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Chapter 9

Hypothesis Tests

Learning Objectives

1. Learn how to formulate and test hypotheses about a population mean and/or a population proportion.

2. Understand the types of errors possible when conducting a hypothesis test.

3. Be able to determine the probability of making various errors in hypothesis tests.

4. Know how to compute and interpretp-values.

5. Be able to use critical values to draw hypothesis testing conclusions.

6. Be able to determine the size of a simple random sample necessary to keep the probability ofhypothesis testing errors within acceptable limits.

7. Know the definition of the following terms:

null hypothesis two-tailed testalternative hypothesis p-valueType I error level of significanceType II error critical valueone-tailed test power curve

Solutions:

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Chapter 9

1. a. H0: 600 Managers claim.

Ha: > 600

b. We are not able to conclude that the managers claim is wrong.

c. The managers claim can be rejected. We can conclude that > 600.

2. a. H0: 14

Ha: > 14 Research hypothesis

b. There is no statistical evidence that the new bonus plan increases sales volume.

c. The research hypothesis that > 14 is supported. We can conclude that the new bonus planincreases the mean sales volume.

3. a. H0: = 32 Specified filling weight

Ha: 32 Overfilling or underfilling exists

b. There is no evidence that the production line is not operating properly. Allow the productionprocess to continue.

c. Conclude 32 and that overfilling or underfilling exists. Shut down and adjust the productionline.

4. a. H0: 220

Ha: < 220 Research hypothesis to see if mean cost is less than $220.

b. We are unable to conclude that the new method reduces costs.

c. Conclude < 220. Consider implementing the new method based on the conclusion that it lowersthe mean cost per hour.

5. a. The Type I error is rejecting H0when it is true. In this case, this error occurs if the researcherconcludes that the mean newspaper-reading time for individuals in management positions is greaterthan the national average of 8.6 minutes when in fact it is not.

b. The Type II error is accepting H0when it is false. In this case, this error occurs if the researcherconcludes that the mean newspaper-reading time for individuals in management positions is lessthan or equal to the national average of 8.6 minutes when in fact it is greater than 8.6 minutes.

6. a. H0: 1 The label claim or assumption.

Ha: > 1

b. Claiming > 1 when it is not. This is the error of rejecting the products claim when the claim istrue.

c. Concluding 1 when it is not. In this case, we miss the fact that the product is not meeting itslabel specification.

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Hypothesis Testing

7. a. H0: 8000

Ha: > 8000 Research hypothesis to see if the plan increases average sales.

b. Claiming > 8000 when the plan does not increase sales. A mistake could be implementing the

plan when it does not help.

c. Concluding 8000 when the plan really would increase sales. This could lead to notimplementing a plan that would increase sales.

8. a. H0: 220

Ha: < 220

b. Claiming < 220 when the new method does not lower costs. A mistake could be implementingthe method when it does not help.

c. Concluding 220 when the method really would lower costs. This could lead to notimplementing a method that would lower costs.

