2.7 -0.6 3.2 1.3 2.2 2.0 -0.2 3.1

20 3 15.5 24 14 12 8.5 13 3 10 3

18.5 18.5 8.5 21 7 23 11 17 15.5 3 22 3 -0.2 4.0 25 Negative rank sum r_=50.5 Positive rank sum TV = 274.5

To determine if the treatment for tendonitis tends to reduce the thickness of tendons, we test: Hq. The distribution of the thickness before treatment has the same location as the distribution of the thickness after the treatment //a: The distribution of the thickness before the treatment is shifted to the right of that for the thickness after the treatment The test statistics '\sT= smaller of T_ and TV which is T_ = 50.5. Reject Hq if T_ < Tq where TV is based on a = .10 and n = 25 (one-tailed). Reject Ho if T_< 101 (From Table VI, Appendix A) (Note: There is no value for or = .10 for a one-tailed test. However, if we reject Hq for a= .05, we would also reject Hq for a= .10.) Since thfe observed value of the test statistic falls in the rejection region (r_ = 50.5 < 101), Ho is rejected. There is sufficient evidence to indicate the treatment for tendonitis tends to reduce the thickness of tendons at a = . 10.

Comparing Population Means

213

7.91

Some preliminary calculations are:

Bowler 1 2 3 4

a.

After 4 Strikes .683 .684 .632 .610

After 4 NonStrikes .432 .400 .421 .529

Rank of Absolute difference

Difference .251 . 3 .284 4 2 .211 1 .081 Negative rank sum r_=0 Positive rank sum TV = 10

To determine if the data support the “hot hand” theory in bowling, we test: Hq. The distribution of the proportion of strikes after rolling 4 strikes has the same location as the distribution of the proportion of strikes after rolling 4 non-strikes The distribution of the proportion of strikes after rolling 4 strikes is shifted to the right of the distribution of the proportion of strikes after rolling 4 non-strikes The test statistic is the smaller of T_ and TV which is T_ = 0. There are no table values in Table VI when n = A. Even if all the observations support rejecting the /?-value is greater than .05. Thus, we are unable to reject Hq. There is insufficient evidence to support the “hot hand” theory at

b.

.05.

When all 43 bowlers are included, the p-value is/? = 0. Since this /7-value is so small, we would reject T/o for any reasonable value of a. There is sufficient evidence to support the “hot hand” theory at any reasonable value of a.

7.93

7.95

214

The conditions required for a valid ANOVA F-test are: 1.

The samples are randomly selected in an independent manner from k treatment populations.

2.

All k sampled populations have distributions that are approximately normal.

3.

The k population variances are equal (i.e.,

a.

UsingTable VIII, Fo5 = 6.59 with vi =3, V2 = 4.

b.

Using Table X, Foi = 16.69 with V| = 3, V2 = 4.

c.

Using Table VII, F io = 1.61 with Vj = 20, V2 = 40.

d.

Using Table IX, F025 = 3.87 with v, = 12, V2 = 9.

= 4.96. For diagram A, the observed value of the test statistic falls in the rejection region (F= 37.5 > 4.96). Thus, Ho is rejected. There is sufficient evidence to indicate the samples were drawn from populations with different means aX a = .05. For diagram B, the observed value of the test statistic falls in the rejection region (F=5.21 > 4.96). Thus, Ho is rejected. There is sufficient evidence to indicate the samples were drawn from populations with different means at a= .05.

h.

216

We must assume both populations are normally distributed with common variances.

Chapter 7

7.99

Refer to Exercise 7.97, the ANOVA table is: For diagram A:

Source

Df

ss

MS

Treatment Error

1 10

75 20

75 2

Total

11

95

F 37.5

For diagram B:

7.101

Source

Df

SS

MS

F

Treatment Error

1 10

75 144

75 14.4

5.21

Total

11

219

a. The experimental units are the NCAA tennis coaches. The dependent variable is the rating on a 7-point scale of how important the coaches think the web site was. The 3 treatments are Division I, Division II, and Division III. b. To determine if the mean ratings of the web sites are different for the different levels of coaches, we test: Hq\ fi\~ fk\ — /Aw At least one mean differs c. Since the p-value is less than .05 (/? < .003), Hq is rejected. There is sufficient evidence to indicate a difference in mean responses among the different divisions of the colleges and universities at

7.103

.05.

a.

The experimental design used is a one-way ANOVA.

b.

There are 4 treatments in this experiment. The four treatments are the 4 “colonies” to which the robots were assigned - 3, 6, 9, or 12 robots per colony.

c.

To determine if the mean energy expended (per robot) of the four different colony sizes differed, we test: Hq. fj.1— p,2~ /a,— /^a

Ha'. At least two treatment means differ d.

The test statistic isF= 7.70. The p-value is p < .001. Since the /7-value is less than a~ .05, Ho is rejected. There is sufficient evidence to indicate that mean energy expended (per robot) of the four different colony sizes differed at a = .05

7.105

a.

To determine if the average heights of male Australian school children differ among the age groups,'we test: ^O'./^l =/l2=fi3

Ha. At least two treatment means differ

Comparing Population Means

217

b.

The test statistic is 4.57. Thep-value is/? — .01. Since the p-value is less than oc{p = .01 < .05), Hq is rejected. There is sufficient evidence to indicate the average heights of male Australian school children differ among the age groups at Gf= .05.

c.

To determine if the average heights of female Australian school children differ among the age groups, we test:

,

‘

Hq’. ij.x= ^2 = Hi 74: At least two treatment means differ The test statistic is F = 0.85. The p-value is p == .43. Since the p-value is not less than a (p = .43 > .05), //o is not rejected. There is insufficient evidence to indicate the average d.

heights of female Australian school children differ among the age groups at a = .05. For the boys, differences were found in the mean standardized heights among the three tertiles. For the girls, no differences were found in the mean standardized heights among the three tertiles.

7.107

To determine if a driver’s propensity to engage in road rage is related to his/her income, we test: 74: Pi=P2 = F3 74: At least two means differ The test statistic is F = 3.90. The p-value is p < .01. Since the p-value is so small, 7/o would be rejected for any reasonable value of a. There is sufficient evidence to indicate that a driver’s propensity to engage in road rage is related to his/her income at a > .01. We have just concluded that the mean road rage score is related to income group. For the Under $30,000 group, the mean road rage score is 4.60. For the $30,000 to $60,000 group, the mean road rage score is 5.08. For the Over $60,000 group, the mean road rage score is 5.15. Thus, as the income increases, the mean road rage score also increases.

7.109

218

a.

The subjects were randomly divided into 3 treatment groups with one group receiving an injection of scopolamine, another group receiving an injection of giycopyrrolate, and the third group receiving nothing. Thus, this is an independent samples design.

b.

There are 3 treatments: injection of scopolamine, injection of giycopyrrolate, and nothing. The dependent variable is the number of pairs recalled.

Chapter 7

c.

Using MINITAB, the descriptive statistics are:

Descriptive Statistics: Scopolamine, Placebo, No Drug Variable Scopolamine Placebo No Drug

N 12 8 8

Variable Scopolamine Placebo No Drug -

Maximum 8 .000 12 .000 12 . 000

N* 0 0 0

Mean 6.167 9.375 10.625

StDev 1.267 1.506 1.506

SE Mean 0.366 0.532 0.532

Minimum 4.000 7.000 8.000

Q1 5.250 8.250 9.250

Median 6.000 9.500 11.000

Q3 7.500 10.000 12.000

The sample means for the 3 groups are: 6.167, 9,375, and 10.625. There is not sufficient information to support the researcher’s theory. We need to take into account the variability within each group. d.

Using MINITAB, the ANOVA table is: One-way ANOVA: Scopolamine, Placebo, No Drug Source Factor Error Total

DF 2 25 27

SS 107.01 49.42 156.43

S = 1. 406

MS 53.51 1.98

R-Sq = 68.41%

F 27.07

P 0.000

R-Sq(adj)

=

Individual 95% CIs For Mean Based on Pooled StDev Level Scopolamine Placebo No Drug

N 12 8 8

Mean 6.167 9.375 10.625

StDev 1.267 1.506 1.506

(-*-

6.4

8.0

9.6

11.2

Pooled StDev,= 1.406

To determine if the mean number of words recalled differed among the 3 groups, we test: Hq. /7l =

= A3

H^;. At least two means differ The test statistic is F = 27.07. The /7-value is/7 = .000. Since the /7-value is so small, Hq would be rejected for any reasonable value of a. There is sufficient evidence to indicate the mean number of words recalled differed among the 3 groups at a= .05. Prior to the experiment, the researchers theorized that the mean number of word pairs recalled for the scopolamine group would be less that the corresponding means for the other 2 groups. From the printout, the sample mean for the scopolamine group is 6.167. The meansTor the placebo group and the nothing group are 9.375 and 10.625, *^ince the sample mean for the scopolamine group is much smaller than the other two, it appears that the researchers’ theory was correct.

