chapter 8 - hypothesis testingwebsites.rcc.edu/pell/files/2015/09/chapter-8-solutions.pdfchapter 8 -...

Chapter 8 - Hypothesis Testing

138

Note: Graphs are not to scale and areintended to convey a general idea. Answersmay vary due to rounding.

EXERCISE SET 8-1

1.The null hypothesis is a statistical hypothesisthat states there is no difference between aparameter and a specific value or there is nodifference between two parameters. Thealternative hypothesis specifies a specificdifference between a parameter and aspecific value, or that there is a differencebetween two parameters. Examples willvary.

2.A type I error occurs by rejecting the nullhypothesis when it is true. A type II erroroccurs when the null hypothesis is notrejected and it is false. They are related inthat decreasing the probability of one type oferror increases the probability of the othertype of error.

3.A statistical test uses the data obtained froma sample to make a decision as to whether ornot the null hypothesis should be rejected.

4.A one-tailed test indicates the nullhypothesis should be rejected when the teststatistic value is in the critical region on oneside of the mean. A two-tailed test indicatesthe null hypothesis should be rejected whenthe test statistic value is in either criticalregion on both sides of the mean.

5.The critical region is the region of values ofthe test-statistic that indicates a significantdifference and the null hypothesis should berejected. The non-critical region is theregion of values of the test-statistic thatindicates the difference was probably due tochance, and the null hypothesis should notbe rejected.

6."H " represents the null hypothesis. "H "! "

represents the alternative hypothesis.

7.Type I is represented by , type II is!represented by ."

8.When the difference between the samplemean and the hypothesized mean is large,then the difference is said to be significantand probably not due to chance.

9.A one-tailed test should be used when aspecific direction, such as greater than orless than, is being hypothesized, whereaswhen no direction is specified, a two-tailedtest should be used.

10.Hypotheses can only be proved true whenthe entire population is used to compute thetest statistic. In most cases, this isimpossible.

11.a. 1.96„

1.96 0 1.96

b. 2.33

2.33 0

c. 2.58

0 2.58


139

11. continuedd. 2.33

0 2.33

e. 1.65

1.65 0

12.a. 2.05

2.05 0

b. 1.65

0 1.65

c. 2.58„

2.58 0 2.58

12. continuedd. 1.75

1.75 0

e. 2.05

0 2.05

13.a. H : 24.6! . œ H : 24.6" . Á

b. H : $51,497! . œ H : $51,497" . Á

c. H : 25.4! . œ H : 25.4" .

d. H : 88! . œ H : 88" .

14.a. H : 27! . œ H : 27" .

b. H : 4.71! . œ H : 4.71" . Á

c. H : 36! . œ H : 36" .

d. H : $79.95! . œ H : $79.95" . Á

e. H : 8.2! . œ H : 8.2" . Á

EXERCISE SET 8-2

1.H : 305! . œ


140

1. continuedH : 305 (claim)" .

C. V. 1.65 3.6 1.897œ œ œ5 Èz 4.69œ œ œX 306.2 305 .

5È Èn

1.897

55

0 1.65 Å 4.69

Reject the null hypothesis. There is enoughevidence to support the claim that theaverage depth has increased. Many factorscould contribute to the increase, includingwarmer temperatures and higher than usualrainfall.

2.H : 130! . œH : 130 (claim)1 .

C. V. 1.65œ

z 5.92œ œ œX 162 130 .5È Èn

38.2

50

0 1.65 Å 5.92Reject the null hypothesis. There is enoughevidence to support the claim that theaverage number of friends is more than 130.

3.H : $24 billion! . œH1: $24 billion (claim).

C. V. 1.65 X $31.5 s $28.7œ œ œ

z 1.85œ œ œX 31.5 24 .5È Èn

28.7

50

3. continued

0 1.65 Å 1.85

Reject the null hypothesis. There is enoughevidence to support the claim that theaverage revenue exceeds $24 billion.

4.H : 8.5! . œH : 8.5 (claim)" . Á

C. V. 1.96œ „

z 2.17œ œ œX 9.6 8.5 .5È Èn

3.2

40

0 1.96 Å 2.17

Reject the null hypothesis. There is enoughevidence to support the claim that there is adifference in the average number of movies.

5.H : 30.9! . œH : 30.9 (claim)" . Á

C. V. 2.58œ „

z 1.89œ œ œX 32.1 30.9 .sn

3.6

32È È

2.58 0 2.58 Å 1.89

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat average number of hours differs from30.9.


141

6.H : $117.91! . œH : $117.91 (claim)" . Á

C. V. 1.65œ „

z 1.40œ œ œX $122.57 $117.91 .5È Èn

$20

36

1.65 0 1.65 Å 1.40

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the cost per square foot differs from$117.91.

7.H : 29! . œH : 29 (claim)" . Á

C. V. 1.96 X 29.45 s 2.61œ „ œ œ

z 0.944œ œ œX 29.45 29 .5È Èn

2.61

30

1.96 0 1.96Å 0.944

Do not reject the null hypothesis. There isenough evidence to reject the claim that theaverage height differs from 29 inches.

8.H : $59,593! . œH : $59,593 (claim)" .

C. V. 2.33œ

z 2.90œ œ œ X $58,800 $59,593 .sn

$1500

30È È

8. continued

2.33 0Å 2.90

Reject the null hypothesis. There is enoughevidence to support the claim that theaverage income is less than $59,593.

9.H : $8121! . œH : $8121 (claim)" .

C. V. 2.33œ

z 1.93œ œ œX $8350 $8121 .5È Èn

750

40

0 2.33Å 1.93

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the average operating cost hasincreased.

10.H : 6.698! . œH : 6.698 (claim)" . Á

= œ œ5.5 inches 0.458 feetC. V. 1.96œ „

z 0.62œ œ œX 6.75 6.698 .5È Èn

0.458

30

1.96 0 1.96 Å 0.62

Do not reject the null hypothesis. There isnot enough evidence to support the claim


142

10. continuedthat the average height differs from 6.698feet.

11.H : 150! . œH : 150 (claim)" .

C. V. 2.33œ

z 1.48œ œ œX $152.59 $150 .5È Èn

$10.78

38

0 2.33Å 1.48

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the average cost is greater than $150.

