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Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 1

Chapter 9

Statistical Inference:

Hypothesis Testing for Single Populations

LEARNING OBJECTIVES

The main objective of Chapter 9 is to help you to learn how to test hypotheses on singlepopulations, thereby enabling you to:

1. Understand the logic of hypothesis testing and know how to establish null andalternate hypotheses.

2. Understand Type I and Type II errors and know how to solve for Type II errors.

3. Know how to implement the HTAB system to test hypotheses.

4. Test hypotheses about a single population mean when is known.

5. Test hypotheses about a single population mean when is unknown.

6. Test hypotheses about a single population proportion.

7. Test hypotheses about a single population variance.

CHAPTER TEACHING STRATEGY

For some instructors, this chapter is the cornerstone of the first statistics course.Hypothesis testing presents the logic in which ideas, theories, etc., are scientifically

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examined. The student can be made aware that much of the development of concepts tothis point including sampling, level of data measurement, descriptive tools such as meanand standard deviation, probability, and distributions pave the way for testing hypotheses.Often students (and instructors) will say "Why do we need to test this hypothesis whenwe can make a decision by examining the data?" Sometimes it is true that examining the

data could allow hypothesis decisions to be made. However, by using the methodologyand structure of hypothesis testing even in "obvious" situations, the researcher has addedcredibility and rigor to his/her findings. Some statisticians actually report findings in acourt of law as an expert witness. Others report their findings in a journal, to the public,to the corporate board, to a client, or to their manager. In each case, by using thehypothesis testing method rather than a "seat of the pants" judgment, the researcherstands on a much firmer foundation by using the principles of hypothesis testing andrandom sampling. Chapter 9 brings together many of the tools developed to this pointand formalizes a procedure for testing hypotheses.

The statistical hypotheses are set up as to contain all possible decisions. The

two-tailed test always has = and in the null and alternative hypothesis. One-tailedtests are presented with = in the null hypothesis and either > or < in the alternativehypothesis. If in doubt, the researcher should use a two-tailed test. Chapter 9 beginswith a two-tailed test example. Usually, that which the researcher wants to demonstratetrue or prove true is usually set up as an alternative hypothesis. The null hypothesis isthat the new theory or idea is not true, the status quo is still true, or that there is nodifference. The null hypothesis is assumed to be true before the process begins. Someresearchers liken this procedure to a court of law where the defendant is presumedinnocent (assume null is true - nothing has happened). Evidence is brought before thejudge or jury. If enough evidence is presented, the null hypothesis (defendant innocent)can no longer be accepted or assume true. The null hypothesis is rejected as not true and

the alternate hypothesis is accepted as true by default. Emphasize that the researcherneeds to make a decision after examining the observed statistic.

Some of the key concepts in this chapter are one-tailed and two-tailed test andType I and Type II error. In order for a one-tailed test to be conducted, the problem mustinclude some suggestion of a direction to be tested. If the student sees such words asgreater, less than, more than, higher, younger, etc., then he/she knows to use a one-tailtest. If no direction is given (test to determine if there is a "difference"), then a two-tailedtest is called for. Ultimately, students will see that the only effect of using a one-tailedtest versus a two-tailed test is on the critical table value. A one-tailed test uses all of thevalue of alpha in one tail. A two-tailed test splits alpha and uses alpha/2 in each tail thus

creating a critical value that is further out in the distribution. The result is that (all thingsbeing the same) it is more difficult to reject the null hypothesis with a two-tailed test.Many computer packages such as MINITAB include in the results a p-value. If youdesignate that the hypothesis test is a two-tailed test, the computer will double the p-valueso that it can be compared directly to alpha.

In discussing Type I and Type II errors, there are a few things to consider. Oncea decision is made regarding the null hypothesis, there is a possibility that the decision is

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correct or that an error has been made. Since the researcher virtually never knows forcertain whether the null hypothesis was actually true or not, a probability of committingone of these errors can be computed. Emphasize with the students that a researcher cannever commit a Type I error and a Type II error at the same time. This is so because aType I error can only be committed when the null hypothesis is rejected and a Type II

error can only be committed when the decision is to not reject the null hypothesis. Type Iand Type II errors are important concepts for managerial students to understand evenbeyond the realm of statistical hypothesis testing. For example, if a manager decides tofire or not fire an employee based on some evidence collected, he/she could becommitting a Type I or a Type II error depending on the decision. If the productionmanager decides to stop the production line because of evidence of faulty raw materials,he/she might be committing a Type I error. On the other hand, if the manager fails toshut the production line down even when faced with evidence of faulty raw materials,he/she might be committing a Type II error.