Comparing Population Means

219

7.111

Using MINITAB, the results of the analysis are: One-way ANOVA: UMRB-1, UMRB-2, UMBR-3, SWRA, SD Source Factor Error Total

DF 4 21 25

S = 0. 4486

SS 5.836 4.225 10.061

F 7.25

MS 1.459 0.201

P 0.001

R-Sq(adj)

R-Sq = 58.00%

Individual 95% CIs For Mean Based on Pooled StDev Level UMRB-1 UMRB-2 UMRB-3 SWRA SD

N 7 6 7 3 3

Mean 3.4971 4.0167 3.7614 2.6433 2.7800

StDev 0.3640 0.5726 0.4832 0.3580 0.2587

*

\ )

*

\

* -

{ \

\ ) )

1 ( — —“—+— 2.40

•' ' 3.00

/

4.20

3.60

Pooled StDev = 0.4486

To determine if there are differences among the mean AI/Be ratios for the 5 boreholes, we test: Hq-. //, = //2 = = a = /is 7/a- At least 2 means differ From the printout, the test statistic is F = 7.25 and the /?-value is /> = 0.001. Since the /»-value is less than «= .10 (p = .001 < .10), T/o is rejected. There is sufficient evidence to indicate a difference in the mean AI/Be ratios among the 5 boreholes at «= .10. 7.113

a.

The 2 samples are randomly selected in an independent manner from the two populations. The sample sizes, n\ and «2» are large enough so that approximately normal sampling distributions and so that sf and approximations to

b.

1.

. This will be true if

provide good

> 30 and n2> 30.

2. 3.

Both sampled populations have relative frequency distributions that are approximately normal. The population variances are equal. The samples are randomly and independently selected from the populations.

c.

1. 2.

The relative frequency distribution of the population of differences is normal. The sample of differences are randomly selected from the population of differences.

d.

1.

The samples are randomly selected in an independent manner from the k treatment populations. All A: sampled populations have distributions that are approximately normal. The k population variances are equal.

2. 3.

220

and

and X2 each have

Chapter 7

7.115

a.

To compare two populations with independent samples, one would use the Wilcoxon Rank Sum test. I

7.117

b.

To compare two populations with matched pairs, one would use the Wilcoxon Signed Rank Test.

a.

For confidence coefficient .90, a = .10 and all = .05. From Table III, Appendix A, Zos = 1.645. The confidence interval is; ?

.(Xi -X2) + 2,05

-

2

2 1 3 0 (12.2-8.3) ± 1.645J—+ — V135 148

+—

«2

• 3.90+ .31 ^(3.59,4.21) b.

Hq.

- //2

0

//i - //2 ^ 0 The test statistic is 2 =

(Xi-X2)-0

=20.60 2.1

n,

n.

3.0

V 135 "^148

The rejection region requires a/2 = .01/2 = .005 in each tail of the z distribution. From Table III, Appendix A, zqos = 2.58. The rejection region is z < -2.58 or z > 2.58. Since the observed value of the test statistic falls in the rejection region (20.60 > 2.58), Ho is rejected. There is sufficient evidence to indicate that c.

For confidence coefficient .90, 2,05= 1.645. „ -n

(o-' + o-O

at a - .01.

.10 and a/l = .05. From Table III, Appendix A,

(1.645)^(2.1 + 3.0)

-i— -=-1--^- 345.02 « 346 , (SEf .2'

Comparing Population Means

221

7.119

Use the Wilcoxon signed rank test. Some preliminary calculations are: Pair

X

Y

Difference

Rank of Absolute Difference

1 2 3 4 5 6 7 8 9

19 27 15 35 13 29 16 22 16

12 19 7 25 11 10 16 10 18

7 8 8 10 2 19 0 12 -2

3 ‘ 4.5 4.5 6 1.5 8 (eliminated) 7 II

To determine if the probability distribution of x is shifted to the right of that fory, we test: Hq: Ha.

The probability distributions are identical for the two variables The probability distribution of jc is shifted to the right of the probability distribution ofy

The test statistic is 7= r_= 1.5 Reject HoifT< Tq where Tq is based on or = .05 and « = 8 (one-tailed): Reject Ho if T< 6 (from Table VI, Appendix A). Since the observed value of the test statistic falls in the rejection region (T = 1.5 < .6), reject Ho at a = .05. There is sufficient evidence to conclude that the probability distribution of x is shifted to the right of that fory. 7.121

a.

SSE = SSTot-SST = 62.55 -36.95 =25.60 df Treatment = A:-1=4-1=3 dfError = «-A: = 20-4= 16 df Total = « - 1 = 20 - 1 = 19 MST = SST/df =

= 12.32 3

MSE = SSE/df =

MSE

= 1.60

1.60 •

The ANOVA table: Source_df_SS_MS_F Treatment 3 Error_16 Total 19

222

36.95 25.60 62.55

12.32 1.60

7?70

Chapter 7

b. To determine if there is a difference in the treatment means, we test: Hq. /^i = a = A3

>^4

//a: At least two of the means differ where the

represents the mean for the /th treatment.

The test statistic is F= 7.70

The rejection region requires a =. 10 in the upper tail of the F distribution with 1) = (4 - 1) = 3 and Vi = {n-K) = (20 - 4) = 16. From Table VII, Appendix A, Fjo = 2.46. The rejection region is F> 2.46. V\-{k-

Since the observed value of the test statistic falls in the rejection region (F = 7.70 > 2.46), Hq is rejected. There is sufficient evidence to conclude that at least two of the means differ at a = . 10. 7.123

a.

Let /ii = mean trait for senior year in high school and = mean trait during sophomore year of college. Also, let n^ = ijl\- fUi. To determine if there is a decrease in the mean self-concept of females between the senior year in high school and the sophomore year in college, we test: Hq'. //j = 0

Fa'- Ai>0 b.

Since the same female students were used in each study, the data should be analyzed as paired differences. The samples are not independent.

c.

Since the number of pairs is Wd 133, no assumptions about the population of differences is necessary. We still assume that a random sample of differences is selected.

d.

Leadership: The/7-value > .05. There is no evidence to reject//o for a = .05. There is no evidence to indicate that there is a decrease in mean self-concept of females on leadership between the senior year in high school and the sophomore year in college for a= .05. Popularity: The /7-value < .05. There is evidence to reject Hq for a= .05. There is evidence to indicate that there is a decrease in mean self-concept of females on popularity between the senior year in high school and the sophomore year in college for a= .05. Intellectual self-confidence: The /7-value < .05. There is evidence to reject

for

a = .05. There is evidence to indicate that there is a decrease in mean self-concept of females on intellectual self-concept between the senior year in high school and the sophomore year in college for a - .05.

Comparing Population Means

223

7.125

To determine if differences in the mean percent of pronoun errors exist among the three groups, we test: Hq:

^2 = >U3

/4: At least two treatment means differ i

where /// represents the mean percent of pronoun errors for the ith group. The test statistic is F = 8.295. Since thep-value is so small {p = .0016), Hq is rejected for any a > .0016. There is sufficient evidence to indicate a difference in the mean percent of pronoun errors among the three groups. A multiple comparison was run on the three group means. A line connects the means of the SLI and the YND groups. The mean percent of pronoun errors is not different for these two groups. The mean percent of pronoun errors for the OND group is significantly smaller than the mean percent of pronoun errors for both the SLI and the YND groups. 7.127

a.

First, some preliminary calculations:

After 6 months of Depro-Provera

1 2 3 4 5 6 7 8

849 903 890 1,092 362 900 1,006 672

96 41 31 124 46 53 113 174

V,

=

V' 2

224

753 862 859 968 316 847 893 498

=749.5

--«d-l

1 >

Pretreatment

II

Patient

O

Testosterone Levels

5 996^

4,847,676-^^^

-=-—-^ = 50,524.857

8-1

Chapter 7

For confidence coefficient .99, a= .01 and all Appendix A, with df =

.0112 = .005. From Table IV,

- 1 = 8 - 1 = 7, foos = 3.499. The confidence interval is:

749.5 + 3.499^^^’-^—

749.5 ± 278.068

V8 (471.432,1027.568) We are 99% confident that the true mean difference between the pretreatment and aftertreatment testosterone levels of young male TBI patients is between 471.432 and 1,027.568. b.

Since 0 is not in the confidence interval, there is evidence that there is a difference in the testosterone levels pre and post treatment. Thus, there is evidence that DepoProvera is effective in reducing testosterone levels in young male TBI patients.

Pair 1 2 3 4 5 6 7 8 9 10 11 12 13

Attitude toward Father 2 5 4 4 3 5 4 2 4 5 4 5 3

Attitude toward Mother 3 5 3 5 4 4 5 4 5 4 5 4 3

Rank of Absolute Difference Difference -1 5.5 (eliminated) 0 1 5.5 -1 5.5 -1 5.5 5.5 1 -1 5.5 -2 11 -1 5.5 1 5.5 -1 5.5 1 5.5 0 (eliminated) Positive rank sum T+ = 22

Hq: The probability distributions of the two populations are identical Ha. The probability distribution of the attitude toward Father is shifted to the right or left of that of the attitude toward Mother The test statistic is T+ = 22 Reject Ho if T+ < To where To is based on a= .05 and « = 11 (two-tailed): Reject //o if T+ < 11 (from Table VI, Appendix A). Since the observed value of the test statistic does not fall in the rejection region {T+ = 22 ^ 11), i/o is not rejected. There is insufficient evidence to indicate the male students’ attitudes toward their fathers differ from their attitudes toward their mothers at a= .05.