12.H : $10,337! . œH : $10,337 (claim)" . Á

C. V. 1.65 at 0.10, 1.96 at 0.05,œ „ „and 2.58 at 0.01„

z 3.62œ œ œX $10,798 $10,337 .5È Èn

1560

150

Since 3.62 is outside the critical values of! œ 0.10, 0.05, and 0.01, reject the nullhypothesis. There is a significant differencein student expenditures.

13.H : 60.350 . œH : 60.35 (claim)" .

C. V. 1.65œ

z 4.82œ œ œ X 55.4 60.35 .sn

6.5

40È È

Å 1.65 0 4.82

13. continuedReject the null hypothesis. There is enoughevidence to support the claim that statesenators are younger than U. S. Senators.

14.H : 34.9! . œH : 34.9 (claim)" . Á

C. V. 2.05œ „

z 4.55œ œ œ X 28.5 34.9 .5È Èn

8.9

40

1.96 0 1.96Å 4.55

Reject the null hypothesis. There is enoughevidence to support the claim that theaverage stay differs from 34.9 months.

15.a. Do not reject.b. Reject.c. Do not reject.d. Reject.e. Reject.

16.H : 52 (claim)0 . œH : 521 . Á

z 8.69œ œ œX 56.3 52 .sn

3.5

50È È

The area corresponding to z 8.69 isœ 0.9999. Then P-value 0.01. Hence,the null hypothesis should be rejected.There is enough evidence to reject the claimthat the mean is 52. The researcher's claimis not valid.

17.H : 264! . œH : 264 (claim)" .

z 2.53œ œ œ X 262.3 264 .5È Èn

3

20

The area corresponding to z 2.53 isœ0.9943. The P-value is 1 0.9943 œ0.0057. The decision is to reject the nullhypothesis since 0.0057 0.01. There is


143

17. continuedenough evidence to support the claim thatthe average stopping distance is less than264 feet. (TI: P-value 0.0056)œ

18.H : 40! . œH : 40 (claim)" . X 29.26 30.9œ œ5

z 2.458œ œ œ X 29.3 40 .5È Èn

30.9

50

The area corresponding to z 2.458 isœ 0.0069. The decision is reject the nullhypothesis since 0.0069 0.01. There isenough evidence to support the claim thatthe average number of copies is less than 40.(TI: P-value 0.00699)œ

19.H : 546! . œH : 546 (claim)" .

z 2.40œ œ œ X 544.8 546 .5È Èn

3

36

The area corresponding to z 2.40 isœ 0.0082. Thus, P-value 0.0082. Theœdecision is to reject the null hypothesis since0.0082 0.01. There is enough evidence tosupport the claim that the number of caloriesburned is less than 546.(TI: P-value 0.0082)œ

20.H : 800 (claim)! . œH : 800" . Á

z 2.61œ œ œ X 793 800 .5È Èn

12

200

The area corresponding to z 2.61 isœ 0.0045. The P-value is found by multiplyingby 2 since this is a two-tailed test. Hence,2(0.0045) 0.0090. The decision is toœreject the null hypothesis since0.009 0.01. There is enough evidence toreject the null hypothesis that the breakingstrength is 800 pounds.

21.H : 444! . œH : 444" . Á

z 1.70œ œ œ X 430 444 .5È Èn

52

40

The area corresponding to z 1.70 isœ 0.0446. The P-value is 2(0.0446) 0.0892.œ

21. continuedThe decision is do not reject the nullhypothesis since P-value 0.05. There isnot enough evidence to support the claimthat the mean differs from 444.(TI: P-value 0.0886)œ

22.H : 65 (claim)! . œH : 65" . Á

z 1.21œ œ œ X 63.2 65 .5È Èn

7

22

The area corresponding to z 1.21 isœ 0.1131. The P-value is 2(0.1131) 0.2262.œThe decision is do not reject the nullhypothesis since 0.2262 0.10. Hence,there is not enough evidence to reject theclaim that the average is 65 acres.

23.H : 30,000 (claim)! . œH : 30,000" . Á

z 1.71œ œ œX 30,456 30,000 .sn

1684

40È È

The area corresponding to z 1.71 isœ0.9564. The P-value is 2(1 0.9564) œ2(0.0436) 0.0872. The decision is toœreject the null hypothesis at 0.10 since! œ0.0872 0.10. The conclusion is that thereis enough evidence to reject the claim thatcustomers are adhering to therecommendation. Yes, the 0.10 significancelevel is appropriate. (TI: P-value 0.0868)œ

24.H : 60 (claim)0 . œH : 601 . Á

X 59.93 s 13.42œ œ

z 0.03œ œ œ X 59.93 60 .sn

13.42

30È È

The P-value is 2(0.4880) 0.9760. (TI: P-œvalue 0.9783). Since 0.9760 > 0.05, theœdecision is do not reject the null hypothesis.There is not enough evidence to reject theclaim that the average number of speedingtickets is 60.

25.H : 100 . œH : 10 (claim)" .


144

25. continuedX 5.025 s 3.63œ œ

z 8.67œ œ œ X 5.025 10 .sn

3.63

40È È

The area corresponding to 8.67 is lessthan 0.0001. Since 0.0001 < 0.05, thedecision is to reject the null hypothesis.There is enough evidence to support theclaim that the average number of daysmissed per year is less than 10.

26.Reject the claim at 0.05 but not at! œ! œ 0.01. There is no contradiction sincethe value of should be chosen before the!test is conducted.

27.The mean and standard deviation are foundas follows:

f X f X f X

8.35 - 8.43 2 8.39 16.78 140.7842

8.44 - 8.52 6 8.48 50.88 431.4624

8.53 - 8.61 12 8.57 102.84 881.3388

8.62 - 8.70 18 8.66 155.88 1349

m m m† † #

.9208

8.71 - 8.79 10 8.75 87.5 765.625

8.80 - 8.88 8.84

50 431.56 3725.4224

2 17.68 156.2912

X 8.63œ œ œ!f X

n 50431.56† m

= œ œÉ É!f X 3725.4224

n 1 49

†

m

( f X )mn 50

(431.56)# † # #!

= œ 0.105

H : 8.65 (claim)0 . œH : 8.651 . Á

C. V. 1.96œ „

z 1.35œ œ œ X 8.63 8.65 .sn

0.105

50È È

Do not reject the null hypothesis. There isnot enough evidence to reject the claim thatthe average hourly wage of the employees is$8.65.