The student can be told that there are some widely accepted values for alpha

(probability of committing a Type I error) in the research world and that a value isusually selected before the research begins. On the other hand, since the value of Beta(probability of committing a Type II error) varies with every possible alternate value ofthe parameter being tested, Beta is usually examined and computed over a range ofpossible values of that parameter. As you can see, the concepts of hypothesis testing aredifficult and represent higher levels of learning (logic, transfer, etc.). Studentunderstanding of these concepts will improve as you work your way through thetechniques in this chapter and in chapter 10.

CHAPTER OUTLINE

9.1 Introduction to Hypothesis TestingTypes of HypothesesResearch HypothesesStatistical HypothesesSubstantive HypothesesUsing the HTAB System to Test HypothesesRejection and Non-rejection RegionsType I and Type II errors

9.2 Testing Hypotheses About a Population Mean Using the z StatisticUsing a Sample Standard DeviationTesting the Mean with a Finite PopulationUsing thep-Value Method to Test HypothesesUsing the Critical Value Method to Test HypothesesUsing the Computer to Test Hypotheses about a Population Mean Using

the z Test9.3 Testing Hypotheses About a Population Mean Using the tStatistic

Using the Computer to Test Hypotheses about a Population Mean Usingthe tTest

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9.4 Testing Hypotheses About a ProportionUsing the Computer to Test Hypotheses about a Population Proportion

9.5 Testing Hypotheses About a Variance

9.6 Solving for Type II ErrorsSome Observations About Type II ErrorsOperating Characteristic and Power CurvesEffect of Increasing Sample Size on the Rejection Limits

KEY TERMS

Alpha( ) One-tailed Test

Alternative Hypothesis Operating-Characteristic Curve (OC)Beta( ) p-Value MethodCritical Value Power Critical Value Method Power CurveHypothesis Testing Rejection RegionLevel of Significance Research HypothesisNonrejection Region Statistical HypothesisNull Hypothesis Substantive ResultObserved Significance Level Two-Tailed TestObserved Value Type I Error

Type II Error

SOLUTIONS TO PROBLEMS IN CHAPTER 9

9.1 a) Ho: = 25

Ha: 25

x = 28.1 n = 57 s = 8.46 = .01

For two-tail, /2 = .005 zc = 2.575

z=

57

46.8251.28

=

n

s

x

= 2.77

observedz= 2.77 >zc = 2.575

Reject the null hypothesis

b) from Table A.5, inside area betweenz= 0 andz= 2.77 is .4972

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p-value = .5000 - .4972 = .0028

Since thep-value of .0028 is less than2

= .005, the decision is to:


c) critical mean values:

zc =

n

s

xc

2.575 =

57

46.8

25cx

x c = 25 2.885

x c = 27.885 (upper value)

x c = 22.115 (lower value)

9.2 Ho: = 7.48Ha: < 7.48

x = 6.91 n = 96 s = 1.21 =.01

For one-tail, = .01 zc = -2.33

z =

96

21.1

48.791.6 =

n

s

x

= -4.62

observedz= -4.62 1,200

x = 1,215 n = 113 s = 100 = .10

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For one-tail, = .10 zc = 1.28

z =

113

100

200,1215,1 =

n

s

x

= 1.59

observedz= 1.59 >zc = 1.28


b) Probability > observedz= 1.59 is .0559 which is less than = .10.Reject the null hypothesis.

c) Critical mean value:

zc =

n

sx c

1.28 =

113

100

200,1cx

x c = 1,200 + 12.04

Since calculated x = 1,215 which is greater than the critical x = 1212.04, rejectthe null hypothesis.