Comparing Population Means

225

7.129

a.

Since we want to compare the means of two independent groups, we would use a two sample test of hypothesis for independent samples. Since the sample sizes are small (21 for each), we would use a small sample test.

b.

We must assume that both populations being sampled from are approximately normally distributed and that the population variances of the two groups are equal.

c.

No. We need to know the common variance of the two groups or an estimate of it. \

d.

Since the /j-value was so large {p = .79), Hq would not be rejected for any reasonable value of a. There is insufficient evidence to indicate a difference in the mean scores for the two groups on the question dealing with clean/sterile gloves at a< .10.

e.

7.131

Since the /7-value was so small {p = .02), Hq is rejected for any a> .02. There is sufficient evidence to indicate a difference in the mean scores for the two groups on the question dealing with a stethoscope at a> .02.

Using MINITAB, the ANOVA table is: Source_df Expressions 5 Error_30 Total 35

SS 23.09 34.99 58.07

MS 4.62 1.17

F

P

3.96

0.007

To determine if differences exist among the mean dominance ratings of the six facial expressions, we test:

Hq. P\= IU2=^ IU2= lLk = il5= At least two treatment means differ where jUi represents the mean dominance rating for facial expression /. The test statistic is F = 3.96. The rejection region requires « = .10 in the upper tail of the F distribution with Vi = A:-l = 6 - 1 = 5 and V2 = « - A:= 36 - 6 = 30. Using Table VII, Appendix A, F.io - 2.05. The rejection region is F> 2.05. Since the observed value of the test statistic falls in the rejection region (F= 3.96 > 2.05), Ho is rejected. There is sufficient evidence to indicate that differences exists among the mean dominance rating for facial expressions at a = .10.

226

Chapter 7

7.133

a.

To determine whether the median is less than $2.25, we test:

Ho. 77-2.25 H^\ 77 2.12). From Table III, Appendix A,p = P(z> 2.12) = .5 - .4830 = .0170. Therefore, reject Hq for a > .017 and conclude that the median amount spent for hamburgers at lunch at McDonald's is less than $2.25.

7.135

b.

No; The sample was taken from only two restaurants in Boston.

c.

The distribution of the prices is continuous. Nothing is assumed about the shape of the probability distribution.

Let jui = mean number of timeout incidents for Option II students and pi ~ mean number of timeout incidents for Option III students. To detemiine if the mean number of timeout incidents for students in Option III is greater than the mean number of timeout incidents for students in Option II, we test: Ho'. P\

-

pi

Ho'. P\ < (T,-T,)-0

(78.67-102.87)-0

The test statistic is z

Since no value of « was given, we will use a= .05. The rejection region requires a = .05 in the lower tail of the z distribution. From Table III, Appendix A, zqs = 1.645. The rejection region is z 1.96), Hq is not rejected. There is insufficient evidence to indicate a difference in the mean task identity between the positive spillover group and the no spillover group at Qr=.05. Of the 5 job-related characteristics, there are only 2 that have significantly different means for the 2 groups. They are “use of creative ideas” and “good use of skills”. For both of these characteristics, the mean for the positive spillover group was significantly higher than the mean for the no spillover group.

230

Chapter 7

Chapter

Comparing Population Proportions 8.1

a. The distribution ofxi is binomial with rii trials and the probability of success, The distribution of X2 is binomial with nj trials and the probability of success, pi. b. For large samples, the sampling distribution of (p, - p^) is approximately normal with mean {p^ ~ P2), and standard deviation cr'

8.3

]j «,

«2

The conditions required for a large-sample inference about (pj - P2) are: 1. 2.

The two samples are randomly selected in an independent manner from the two target populations. The sample sizes, «i and are both large so that the sampling distribution of (A ~ P2)

t)e approximately normal. (This condition will be satisfied if all of

the following are true: 8.5

'

>15,

>15 and n^P2 ^15,

The sample sizes are large enough if pp, > 15, a.

«,p, =10(.5) = 5^15, /72P2 =12(.5) = 6^15,

^15.)

>15 and n^p^ >15 , n^q^, > 15 .

=10(.5) = 5 3^15 «2^2 =12(.5) = 6 5^15

Since none of the products are greater than or equal to 15, the sample sizes are not large enough to assume normality. b.

n,p, =10(.1) = 15^15,

=10(.9) = 93^15

«2A =12(.08)=.96 5^15,

^2^2 =12(.92) = 11.04 3^15

Since none of the products are greater than or equal to 15, the sample sizes are not large enough to assume normality. c.

«,p, =30(.2)-6 3^15,

n,(7, =30(.8) = 24>15

n^P2 =30(.3) = 9^15,

n^q2 =30(.7) = 21>,15

Since two of the products are not greater than or equal to 15, the sample sizes are not large enough to assume normality. d.

«,p,

=100(.05) = 5 3^15,

«2A = 200009) = 18 > 15,

=100(.95) = 95 >15 «2^2 = 200(.9 1) = 182 > 15

Since one of the products is not greater than or equal to 15, the sample sizes are not large enough to assume normality.

Comparing Population Proportions

231

e.

=100(.95) = 95>15,

=100(.05) = 5 5^15

n^A =200(.91) = 182>15,

= 200(.09) = 18 > 15

Since one of the products is not greater than or equal to 15, the sample sizes are not large enough to assume normality. 8.7

For confidence coefficient .95, a= 1 - .95 == .05 and aJl = .05/2 = .025. From Table III, Appendix A, zo25 = 1.96. The 95% confidence interval forp\ -pi is approximately: A

a.

A.

A

(A-A)±^«/2j—+ «i

A

«2

(.65~.58)± 1.96=>.07+ .067 =>(.003, .137) 400 400

b.

(p^-p^)±zA^^Mi. «1

^2

(.31-.25)+

+ 180 A

C.

A

A

(A-A)±^«/2J—+ «1

A

«2

(.46-.61) +

8.11

a.

=^.06 + .086 => (-026,. 146) 250

100

120

=>-.15+ .131 => (-.281,-.019)

The observed proportion of St. John’s wort patients who were in remission is: X] 14 /7i =-!- = — = .143 «i 98

b.

The observed proportion of placebo patients who were in remission is: ^2 5 P2=~=- =-049 «2 102

c.

Some preliminary calculations are: .

X,+^2

14 + 5

19

n,+«2

98 + 102

200

P = --- =-=-=

232

.095

^ = l-^ = l-.095 = .905

Chapter 8

To detemiine if the proportion of St. John wort patients in remission exceeds the proportion of placebo patients in remission, we test: Hq. p\ =P2

P\ > Pi The test statistic is z =

P\-Pi f 1 1 +Pq n2 y V"i

.143-.049 .095(.905)

2.27

f 1

1 ■ + -

1^98

102 j

The rejection region requires .01 in the upper tail of the z distribution. From Table III, Appendix A, z.oi = 2.33. The rejection region is z > 2.33. Since the observed value of the test statistic does not fall in the rejection region (z = 2.27 2.33), //o is not rejected. There is insufficient evidence to indicate the proportion of St. John wort patients in remission exceeds the proportion of placebo patients in remission at a = .01. d.

The hypotheses and test statistic are the same as in part c. The rejection region requires = .10 in the upper tail of the z distribution. From Table III, Appendix A, z io = 1.28. The rejection region is z > 1.28. Since the observed value of the test statistic falls in the rejection region (z = 2.27 > 1.28), //o is rejected. There is sufficient evidence to indicate the proportion of St. John wort patients in remission exceeds the proportion of placebo patients in remission a = .10.

e.

By changing the value of a in this problem, the conclusion changes. When a is small («= .01), we did not reject//q- When ^increases {a- .10), we rejectedAs a increases, the chance of making a Type I error increase, but the chance of a Type II error decrease.

8.13

a.

Let p\ = proportion of African-American drivers who were searched and pi = proportion of White drivers who were searched. The parameter of interest is p\ -pj, or the difference in the proportions of drivers searched between the two ethnic groups. Some preliminary calculations are: 5,312

X, _ 12,016

P=

X,

+X2

n^ +

12,016 + 5,312

= .050

106,892

61,688

= .103

61,688 + 106,892

To determine if there is a difference in the proportions of African-American and White drivers who are searched by the LA police, we test: Hq\ p\-p2 = ^

P\ -pi ^ 0

Comparing Population Proportions

233

The test statistic is

iPx -A)-Q 1

A

1

\P(1

--

(.195-.050)-0

94.35

.103(.897)

^2)

The rejection region requires all = .05/2 = .025 in each tail of the z distribution. From Table III, Appendix A, Z025 = 1-96. The rejection region is z < -1.96 or z > 1.96. Since the observed value of the test statistic falls in the rejection region (z = 94.35 > 1.96), Hq is rejected. There is sufficient evidence to indicate a difference in the proportions of African-American and White drivers who are searched by the LA police at «= .05. b.