EXERCISE SET 8-3

1.It is bell-shaped, symmetric about the mean,and it never touches the x axis. The mean,

1. continuedmedian, and mode are all equal to 0 and theyare located at the center of the distribution.The t distribution differs from the standardnormal distribution in that it is a family ofcurves, the variance is greater than 1, and asthe degrees of freedom increase the tdistribution approaches the standard normaldistribution.

2.The degrees of freedom are the number ofvalues that are free to vary after a samplestatistic has been computed. They tell theresearcher which specific curve to use whena distribution consists of a family of curves.

3.a. d. f. 9 C. V. 1.833œ œ b. d. f. 17 C. V. 1.740œ œ c. d. f. 5 C. V. 3.365œ œ d. d. f. 8 C. V. 2.306œ œ

4.a. d. f. 14 C. V. 1.761œ œ „b. d. f. 22 C. V. 2.819œ œ c. d. f. 27 C. V. 2.771œ œ „d. d. f. 16 C. V. 2.583œ œ „

5.a. 0.01 < P-value < 0.025 (0.018)b. 0.05 < P-value < 0.10 (0.062)c. 0.10 < P-value < 0.25 (0.123)d. 0.10 < P-value < 0.20 (0.138)

6.a. P-value < 0.005 (0.003)b. 0.10 < P-value < 0.25 (0.158)c. P-value 0.05 (0.05)œd. P-value > 0.25 (0.261)

7.H : 2000 . œH : 200 (claim)" .

C. V. 1.833 d. f. 9œ œ

t 4.680œ œ œ X 185.2 200 .sn

10

10È È


145

7. continued

Å 1.833 0 4.680

Reject the null hypothesis. There is enoughevidence to support the claim that theaverage number of seeds is less than 200.

8.H : $5400! . œH : $5400 (claim)" .

C. V. 2.052 d. f. 27œ œ

t 1.262œ œ œ X $5250 $5400 .sn

$629

28È È

2.052 0 Å 1.262

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the average cost is less than $5400.

9.H : 700 (claim)0 . œH : 700" . X 606.5 s 109.1œ œ

C. V. 2.262 d. f. 9œ œ

t 2.710œ œ œ X 606.5 700 .sn

109.1

10È È

2.262 0Å 2.710

9. continuedReject the null hypothesis. There is enoughevidence to reject the claim that the averageheight of the buildings is at least 700 feet.

10.H : 50,000! . œH : 50,000 (claim)" .

C. V. 1.440 d. f. 6œ œX 50,363.57 1113.16œ = œ

t 0.864œ œ œX 50,363.57 50,000 .sn

1113.16

7È È

0 1.440Å 0.864

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the average number of words is greaterthan 50,000.

11.H : 73! . œH : 73 (claim)" .

C. V. 2.821 d. f. 9œ œX 123.5 s 39.303œ œ

t 4.063œ œ œX 123.5 73 .sn

39.303

10È È

0 2.821 Å 4.063

Reject the null hypothesis. There is enoughevidence to support the claim that theaverage viewing time is longer than 73minutes.

12.H : 110! . œH : 110 (claim)" .

C. V. 2.624 d. f. 14œ œ


146

12. continuedX 137.333 24.118œ = œ

t 4.389œ œ œX 137.333 110 .sn

24.118

15È È

0 2.624 Å 4.389

Reject the null hypothesis. There is enoughevidence to support the claim that theaverage number of calories is greater than110.

13.H : $54.8! . œH : $54.8 (claim)1 .

C. V. 1.761 d. f. 14œ œ

t 3.058œ œ œX $62.3 $54.8 .sn

$9.5

15È È

0 1.761 Å 3.058

Reject the null hypothesis. There is enoughevidence to support the claim that the cost toproduce an action movie is more than $54.8million.

14.H : 36! . œH : 36 (claim)" . Á

C. V. 2.807 d. f. 23œ „ œ

t 5.638œ œ œX 42.1 36 .sn

5.3

24È È

2.807 0 2.807 Å 5.638

14. continuedReject the null hypothesis. There is enoughevidence to support the claim that theaverage usage differs from 36.

15.H : $50.07! . œH : $50.07 (claim)" .

C. V. 1.833 d. f. 9œ œX $56.11 s $6.97œ œ

t 2.741œ œ œX $56.11 $50.07 .sn

6.97

10È È

0 1.833 Å 2.741

Reject the null hypothesis. There is enoughevidence to support the claim that theaverage bill has increased.

16.H : $15,0000 . œH : $15,000 (claim)" . Á

C. V. 2.201 d. f. 11œ „ œX $14,347.17 $2048.54œ = œ

t 1.104œ œ œ X $14,347.17 $15,000 .sn

$2048.54

12È È

2.201 0 2.201 Å 1.104

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the average stipend differs from$15,000.

17.H : $7.89! . œH : $7.89 (claim)" .

C. V. 2.624 d. f. 14œ œ


147

17. continued

t 2.550œ œ œX $11.09 $7.89 .sn

$4.86

15È È

0 2.624Å 2.550

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the average cost is greater than $7.89

18.H : 2.270 . œH : 2.27 (claim)1 . Á

C. V. 2.093 d. f. 19œ „ œ

t 3.240œ œ œX 2.98 2.27 .sn

0.98

20È È

Å2.093 0 2.093 3.240

Reject the null hypothesis. There is enoughevidence to support the claim that theaverage call differs from 2.27 minutes.

19.H : 25.40 . œH : 25.4 (claim)1 .

C. V. 1.318 d. f. 24œ œ

t 3.11œ œ œ X 22.1 25.4 .sn

5.3

25È È

Å 1.318 0 3.11

19. continuedReject the null hypothesis. There is enoughevidence to support the claim that thecommute time is less than 25.4 minutes.

20.H : 3.180 . œH : 3.18 (claim)" . Á

X 3.833 s 1.4346œ œd. f. 23 C. V. 2.069œ œ „

t 2.2299 or 2.23œ œ œX 3.833 3.18 .sn

1.4346

24È È

(TI answer 2.231)œ

2.069 0 2.069 Å 2.23

Reject the null hypothesis. There is enoughevidence to support the claim that theaverage family size differs from 3.18.

21.H : 5.8! . œH : 5.8 (claim)" . ÁX 3.85 s 2.519œ œd. f. 19 0.05œ œ!P-value < 0.01 (0.0026)

t 3.462œ œ œ X 3.85 5.8 .sn

2.519

20È È

Since P-value < 0.01, reject the nullhypothesis. There is enough evidence tosupport the claim that the mean is not 5.8.