9. a. 0 19.4 20 2.12

/ 2 / 50

xz

n

= = =

b. Area = .4830

p-value = .5000 - .4830 = .0170

c. p-value .05, reject H0

d. Reject H0ifz-1.645

-2.12 -1.645, reject H0

10. a. 0 26.4 25 1.48

/ 6 / 40

xz

n

= = =

b. Area = .4306

p-value = .5000 - .4306 = .0694

c. p-value >.01, do not reject H0

d. Reject H0ifz2.33

1.48 < 2.33, do not reject H0

11. a. 0 14.15 15 2.00

/ 3 / 50

xz

n

= = =

b. Area = .4772

p-value = 2(.5000 - .4772) = .0456

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Chapter 9

c. p-value .05, reject H0

d. Reject H0ifz-1.96 or z 1.96

-2.00 -1.96, reject H0

12. a. 0 78.5 80 1.25

/ 12 / 100

xzn

= = =

p-value = .5000 - .3944 = .1056

p-value >.01, do not reject H0

b. 0 77 80 2.50

/ 12 / 100

xz

n

= = =

p-value = .5000 - .4938 = .0062

p-value .01, reject H0

c. 0 75.5 80 3.75

/ 12 / 100

xz

n

= = =

p-value 0


d. 0 81 80 .83

/ 12 / 100

xz

n

= = =

Area to left ofz= .83

p-value = .5000 + .2967 = .7967


13. Reject H0ifz1.645

a. 0 52.5 50 2.42

/ 8 / 60

xz

n

= = =

2.42 1.645, reject H0

b. 0 51 50 .97

/ 8 / 60

xzn

= = =

.97 < 1.645, do not reject H0

c. 0 51.8 50 1.74

/ 8 / 60

xz

n

= = =

1.74 1.645, reject H0

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Hypothesis Testing

14. a. 0 23 22 .87

/ 10 / 75

xz

n

= = =

p-value = 2(.5000 - .3078) = .3844


b. 0 25.1 22 2.68

/ 10 / 75

xz

n

= = =

p-value = 2(.5000 - .4963) = .0074


c. 0 20 22 1.73

/ 10 / 75

xz

n

= = =

p-value = 2(.5000 - .4582) = .0836


15. a. H0: 1056

Ha: < 1056

b. 0 910 1056 1.83

/ 1600 / 400

xz

n

= = =

p-value = .5000 - .4664 = .0336

c. p-value .05, reject H0. Conclude the mean refund of last minute filers is less than $1,056.

d. Reject H0ifz-1.645

-1.83 -1.645, reject H0

16. a. H0: 10994

Ha: > 10994

b. 19.1

180/2770

1099411240

/

0=

=

=

n

xz

Area = .3830

p-value = .5000 - .3830 = .1170

c. Do not reject H0. We cannot conclude the rental rates have increased.

d. Recommend withholding judgment and collecting more data on apartment rental rates beforedrawing a final conclusion.

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Chapter 9

17. a. H0: =44.75

Ha: 44.75

b. 54.1111/48.5

75.4495.43

/

0=

=

=

n

x

z

p-value = 2(.5000 - .4382) = .1236

c. p-value > .05, do not reject H0. We cannot conclude that the mean length of a work week haschanged.

d. Reject H0ifz-1.96 orz1.96

z= -1.54; cannot reject H0

18. a. H0: =36933

Ha: 36933

b. 09.2516/10740

3693335935

/

0=

=

=

n

xz

p-value = 2(.5000 - .4817) = .0366

c. p-value .05.

Reject H0; Taipei County has a mean monthly wage different from the nationwide mean monthlywage.

19. H0: 45064

Ha: > 45064

15.275/96.4562

4506446196

/

0=

=

=

n

xz

Area = .4842

p-value = .5000 - .4842 = .0158

p-value .05, reject H0. Conclude that there has been an increase in the mean monthly earnings of

service industry workers.

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Hypothesis Testing

20. a. H0: 181,900

Ha: < 181,900

b.166, 400 181,900

2.93/ 33,500 / 40

xz

n

= = =

c. p-value = .5000 - .4983 = .0017

d. p-value .01; reject H0. Conclude mean selling price in South is less than the national mean selling

price.

21. a. H0: 15

Ha: > 15

b.17 15

2.96/ 4 / 35

xz

n

= = =

c. p-value = .5000 - .4985 = .0015

d. p-value .01; reject H0; the premium rate should be charged.

22. a. H0: =8

Ha: 8

b. 0 8.4 8 1.37

/ 3.2 / 120

xz

n

= = =

p-value = 2(.5000 - .4147) = .1706

c. Do not reject H0. Cannot conclude that the population mean waiting time differs from 8 minutes.

d. .025 ( / )x z n

8.4 1.96 (3.2/ 120)

8.4 .57 (7.83 to 8.97)

Yes; =8 is in the interval. Do not reject H0.

23. a.

0 14 12

2.31/ 4.32 / 25

x

t s n

= = =

b. Degrees of freedom = n 1 = 24

Using ttable,p-value is between .01 and .025.

Actualp-value = .0147

c. p-value .05, reject H0.