Let p\ = “hit rate” of African-American drivers and pj = “hit rate” of White drivers. The parameter of interest is p\ -pj, or the difference in the “hit rates” for the two ethnic groups. Some preliminary calculations are:

M 5,312

A =^ = ^^ = .427 n,

51

12,016

=

,566

For confidence coefficient .95, a= .05 and all = .05/2 = .025. From Table III, Appendix A, Z025 = L96. The 95% confidence interval is:

(.427-.566) ±1.96

427(.573)

.566(.434)

12,016

5,312

=>-.139±.016ii5>(-.155.

-.123)

V

We are 95% confident that the difference in “hit rates” between African-American and White drivers searched by the LA police is between -.155 and -.123. Since both end points of the interval are negative, there is evidence that the “hit rate” for AfricanAmerican drives is less than that for White drivers. 8.15

Let p\ = proportion of rice weevils found dead after 4 days of exposure to nitrogen gas and P2 = proportion of rice weevils found dead after 3.5 days of exposure to nitrogen gas. The parameter of interest is/^i —pi, or the difference in the proportions of rice weevils found dead between the 2 different exposures to nitrogen gas. Some preliminary calculations are:

234

31,386

23,516

31,421

23,676

^1+^2

31,386 + 23,516

«, + «2

31,421 + 23,676

= .993

Chapter 8

To determine if there is a difference in the proportions of rice weevils found dead between the 2 exposure times, we test; Ho. Pi-P2 = 0

H^'. P\-P2^^ The test statistic is z =

(a -A)-o 11

/V

A

^- + -1 [n^

nj

(.999-.993)-0

11.05

1 f 1 1 J.996(.004) ■f ■ V V 31,421 23,676

The rejection region requires ccH = .10/2 = .05 in each tail of the z distribution. From Table III, Appendix A, Zqs = 1.645. The rejection region is z < -1.645 or z > 1.645. Since the observed value of the test statistic falls in the rejection region (z = 11.05 > 1.645), Ho is rejected. There is sufficient evidence to indicate a difference in the proportions of rice weevils found dead between the 2 exposure times at a= .10. 8.17

Let p\ - recall rate of the 60 to 72 year-old seniors and pj = recall rate of the 73 to 83 yearold seniors. Some preliminary calculations are: 31 + 22

^1+^2 n, + «2

2* =.775 40

= .6625

40 + 40

To determine if the recall rates of the two groups differ, we test: Hq\ Pi =P2

H^'. p\ ^P2

The test statistic is

(a-A)-Q —+—■

(.775-.55)-0

r .6625(.3375)

1

P

2.13

Tlie rejection region requires a/2 = .05/2 = .025 in each tail of the z distribution. From Table III, Appendix A, zo25 = 1.96. The rejection region is z < -1.96 or z > 1.96. Since the observed value of the test statistic falls in the rejection region (z = 2.13 > 1.96), Ho is rejected. There is sufficient evidence to indicate the recall rates of the two groups differ at a= .05.

Comparing Population Proportions

235

8.19

Let p\ = proportion of volunteers who figured out the third rule in the group that slept and p2 = proportion of volunteers who figured out the third rule in the group that stayed awake all night. The parameter of interest is p\ —pi, or the difference in the proportions of volunteers who figured out the third rule between the 2 groups. Some preliminary calculations are:

For confidence coefficient .90, a= AQ and all = .10/2 = .05. From Table III, Appendix A, zo5 = 1.645. The 90% confidence interval is:

(.78-.30) ±1.645^- -1-^.48 i. 144=::>(.336, .624)

We are 90% confident that the difference in the proportions of volunteers who figured out the third rule between the 2 groups is between .336 and .624. Since both end points are greater than 0, there is evidence that the proportion of those who slept who figured out the third rule is greater than the proportion of those who stayed awake all night who figured out the third rule. 8.21

a.

Let p\ = true proportion of women who acknowledge having food cravings and P2 = true proportion of men who acknowledge having food cravings. To determine if the true proportion of women who acknowledge having food cravings exceeds the corresponding proportion of men, we test: Ho’. p\ = Pi P\ > Pi

(A-A)-Q

The test statistic would be z

/

A A

f -1 1

i] --

i The rejection region requires a= .01 in the upper tail of the z distribution. From Table III, Appendix A, zqi = 2.33. The rejection region is z > 2.33. We need to find «i = «2 so that the test statistic falls in the rejection region or z > 2.33. We know that p, = .97 and p2 = .67. Assuming «, = «2 = n. g . A+P2

^2

236

•97 + .67

2

■

Chapter 8

__ (A-A)-O pq —+ — ^2)

(.97-.67)-0

2.33

. |.82(.18) ——— \n n)

c

V

'1 n

=>.30 = 2.33^.82(.18) —+ — n) .30 = 2.33. .1476

^2"

UJ

.2952 =>.30-1.33,1 V n

r^.30'« = 2.33'(.2952) =i>« =

2.33'(.2952)

= 17.807«18

.3' We would need at least 18 observations from each group. b.

This study involved Students at McMaster University students. It is very dangerous to generalize the results of this study to the general adult population of North America. The sample of students used may not be representative of the population of interest.

8.23

If the sample size calculation yields a value of n that is too large to be practical, we might decide to use a large sampling error (SE) in order to reduce the sample size, or we might decrease the confidence coefficient.

8.25

For confidence coefficient .90, « = 1 - .90 = .10 and all = .10/2 = .05. From Table III, Appendix A, zqs = 1.645. For width = . 1, the standard error is SE -AH - .05. ! (A?, H- PA) ^ 1,645^.6(.4) + .6(.4)) ^ ^

{SEf

^^

.05'

In order to estimate the difference in proportions using a 90% confidence interval of width .1, we would need to sample 520 observations from each population. Since only enough money was budgeted to sample 100 observations from each population, insufficient funds have been allocated. 8.27

For confidence coefficient .90, «2 =

J

2(5'^)"

_ 1.645'(.2(.8) + 2(.2)(.8)) = 259.8 « 260 «2 =

Thus, the manufacturer would sample 2(260) = 520 buyers of its sets and 260 buyers of its competitors sets. 8.31

The characteristics of the multinomial experiment are: 1. 2. 3.

The experiment consists of n identical trials. There are k possible outcomes to each trial. The probabilities of the k outcomes, denoted p\,p2,... ,pk, remain the same from trial to

4. 5.

trial, where/?i +/72 + •" + A = I ■ The trials are independent. The random variables of interest are the counts «i,

•••, «* in each of the k cells.

The characteristics of the binomial are the same as those for the multinomial with k=2. 8.33

8.35

a.

With df= 10, ;jrj5 = 18.3070

b.

With df = 50, zlgo = 29.7067

c.

With df= 16, ;jrjo =23.5418

d.

With df = 50, zlos = 79.4900

a.

The rejection region requires a = .05 in the upper tail of the ^ distribution with df= A: - 1 = 3 - 1 = 2. From Table XI, Appendix A, region is

b.

> 5.99147.

The rejection region requires a =. 10 in the upper tail of the df=A:-l = 5- l=4. From Table XI, Appendix A, region is

c.

= 5.99147. The rejection

distribution with

1.119AA. The rejection

> 1.119AA.

The rejection region requires a = .01 in the upper tail of the ^ distribution with df = A: - 1 = 4 - 1 = 3. From Table XI, Appendix A, x]^, = 11.3449. The rejection region is

238

> 11.3449.

Chapter 8

8.37

Some preliminary calculations are:

If the probabilities are the same, /?i,o = Pifi

~

Pi,Q = P4,o = -25

■£i = «/?i.o = 205(.25)= 51.25 £:2 = £3 = £’4 = 205(.25)- 51.25 a. To determine if the multinomial probabilities differ, we test:

Ho: pi=p2=P2=P4 = -25 Ha'. At least one of the probabilities differs from .25 "

r

The test statistic is ;j^ = ^

12

^ '

_ (43-51.25r ^ (56-51.25)^ ^ (59-51.25^ ^ (47-51.25y 51.25

51.25

51.25

= 3.293

51.25

The rejection region requires a = .05 in the upper tail of the ‘2’ distribution with df = ^ 1 = 4 - 1 = 3. From Table XI, Appendix A, Jqj = 7.81473. The rejection region is

>

7.81473. Since the observed value of the test statistic does not fall in the rejection region

-■ 3.293 'i> 7.81473), Ho is not rejected. There is insufficient evidence to indicate the multinomial probabilities differ at a= .05. b. The Type I error is concluding the multinomial probabilities differ when, in fact, they do not. The Type II error is concluding the multinomial probabilities are equal, when, in fact, they are not. c. For confidence coefficient .95, a - .05 and a/2 = .05/2 = .025. From Table III, Appendix A, z.025 = 1.96. ^3 = 59/205 = .288 The confidence interval is:

P±Zq25^ ^ .288 + 1.96^^^^^^^^^ =i> .288 ± .062

8.39

(.226, .350)

a.