22.H : 9.2 (claim)0 . œH : 9.2" . Á

X 8.25 s 5.06œ œd. f. 7œP-value > 0.50 (0.6121)

t 0.531œ œ œ X 8.25 9.2 .sn

5.06

8È È

Since P-value > 0.50, do not reject the nullhypothesis. There is enough evidence tosupport the claim that the mean number ofjobs is 9.2. One reason why a person maynot give the exact number of jobs is that he


148

22. continuedor she may have forgotten about a particularjob.

23.H : 1230 . œH : 123 (claim)" . Ád. f. 15œP-value < 0.01 (0.0086)

t 3.019œ œ œ X 119 123 .sn

5.3

16È È

Since P-value < 0.05, reject the nullhypothesis. There is enough evidence tosupport the claim that the mean is not 123gallons.

EXERCISE SET 8-4

1.Answers will vary.

2.The proportion of A items can be considereda success whereas the proportion of itemsthat are not included in A can be considereda failure.

3.np 5 and nq 5� �

4.

. 5œ : œ :;Î8 È

5.H : p 0.456! œH : p 0.456 (claim)" Á

s œ œ œ œp 0.545 p 0.456 q 0.54460110

C. V. 1.65œ „

z 1.87œ œ œs p p 0.545 0.456

É Épqn

(0.456)(0.544)110

(TI: z 1.884)œ

1.65 0 2.58 Å 1.87

5. continuedDo not reject the null hypothesis. There isnot enough evidence to support the claimthat the proportion of accidents involvingimproper driving differs from 45.6%.

6.H : p 0.5030 œH : p 0.503 (claim)" Á

s œ œ œ œp 0.57 p 0.503 q 0.497171300

z 2.32œ œ œs p p 0.57 0.503

É Épqn

(0.503)(0.497)300

Therefore, reject H at any > 0.025 since! !the critical value will be greater than 2.33(or any P-value 0.025).Ÿ

7.H : p 0.36! œH : p 0.36 (claim)"

s œ œ œ œp 0.44 p 0.36 q 0..6488200

C. V. 2.05œ

z 2.357œ œ œs p p 0.44 0.36

É Épqn

(0.36)(0.64)200

0 2.05 Å 2.357

Reject the null hypothesis. There is enoughevidence to support the claim that theproportion of baseball fans is greater than36%.

8.H : p 0.279! œH : p 0.279 (claim)"

s œ œ œ œp 0.375 p 0.279 q 0.72145120

C. V. 1.65œ

z 2.35œ œ œs p p 0.375 0.279

É Épqn

(0.279)(0.721)120


149

8. continued

0 1.65 Å 2.35

Reject the null hypothesis. There is enoughevidence to conclude that the proportion offemale physicians is higher than 27.9%.

9.H : p 0.30! œH : p 0.30 (claim)" Á

s œ œ œ œp 0.346 p 0.30 q 0.7045130

C. V. 1.96œ „

z 1.145œ œ œs p p 0.346 0.3

É Épqn

(0.3)(0.7)130

(TI: 1.148)D œ

1.96 0 1.96 Å 1.145

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the proportion of open or unlockedwindow or door burglaries differs from 30%.

10.H : p 0.856! œH : p 0.856 (claim)" Á

s œ œ œ œp 0.84 p 0.856 q 0.144420500

C. V. 1.96œ „

z 1.02œ œ œ s p p 0.84 0.856

É Épqn

(0.856)(0.144)500

1.96 0 1.96 Å 1.02

10. continuedDo not reject the null hypothesis. There isnot enough evidence to support the claimthat the percentage differs from the nationalrate.

11.H : p 0.320 œH : p 0.32 (claim)" Á

s œ œ œ œp 0.4019 p 0.32 q 0.68170423

C. V. 2.58œ „

z 3.61œ œ œs p p 0.4019 0.32

É Épqn

(0.32)(0.68)423

2.58 0 2.58 Å 3.61

Reject the null hypothesis. There is enoughevidence to support the claim that theproportion of 40- to 59-year-old moviegoersdiffers from 32%.

12.H : p 0.83! œH : p 0.83 (claim)"

p 0.8 p 0.83 q 0.17^ œ œ œ œ240300

C. V. 1.65œ

z 1.38œ œ œ s p p 0.8 0.83

É Épqn

(0.83)(0.17)300

1.65 0 Å 1.38

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the percentage is less than 83%.

13.H : p 0.54 (claim)! œH : p 0.54" Á

s œ œ œ œp 0.6 p 0.54 q 0.463660


150

13. continued

z 0.93œ œ œs p p 0.6 0.54

É Épqn

(0.54)(0.46)60

Area 0.8238œP-value 2(1 0.8238) 0.3524œ œSince P-value > 0.01, do not reject the nullhypothesis. There is enough evidence tosupport the claim that 54% of kids had asnack after school. Yes, a healthy snackshould be made available for children to eatafter school. (TI: P-value 0.3511)œ

14.H : p 0.517 (claim)! œH : p 0.517" Á

s œ œ œ œp 0.575 p 0.517 q 0.483115200

z 1.64œ œ œs p p 0.575 0.517

É Épqn

(0.517)(0.483)200

Area 0.9495œP-value 2(1 0.9495) 0.101œ œSince P-value > 0.05, do not reject the nullhypothesis. There is not enough evidence toreject the claim that 51.7% of homes inAmerica were heated by natural gas. Theevidence supports the claim. The conclusioncould be different if the sample is taken in anarea where natural gas is not commonly usedto heat homes.

15.H : p 0.18 (claim)! œH : p 0.18"

s œ œ œ œp 0.1667 p 0.18 q 0.8250300

z 0.60œ œ œ s p p 0.1667 0.18

É Épqn

(0.18)(0.82)300

P-value 0.2743 (TI: P-value 0.2739)œ œSince P-value > 0.05, do not reject the nullhypothesis. There is not enough evidence toreject the claim that 18% of all high schoolstudents smoke at least a pack of cigarettes aday.

16.H : p 0.14 (claim)0 œH : p 0.14" Á

s œ œ œ œp 0.10 p 0.14 q 0.8610100

z 1.15œ œ œ s p p 0.10 0.14

É Épqn

(0.14)(0.86)100

Area 0.1251œP-value 2(0.1251) 0.2502œ œ

16. continuedSince P-value > 0.10, do not reject the nullhypothesis. There is not enough evidence toreject the claim that 14% of men useexercise to relieve stress. The results cannotbe generalized to all adults since only menwere surveyed.