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Chapter 9

d. With df= 24, t.05 = 1.711

Reject H0 if t1.711

2.31 > 1.711, reject H0.

24. a. 0 17 18 1.54

/ 4.5 / 48

xt

s n

= = =

b. Degrees of freedom = n 1 = 47

Area in lower tail is between .05 and .10

Using ttable,p-value (two-tail) is between .10 and .20


c. p-value >.05, do not reject H0.

d. With df= 47, t.025 = 2.012

Reject H0 if t-2.012 or t2.012

t= -1.54; do not reject H0

25. a. 0 44 45 1.15

/ 5.2 / 36

xt

s n

= = =

Degrees of freedom = n 1 = 35

Using ttable,p-value is between .10 and .20



b. 0 43 45 2.61

/ 4.6 / 36

xt

s n

= = =




c. 0 46 45 1.20

/ 5 / 36

xt

s n

= = =

Using ttable, area in upper tail is between .10 and .20

p-value (lower tail) is between .80 and .90Actualp-value = .8809

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Hypothesis Testing


26. a. 0 103 100 2.10

/ 11.5 / 65

xt

s n

= = =


Using ttable, area in tail is between .01 and .025

p-value (two tail) is between .02 and .05



b. 0 96.5 100 2.57

/ 11/ 65

xt

s n

= = =





c. 0 102 100 1.54

/ 10.5 / 65

xt

s n

= = =





27. a. H0: 238

Ha: < 238

b. 0 231 238 .88

/ 80 / 100

xt

s n

= = =




c. p-value > .05; do not reject H0. Cannot conclude mean weekly benefit in Virginia is less than thenational mean.

d. df= 99 t.05= -1.66

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Chapter 9

Reject H0if t-1.66

-.88 > -1.66; do not reject H0

28. a. H0: 3530

Ha: > 3530

b. 0 3740 3530 2.49

/ 810 / 92

xt

s n

= = =




c. p-value .01; reject H0. The mean attendance per game has increased. Anticipate a new all-time

high season attendance during the 2002 season.

29. a. H0: = 5600

Ha: 5600

b. 0 5835 5600 2.26

/ 520 / 25

xt

s n

= = =




c. p-value .05; reject H0. The mean diamond price in New York City differs.

d. df= 24 t.025= 2.064

Reject H0if t< -2.064 or t> 2.064

2.26 > 2.064; reject H0

30. a. H0: = 600

Ha: 600

b. 0 612 600 1.17

/ 65 / 40

xt

s n

= = =

df= n- 1 = 39


p-value is between .20 and .40

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Chapter 9



c. p-value .05; reject H0. The population mean distance for the new driver is greater than the USGA

approved driver.

34. a. H0: = 2

Ha: 2

b. 22

2.210

ix

xn

= = =

c. ( )2

.5161

ix x

sn

= =

d. 0 2.2 2 1.22

/ .516 / 10

xt

s n

= = =

Degrees of freedom = n- 1 = 9




e. p-value > .05; do not reject H0. No reason to change from the 2 hours for cost estimating purposes.

35. a.

0

0 0

.175 .20 1.25(1 ) .20(1 .20)

400

p pz

p p

n

= = =

b. Area in tail = (.5000 - .3944) = .1056

p-value = 2(.1056) = .2112

c. p-value > .05; do not reject H0

d. z.025= 1.96

Reject H0ifz-1.96 orz1.96

z= 1.25; do not reject H0

36. a.

0

0 0

.68 .752.80

(1 ) .75(1 .75)

300

p pz

p p

n

= = =

p-value = .5000 - .4974 = .0026

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Hypothesis Testing

p-value .05; reject H0

b.

.72 .751.20

.75(1 .75)

300

z

= =

p-value = .5000 - .3849 = .1151

p-value > .05; do not reject H0

c.

.70 .752.00

.75(1 .75)

300

z

= =

p-value = .5000 - .4772 = .0228

p-value .05; reject H0

d.