The qualitative variable of interest is nonfunctional jaw habits. It has 4 levels: bruxism, teeth clenching, both bruxism and clenching, and neither.

b.

A one-way table of the data is: Cell #

tlj

Bruxism

Clench

3

11

Comparing Population Proportions

Bruxism & Clench 30

Neither 16

239

c.

To determine whether the percentages associated with the admitted habits differ, we test: Hq: Pi =P2 =P3

= A " -25

At least one pt differs from .25, d.

E\= E2==E3 = E4 = npifl = 60(.25)= \5.

e.

The test statistic is

/ = 1, 2, 3, 4

[3-15f ^ [ll-isr ^ [30-l5f ^[16-15]l

^ f.

E,

15

15

15

15

The rejection region requires a= .05 in the upper tail of the ^ distribution with df = 1 = 4 - 1 = 3. From Table XI, Appendix A, zls^ 7.81473. The rejection region is

/>7.m73. g.

Since the observed value of the test statistic falls in the rejection region (j^ = 25.733 > 7.81473), Hq is rejected. There is sufficient evidence to indicate the percentages associated with the admitted habits differ at a= .05.

For confidence coefficient .95, a= .05 and af2 = .05/2 = .025. From Table III, Appendix A, 2,025 - 1 -96. The 95% confidence interval is:

I

A ±2. 025'

\Ml

.5 + 1.96

n

.5(.5)

.5±.127=>(.373, .627)

V 60

We are 95% confident that the true proportion of dental patients who admit to both habits is between .373 and .627. 8.41

a.

The qualitative variable of interest in this problem is the type of pottery found. There are 4 levels of the variable: burnished, monochrome, painted, and other.

b.

If all 4 types of pottery occur with equal probabilities, the values of p\,p2, ps, and p4 are all .25. «

c.

To determine if one type of pottery is more likely to occur at the site than any other, we test:

Ho: P\ =P2=P3=Pa = -25 Ha: At least one/?, ^ .25 for / = 1, 2, 3, 4

240

Chapter 8

d.

Some preliminary calculations are:

Ei=E2 = Ei^E^ = npi,o = 837(.25) = 209.25.

[n.-E,f

[133-209.25f

[460-209.25f

[183-209.25f

[61-209.25f

209.25

209.25

209.25

209.25

436.591 e.

Thep-value \sp =

> 436.591). Using Table XI, Appendix A, with df = A: - 1 = 4 - I = lt,p = P{^ > 436.591) < .005. Since the p-value is less than a= .10, Hq is rejected. There is sufficient evidence to indicate at least one type of pottery is more likely to occur at the site than another at a = .10.

8.43

a.

Some preliminary calculations are:

£3

1,000(.50) = 500 - wpa’o = 1,000(. 11) = 110

£2 £4

= «P2o = 1,000(.22) - 220 = npisi = 1,000(. 17) = 170

To determine if the data disagree with the percentages reported by Nielson/NetRatings, we test: Hq.

Pi = .50,^2 = •22,/73 = .11, and/74 = .17 At least one pi differs from it hypothesized value

The test statistic is

Ei

_[487-500f ^ [245-220f ^ [121-llOf ^ [147-170f _ ^ 500

^

220

110

^

170

The rejection region requires a = .05 in the upper tail of the ^ distribution with df = A:1 =4-1 =3. From Table XI, Appendix A, Zos ~ 7.81473. The rejection region is /> 7.81473. Since the observed value of the test statistic falls in the rejection region

= 7.39 > 7.81473), Hq is rejected. There is sufficient evidence to indicate the data disagree with the percentages reported by Nielson/NetRatings at a= .05.

b.

487

P\ =

1000

= .487

For confidence coefficient .95, a = .05 and all - .05/2 = .025. From Table III, Appendix A, Z025 = R96. The 95% confidence interval is:

p, ± z025

n

=> .487 ± 1.487 ±.031=^ (.456, .518)

M

1000

X*'

We are 95% confident that the true proportion of all Internet searches that use the Google Search engine is between .456 and .518.

Comparing Population Proportions

241

8.45

a.

There were 1,470 responses that were missing. In addition, 14 responses were 8 = Don’t know and 7 responses were 9 = Missing. Those responding with 8 or 9 were not included. The frequency table is: Response 1 2 3 4 Totals

b.

Frequency 450 627 219 23 1319

To determine if the true proportions in each category are equal, we test: Hq: P\=P2=Pi=p^^ 2S Ha'. At least one pi differs

c. Ei=E2 = E2=E4=^ npi,o = 1,319(.25) = 329.75 d. The test statistic is

[450-329.75f 329.75

+

[627-329.75f 329.75

[219-329.75f

[23-329.75]

329.75

329.75

+ ■

e. The rejection region requires a = .10 in the upper tail of the 1 = 4 - 1 = 3. From Table XI, Appendix A,

= 634.36

distribution with df=

^ 6.25139. The rejection region is

>6.25139. Since the observed value of the test statistic falls in the rejection region = 634.36 > 6.25139), is rejected. There is sufficient evidence to indicate at least one proportion differs at cr = . 10. f

To determine if the true proportions follow the proportions given, we test: Ho'. Pi = .30,/?2 = .50,= .15,^4 = -05 Ha'. At least one pi differs from its hypothesized value El = npi^o = 1,319(.30) = 395.7 E2 = np2fi= 1,319(. 15)= 197.85

E2 = «/?2.o = 1,319(.50) = 659.5 £'4 = «p4’o = 1,319(.05) = 65.95

The test statistic is

[n-E,f

E. [450-395.7f , [627-659.5f [219-197.85f [23-65.951^ =-+-+ + i-1- = 39 29 395.7 659.5 197.85 65.95

242

Chapter 8

The rejection region requires a = .10 in the upper tail of the

distribution with

df = A: - 1= 4 - 1 = 3. From Table XI, Appendix A, ^'lo “ 6.25139. The rejection region is

>6.25139.

Since the observed value of the test statistic falls in the rejection region Oj^ = 39.29 > 6.25139), //q is rejected. There is sufficient evidence to indicate at least one proportion differs from its hypothesized value at (.165, .223)

700

n

We are 95% confident that the true proportion of vowels drawn by the electronic game is between .165 and .223. The actual proportion of vowels drawn in the board game is .42. Since this value is not contained in the 95% confidence interval, one can conclude that the ScrabbleExpress™ game produces too few vowels in the 7-letter draws. 8.49 8.51

A two-way contingency table presents multinomial count data classified on two scales of classification. The conditions that are required for a valid chi-square test of data from a contingency table are: 1. 2.

8.53

The n observed counts are a random sample from the population of interest. The sample size, n, will be large enough so that, for every cell, the expected count, Ey, will be equal to 5 or more.

a.

Ho".

The row and column classifications are independent The row and column classifications are dependent

b.

The test statistic is

- XI

\yi E 1^ —---—

The rejection region requires or = .01 in the upper tail of the

distribution with df=

(r- l)(c - 1) = (2 - 1)(3 - 1) = 2. From Table XI, Appendix A, x\\ = 9.21034. The rejection region is 2^ > 9.21034. c.

The expected cell counts are: /?,C, ^ 96(25) ^11

n

E\2

n ^13

244

“ 167 _ 96(64) ~

167

^ 96(78) n

~

167

= 14.37 = 36.79

■£^21

^71(25) n

167 _ 71(64)

E22 n

= 44.84

167

RjC, ^71(78) £^23

n

~

167

10.63 27.21 33.16

Chapter 8

d.

The test statistic is

Fw E 1^ —---—

~

^ij

^ (9-14.37)^

(34-36.79)^

(53-44.84)^

(16-10.63)^

36.79

44.84

10.63

14.37

^ (30-27.2tf ^ (25-33.16)^ 27.21

33.16

Since the observed value of the test statistic does not fall in the rejection region = 8.71

9.21034), Ho is not rejected. There is insufficient evidence to indicate the

row and column classifications are dependent at a = .01. 8.55

Some preliminary calculations are: /^C, J54(134) ^11 =

En ~

n

E2X =

439 =57.180

E22 —

439

E\o -

=49.813

E23 =

439

En =

=30.219

439 =69.062 439 =60.164 439 =32.023

11

En =

186(134)

439

439 =36.759

439

To determine if the row and column classifications are dependent, we test: Ho'. The row and column classifications are independent Ha'. The row and column classifications are dependent

The test statistic is

~

-(40-47.007)^

(72-57.180)^

(42-49.813)^

47.007

57.180

49.813

^ (53-69.062)^ 69.062

{63-56.11

(70-60.164)^

(31-30.219)^

60.164

30.219

56.774

, (38-36.759)2 (30-32.023)2 _ +-1-12.36 36.759 32.023 The rejection region requires a = .05 in the upper tail of the ^ distribution with df= (r - l)(c - 1) = (3 - 1)(3 - 1) = 4. From Table XI, Appendix A, jo5= 9.48773. The rejection region is

> 9.48773.