17.H : p 0.67! œH : p 0.67 (claim)" Á

p 0.82 p 0.67 q 0.33^ œ œ œ œ82100

C. V. 1.96œ „

z 3.19œ œ œs p p 0.82 0.67

É Épqn

(0.67)(0.33)100

1.96 0 1.96 Å 3.19

Reject the null hypothesis. There is enoughevidence to support the claim that thepercentage is not 67%.

18.H : p 0.6! œH : p 0.6 (claim)"

p 0.52 p 0.6 q 0.4^ œ œ œ œ2650

C. V. 1.65œ

z 1.15œ œ œ s p p 0.52 0.6

É Épqn

(0.6)(0.450

1.65 0 Å 1.15

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the percentage of paid assistantships isless than 60%.

19.H : p 0.5760 œH : p 0.576 (claim)1


151

19. continued

p 0.472 p 0.576 q 0.424^ œ œ œ œ1736

C. V. 1.65œ

z 1.26œ œ œ s p p 0.472 0.576

É Épqn

(0.576)(0.424)36

1.65 0 Å 1.26

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the percentage of injuries duringpractice is less than 57.6%.

20.H : p 0.7! œH : p 0.7 (claim)"

p 0.816 p 0.7 q 0.3^ œ œ œ œ204250

C. V. 2.33œ

z 4.00œ œ œs p p 0.816 0.7

É Épqn

(0.7)(0.3)250

0 2.33 Å 4.00

Reject the null hypothesis. There is enoughevidence to support the claim that thepercentage of college students who recycleis greater than 70%.

21.This represents a binomial distribution withp 0.50 and n 9. The P-value isœ œ2 P(X 3) 2(0.254) 0.508.† Ÿ œ œ

Since P-value 0.10, the conclusion thatthe coin is not balanced is probablyfalse. The answer is no.

22.This represents a binomial distribution withp 0.20 and n 15. The P-value isœ œ2 ) 2(0.061) 0.122, which is† T Ð\ & œ œ

22. continuedgreater than 0.10. TDo not reject the! œnull hypothesis. here is not enough evidenceto conclude that the proportions havechanged.

23.

z œ H.5

z œ HnpnpqÈ

z œH

n nnp

1n

Ènpq

z œH

n nnp

npq

n

É #

z œ sp p

É pqn

EXERCISE SET 8-5

1.a. H : 225!

#5 œ H : 225"

#5

C. V. 27.587 d. f. 17œ œ

0 27.587

b. H : 225!#5 œ

H : 225"#5

C. V. 14.042 d. f. 22œ œ

0 14.042

c. H : 225!#5 œ

H : 225"#5 Á

C. V. 5.629, 26.119 d. f. 14œ œ


152

1c. continued

0 5.629 26.119

d. H : 225!#5 œ

H : 225"#5 Á

C. V. 2.167, 14.067 d. f. 7œ œ

0 2.167 14.067

2.a. H : 225!

#5 œ H : 225"

#5

C. V. 32.000 d. f. 16œ œ

0 32.000

b. H : 225!#5 œ

H : 225"#5

C. V. 8.907 d. f. 19œ œ

0 8.907

c. H : 225!#5 œ

H : 225"#5 Á

C. V. 3.074, 28.299 d. f. 12œ œ

2c. continued

0 3.074 28.299

d. H : 225!#5 œ

H : 225"#5

C. V. 15.308 d. f. 28œ œ

0 15.308

3.a. 0.01 < P-value < 0.025 (0.015)b. 0.005 < P-value < 0.01 (0.006)c. 0.01 < P-value < 0.02 (0.012)d. P-value < 0.005 (0.003)

4a. 0.02 < P-value < 0.05 (0.037)b. 0.05 < P-value < 0.10 (0.088)c. 0.05 < P-value < 0.10 (0.066)d. P-value < 0.01 (0.007)

5.H : 15! 5 œH : 15 (claim)" 5

C. V. 4.575 0.05 d. f. 11œ œ œ!

;# œ œ œ(n 1)s (12 1)(13.6)15

# #

# #59.0425

0 4.575 Å 9.0425

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the standard deviation is less than 15.


153

6.H : 100!

#5 œH : 100 (claim)"

#5 Á

s 135.4333# œC. V. 2.700, 19.023 0.05 d. f. 9œ œ œ!

;# œ œ œ(n 1)s (10 1)(135.4333)100

#

#512.189

0 2.700 19.023Å 12.189

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the variance differs from 100.

7.H : 1.2 (claim)! 5 œH : 1.2" 5

! œ œ0.01 d. f. 14

;# œ œ œ(n 1)s (15 1)(1.8)(1.2)

# #

# #531.5

P-value < 0.005 (0.0047)Since P-value < 0.01, reject the nullhypothesis. There is enough evidence toreject the claim that the standard deviation isless than or equal to 1.2 minutes.

8.H : 0.03 (claim)! 5 œH : 0.03" 5

s 0.043œ! œ œ0.05 d. f. 7

;# œ œ œ(n 1)s (7(0.043)0.03

# #

# #514.381

0.025 < P-value < 0.05 (0.045)Since P-value < 0.05, reject the nullhypothesis. There is enough evidence toreject the claim that the standard deviation isless than or equal to 0.03 ounce.

9.H : 100! 5 œH : 100 (claim)" 5

9. continueds 126.721œC. V. 12.017 0.10 d. f. 7œ œ œ!

;# œ œ œ(n 1)s (8 1)(126.721)100

# #

# #511.241

0 12.017Å 11.241

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the standard deviation is more than 100mg.

10.H : 100!

#5 œH : 100 (claim)"

#5

s 183.7755# œC. V. 23.685 0.05 d. f. 14œ œ œ!

;# œ œ œ(n 1)s (15 1)(183.7755)100

#

#525.729

0 23.685 Å 25.729

Reject the null hypothesis. There is enoughevidence to support the claim that thevariance is more than 100.

11.H : 35! 5 œH : 35 (claim)" 5

C. V. 3.940 0.05 d. f. 10œ œ œ!

;# œ œ œ(n 1)s (11 1)(32)35

# #

# #58.359


154

11. continued

0 3.940 Å 8.359

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the standard deviation is less than 35.

