Ken Black QA 5th chapter 9 Solution

8/3/2019 Ken Black QA 5th chapter 9 Solution

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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 and

alternate 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 scientificallyexamined. 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 thedata could allow hypothesis decisions to be made. However, by using the methodologyand structure of hypothesis testing even in "obvious" situations, the researcher has added

credibility 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.


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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. Often that which the researcher wants to demonstrate true

or prove true is set up as the alternative hypothesis. The null hypothesis is that the newtheory or idea is not true, the status quo is still true, or that there is no difference. Thenull hypothesis is assumed to be true before the process begins. Some researchers likenthis procedure to a court of law where the defendant is presumed innocent (assume null istrue - nothing has happened). Evidence is brought before the judge or jury. If enoughevidence is presented, then the null hypothesis (defendant innocent) can no longer beaccepted or assumed true. The null hypothesis is rejected as not true and the alternatehypothesis is accepted as true by default. Emphasize that the researcher needs to make adecision after examining the observed statistic.

Some of the key concepts in this chapter are one-tailed and two-tailed test and

Type 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 thuscreating 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 thep-value

so 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 iscorrect 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 IIerror 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 even

beyond 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.


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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 of

possible 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 Hypotheses

Research 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 thezStatistic ( known)Testing the Mean with a Finite PopulationUsing thep-Value Method to Test HypothesesUsing the Critical Value Method to Test Hypotheses

Using the Computer to Test Hypotheses about a Population Mean UsingthezStatistic

9.3 Testing Hypotheses About a Population Mean Using the tStatistic ( unknown)Using the Computer to Test Hypotheses about a Population Mean Using

the tTest

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


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KEY TERMS

Alpha( ) One-tailed TestAlternative Hypothesis Operating-Characteristic Curve (OC)

Beta( ) p-Value MethodCritical Value Power Critical Value Method Power CurveHypothesis Rejection RegionHypothesis Testing Research HypothesisLevel of Significance Statistical HypothesisNonrejection Region Substantive ResultNull Hypothesis Two-Tailed TestObserved Significance Level Type I Error Observed Value Type II Error

SOLUTIONS TO PROBLEMS IN CHAPTER 9

9.1 a) Ho: = 25

Ha: 25

x = 28.1 n = 57 = 8.46 = .01

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

z=

57

46.8

251.28 =

n

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

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

Since thep-value of .0028 is less than /2 = .005, the decision is to:Reject the null hypothesis


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c) critical mean values:

zc =

n

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 = 24 = 1.21 =.01

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

z =

24

21.1

48.791.6 =

n

x

= -2.31

observedz= -2.31 >zc = -2.33

Fail to reject the null hypothesis


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9.3 a) Ho: = 1,200Ha: > 1,200

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

For one-tail, = .10 zc = 1.28

z =

113

100

200,1215,1 =

n

x

= 1.59

observedz= 1.59 >zc = 1.28

Reject the null hypothesis

b) Probability > observedz= 1.59 is .5000 - .4441 = .0559 (thep-value) which is

less than = .10.Reject the null hypothesis.

c) Critical mean value:

zc =

n

x c

1.28 =113

100

200,1cx

x c = 1,200 + 12.04

Since the observed x = 1,215 is greater than the critical x = 1212.04, thedecision is to reject the null hypothesis.


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9.4 Ho: = 82Ha: < 82

x = 78.125 n = 32 = 9.184 = .01

z.01 = -2.33

z=

32

184.9

82125.78 =

n

x

= -2.39

Since observedz= -2.39


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9.6 H0: = $62,600

Ha: < $62,600

x = $58,974 n = 18 = $7,810 = .01

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

z =

18

810,7

600,62974,58 =

n

x

= -1.97

Since the observedz= -1.97 >z.01 = -2.33, the decision is to fail to reject thenull hypothesis.

9.7 H0

:

= 5

Ha: 5

x = 5.0611 n = 42 N= 650 = 0.2803 = .10

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

z =

1650

42650

42

2803.0

50611.5

1

=

N

nN

n

x

= 1.46

Since the observedz= 1.46


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x = $4,008 n = 55 = $386 = .01

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

z =

55

386$292,4$008,4$ =

n

x

= -5.46

Since the observedz= -5.46 123

= .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.


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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 =

20

59.3

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 = 51 x = 58.42 s2 = 25.68 df = 51 - 1 = 50 = .01

Ho: = 60Ha: < 60

For one-tail test, = .01 critical t.01,50 = -2.403

t =

51

68.25

6042.58 =

n

s

x

= -2.23

Observed t= -2.23 > t.01,7 = -2.403



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

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

or 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 = .1895 df = 20-1 = 19 = .01

Ho: = 8.3

Ha: 8.3

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

t =

20

1895.

3.837.8 =

n

s

x

= 1.65

Observed t= 1.65 < 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

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

t = 1202353.