Since the observed value of the test statistic falls in the rejection region = 12.36 > 9.487^), Ho is rejected. There is sufficient evidence to indicate the row and column classification are dependent at a= .05.

Comparing Population Proportions

245

c.

The proportion of girls drawing a dog is almost 2 times the proportion of boys drawing a dog.

d.

To determine if the likelihood of drawing a dog depends on gender, we test. Hq\ Presence of a dog and gender are independent Presence of a dog and gender are dependent

e. From the printout, the test statistic is 2^ = 2.250 and the p-value is /? = 0 J 34. Since the p-value is not less than a — .05 (p = .134

.05), //o is not rejected. There is

insufficient evidence to indicate the likelihood of drawing a dog depends on gender at a= .05. f. Using MINITAB, the results are: Tabulated statistics; TV, GENDER Using frequencies in NUMBER

Rows:

TV

Columns: GENDER Boy

Girl

All

No

66 66.52

61 60.48

127 127.00

Yes

11 10.48

9 9.52

20 20.00

All

77 77.00

70 70.00

147 147.00

Cell Contents ;

Count Expected count

Pearson Chi-Square = 0.064, DF = 1, P-Value = 0.801 Likelihood Ratio Chi-Square = 0.064, DF = 1, P-Value = 0.801

To determine if the likelihood of drawing a TV in the bedroom depends on gender, we test: Hq. Presence of TV and gender are independent 74: Presence of TV and gender are dependent From the printout, the test statistic is

2^

= .064 and the /?-value \sp = 0.801.

Since the/?-value is not less than «= .05 (p = .801 ft .05), Hq is not rejected. There is insufficient evidence to indicate the likelihood of drawing TV depends on gender at a= .05.

246

Chapter 8

8.59

a.

The two categorical variables are Risk of masculinity and type of event. Risk of masculinity has two levels: High and Low. Type of event has two levels: Violent event and Avoided-Violent event. ‘

b.

The experimental units are 1,507 newly incarcerated men.

n

d.

1,507

=

n

= n 8.202

£,,=:^ = MmZ) = 776.202

1,507

n

n

1,507

2

_[236-260.798f

X -LL

-

[801-776.202f

260.798

[143-118.202f ^

118.202

[327-351.798]"

+ --- + --—

776.202 f.

1,507

=

,, ,,, 10.101

351.798

To determine whether event type depends on high/low risk masculinity, we test:

Hq: Event Type and High/Low Risk Masculinity are independent 7/a: Event Type and High/Low Risk Masculinity are dependent The test statistic is

= 10.101.

The rejection region requires a= .05 in the upper tail of the 2^ distribution with df=(r-l)(c-l) = (2-l)(2-l)= 1. From Table XI, Appendix A,

2r"o5=

3.84146.

The rejection region is 2^ > 3.84146. Since the observed value of the test statistic falls in the rejection region (f =iO.IOI > 3.84146), Ho is rejected. There is sufficient evidence that event type depends on high/low risk masculinity at «:= .05. 8.61

Some preliminary calculations are: ^ R,C, 192(115) Ai, =---- 41.98 n 526 E21

R^C,

^31 =

157(115)

n

526

R,C,

177(115)

n

526

£’,2

_R,C^ _ 192(411)_ = 150.02 n 526

—

= 34.33

= 38.70

122,67

=

^

n

=

526

£3.=^ = ™) =138.30 ’

n

526

To determine if the proportion of students diagnosed with MR depends on the IQ test/retest methoij, we test: Ho'. MR diagnosis and Test/Retest are independent H^'. MR diagnosis and Test/Retest are dependent

Comparing Population Proportions

247

The test statistic is ,

r25-41.98f

r =

41.98

[54-34.33]' [36-38.70]' [167-150.02]' + ---— + ' 150.02 34.33 38.70

[103-122.67]'

[141-138.30]'

+ ■

122.67

^

138.30

,

The rejection region requires (X= .01 in the upper tail of the x

’ ‘ distribution with df

= (r- i)(c_ 1) = (3 - 1)(2 - 1) = 2. From Table XI, Appendix A, rejection region is

xl\

= 9.21034. The

> 9.21034.

Since the observed value of the test statistic falls in the rejection region (/ = 23.46 > 9.21034), Hq is rejected. There is sufficient evidence to indicate the proportion of students diagnosed with MR depends on the IQ test/retest method at or = .01. 8.63

Some preliminary calculations are:

e,=:^ = ^ = 23.90

^12

275

n

59(106)

£,,=^ = ^=12.85 ”

n

^21

275

■^23

= 24.03

■‘22

275

n

£,,=^=*=22.74 ^31 n

n

~

n

= 12.23

^41

275 95(112)

E,2 =

^32

275

_ 59(57) ^33 =

275

n

= 38.69

275

R^C^

59(57)

n

275

^3^2 n

59(112)

.RA

95(106)

= 24.03

275

n

^43

275

n

275

RA2

95(57) _ .

n

275

To determine if political strategy of ethnic groups depends on world region, we test: Hq. Political strategy and world region are independent Ha'. Political strategy and world region are dependent ... 2 (24-23.90)' The test statistic is /' = J ^--- = -23.90 2

(31-25.25)' 25.25

(7-12.85)"

(32-22.74)"

(23-24.03)"

(4-12.23)"

12.85

22.74

24.03

12.23

H-1-1-1-h ■

(11-22.74)^ .74

(22-24.03)' (26-12.23)' (39-36.62)' (36-38.69)' (20-19.69)' H-1-1-1-1- ■ 24.03 12.23 36.62 38.69 19.69 = 35.409

248

Chapter 8

The rejection region requires 10.6446. Since the observed value of the test statistic falls in the rejection region (j^ = 35.409 > 10.6446), Ho is rejected. There is sufficient evidence to indicate that the political strategy of the ethnic groups depends on world region at a = .10. To graph the data, we will first compute the percent of observations in each category of Political Strategy for each World Region. To do this, we divide each cell frequency by the row total and then multiply by 100%. Using SAS, a graph of the data is: PER Sum •k "k

■k -k kk

k k

kk

kk

kk

kk

* ★

kk

kk

k k

kk

kk

kk

kk

kk

k k

0.5

0.4 **

kk

kk

kk

k k

kk

kk

kk

kk

kk

kk

k k

k k

kk

★

kk

kk

k k

kk

kk

kk

kk

•k ★

kk

kk

kk

k*

kk

kk

kk

* ★

kk

kk

kk

★★

kk

kk

k k

•k "k

kk

•* *

kk

k

kk

kk

k★

★k

kk

**

kk

kk

k k

kk

*

kk

kk

★★

kk

kk

kk

kk

**

kk

k k

kk

★k

kk

kk

kk

kk

★k

kk

kk

kk

kk

kk

kk

kk

k k

kk

k k

k *

kk

kk

kk

kk

kk

kk

kk

kk

k *•

k k

kk

kk

k k

kk

kk

kk

kk

kk

kk

k k

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

★k

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

k k

k k

kk

kk

kk

kk

kk

k k

kk

k k

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

” kk

kk

kk

kk

kk

kk

kk

kk

kk

kk

M

N

T

M

N

T

M

N

T

M

N

T

0

0

E

0

0

E

0

E

0

0

E

B

N

R

B

N

R

B

0 N

R

B

N

R

I

E

R

I

E

R

I

E

R

I

E

R

0

L

0

L

0

L

*

0.3

0.2

0.1

kk

kk

L

R

R AFRICA

EAST ASIA

R LATIN AM.

*

STRAT

0 R

POST--COMMON

REGION

The graph supports the results of the test above. The percents in each level of Political Strategy differs for the different values of World Region.

Comparing Population Proportions

249

8.65

Using MINITAB, the results of the analyses are: Results for: AIRTHR~1.MTP Tabulated statistics: Instruction, Strategy Columns :

Strategy

Guess

Other

TTBC

All

Cue

5 7.000

6 8.500

13 8.500

24 24.000

Pattern

9 7.000

11 8.500

4 8.500

24 24.000

14 14.000

17 17.000

17 17.000

48 48.000

Rows:

All

Instruction

Cell Contents:

Count Expected count

Pearson Chi-Square = 7.378, DF = 2, P-Value = 0.025 Likelihood Ratio Chi-Square = 7.668, DF = 2, P-Value = 0.022

To determine if the choice of heuristic strategy depends on type of instruction provided, we test: Hq\ Choice of heuristic strategy and instruction provided are independent Hq. Choice of heuristic strategy and instruction provided are dependent From the printout, the test statistic ’\s ^ = 7.378 and the

value \sp = .025.

Since the p-value is less than a= .05 (p = .025 < .05), Hq is rejected. There is sufficient evidence to indicate the choice of heuristic strategy depends on type of instruction provided ata= .05.

250

Chapter 8

8.67

a.