.77 .75.80

.75(1 .75)

300

z

= =

p-value = .5000 + .2881 = .7881

p-value > .05; do not reject H0

37. a. H0: p.40

Ha: p> .40

b. 189

.45425

p= =

0

0 0

.45 401.88

(1 ) .40(1 .40)

425

p pz

p p

n

= = =

Area = .4699

p-value = .5000 - .4699 = .0301

c. p-value .05; reject H0. Conclude more than 40% access less than seven hours per week.

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Chapter 9

38. a. H0: p=.64

Ha: p.64

b. 52

.52100

p= =

0

0 0

.52 .642.50

(1 ) .64(1 .64)

100

p pz

p p

n

= = =

Area = .4938

p-value = 2(.5000 - .4938) = .0124

c. p-value .05; reject H0. Proportion differs from the reported .64.

d. Yes. Since p = .52, it indicates that fewer than 64% of the shoppers believe the supermarket brand

is as good as the name brand.

39. a. H0: p=.70

Ha: p.70

b. Wisconsin252

.72350

p= =

0

0 0

.72 .70.82

(1 ) .70(1 .70)

350

p pz

p p

n

= = =

Area = .2939

p-value = 2(.5000 - .2939) = .4122

Cannot reject H0.

California189

.63300

p= =

.63 .702.65

.70(1 .70)

300

z

= =

Area = .4960

p-value = 2(.5000 - .4960) = .0080

Reject H0. California has a different (lower) percentage of adults who do not exercise regularly.

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Hypothesis Testing

40. a. 414

.27021532

p= = (27%)

b. H0: p.22

Ha: p> .22

0

0 0

.2702 .224.75

(1 ) .22(1 .22)

1532

p pz

p p

n

= = =

p-value 0

Reject H0; There has been a significant increase in the intent to watch the TV programs.

c. These studies help companies and advertising firms evaluate the impact and benefit of commercials.

41. a. H0: p.75

Ha: p< .75

b.

0

0 0

.72 .751.20

(1 ) .75(1 .75)

300

p pz

p p

n

= = =

Area = .3849

p-value = .5000 - .3849 = .1151

c. p-value > .05; do not reject H0. The executive's claim cannot be rejected.

42. H0: p.31

Ha: p> .31

38.500

190==p

38.3

500

)31.1(31.

31.38.

)1( 00

0=

=

=

n

pp

ppz

Area = .4996

p-value = .5000 - .4996 = .0004

p-value .05; reject H0. This is an increase compared to 2003.

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Chapter 9

43. a. H0: p=.48

Ha: p.48

b. 360

.45800

p= =

0

0 0

.45 .481.70

(1 ) .48(1 .48)

800

p pz

p p

n

= = =

Area = .4554

p-value = 2(.5000 - .4554) = .0892

c. p-value > .05; do not reject H0. There is no reason to conclude the proportion has changed.

44. a. H0: p.50

Ha: p> .50

b. 285

.57500

p= =

0

0 0

.57 .503.13

(1 ) .50(1 .50)

500

p pz

p p

n

= = =

p-value 0

c. p-value .01; reject H0. Conclude Burger King fries are preferred by more than .50 of thepopulation.

d. Yes; statistical evidence shows Burger King fries are preferred. The give-away was a good way toget customers to try the new fries.

45. a. H0: p= .44

Ha: p.44

b. 205

.41500

p= =

0

0 0

.41 .44 1.35(1 ) .44(1 .44)

500

p pz

p p

n

= = =

Area = .4115

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Hypothesis Testing

p-value = 2(.5000 - .4115) = .1770

p-value > .05; do not reject H0. No change.

c. 245

.49

500

p= =

.49 .442.25

.44(1 .44)

500

z

= =

Area = .4878

p-value = 2(.5000 - .4878) = .0244

p-value .05; reject H0. There has been a change: an increase in repeat customers.