84.185083.1 =

ns

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 = 3.1948 s = .0889 df = 25 - 1 = 24 = .01


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Ho: = $3.16Ha: > $3.16

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

t =

25

0889.16.31948.3

=

n

sx

= 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 = 61 x = 3.72 s = 0.65 df = 61 1 = 60 = .01

H0: = 3.51

Ha: > 3.51

One-tailed test, = .01 t.01,60 = 2.390

t =

61

65.0

51.372.3 =

n

s

x

= 2.52

The observed t= 2.52 > t.01,60 = 2.390,The decision is to reject the null hypothesis


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9.19 n = 22 x = 1031.32 s = 240.37 df = 22 1 = 21 = .05

H0: = 1135

Ha: 1135

Two-tailed test, /2= .025 t.025,21 = +2.080

t =

22

37.240

113532.1031 =

n

s

x

= -2.02

The observed t= -2.02 > t.025,21 = -2.080,The decision is to fail toreject the null hypothesis

9.20 n = 12 x = 42.167 s = 9.124 df = 12 1 = 11 = .01

H0: = 46

Ha: < 46

One-tailed test, = .01 t.01,11 = -2.718

t =

12

124.9

46167.42 =

n

s

x

= -1.46

The observed t= -1.46 > t.01,11 = -2.718,The decision is to fail toreject the null hypothesis

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

H0: = 19

Ha: 19

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.


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9.22 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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9.24 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 which is thep-value

Since thep-value = .2743 > /2 = .025, the decision is to Fail to reject the nullhypothesis

Solving for critical values:

z =

n

qp

ppc

1.96 =

740

)71)(.29(.

29. cp

cp = .29 .033

.257 and .323 are the critical values

Since p = .28 is not outside critical values in tails, the decision is to Fail to rejectthe null hypothesis


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9.27 Ho:p = .31

Ha:p .31

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

600200 ==

nxp = .3333

z =

600

)69)(.31(.

31.3333. =

n

qp

pp

= 1.23

Since the observedz= 1.23 is less thanz.005= 1.645, The decision is to fail toreject the null hypothesis. There is not enough evidence to declare that theproportion is any different than .31.

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.


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19/44


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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 nullhypothesis.

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

Ha: 2 5

2.025,10 = 20.4832 2.975,10 = 3.24696

2 =

5

)2.1)(111( = 2.4

Since 2 = 2.4 < 2.975,10 = 3.24696, the decision is to reject the nullhypothesis.

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

30.0833

Ha: 2 14

2.025,11 = 21.9200

2.975,11 = 3.81574

2 =

14

)0833.30)(112( = 23.64

Since 2 = 23.64 > 2.025,11 = 21.9200, the decision is to reject the nullhypothesis.

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

Ha: 2 > .001

2.01,15 = 30.5780

2 =

001.)00144667)(.116(

= 21.7

Since 2 = 21.7 < 2.01,15 = 30.5780, the decision is to fail to reject the nullhypothesis.


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22/44


z =

n

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

x c

-1.645 =

48

14100

cx

x c = 96.68

z =

n

x c

=

48

14

9968.96

= -1.15


= .3749 + .5000 = .8749

c) = .01 z.01 = -2.33

zc =

n

x c

-2.33 =4814

100cx

x c = 95.29


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z =

n

x c

=

48

14

9929.95

= -1.84


= .4671 + .5000 = .9671

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

9.37 = .05 = 100 n = 48 = 14

a) a = 98.5 zc = -1.645

zc =

n

x c

-1.645 =

48

14

100cx

x c = 96.68

z =

n

x c

=

48

14

5.9868.96

= -0.90


= .3159 + .5000 = .8159


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b) a = 98 zc = -1.645

x c = 96.68

zc =n

x c

=4814

9868.96

= -0.65


= .2422 + .5000 = .7422

c) a = 97 z.05 = -1.645

x c = 96.68

z =

n

x c

=

48

149768.96

= -0.16


= .0636 + .5000 = .5636

d) a = 96 z.05 = -1.645

x c = 96.68

z =

n

x c

=

48

14

9668.96

= 0.34

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

= .5000 - .1331 = .3669

e) As the alternative value gets farther from the null hypothesized value, theprobability of committing a Type II error reduces (all other variables being heldconstant).


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9.38 Ho: = 50

Ha: 50

a= 53 n = 35 = 7 = .01

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

zc =

n

x c

2.575 =

35

7

50cx

x

c = 50 3.05

46.95 and 53.05

z =

n

x c

=

35

7

5305.53

= 0.04

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

Other end:

z =

n

x c

=

35

7

5395.46

= -5.11

Area associated withz= -5.11 is .5000

= .5000 + .0160 = .5160


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

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

zc =

n

qpppc

-1.645 =

360

)35)(.65(.

65. cp

p c = .65 - .041 = .609

z =n

qp

ppc

=360

)40)(.60(.

60.609.

= 0.35


= .5000 - .1368 = .3632

b) pa = .55 z.05 = -1.645 p c = .609

z =

nqp

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

= .5000 - .5000 = .0000


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9.40 n = 58 x = 45.1 = 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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For 47 years:

z =587.8

47239.46

= -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

48239.46

= 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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9.41 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.

Type II error:

Solving for the critical proportion, p c:

zc =

n

qp

ppc

-1.28 =

463

)29)(.71(.