12.H : 8! 5 œH : 8 (claim)" 5

C. V. 55.758 0.05 d. f. 49œ œ œ!

;# œ œ œ(n 1)s (50 1)(10.5)8

# #

# #584.410

0 55.758 Å 84.410

Reject the null hypothesis. There is enoughevidence to support the claim that thestandard deviation is more than 8.

13.H : 150! 5 œH : 150 (claim)" 5

s 162.09œC. V. 21.026 0.05 d. f. 12œ œ œ!

;# œ œ œ(n 1)s (13 1)(162.09)150

# #

# #514.012

0 21.026Å 14.012

13. continuedDo not reject the null hypothesis. There isnot enough evidence to support the claimthat the standard deviation is greater than150 mg of sodium.

14.H : 12! 5 œH : 12 (claim)" 5 Á

s 14.608œC. V. 5.226, 21.026 0.10 d.f. 12œ œ œ!

;# œ œ œ(n 1)s (13 1)(14.608)12

# #

# #517.783

0 5.226 21.026Å 17.783

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the standard deviation differs from 12mg.

15.H : 0.52! 5 œH : 0.52 (claim)" 5

C. V. 30.144 0.05 d. f. 19œ œ œ!

;# œ œ œ(n 1)s (20 1)(0.568)(0.52)

# #

# #522.670

0 30.144Å 22.670

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the standard deviation is more than 0.52mm.

16.H : 9!

#5 œH : 9 (claim)"

#5


155

16. continued= œ# 9.344C. V. 14.684 0.10 d. f. 9œ œ œ!

;# œ œ œ(n 1)s (10 1)(9.344)9

#

#59.344

0 14.684Å 9.344

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the variance is more than 9.

17.H : 60 (claim)! 5 œH : 60" 5 ÁC. V. 8.672, 27.587 0.10œ œ!d. f. 17œs 64.6œ

;# œ œ œ(n 1)s (18 1)(64.6)(60)

# #

# #519.707

0 8.672 27.587Å 19.707

Do not reject the null hypothesis. There isnot enough evidence to reject the claim thatthe standard deviation is 60.

18.H : 8! 5 œH : 8 (claim)" 5 C. V. 30.144 0.05 d. f. 19œ œ œ!X 46.3 s 11.017œ œ

;# œ œ œ(n 1)s (20 1)(11.017)8

# #

# #536.033

0 30.144 Å 36.033

18. continuedReject the null hypothesis. There is enoughevidence to support the claim that thestandard deviation is higher than 8 degrees.

19.

5 ¸ Range4

5 ¸ œ$9500 $67824 $679.50

H : $679.50! 5 œH : $679.50 (claim)" 5 Ás 770.67œC. V. 5.009, 24.736 0.05 d. f. 13œ œ œ!

;# œ œ œ(n 1)s (14 1)(770.67)679.5

# #

# #516.723

0 5.009 24.736Å 16.723

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the standard deviation differs from$679.50.

20.H : 2385.9! 5 œH : 2385.9 (claim)" 5 s 2194.845œC. V. 1.145 0.05 d. f. 5œ œ œ!

;# œ œ œ(n 1)s (6 1)(2194.845)2385.9

# #

# #54.231

0 1.145 Å 4.231

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the standard deviation is less than2385.9 feet.

EXERCISE SET 8-6

1.H : 25.2! . œH : 25.2 (claim)" . Á


156

1. continuedC. V. 2.032œ „

> œ œ œX 28.7 25.2 .=

È Èn4.6

35

4.50

2.032 0 2.032 Å 4.50

Reject the null hypothesis. There is enoughevidence to support the claim that theaverage age differs from 25.2.

The 95% confidence interval of the mean is:X X > >! !

2 2

= =È Èn n

.

28.7 2.032 ) ˆ 4.6

35È .

28.7 2.032 ) ˆ 4.6

35È27.1 30.3 .(TI: 27.2 30.2) .The confidence interval does not contain thehypothesized mean age of 25.2.

2.H : $236! . œH : $236 (claim)" . Á

C. V. 2.539 d.f. 19œ „ œ

t 2.704œ œ œ X 210 236 .sn

43

20È È

2.539 0 2.539Å 2.704

Reject the null hypothesis. There is enoughevidence to support the claim that theaverage airfare differs from $236.

The 98% confidence interval of the mean is:X z X z ! !

2 2

s sn nÈ È.

210 2.539 210 2.539 † †43 43

20 20È È.

$185.59 $234.41 .

2. continuedThere is agreement between the z-test andthe confidence interval because the intervaldoes not contain the mean of $236.

3.H : $19,150! . œH : $19,150 (claim)" . Á

C. V. 1.96œ „

z 3.69œ œ œ X $17,020 $19,150 .5È Èn

4080

50

1.96 0 1.96Å 3.69

Reject the null hypothesis. There is enoughevidence to support the claim that the meanis not $19,150.

X z X z ! !2 2

5 5È Èn n

.

$17,020 1.96 Ð Ñ 4080

50È .

$17,020 1.96 Ð Ñ4080

50È$15,889 $18,151 .The 95% confidence interval supports theconclusion because it does not contain thehypothesized mean.

4.H : 0.694! : œH : 0.694 (claim)" : Á

C. V. 1.96œ „

p 0.727 p 0.694 q 0.306^ œ œ œ œ120165

z 0.920œ œ œs p p 0.727 0.694

É Épqn

(0.694)(0.306)165

1.96 0 1.96 Å 0.920


157

4. continuedThe 95% confidence interval of theproportion is:

p (z < p < p (z^ ^ Ñ Ñ! !# #É Ép q p q^ ^ ^ ^

n n

0.727 1.96 † : É (0.727)(0.273)165

0.727 1.96 †É 0.727)(0.273)165

0.659 0.795 :

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the proportion of white collar criminalswho serve time differs from 69.4%. Theconfidence interval does contain thehypothesized percentage of 69.4%.

5.H : 19! . œH : 19 (claim)" . Á

C. V. 2.145œ „

t 1.37œ œ œX 21.3 19 .sn

6.5

15È È

2.145 0 2.145 Å 1.37

The 99% confidence interval of the mean is:

X z X z ! !2 2

5 5È Èn n

.

21.3 2.145 21.3 2.145 † †6.5 6.5

15 15È È.

17.7 24.9 .

The decision is do not reject the nullhypothesis since 1.37 2.145 and the 99%confidence interval does contain thehypothesized mean of 19. The conclusion isthat there is not enough evidence to supportthe claim that the average time worked athome is not 19 hours per week.