9.4 Ho: = 82Ha: < 82

x = 78.125 n = 32 s = 9.184 = .01

z.01 = -2.33

z=32184.9

82125.78 =

n

s

x

= -2.39

Since observedz= -2.39

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Statistically, we can conclude that urban air soot is significantly lower. From abusiness and community point-of-view, assuming that the sample result isrepresentative of how the air actually is now, is a reduction of suspended particlesfrom 82 to 78.125 really an important reduction in air pollution? Certainly itmarks an important first step and perhaps a significant start. Whether or not it

would really make a difference in the quality of life for people in the city of St.Louis remains to be seen. Most likely, politicians and city chamber of commercefolks would jump on such results as indications of improvement in cityconditions.

9.5 H0: = $424.20

Ha: $424.20

x = $432.69 n = 54 s = $33.90 = .05

2-tailed test, /2 = .025 z.025 = + 1.96

z =

54

90.33

20.42469.432 =

n

s

x

= 1.84

Since the observedz= 1.85 $62,600

x = $64,820 n = 48 s = $7,810 = .01

1-tailed test, = .01 z.01 = 2.33

z =

48

810,7

600,62820,64 =

n

s

x

= 1.97

Since the observedz= 1.97

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9.7 H0

: = 5

Ha: 5

x = 5.0611 n = 42 s = 0.2803 = .10

2-tailed test, /2 = .05 z.05 = + 1.645

z =

42

2803.0

50611.5 =

n

s

x

= 1.41


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The CEO could use this information as a way of discrediting the Runzheimerstudy and using her own figures in recruiting people and in discussing relocationoptions. In such a case, this could be a substantive finding. However, one mustask if the difference between $4,292 and $4,008 is really an important differencein monthly rental expense. Certainly, Paris is expensive either way. However, an

almost $300 difference in monthly rental cost is a non trivial amount for mostpeople and therefore might be considered substantive.

9.10 Ho: = 123Ha: > 123

= .05 n = 40 40 people were sampled

x = 132.36 s = 27.68

This is a one-tailed test. Since thep-value = .016, we

reject the null hypothesis at = .05.The average water usage per person is greater than 123 gallons.

9.11 n = 20 x = 16.45 s = 3.59 df = 20 - 1 = 19 = .05

Ho: = 16

Ha: 16

For two-tail test, /2 = .025, critical t.025,19 = 2.093

t =2059.3

1645.16 =

n

s

x

= 0.56

Observed t= 0.56 < t.025,19 = 2.093

The decision is to Fail to reject the null hypothesis

9.12 n = 8 x = 58.42 s2 = 25.68 df = 8 - 1 = 7 = .01

Ho: = 60

Ha: < 60

For one-tail test, = .01 critical t.01,7 = -2.998

t =

8

68.25

6042.58 =

n

s

x

= -0.88

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Observed t= -0.88 > t.01,7 = -2.998


9.13 n = 11 x = 1,235.36 s = 103.81 df = 11 - 1 = 10 = .05

Ho: = 1,160Ha: > 1,160

For one-tail test, = .05 critical t.05,10 = 1.812

t =

11

81.103

160,136.236,1 =

n

s

x

= 2.44

Observed t= 2.44 > t.05,10 = 1.812

The decision is to Reject the null hypothesis

9.14 n = 20 x = 8.37 s = .189 df = 20-1 = 19 = .01

Ho: = 8.3

Ha: 8.3For two-tail test, /2 = .005 critical t.005,19 = 2.861

t =

20

189.

3.837.8 =

n

s

x

= 1.66

Observed t= 1.66 < t.005,19 = 2.861


9.15 n = 12 x = 1.85083 s = .02353 df = 12 - 1 = 11 = .10

H0: = 1.84

Ha: 1.84

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For a two-tailed test, /2 = .05 critical t.05,11 = 1.796

t =

12

02353.

84.185083.1 =

n

s

x

= 1.59

Since t= 1.59 < t11,.05 = 1.796,The decision is to fail to reject the null hypothesis.

9.16 n = 25 x = 1.1948 s = .0889 df = 25 - 1 = 24 = .01

Ho: = $1.16Ha: > $1.16

For one-tail test, = .01 Critical t.01,24 = 2.492

t =

25

0889.

16.11948.1 =

n

s

x

= 1.96

Observed t= 1.96 < t.01,24 = 2.492The decision is to Fail to reject the null hypothesis

9.17 n = 19 x = $31.67 s = $1.29 df = 19 1 = 18 = .