71. cp

p = .683


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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

From Table A.5, the area forz= 1.04 is .3508

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

Forpa = .60

z =

463

)40)(.60(.

60.683.

= 3.65

From Table A.5, the area forz= 3.65 is essentially, .5000

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


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9.42 HTAB steps:

1) Ho: = 36

Ha: 36

2) z =

n

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, = 5.93

6) z =

n

x

=

63

93.5364.38

= 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.

9.43 HTAB steps:

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


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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

is to reject the null hypothesis.

8) The population mean is significantly less than 7.82.

9.44 HTAB steps:

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

2) z =

n

qppp

3) = .10

4) This is a one-tailed test,z.10 = 1.28. If the observed value ofzis greaterthan 1.28, 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 thenull hypothesis.

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


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b. 1) Ho:p = .61

Ha:p .61

2) z =nqp

pp

3) = .05

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

5) n = 401 p = .56

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 nullhypothesis.

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

9.45 HTAB steps:

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


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7) Since the observed 2 = 32.675 is less than 33.4087, the decision is tofail

to reject the null hypothesis.

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

9.46 a) H0: = 130Ha: > 130

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

Solving for x c:

zc =

n

x c

2.33 =

75

12

130cx

x c = 133.23

z =

75

1213523.133

= -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


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-1.645 =

1095

)56)(.44(.

44. cp

cp = .4153

z =

1095

)58)(.42(.

42.4153.

= -0.32

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

= .5000 + .1255 = .6255

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

n = 80 = .01 p = .39

z.01 = 2.33

z =

80

)68)(.32(.

32.39. =

n

qp

pp

= 1.34



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9.49 n = 210 x = 93 = .10210

93 ==

n

xp = .443

Ho: p = .57Ha: p < .57

For one-tail, = .10 zc = -1.28

z =

210

)43)(.57(.

57.443. =

n

qp

pp

= -3.72

Since the observedz= -3.72 16

s = 0.4987864 ft. = 5.98544 in.

2.05,11 = 19.6752

2 =

16

)98544.5)(112( 2= 24.63

Since 2 = 24.63 > 2.05,11 = 19.6752, the decision is to reject the nullhypothesis.

9.51 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.


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9.52 x = $26,650 n = 100 = $12,000

a) Ho: = $25,000

Ha: > $25,000 = .05

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

z =

n

x

=

100

000,12

000,25650,26

= 1.38



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

Ha: 2 > 4

2.10,7 = 12.0170

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.

9.54 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

Since the observed value ofz= 1.53


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H0: = 185Ha: < 185

t.05,15 = - 1.753

t =

n

s

x

=

16

28286.14

185175

= -2.80

Since observed t= - 2.80 < t.05,15 = - 1.753, the decision is to reject the nullhypothesis.

9.56 H0: p = .182

Ha: p > .182

n = 428 x = 84 = .01428

84 ==

n

xp = .1963

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

z =

428

)818)(.182(.

182.1963. =

n

qp

pp

= 0.77



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z =

428

)79)(.21(.

21.2255. =

n

qp

pp

aa

ac

= 0.79

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

= .5000 + .2852 = .7852

9.57 Ho: = $15Ha: > $15

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

For one-tail and = .10 zc = 1.28

z =

n

s

x

=

35

52.4

1534.19

= 5.68

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

9.58 H0: 2

= 16 n = 22 df = 22 1 = 21 s = 6 = .05Ha:

2 > 16

2.05,21 = 32.6706

2 =

16

)6)(122( 2= 47.25

Since the observed 2 = 47.25 > 2.05,21 = 32.6706, the decision is to reject thenull hypothesis.


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9.59 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

sx

=

9

6.05.24.3

= 4.50

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

9.60 a) Ho: = 23.58

Ha: 23.58

n = 95 x = 22.83 = 5.11 = .05

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

z =

n

x

=

95

11.5

58.2383.22

= -1.43

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

b) zc =

n

x c

+ 1.96 =

95

11.5

58.23cx

cx = 23.58 + 1.03

cx = 22.55, 24.61

for Ha: = 22.30


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z =

n

x ac

=

95

11.5

30.2255.22

= 0.48

z =

n

x ac

=95

11.5

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 .

9.61 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 = 21.9200

2..975,11 = 3.81574

If the observed 2 is greater than 21.9200 or less than 3.81574, 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.


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9.62 H0: = 23 x = 18.5 s = 6.91 = .10 n = 16 df = 16 1 =15

Ha: < 23

t.10,15 = -1.341

t =

n

s

x

=

16

91.6

235.18

= -2.60

Since the observed t= -2.60 < t.10,15 = -1.341, the decision is to reject the nullhypothesis.

9.63 The sample size is 22. x is 3.969 s = 0.866 df = 21

The test statistic is:

t =

n

s

x

The observed t= -2.33. The p-value is .015.

The results are statistical significant at = .05.

The decision is to reject the null hypothesis.

9.64 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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9.65 H0: = 2.51

Ha: > 2.51

This is a one-tailed test. The sample mean is 2.55 which is more than thehypothesized value. The observed tvalue is 1.51 with an associated

p-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.

9.66 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

Ken Black QA 5th chapter 9 Solution

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