46.5

.46120

x

n

= = =

x

10

.05

c

H0:10

Ha:< 10

c = 10 - 1.645 (5 / 120 ) = 9.25

Reject H0if x 9.25

a. When = 9,

9.25 9.55

5/ 120z

= =

Prob (H0) = (.5000 - .2088) = .2912

b. Type II error

c. When = 8,

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Chapter 9

9.25 82.74

5/ 120z

= =

= (.5000 - .4969) = .0031

47. Reject H0ifz -1.96 or ifz 1.96

10.71

200x

n

= = =

xx

c1 c220

Ha:20 Ha:20H0:= 20

.025 .025

c1 = 20 - 1.96 (10 / 200 ) = 18.61

c2 = 20 + 1.96 (10 / 200 ) = 21.39

a. = 18

18.61 18.86

10/ 200z

= =

= .5000 - .3051 = .1949

b. = 22.5

21.39 22.51.57

10/ 200z

= =

= .5000 - .4418 = .0582

c. = 21

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Hypothesis Testing

21.39 21.55

10/ 200z

= =

= .5000 +.2088 = .7088

48. a. H0: 15

Ha: > 15

Concluding 15 when this is not true. Fowle would not charge the premium rate even thoughthe rate should be charged.

b. Reject H0ifz 2.33

0 15 2.33/ 4 / 35

x xz

n

= = =

Solve forx = 16.58

Decision Rule:

Accept H0if x< 16.58

Reject H0if x 16.58

For = 17,

16.58 17.62

4 / 35z

= =

= .5000 -.2324 = .2676

c. For = 18,

16.58 182.10

4 / 35z

= =

= .5000 -.4821 = .0179

49. a. H0: 25

Ha: < 25

Reject H0ifz -2.05

0 25 2.05/ 3 / 30

x xz

n

= = =

Solve for x = 23.88Decision Rule:

Accept H0if x> 23.88

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Chapter 9

Reject H0if x 23.88

b. For = 23,

23.88 231.61

3 / 30

z

= =

= .5000 -.4463 = .0537

c. For = 24,

23.88 24.22

3 / 30z

= =

= .5000 +.0871 = .5871

d. The Type II error cannot be made in this case. Note that when = 25.5, H0is true. The Type II

error can only be made when H0is false.

50. a. Accepting H0 and concluding the mean average age was 28 years when it was not.

b. Reject H0ifz -1.96 or ifz 1.96

0 28

/ 6 / 100

x xz

n

= =

Solving forx , we find

at z = -1.96, x = 26.82

at z = +1.96, x = 29.18

Decision Rule:

Accept H0if 26.82 < x

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Hypothesis Testing

At = 29,

29.18 29.30

6/ 100z

= =

= .5000 +.1179 = .6179

At = 30,

29.18 301.37

6/ 100z

= =

= .5000 -.4147 = .0853

c. Power = 1 -

at = 26, Power = 1 - .0853 = .9147

When = 26, there is a .9147 probability that the test will correctly reject the null hypothesis that = 28.

51. a. Accepting H0and letting the process continue to run when actually over - filling or under - fillingexists.

b. Decision Rule: Reject H0ifz -1.96 or ifz 1.96 indicates

Accept H0if 15.71 < x

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Chapter 9

x

16.516.29

c

c. Power = 1 - .0749 = .9251

d. The power curve shows the probability of rejecting H0for various possible values of. Inparticular, it shows the probability of stopping and adjusting the machine under a variety ofunderfilling and overfilling situations. The general shape of the power curve for this case is

.00

.25

.50

.75

1.00

15.6 15.8 16.0 16.2 16.4

Possible Vl!es ofu

Po"er

52. 0 .014

15 2.33 16.3250

c zn

= + = + =

At = 1716.32 17

1.204 / 50

z

= =

= .5000 - .3849 = .1151

At = 1816.32 18

2.974 / 50

z

= =

= .5000 - .4985 = .0015

Increasing the sample size reduces the probability of making a Type II error.

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Hypothesis Testing

53. a. Accept 100 when it is false.

b. Critical value for test:

0 .05

75100 1.645 119.51

40c z

n

= + = + =

At = 120119.51 120

.0475/ 40

z

= =

= .5000 - .0160 = .4840

c. At = 13119.51 130

.8875/ 40

z

= =

=.5000 - .3106 = .1894

d. Critical value for test:

0 .05

75100 1.645 113.79

80c z

n

= + = + =

At = 120113.79 120

.7475/ 80

z

= =

= .5000 - .2704 = .2296

At = 130113.79 130

1.9375/ 80

z

= =

= .5000 - .4732 = .0268

Increasing the sample size from 40 to 80 reduces the probability of making a Type II error.