05

H0: = $32.28

Ha: $32.28

Two-tailed test, /2 = .025 t.025,18 = + 2.101

t =

19

29.1

28.3267.31 =

n

s

x

= -2.06

The observed t= -2.06 > t.025,18 = -2.101,The decision is to fail to reject the null hypothesis

9.18 n = 26 x = 19.534 minutes s = 4.100 minutes = .05

H0: = 19

Ha: 19

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Two-tailed test, /2 = .025, critical tvalue = + 2.06

Observed tvalue = 0.66

Since the observed t= 0.66 < critical tvalue = 2.06,

The decision is to fail to reject the null hypothesis.

Since the Excelp-value = .256 > /2 = .025 and MINITABp-value =.513 > .05,the decision is to fail to reject the null hypothesis.

She would not conclude that her city is any different from the ones in the

national survey.

9.19 Ho: p = .45Ha: p > .45

n = 310 p = .465 = .05

For one-tail, = .05 z.05 = 1.645

z =

310

)55)(.45(.

45.465. =

n

qp

pp

= 0.53

observedz= 0.53 zc = -2.33

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9.21 Ho: p = .29

Ha: p .29

n = 740 x = 207740

207 ==

n

xp = .28 = .05

For two-tail, /2 = .025 z.025 = 1.96

z =

740

)71)(.29(.

29.28. =

n

qp

pp

= -0.60

observedz= -0.60 >zc = -1.96


p-Value Method:

z= -0.60

from Table A.5, area = .2257

Area in tail = .5000 - .2257 = .2743

.2743 > .025

Again, the decision is to Fail to reject the null hypothesis

Solving for critical values:

z =

n

qp

ppc

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1.96 =

740

)71)(.29(.

29. c

p

cp

= .29 .033

.257 and .323

Samplep = p = .28 not outside critical values in tails

Again, the decision is to Fail to reject the null hypothesis

9.22 Ho:p = .48

Ha:p .48

n = 380 x = 164 = .01 /2 = .005 z.005 = +2.575

380

164 ==

n

xp = .4316

z =

380

)52)(.48(.

48.4316. =

n

qp

pp

= -1.89

Since the observedz= -1.89 is greater thanz.005= -2.575, The decision is to fail toreject the null hypothesis. There is not enough evidence to declare that theproportion is any different than .48.

9.23 Ho:p = .79Ha:p < .79

n = 415 x = 303 = .01 z.01 = -2.33

415

303 ==

n

xp = .7301

z =

415

)21)(.79(.

79.7301 =

n

qp

pp

= -3.00

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Since the observedz= -3.00 is less than z.01= -2.33, The decision is to reject thenull hypothesis.

9.24 Ho:p = .31

Ha:p .31

n = 600 x = 200 = .10 /2 = .05 z.005 = +1.645

600

200 ==

n

xp = .3333

z =

600

)69)(.31(.

31.3333. =

n

qp

pp

= 1.23

Since the observed z= 1.23 is less than z.005= 1.645, The decision is to fail to

reject the null hypothesis. There is not enough evidence to declare that theproportion is any different than .48.

Ho:p = .24Ha:p < .24

n = 600 x = 130 = .05 z.05 = -1.645

600

130 ==

n

xp = .2167

z =

600

)76)(.24(.

24.2167. =

n

qp

pp

= -1.34

Since the observedz= -1.34 is greater than z.05= -1.645, The decision is to fail toreject the null hypothesis. There is not enough evidence to declare that theproportion is less than .24.

9.25 Ho:p = .18Ha:p > .18

n = 376 p = .22 = .01

one-tailed test, z.01 = 2.33

z =

376

)82)(.18(.

18.22. =

n

qp

pp

= 2.02

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Since the observedz= 2.02 is less thanz.01= 2.33, The decision is to fail to rejectthe null hypothesis. There is not enough evidence to declare that the proportionis greater than .18.

9.26 Ho:p = .32Ha:p < .32

n = 118 x = 22118

22 ==

n

xp = .186 = .01

For one-tailed test, z.05 = -1.645

z =

118

)68)(.32(.