6.H : 10.8 (claim)! . œH : 10.8" . Á

C. V. 2.33œ „

z 2.80œ œ œX 12.2 10.8 .5È Èn

3

36

6. continued

2.33 0 2.33 Å 2.80

X z X z ! !2 2

5 5È Èn n

.

12.2 2.33 12.2 2.33 † †3 3

36 36È È.

11.0 13.4 .

The decision is to reject the null hypothesissince 2.80 2.33 and the confidenceinterval does not contain the hypothesizedmean of 10.8. The conclusion is that there isenough evidence to reject the claim that theaverage time a person spends reading anewspaper is 10.8 minutes.

7.The power of a statistical test is theprobability of rejecting the null hypothesiswhen it is false.

8.The power of a test is equal to 1 where "" is the probability of a type II error.

9.The power of a test can be increased byincreasing or selecting a larger sample!size.

REVIEW EXERCISES - CHAPTER 8

1.H : 18.3! . œH : 18.3 (claim)" . Á

C. V. 2.33œ „

= œ œÈ32.49 5.7

z 3.16œ œ œX 20.9 18.3 .sn

5.7

48È È


158

1. continued

2.33 0 2.33 Å 3.16

Reject the null hypothesis. There is enoughevidence to support the claim that theaverage time spent online differs from 18.3hours.

2.H : 25.3! . œH : 25.3 (claim)" .

C. V. 2.33œ

z 2.19œ œ œ X 23.9 25.3 .5È Èn

6.39

100

2.33 0 Å 2.19

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the average commute time is less than25.3 minutes.

3.H : 18,000! . œH : 18,000 (claim)" .

X 16,298.37 s 2604.82œ œ

C. V. 2.33œ

z 3.58œ œ œ X 16,298.37 18,000 .sn

2604.82

30È È

2.33 0Å 3.58

3. continuedReject the null hypothesis. There is enoughevidence to support the claim that averagedebt is less than $18,000.

4.H : 10! . œH : 10 (claim)" .

z 2.22œ œ œ X 9.25 10 .5È Èn

2

35

P-value 0.0132œSince 0.0132 0.05, reject the nullhypothesis. The conclusion is that there isenough evidence to support the claim thatthe average time is less than 10 minutes.

5.H : $1229! . œH : $1229 (claim)" . Á

C. V. 1.96œ „

t 1.875œ œ œX $1350 $1229 .sn

$250

15È È

1.96 0 1.96 Å 1.875

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the average rent differs from $1229.

6.H : $50! . œH : $50 (claim)" .

C. V. 1.356 d. f. 12œ œX 64.962 51.929œ = œ

t 1.04œ œ œX 64.962 50 .sn

51.929

13È È

0 1.356Å 1.04


159

6. continuedDo not reject the null hypothesis. There isnot enough evidence to support the claimthat the average winnings exceed $50.

7.H : 10! . œH : 10 (claim)" .

C. V. 1.782 X 9.6385 s 0.5853œ œ œ

t 2.227 or 2.230œ œ œ X 9.6385 10 .sn

0.5853

13È È

1.782 0Å 2.230

Reject the null hypothesis. There is enoughevidence to support the claim that averageweight is less than 10 ounces.

8.H : 208! . œH : 208 (claim)" .

C. V. 2.896 d. f. 8œ œX 209.74 s 1.67œ œ

t 3.13œ œ œX 209.74 208 .sn

1.67

9È È

0 2.896 Å 3.13

Reject the null hypothesis. There is enoughevidence to support the claim that weight isgreater than 208 grams.

9.H : p 0.17! œH : p 0.17 (claim)"

C. V. 1.65œ „s œ œ œp 0.28 p 0.17 q 0.83

9. continued

z 4.34œ œ œs p p 0.28 0.17

É Épqn

(0.17)(0.83)220

1.65 0 1.65 Å 4.34

Reject the null hypothesis. There is enoughevidence to support the claim that thepercentage is greater than 17%.

10.H : p 0.602! œH : p 0.602 (claim)"

C. V. 1.65œs œ œ œp 0.65 p 0.602 q 0.398

z 1.96œ œ œs p p 0.65 0.602

É Épqn

(0.602)(0.398)400

0 1.65 Å 1.96

Reject the null hypothesis. There is enoughevidence to support the claim that thepercentage of drug offenders is higher than60.2%.

11.H : p 0.593! œH : p 0.593 (claim)"

C. V. 2.33œ s œ œ œ œp 0.52 p 0.593 q 0.407156

300

z 2.57œ œ œ s p p 0.52 0.593

É Épqn

(0.593)(0.407)300

2.33 0Å 2.57


160

11. continuedReject the null hypothesis. There is enoughevidence to support the claim that less than59.3% of school lunches are free or at areduced price.

12.H : p 0.65 (claim)! œH : p 0.65" Á

s œ œ œ œp 0.7125 p 0.65 q 0.355780

z 1.17œ œ œs p p 0.7125 0.65

É Épqn

(0.65)(0.35)80

Area 0.8790œPvalue 2(1 0.8790) 0.242 (0.2412)œ œSince P-value > 0.05, do not reject the nullhypothesis. There is not enough evidence toreject the claim that 65% of the teenagersown their own MP3 players.

13.H : p 0.204! œH : p 0.204 (claim)" Á

p 0.18 p 0.204 q 0.796^ œ œ œC. V. 1.96œ „

z 1.03œ œ œ s p p 0.18 0.204

É Épqn

(0.204)(0.796)300

1.96 0 1.96 Å 1.03

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the proportion of high school smokersdiffers from 20.4%.

14.H : p 0.205! œH : p 0.205 (claim)"

C. V. 1.28œs œ œ œp 0.3167 p 0.20538

120

q 0.795œ

z 3.03œ œ œs p p 0.3167 0.205

É Épqn

(0.205)(0.795)120

14. continued

0 1.28 Å 3.03

Reject the null hypothesis. There is enoughevidence to support the claim that thepercentage of men over the age of 65 stillworking is greater than 20.5%.

15.H : 4.3 (claim)! 5 œH : 4.3" 5

d. f. 19œ

;# œ œ œ(n 1)s (20 1)(2.6)(4.3

# #

# #56.95

0.005 < P-value < 0.01 (0.006)Since P-value < 0.05, reject the nullhypothesis. There is enough evidence toreject the claim that the standard deviation isgreater than or equal to 4.3 miles per gallon.