54.

2 2 2 2

2 2

0

( ) (1.645 1.28) (5)214

( ) (10 9)a

z zn

+ += = =

55.

2 2 2 2

2 2

0

( ) (1.96 1.645) (10)325

( ) (20 22)a

z zn

+ += = =

56. At0 = 3, = .01. z.01 = 2.33

At a = 2.9375, = .10. z.10 = 1.28

= .18

2 2 2 2

2 2

0

( ) (2.33 1.28) (.18)108.09

( ) (3 2.9375)a

z zn

+ += = =

Use 109

57. At0 = 400, = .02. z.02 = 2.05

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Chapter 9

Ata = 385, = .10. z.10 = 1.28

= 30

2 2 2 2

2 20

( ) (2.05 1.28) (30)

44.4( ) (400 385)a

z z

n

+ +

= = = Use 45

58. At0 = 28, = .05. Note however for this two - tailed test,z/ 2 = z.025 = 1.96

Ata = 29, = .15. z.15 = 1.04

= 6

2 2 2 2/ 2

2 2

0

( ) (1.96 1.04) (6)324

( ) (28 29)a

z zn

+ += = =

59. At0 = 25, = .02. z.02 = 2.05

Ata = 24, = .20. z.20 = .84

= 3

2 2 2 2

2 2

0

( ) (2.05 .84) (3)75.2

( ) (25 24)a

z zn

+ += = =

Use 76

60. a. H0: = 16

Ha: 16

b. 0 16.32 16 2.19

/ .8 / 30

xz

n

= = =

Area = .4857

p-value = 2(.5000 - .4857) = .0286

p-value .05; reject H0. Readjust production line.

c. 0 15.82 16 1.23

/ .8 / 30

xz

n

= = =

Area = .3907

p-value = 2(.5000 - .3907) = .2186

p-value > .05; do not reject H0. Continue the production line.

d. Reject H0ifz-1.96 orz1.96

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Hypothesis Testing

For x = 16.32, z= 2.19; reject H0

For x = 15.82, z= -1.23; do not reject H0

Yes, same conclusion.

61. a. H0: = 46.69

Ha: 46.69

b. .025x zn

200

65.1096.162.48

48.62 1.48 (47.14 to 50.10)

c. Reject H0because= 46.69 is not in the interval.

d. 56.2200/65.10

69.4662.48

/

0=

=

=

n

xz

p-value = 2(.5000 - .4948) = .0104

62. a. H0: 48973

Ha: >48973

b. 71.295/6818

4897350867

/

0=

=

=

n

xz

p-value = .5000 - .4966 = .0034

p-value .01; reject H0. Conclude Taipei City has a higher mean salary.

63. a. H0: 37,000

Ha: >37,000

b. 0 38,000 37,000 1.47

/ 5200 / 48

xt

s n

= = =




c. p-value > .05, do not reject H0. Cannot conclude mean greater than $37,000. A larger sample isdesirable.

64. H0: = 6000

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Chapter 9

Ha: 6000

0 5812 6000 .93/ 1140 / 32

xt

s n

= = =





Do not reject H0. There is no evidence to conclude that the mean number of freshman applicationshas changed.

65. a. H0: 30

Ha: < 30

b. 0 29.6 30 1.57

/ 1.8 / 50

xt

s n

= = =

Degrees of freedom = 50 1 = 49



c. p-value > .01; do not reject H0. Cannot conclude miles per gallon is less than 30.

66. H0: 125,000

Ha: > 125,000

0 130,000 125,000 2.26/ 12,500 / 32

xt

s n

= = =

Degrees of freedom = 32 1 = 31



p-value .05; reject H0. Conclude that the mean cost is greater than $125,000 per lot.