32.186. =

n

qp

pp

= -3.12

Observedz= -3.12 20

2.05,14 = 23.6848

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2 =

20

)32)(115( = 22.4

Since 2 = 22.4 < 2.05,14 = 23.6848, the decision is to fail to reject the null

hypothesis.

b) H0: 2 = 8.5 = .10 /2 = .05 n = 22 df = n-1 = 21 s2 = 17

Ha: 2 8.5

2.05,21 = 32.6705

2 =

5.8

)17)(122( = 42

Since 2 = 42 > 2.05,21 = 32.6705, the decision is to reject the null hypothesis.

c) H0: 2 = 45 = .01 n = 8 df = n 1 = 7 s = 4.12

Ha: 2 < 45

2

.01,7 = 18.4753

2 =

45

)12.4)(18( 2= 2.64

Since 2 = 2.64 < 2.01,7 = 18.4753, the decision is to fail to reject the null

hypothesis.

d) H0: 2 = 5 = .05 /2 = .025 n = 11 df = 11 1 = 10 s2 = 1.2

Ha: 2 5

2.025,10 = 20.4831 2.975,10 = 3.24697

2 =

5

)2.1)(111( = 2.4

Since 2 = 2.4 < 2.975,10 = 3.24697, the decision is to reject the nullhypothesis.

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9.29 H0: 2 = 14 = .05 /2 = .025 n = 12 df = 12 1 = 11 s2 =30.0833

Ha: 2 14

2.025,11 = 21.92

2.975,11 = 3.81575

2 =

14

)0833.30)(112( = 23.64

Since 2 = 23.64 < 2.025,11 = 21.92, the decision is to reject the nullhypothesis.

9.30 H0: 2 = .001 = .01 n = 16 df = 16 1 = 15 s2 = .00144667

Ha: 2 > .001

2.01,15 = 30.5779

2 =

001.

)00144667)(.116( = 21.7

Since 2 = 21.7 < 2.01,15 = 30.5779, the decision is to fail to reject the nullhypothesis.

9.31 H0: 2 = 199,996,164 = .10 /2 = .05 n = 13 df =13 - 1 = 12

Ha: 2 199,996,164 s2 = 832,089,743.7

2.05,12 = 21.0261 2.95,12 = 5.22603

2 =

164,996,199

)7.743,089,832)(113( = 49.93

Since 2 = 49.93 > 2.05,12 = 21.0261, the decision is to reject the nullhypothesis. The variance has changed.

9.32 H0: 2 = .04 = .01 n = 7 df = 7 1 = 6 s = .34 s2 = .1156

Ha: 2 > .04

2.01,6 = 16.8119

2 =

04.

)1156)(.17( = 17.34

Since 2 = 17.34 > 2.01,6 = 16.8119, the decision is to reject the nullhypothesis

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9.33 Ho: = 100Ha: < 100

n = 48 = 99 s = 14

a) = .10 z.10 = -1.28

zc =

n

s

x c

-1.28 =

48

14

100cx

x c = 97.4

z =

n

s

x c

=

48

14

994.97

= -0.79

from Table A.5, area forz= -0.79 is .2852

= .2852 + .5000 = .7852

b) = .05 z.05 = -1.645

zc =

n

s

x c

-1.645 =

48

14

100cx

x c = 96.68

z =

n

s

x c

=

48

14

9968.96

= -1.15

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= .3749 + .5000 = .8749

c) = .01 z.01 = -2.33

zc =

n

s

x c

-2.33 =

48

14

100cx

x c = 95.29

z =

n

s

x c

=

48

14

9929.95

= -1.84


= .4671 + .5000 = .9671

d) As gets smaller (other variables remaining constant), beta gets larger.Decreasing the probability of committing a Type I error increases the probabilityof committing a Type II error if other variables are held constant.

9.34 = .05 = 100 n = 48 s = 14

a) a = 98.5 zc = -1.645

zc =

n

s

x c

-1.645 =

48

14

100cx

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x c = 96.68

z =

n

s

x c

=

48

14

9968.96

= -0.90


= .3159 + .5000 = .8159

b) a = 98 zc = -1.645

x c = 96.68

zc =

n

s

x c

=48

14

9868.96

= -0.65


= .2422 + .5000 = .7422

c) a = 97 z.05 = -1.645

x c = 96.68

z =

n

s

x c

=

48

14

9768.96

= -0.16


= .0636 + .5000 = .5636

d) a = 96 z.05 = -1.645

x c = 97.4

z =

n

s

x c

=

48

14

9668.96

= 0.34

from Table A.5, area forz= 0.34 is .1331

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= .5000 - .1331 = .3669

e) As the alternative value get farther from the null hypothesized value, the

probability of committing a Type II error reduces. (All other variables being heldconstant).