16.H : 3.810 5# œH : 3.81 (claim)1 5# Á

s (2.08) 4.3264# #œ œC. V. 5.629, 26.119 d. f. 14œ œ

;# œ œ œ(n 1)s (15 1)(4.3264)3.81

#

#515.898

5.629 26.119Å 15.898

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the variance of admission prices differsfrom 3.81.

17.H : 40!

#5 œH : 40 (claim)"

#5 Á


161

17. continueds 6.6 s (6.6) 43.56œ œ œ# #

C. V. 2.700, 19.023 d. f. 9œ œ

;# œ œ œ(n 1)s (10 1)(43.56)40

#

#59.801

2.700 19.023Å 9.801

Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the variance differs from 40.

18.H : 3.4 (claim)! 5 œH : 3.4" 5 Á

C. V. 11.689, 38.076 d. f. 23œ œ

;# œ œ œ(n 1)s (24 1)(4.2)(3.4)

# #

# #535.097

11.689 38.076Å 35.097

Do not reject the null hypothesis. There isnot enough evidence to reject the claim thatthe standard deviation is 3.4 minutes.

19.H : 4! . œH : 4 (claim)" . Á

C. V. 2.58œ „

z 1.49œ œ œX 4.2 4 .sn

0.6

20È È

The 99% confidence interval of the mean is:

X z X z ! !2 2

5 5È Èn n

.

4.2 2.58 4.2 2.58 † †0.6 0.6

20 20È È.

3.85 4.55 .

19. continuedThe decision is do not reject the nullhypothesis since 1.49 2.58 and theconfidence interval does contain thehypothesized mean of 4. There is notenough evidence to support the claim thatthe growth has changed. Yes, the resultsagree. The hypothesized mean is containedin the interval.

20.H : 35 (claim)! . œH : 35" . Á

C. V. 1.65œ „

z 3.00œ œ œ X 33.5 35 .sn

3

36È È

The 90% confidence interval of the mean is:X z X z ! !

2 2

5 5È Èn n

.

33.5 1.65 33.5 1.65 † †3 3

36 36È È.

32.675 34.325 .

The decision is to reject the null hypothesissince 3.00 1.65 and the 90% confidence interval does not contain thehypothesized mean of 35. The conclusion isthat there is enough evidence to reject theclaim that the mean is 35 pounds.

CHAPTER 8 QUIZ

1. True2. True3. False, the critical value separates thecritical region from the noncritical region.4. True5. False, it can be one-tailed or two-tailed.6. b7. d8. c9. b10. Type I11. "12. Statistical hypothesis13. Right14. n 1

15. H : 28.6 (claim)! . œH : 28.6" . ÁC. V. 1.96œ „z 2.15œ


162

15. continuedReject the null hypothesis. There is enoughevidence to reject the claim that the averageage is 28.6.

16. H : $6,500 (claim)! . œH : $6,500" . ÁC. V. 1.96œ „z 5.27œReject the null hypothesis. There is enoughevidence to reject the agent's claim.

17. H : 8! . œH : 8 (claim)" . C. V. 1.65œz 6.00œReject the null hypothesis. There is enoughevidence to support the claim that theaverage is greater than 8.

18. H : 500 (claim)! . œH : 500" . ÁC. V. 3.707œ „t 0.571œ Do not reject the null hypothesis. There isnot enough evidence to reject the claim thatthe average is 500.

19. H : 67! . œH : 67 (claim)" . t 3.1568œ P-value < 0.005 (0.003)Since P-value < 0.05, reject the nullhypothesis. There is enough evidence tosupport the claim that the average height isless than 67 inches.

20. H : 12.4! . œH : 12.4 (claim)" . C. V. 1.345œ t 2.324œ Reject the null hypothesis. There is enoughevidence to support the claim that theaverage is less than what the companyclaimed.

21. H : 63.5! . œH : 63.5 (claim)" . t 0.47075œP-value > 0.25 (0.322)Since P-value > 0.05, do not reject the nullhypothesis. There is not enough evidence tosupport the claim that the average is greaterthan 63.5.

22. H : 26 (claim)! . œH : 26" . ÁC. V. 2.492œ „t 1.5œ Do not reject the null hypothesis. There isnot enough evidence to reject the claim thatthe average is 26.

23. H : p 0.39 (claim)! œH : p 0.39" ÁC. V. 1.96œ „z 0.62œ Do not reject the null hypothesis. There isnot enough evidence to reject the claim that39% took supplements. The study supportsthe results of the previous study.

24. H : p 0.55 (claim)! œH : p 0.55" C. V. 1.28œ z 0.8989œ Do not reject the null hypothesis. There isnot enough evidence to reject the survey'sclaim.

25. H : p 0.35 (claim)! œH : p 0.35" ÁC. V. 2.33œ „z 0.666œDo not reject the null hypothesis. There isnot enough evidence to reject the claim thatthe proportion is 35%.

26. H : p 0.75 (claim)! œH : p 0.75" ÁC. V. 2.58œ „z 2.6833œReject the null hypothesis. There is enoughevidence to reject the claim.

27. The area corresponding to z 2.15 isœ0.9842.P-value 2(1 0.9842) 0.0316œ œ

28. The area corresponding to z 5.27 isœgreater than 0.9999.Thus, P-value 2(1 0.9999) 0.0002.Ÿ Ÿ(TI: P-value 0.0001)

29. H : 6! 5 œH : 6 (claim)" 5 C. V. 36.415œ;# œ 54


163

29. continuedReject the null hypothesis. There is enoughevidence to support the claim that thestandard deviation is more than 6.

30. H : 8 (claim)! 5 œH : 8" 5 ÁC. V. 27.991, 79.490œ;# œ 33.2Do not reject the null hypothesis. There isnot enough evidence to reject the claim that5 œ 8.

31. H : 2.3! 5 œH : 2.3 (claim)" 5 C. V. 10.117œ;# œ 13Do not reject the null hypothesis. There isnot enough evidence to support the claimthat the standard deviation is less than 2.3.

32. H : 9 (claim)! 5 œH : 9" 5 Á;# œ 13.4P-value > 0.20 (0.291)Since P-value > 0.05, do not reject the nullhypothesis. There is not enough evidence toreject the claim that 9.5 œ

33. 28.9 31.2; no .

34. $6562.81 $6,637.19; no .

chapter 8 - hypothesis testingwebsites.rcc.edu/pell/files/2015/09/chapter-8-solutions.pdfchapter 8 -...

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