67. a. H0: = 3

Ha: 3

b. 2.8i

xx

n

= =

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Hypothesis Testing

c. ( )

2

.701

ix x

sn

= =

d. 0 2.8 3 .90

/ .70 / 10

xt

s n

= = =

Degrees of freedom = 10 - 1 = 9




e. p-value > .05; do not reject H0. There is no evidence to conclude a difference compared to prioryear.

68. a. H0: p.50

Ha: p>.50

b. 64

.64100

p= =

c.

0

0 0

.64 .502.80

(1 ) .50(1 .50)

100

p pz

p p

n

= = =

Area in tail = .4974

p-value = .5000 - .4974 = .0026

p-value .01; reject H0. College graduates have a greater stop-smoking success rate.

69. a. H0: p= .6667

Ha: p.6667

b. 355

.6502546

p= =

c.

0

0 0

.6502 .6667

.82(1 ) .6667(1 .6667)

546

p p

z p p

n

= = =

p-value = 2(.5000 - .2939) = .4122

p-value > .05; do not reject H0; Cannot conclude that the population proportion differs from 2/3.

70. a. H0: p.50

Ha: p> .50

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Chapter 9

b. 67

.6381105

p= = (64%)

c.

0

0 0

.6381 .502.83

(1 ) .50(1 .50)

105

p pz

p p

n

= = =

p-value = .5000 - .4977 = .0023

p-value .01; reject H0. Conclude that the four 10-hour day schedules is preferred by more than50% of the office workers.

71. a. 330

.825400

p= =

b. H0: p= .78

Ha: p.78

0

0 0

.825 .782.17

(1 ) .78(1 .78)

400

p pz

p p

n

= = =

p-value = 2(.5000 - .4850) = .03

c. p-value .05; reject H0. The on-time arrival record has changed. It is improving.

72. H0: p.90

Ha: p< .90

49.8448

58p= =

0

0 0

.8448 .901.40

(1 ) .90(1 .90)

58

p pz

p p

n

= = =

p-value = .5000 - .4192 = .0808

p-value > .05; do not reject H0. Claim of at least 90% cannot be rejected.

73. a. H0: p.47

Ha: p< .47

b. 44

.352125

p= =

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Hypothesis Testing

c.

0

0 0

.352 .472.64

(1 ) .47(1 .47)

125

p pz

p p

n

= = =

p-value = .5000 - .4959 = .0041

d. p-value .01; reject H0. The proportion of foods containing pesticides has declined.

74. a. H0: 72

Ha: > 72

Reject H0ifz 1.645

0 72 1.645/ 20 / 30

x xz

n

= = =

Solve for x = 78

Decision Rule:

Accept H0if x < 78

Reject H0if x 78

b. For = 80

78 80.55

20/ 30z

= =

= .5000 -.2088 = .2912

c. For = 75,

78 75.82

20/ 30z

= =

= .5000 +.2939 = .7939

d. For = 70, H0is true. In this case the Type II error cannot be made.

e. Power = 1 -

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Chapter 9

.2

.4

.6

.8

1.0

72 74 76 78 80 82 84

Ho#lsePossible Vl!es of

P

o

"

er

75. H0: 492622

Ha: < 492622

At0 = 492622, = .02. z.02 = 2.05

Ata = 490622, = .05. z.10 = 1.645

44.218)490622492622(

)8000()645.105.2(

)(

)(2

22

2

0

22

=

+=

+

=

a

zzn

Use 219

76. H0: = 120

Ha: 120

At0 = 120, = .05. With a two - tailed test,z/ 2 = z.025 = 1.96

Ata = 117, = .02. z.02 = 2.05

2 2 2 2/ 2

2 2

0

( ) (1.96 2.05) (5)44.7

( ) (120 117)a

z zn

+ += = =

Use 45

b. Example calculation for = 118.

Reject H0ifz -1.96 or ifz 1.96

0 120/ 5 / 45

x xzn

= =

Solve forx . Atz = -1.96, x = 118.54

Atz = +1.96, x = 121.46

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SBE_SM09