Using MINITAB, the results of the analyses are: Tabulated statistics: SHSS, CAHS Using frequencies in Fr Rows:

SHSS

Columns :

CAHS

H

L

M

V

All

19 10.02

6 15.83

14 14.22

3 1.94

42 42.00

L

2 11.45

32 18.09

14 16.25

0 2.22

48 48.00

M

6 7.39

11 11.68

14 10.49

0 1.43

31 31.00

V

4 2.15

0 3.39

2 3.05

3 0.42

9 9.00

31 31.00

49 49.00

44 44.00

6 6.00

130 130.00

H -

All

Cell Contents

.

Count Expected count

Pearson Chi-Square = 60.103, DF = 9 Likelihood Ratio Chi-Square = 59.638,

DF = 9

* WARNING * 1 cells with expected counts less than 1 * WARNING * Chi-Square approximation probably invalid * NOTE * 7 cells with expected counts less than 5

From the printout, 7 cells have expected counts less than 5. In order for the test to be valid, all of the cells should have expected counts greater than 5. Thus, we should not proceed with the analysis. b.

Combining the High and Very High categories. the new table is:

SHSS: C LEVEL

Comparing Population Proportions

Low Medium HighA^ery High

Low 32 11 6

CAHS LEVEL Medium High/Very High 14 2 14 6 16 29

251

c.

The expected cell counts are: 48(49)

Rf,

= 18.09

130

n

£

n

130

RiC,

48(37)

■'13

130 31(44)

^22

n

130

R,C,

31(37)

n

130

n

130

E23

51(49)

= 19.22

130

n

10.49

—

130

n

'31

= 13.66

=

-

48(44) ■'12

11,68

=

''

''

51(37) ■'33

130

n

Since all of the expected cell counts are now 5 or greater, the assumption is met. To determine whether CARS Levels and SHSS:C Levels are dependent, we test: Hq: cars Levels and SRSS:C Levels are independent Ha. CARS Levels and SRSS:C Levels are dependent

The test statistic is % ~

—---

(32-18.09)^

(14-16.25)^

(2-13.66)^

(11-11.68)^

16.25

13.66

11.68

(14-10.49)^

(6-8.82)^

(6-19.22)^

(16-17.26)^

10.49

8.82

19.22

17.26

18.09 +

^

+ (29-14,52)^ .46.70 14.52 The rejection region requires a - .05 in the upper tail of the

distribution with df=

(r - l)(c - 1) = (3 - 1)(3 - 1) = 4. From Table XI, Appendix A, Jqs ~ 9.48773. The rejection region is

> 9.48773.

Since the observed value of the test statistic falls in the rejection region (2^ = 46.70 > 9.48773), //o is rejected. There is sufficient evidence to indicate that CARS Levels and SRSS:C Levels are dependent at .05.

252

Chapter 8

8.69

a.

Some preliminary calculations are:

=^= n

38

''

= ^ = 2(?1)=4.42 n

31(14)

=19.58 n

38

_ R,C, _ 7(14)

38

n

t

= 11.42

= 2.58

38

To determine whether the vaccine is effective in treating the MN strain of HIV, we test: Hq. Vaccine and MN strain are independent //a’. Vaccine and MN strain are dependent

The test statistic is

(22-19.58)^ ^ (9-11.42)

= ZI

19.58

11.42

(2-4.42)^ (5-2.58)2 , + 1^-- + --— =4.407 4.42

2.58

The rejection region requires a = .05 in the upper tail of the ^ distribution with df= (r- l)(c - 1) = (2 - 1)(2 - 1) = 1. From TableXI, Appendix A, xls = 3.84146. The rejection region is 2^ > 3.84146. Since the observed value of the test statistic falls in the rejection region = 4.407 > 3.84146), Ho is rejected. There is sufficient evidence to indicate that the vaccine is effective in treating the MN strain of HIV at a = .05. b.

The necessaiy assumptions are: 1. The n observed counts are a random sample from the population of interest. 2. The sample size, n, will be large enough so that, for every cell, the expected count, E(nij), will be equal to 5 or more. For this example, the second assumption is violated. Two of the expected cell counts are less than 5. What we have computed for the test statistic may not have a distribution.

c.

For this contingency table:

r7Y3i v22 22 P=

Y8^

I24J

7!

31!

2!5! 22!9! = .0438 38! 24114!

Comparing Population Proportions

253

d.

If vaccine and MN strain are independent, then the proportion of positive results should be relatively the same for both patient groups. In the two tables presented, the proportion of positive results for the vaccinated group is smaller than the proportion for the original table. For the first table,

I

7!

31!

Kh v23y ^il6!a = .0057 38! "38" 24!14!

I24J

For the second table: ("31"|

P=

(A w I24J "38"

7!

^ 0!7! 24!7! = .0003 38!

I24J e.

31!

24!14!

The/?-value for Fisher’s exact test is/? = .0438 + .0057+ .0003 = .0498. Since thepvalue is small, there is evidence to reject Hq. There is sufficient evidence to indicate that the vaccine is effective in treating the MN strain of HIV at a > .0498.

8.71

In a one-way chi-square analysis, one must hypothesize the proportions in each of the cells. In a two-way chi-square analysis, one does not need to specify the proportions in each of the cells. In a two-way chi-square analysis, one tests to see if the two categorical variables are independent or not.

8.73

Some preliminary calculations are: P a.

_ no «i ~ 200

130

.55;

200

.65;

,

X; +?C2

110 + 130 _ 240

A7 +«2

200+200 ~ 400

Hq\ (/?i-/?2) = 0 ^a: iP\-P2) 9.48773), Hq is rejected. There is sufficient evidence to indicate a dependence between rows and columns at

.05.

b.

No, the analysis remains identical.

c.

Yes, the assumptions on the sampling differ.

d.

The percentages are in the table below. -

^

Column

1 1 20

_

2

20

— X 100% = 40% — X 100% = 22.2% 50 90

10 110

X 100%-9.1%

Row 2 10

20 70 — X 100% = 20% — X 100% = 22.2% — X 100% = 63.6% 50

90

3 20 50 — X 100% = 40% — X 100% = 55.6% 50 90

Totals

3 50

100

no

250

30 -X 100% = 37.3%

100

no

X 100% - 20%

250

X 100%-40% X 100%-40%

250

If the rows and columns are independent, then the proportion of observations for each column should be relatively the same. In this graph, the percentages are quite different, supporting the results of the test in part a.

256

Chapter 8

8.77

Some preliminary calculations are: E\= E2= Ei,= £4 = E5- nptfi = 95(.20)

=

19

«

To determine if the true percentages of ADEs in the 5 “cause” categories differ, we test: Ho: Pi ^p2 =Po = A =A == -2 At least one/?, differs from .2, i = 1, 2, 3, 4, 5 The test statistic is [«,-£,f

/='X

[29-19f

E:

+

[17-19f

[13-19f

19

[9-19f

19

19

[27-19]

+ ■

+ •

19

46

19

Since no a was given, we will use a = .05. The rejection region requires a= .05 in the upper tail of the ^ distribution with df=A:-l=5-l=4. From Table XI, Appendix A, x])5~ 9.48773. The rejection region is x^> 9.48773. Since the observed value of the test statistic falls in the rejection region = 16>9 AS113), Ho is rejected. There is sufficient evidence to indicate the true percentages of ADEs in the 5 “cause” categories differ at or = .05. (X"

8.79

a.

b.

The 2x2 contingency table is: LLlD Genetic Trait Yes 21 Yes 150 No Total 171

Total 36 456 492

No 15 306 321

Some preliminary calculations are: ^

/?,C, "

n

36(171)

n

^12=-^

158.49

492

23.49

492

n

A92

Rf, ^ 456(171) ''

36(321)

12.51

R,C,

456(321)

n

492

297.51

To determine if the genetic trait occurs at a higher rate in LDD patients than in the controls, we test: Ho'. Genetic trait and LDD group are independent //a: Genetic trait and LDD group are dependent

The test statistic is

K-4]^

(21-12.51)^

(15-23.49)

12.51

23.49

(150-158.49)^ ^ (306-297.51)^ 158.49

Comparing Population Proportions

297.51

257

The rejection region requires a= .01 in the upper tail of the ^ distribution with df (r - l)(c - 1) = (2 - 1)(2 - 1) = l.From Table XI, Appendix A, x\\ = 6.63490. The rejection region is x^> 6.63491. Since the observed value of the test statistic falls in the rejection region — 9.52 > 6.63491), Ho is rejected. There is sufficient evidence to indicate that the genetic trait occurs at a higher rate in LDD patients than in the controls at a = .01. \

c.

To construct a bar graph, we will first compute the proportion of LDD/Control patients that have the genetic trait. Of the LLD patients, 21 /171 = .123 have the trait. For the Controls, 15/321 = .043. The bar graph is:

0.15

Z

0.1

Iu

0.05

0) 0-

0 LLD

CONTROL

TOTAL

Group

Because the bars are not close to being the same height, it indicates that the genetic trait appears at a higher rate among LLD patients than among the controls. This supports the conclusion of the test in part b. 8.81

a.

No. In a multinomial experiment, each trial results in one of A: = 8 possible outcomes. In this experiment, each trial cannot result in one of all of the eight possible outcomes. The people are assigned to a particular diet before the experiment. Once the people are assigned to a particular diet, there are only two possible outcomes, cancer tumors or no cancer tumors.

b.