9.35 Ho: = 50

Ha: 50

a= 53 n = 35 s = 7 = .01

Since this is two-tailed, /2 = .005 z.005 = 2.575

zc =n

s

x c

2.575 =

35

7

50cx

x c = 50 3.05

46.95 and 53.05

z =

n

s

x c

=

35

7

5305.53

= 0.04

from Table A.5 forz= 0.04, area = .0160

Other end:

z =

n

s

x c

=

35

7

539.46

= -5.11

Area associated withz= -5.11 is .5000

= .5000 + .0160 = .5160

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9.36 a) Ho: p = .65Ha: p < .65

n = 360 = .05 pa = .60 z.05 = -1.645

zc =

n

qpppc

-1.645 =

360

)35)(.65(.

65. c

p

p c = .65 - .041 = .609

z =

n

qpppc

=

360

)40)(.60(.60.609.

= -0.35


= .5000 - .1368 = .3632

b) pa = .55 z.05 = -1.645

p c = .609

z =

n

QP

Ppc

=

360

)45)(.55(.

55.609.

= -2.25


= .5000 - .4878 = .0122

c) pa = .50 z.05 = -1.645

p c = .609

z =

n

qp

ppc

=

360

)50)(.50(.

50.609.

= -4.14

from Table A.5, the area for z = -4.14 is .5000

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= .5000 - .5000 = .0000

9.37 n = 58 x = 45.1 s = 8.7 = .05 /2 = .025

H0: = 44

Ha: 44 z.025 = 1.96

z =

58

7.8

441.45

= 0.96

Sincez= 0.96

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Power = 1 - = 1 - .5832 = .4168

For 47 years:

z =

58

7.8479.46

= -0.67

From Table A.5, the area forz= -0.67 is .2486

= .5000 - .2486 = .2514

Power = 1 - = 1 - .2514 = .7486

For 48 years:

z =

58

7.8

48248.46

= 1.54

From Table A.5, the area forz= 1.54 is .4382

= .5000 - .4382 = .0618

Power = 1 - = 1 - .0618 = .9382

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9.38 H0: p= .71Ha: p < .71

n = 463 x = 324 p =463

324= .6998 = .10

z.10 = -1.28

z =

463

)29)(.71(.

71.6998. =

n

qp

pp

= -0.48

Since the observedz= -0.48 >z.10 = -1.28, the decision is to fail to rejectthe nullhypothesis.

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Type II error:

Solving for the critical proportion, p c:

zc =nqp

ppc

-1.28 =

463

)29)(.71(.

71. c

p

p = .683

Forpa = .69

z =

463

)31)(.69(.

69.683.

= -0.33

From Table A.5, the area forz= -0.33 is .1293

The probability of committing a Type II error = .1293 + .5000 = .6293

Forpa = .66

z =

463

)34)(.66(.66.683.

= 1.04


The probability of committing a Type II error = .5000 - .3508 = .1492

Forpa = .60

z =

493

)40)(.60(.

60.683.

= 4.61


The probability of committing a Type II error = .5000 - .5000 = .0000

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9.391) Ho: = 36

Ha: 36

2) z =n

s

x

3) = .01

4) two-tailed test, /2 = .005, z.005 = + 2.575If the observed value ofz is greater than 2.575 or less than -2.575, the decisionwill be to reject the null hypothesis.

5) n = 63, x = 38.4, s = 5.93

6) z =n

s

x

=6393.5

364.38

= 3.21

7) Since the observed value ofz= 3.21 is greater thanz.005 = 2.575, the decision isto reject the null hypothesis.

8) The mean is likely to be greater than 36.

9.40 1) Ho: = 7.82Ha: < 7.82

2) The test statistic is

t =

n

s

x

3) = .05

4) df = n - 1 = 16, t.05,16 = -1.746. If the observed value oftis less than -1.746, thenthe decision will be to reject the null hypothesis.