The expected cell counts are: 80(30) n R,C, ^12

n

_ ^13

120

^21

80(30)

120

^22

120 ^C^_40(30)_j^ n

n

120

n

120

80(30) n

120

^23

80(30) ^14

258

n

120

•^24

n

10

^=10 120

Chapter 8

c.

The test statistic is ^ij _ (27-20)^

(20-20)^

(19-20)

(14-20)^

(3-10)^

(10-10)^

20

20

20

10

10

20

^(11-10)^ ^(16-10)^

10 d.

10

To determine if diet and presence/absence of cancer are independent, we test: Hq: Diet and presence/absence of cancer are independent /7a'• Diet and presence/absence of cancer are dependent The test statistic is = 12.9 (from part c). The rejection region requires « = .05 in the upper tail of the

distribution with df=

(r - l)(c - 1) = (2 - 1)(4 - 1) = 3. From Table XI, Appendix A,

7.81473.

The rejection region is ;^> 7.81473. Since the observed value of the test statistic falls in the rejection region - 12.9 > 7.81473), Ho is rejected. There is sufficient evidence to indicate that diet and presence/absence of cancer are dependent for a= .05. e.

= 27/30 = .9; pj = 20/30 = .667 For confidence coefficient .95, a= .05 and a/2 = .05/2 = .025. From Table III, Appendix A, Z025 - 1-96. The confidence interval is:

(A-7’2)±^. 0251

IPl^l , P2H n,

(.900-.667)+ 1.96.

.900(.100) ^ .667(.333) 30

n.

30

=> .233 ± .200 => (.033, .433) We are 95% confident that the difference in the percentage of rats on a high fat/no fiber diet with cancer and the percentage of rats on a high fat/fiber diet with cancer is between 3.3% and 43.3%.

8.83

a.

-^-^-.153 p^ =^ = n, 189

b.

X 32 A _±2.=:J^ = .215 " 149 For confidence coefficient .90, a= .10 and a/2 = .10/2- .05. From Table III, Appendix A, z.os = 1.645. The 90% confidence interval is:

-.062±.070=>(-.132,

Comparing Population Proportions

.008)

259

d.

8.85

a.

We are 90% confident that the difference in the true proportion of super-experienced bidders who fall prey to the winner’s curse and the true proportion of lessexperienced bidders who fall prey to the winner’s curse is between -0.132 and 0.008. Let p\ = proportion of female students who switched due to loss of interest in SME and P2 = proportion of male students who switched due to lack of interest in SME. Some preliminary calculations are: 74

jCj _ 72

.43;

«2

172

.44;

^ _ Xi + X2 _ 74 + 72 ^

163

+«2

_

172 + 163

To determine if the proportion of female students who switch due to lack of interest in SME differs from the proportion of males who switch due to a lack of interest, we test: Ho’. pi-p2 = 0

H^- P\ -p2 ^ 0 The test statistic is z =

(.43-.44)-0

(A-P2)-0

f 1 \pq

1

.436(.564)

—+—

( 1

+

U72

= -0.18 _1_^ 163^

The rejection region requires d2 = .10/2 = .05 in each tail of the z distribution. From Table III, Appendix A, z.05 = 1.645. The rejection region is z < -1.645 or z > 1.645. Since the observed value of the test statistic does not fall in the rejection region (z = -0.18 ft -1.645), Ho is not rejected. There is insufficient evidence to indicate the proportion of female students who switch due to lack of interest in SME differs from the proportion of males who switch due to a lack of interest in SME at a = . 10. b.

Let p\ = proportion of female students who switched due to low grades in SME and P2 = proportion of male students who switched due to low grades in SME. Some preliminary calculations are:

il=.19; "1

172

p,=il = ii=.27 n,

163

For confidence coefficient .90, or= .10 and fl/2 = .10/2 = .05. From Table III, Appendix A, Zqs = 1.645. The confidence interval is: ^

(a - P2) ± 2.05

A

+ PjBi =^(.19-.27) ±1.645 /.19(.81) ^ .27(.73)

V

"2 => -.08 + .075

172

163

(-.155, -.005)

We are 90% confident that the difference between the proportions of female and male switchers who lost confidence due to low grades in SME is between -.155 and -.005. Since the interval does not include 0, there is evidence to indicate the proportion of female switchers due to low grades is less than the proportion of male switchers due to low grades.

260

Chapter 8

8.87

First, the values of the variables will be defined. 1. 2. 3. 4. 5.

6.

7.

8.

The variable WAR has values 1, 2, and 9. 1 = Support, 2 = Oppose, and 9 = Don’t know or refused to answer. The variable INTERNET has values 1, 2, and 9. 1 = Yes, 2 = No, and 9 = Don’t know or refused to answer. The variable PARTY has values 1, 2, 3, 4, 5, and 9. 1 = Republican, 2 = Democrat, 3 = Independent, 4 = No preference, 5 = Other, and 9 = Don’t know or refused to answer. The variable VET has values 1, 2, 3, 4, and 9. 1 = Yes, I have, 2 = Yes, other household member, 3 = Yes, both, 4 = No, and 9 = Don’t know or refused to answer. The variable IDEO has values 1, 2, 3,4, 5, and 9. 1 = Very conservative, 2 = Conservative, 3 = Moderate, 4 = Liberal, 5 = Very liberal, and 9 = Don’t know or refused to answer. The variable RACE has values 1, 2, 3, 4, 5, and 9. 1 = White, 2 = Black or African American, 3 = Asian or Pacific Islander, 4 = Mixed, 5 = Native American, 6 = Other and 9 = Don’t know or refused to answer. The variable INCRANGE has values 1, 2, 3, 4, 5, 6, 7, 8, and 9. 1 = 11.0705), reject Hq. There is sufficient evidence to indicate the probabilities differ from their hypothesized values at a = .05.

b.

D, = —= .376 ' 85 For confidence coefficient .95, a= .05 and a/2 = .05/2 = .025. From Table III, Appendix A, z 025 1 -96. The 95% confidence interval is:

Pi

± 2025 yp- =>.376 + 1

=> -376 ±. 103

(.273, .479)

We are 95% confident that the true proportion of Avonex MS patients who are exacerbation-free during a 2-year period is between .273 and .479. c.

8.91

From previous studies, it is known that 26% of the MS patients on a placebo experienced no exacerbations in a 2-year period. Since .26 does not fall in the 95% confidence interval, there is evidence that Avonex patients are more likely to have no exacerbations than the placebo patients at a = .05.

Some preliminary calculations are:

”

n

''

n

^ R,C,

12(14) _

^

n R,C,

24 12(14) _

24 _ 12(10) _

.

^

24

n

24

To determine if a relationship exists between food choice and whether or not chickadees fed on gypsy moth eggs, we test: No: Food choice and whether or not chickadees fed on gypsy moth eggs are independent Nai Food choice and whether or not chickadees fed on gypsy moth eggs are dependent

268

Chapter 8

The test statistic is

-

IE

[n.j-E.f

= (2-5r , (10-7/ , (8-5)^ ^ (4-7)^ 5

7

5

7'

The rejection region requires «= .10 in the upper tail of the 2^ distribution with df= (r - l)(c - 1) = (2 - 1)(2 - 1) = 1. From Table XI, Appendix A,

2-^0554.

The rejection region is ;^> 2.70554. Since the observed value of the test statistic falls in the rejection region = 6.111 > 2.70554), Ho is rejected. There is sufficient evidence to indicate that a relationship exists between food choice and whether or not chickadees fed on gypsy moth eggs at a = .10. 8.93

Let Pi = proportion of patients receiving Zyban who were not smoking one year later and p2 = proportion of patients not receiving Zyban who were not smoking one year later. Some preliminary calculations are; 71

= .230

=

309

X, + Xj

p=-^-^

n, + «2

«2

-^il^=.176

.121

306

q = \ = p^\-.\16 =.824

309 + 306

To determine if the antidepressant drug Zyban helped cigarette smokers kick their habit, we test; Ho\ p\ -P2 = 0 /fa: P\ -pi > 0

The test statistic is z =

(A-A)-o pq

1

1

— + —

V”i

(.230-.121)-0 .176(.824)

n2 y

3.55

1

r 1 ■ + ■

309

306

The rejection region requires a = .05 in the upper tail of the z distribution. From Table III, Appendix A, z.os = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic falls in the rejection region (z = 3.55 > 1.645), Ho is rejected. There is sufficient evidence to indicate that the antidepressant drub Zyban helped cigarette smokers kick their habit at a= .05.

Comparing Population Proportions

269

8.95

a.

(26-23?

(146-136)^

(361-341)^

136

341

(143-136/ , (13-23)^ 136

23

= 9.647 b.

From Table XI, Appendix A, with df = 5, ^qs ~ ^ ^ .0705

c.

No. / = 9.647

11.0705. Do not reject Hq. There is insufficient evidence to indicate

the salary distribution is nonnormal for a= .05. d.

The />-value = P{x^ > 9.647). Using Table XI, Appendix A, with df = 5, .05