5) n = 17 x = 7.01 s = 1.69

6) t =

n

s

x

=

17

69.1

82.701.7

= -1.98

7) Since the observed t= -1.98 is less than the table value oft= -1.746, the decision

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is to reject the null hypothesis.

8) The population mean is significantly less than 7.82.

9.41

a. 1) Ho:p = .28Ha:p > .28

2) z =

n

qp

pp

3) = .10

4) This is a one-tailed test,z.10 = 1.28. If the observed value ofzis greater than1.28, the decision will be to reject the null hypothesis.

5) n = 783 x = 230

783

230 =p = .2937

6) z =

783

)72)(.28(.

28.2937.

= 0.85

7) Sincez= 0.85 is less thanz.10 = 1.28, the decision is to fail to reject the nullhypothesis.

8) There is not enough evidence to declare thatp is not .28.

b. 1) Ho:p = .61

Ha:p .61

2) z =

n

qp

pp

3) = .05

4) This is a two-tailed test,z.025 = + 1.96. If the observed value ofzis greater than1.96 or less than -1.96, then the decision will be to reject the null hypothesis.

5) n = 401 p = .56

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6) z =

401

)39)(.61(.

61.56.

= -2.05

7) Sincez= -2.05 is less thanz.025 = -1.96, the decision is to reject the null

hypothesis.

8) The population proportion is not likely to be .61.

9.42 1) H0: 2 = 15.4

Ha: 2 > 15.4

2) 2 =2

2)1(

sn

3) = .01

4) n = 18, df = 17, one-tailed test

2.01,17 = 33.4087

5) s2 = 29.6

6) 2 =2

2)1(

sn=

4.15

)6.29)(17(= 32.675

7) Since the observed 2 = 32.675 is less than 33.4087, the decision is to fail toreject the null hypothesis.

8) The population variance is not significantly more than 15.4.

9.43 a) H0: = 130Ha: > 130

n = 75 = 12 = .01 z.01 = 2.33 a = 135

Solving for x c:

zc =

n

x c

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2.33 =

75

12

130cx

x c = 133.23

z =

75

12

13523.133

= -1.28

from table A.5, area forz= -1.28 is .3997

= .5000 - .3997 = .1003

b) H0:p = .44Ha:p < .44

n = 1095 = .05 pa = .42 z.05 = -1.645

zc =

n

qp

ppc

-1.645 =

1095

)56)(.44(.

44. c

p

cp = .4153

z =

1095

)58)(.42(.

42.4153.

= -0.32

from table A.5, area forz= -0.32 is .1255

= .5000 + .1255 = .6255

9.44 H0: p = .32Ha: p > .32

n = 80 = .01 p = .39

z.01 = 2.33

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z =

80

)68)(.32(.

32.39. =

n

qp

pp

= 1.34


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9.47 H0: 2 = 16 n = 12 = .05 df = 12 - 1 = 11

Ha: 2 > 16

s = 0.4987864 ft. = 5.98544 in.

2.05,11 = 19.6751

2 =

16

)98544.5)(112( 2= 24.63

Since 2 = 24.63 > 2.05,11 = 19.6751, the decision is to reject the nullhypothesis.

9.48 H0: = 8.4 = .01 /2 = .005 n = 7 df = 7 1 = 6 s =1.3

Ha: 8.4

x = 5.6 t.005,6 = + 3.707

t =

7

3.1

4.86.5

= -5.70

Since the observed t= - 5.70 < t.005,6 = -3.707, the decision is to reject the nullhypothesis.

9.49 x = $26,650 n = 100 s = $12,000

a) Ho: = $25,000

Ha: > $25,000 = .05

For one-tail, = .05 z.05 = 1.645

z =

n

s

x

=

100

000,12

000,25650,26

= 1.38


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Solving for x c:

zc =

n

s

x c

1.645 =

100

000,12

)000,25( cx

x c = 25,000 + 1,974 = 26,974

z =

100

000,12

000,30974,26

= -2.52

from Table A.5, the area forz= -2.52 is .4941

= .5000 - .4941 = .0059

9.50 H0: 2 = 4 n = 8 s = 7.80 = .10 df = 8 1 = 7

Ha: 2 > 4

2.10,7 = 12.017

2 =

4

)80.7)(18( 2= 106.47

Since observed 2 = 106.47 > 2.10,7 = 12.017, the decision is to reject the nullhypothesis.

9.51 H0: p = .46Ha: p > .46

n = 125 x = 66 = .05125

66 ==

n

xp = .528

Using a one-tailed test, z.05 = 1.645

z =

125

)54)(.46(.

46.528. =

n

qp

pp

= 1.53

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Since the observed value ofz= 1.53 .16

n = 428 x = 84 = .01428

84 ==

n

xp = .1963

For a one-tailed test, z.01 = 2.33

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z =

428

)84)(.16(.

16.1963. =

n

qp

pp

= 2.05

Since the observedz= 2.05 $15

x = $19.34 n = 35 s= $4.52 = .10

For one-tail and = .10 zc = 1.28

z =

n

s

x

=

35

52.41534.19

= 5.68

Since the observedz= 5.68 >zc = 1.28, the decision is to reject the nullhypothesis.

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9.55 H0: 2 = 16 n = 22 df = 22 1 = 21 s = 6 = .05

Ha: 2 > 16

2.05,21 = 32.6705

2 =

16)6)(122( 2 = 47.25

Since the observed 2 = 47.25 > 2.05,21 = 32.6705, the decision is to reject thenull hypothesis.

9.56 H0: = 2.5 x = 3.4 s = 0.6 = .01 n = 9 df = 9 1 = 8Ha: > 2.5

t.01,8 = 2.896

t =

n

s

x

=

9

6.0

5.24.3

= 4.50

Since the observed t= 4.50 > t.01,8 = 2.896, the decision is to reject the nullhypothesis.

9.57 a) Ho: = 23.58

Ha: 23.58

n = 95 x = 22.83 s = 5.11 = .05

Since this is a two-tailed test and using /2 = .025: z.025 = + 1.96

z =

n

s

x

=

95

11.5

58.2383.22

= -1.43

Since the observedz= -1.43 >z.025 = -1.96, the decision is to fail to reject thenull hypothesis.

b) zc =

n

s

x c

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+ 1.96 =

95

11.5

58.23c

x

cx = 23.58 + 1.03

cx = 22.55, 24.61

for Ha: = 22.30

z =

n

s

x ac

=

95

11.5

30.2255.22

= 0.48

z =n

s

xac

=9511.5

30.2261.24

= 4.41

from Table A.5, the areas forz= 0.48 andz= 4.41 are .1844 and .5000

= .5000 - .1844 = .3156

The upper tail has no effect on .

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9.58 n = 12 x = 12.333 s2 = 10.424

H0: 2 = 2.5

Ha: 2 2.5

= .05 df = 11 two-tailed test, /2 = .025

2.025,11 = 21.92

2..975,11 = 3.81575

If the observed 2 is greater than 21.92 or less than 3.81575, the decision is toreject the null hypothesis.

2 =

2

2)1(

sn=

5.2

)424.10(11= 45.866

Since the observed 2 = 45.866 is greater than 2.025,11 = 21.92, the decision is toreject the null hypothesis. The population variance is significantly more than2.5.

9.59 The sample size is 22. x is 3.967 s = 0.866 df = 21

The test statistic is:

t =n

s

x

The observed t= -2.34. The p-value is .015.

The results are statistical significant at = .05.

The decision is to reject the null hypothesis.

9.60 H0: p = .25

Ha: p .25

This is a two-tailed test with = .05. n = 384.

Since thep-value = .045 < = .05, the decision is to reject the null hypothesis.

The sample proportion, p = .205729 which is less than the hypothesizedp = .25.One conclusion is that the population proportion is lower than .25.

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9.61 H0: = 2.51

Ha: > 2.51

This is a one-tailed test. The sample mean is 2.555 which is more than the

hypothesized value. The observed tvalue is 1.51 with an associatedp-value of .072 for a one-tailed test. Because thep-value is greater than

= .05, the decision is to fail to reject the null hypothesis.There is not enough evidence to conclude that beef prices are higher.

9.62 H0: = 2747

Ha: < 2747

This is a one-tailed test. Sixty-seven households were included in this study.The sample average amount spent on home-improvement projects was 2,349.Sincez= -2.09

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