ch9 solution

9 - 1

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 interpret p-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 of

hypothesis testing errors within acceptable limits. 7. Know the definition of the following terms: null hypothesis two-tailed test alternative hypothesis p-value Type I error level of significance Type II error critical value one-tailed test power curve

Chapter 9

9 - 2

Solutions: 1. a. H0: µ ≤ 600 Manager’s claim. Ha: µ > 600 b. We are not able to conclude that the manager’s claim is wrong. c. The manager’s claim can be rejected. We can conclude that µ > 600. 2. a. H0: µ ≤ 14 Ha: µ > 14 Research hypothesis b. There is no statistical evidence that the new bonus plan increases sales volume. c. The research hypothesis that µ > 14 is supported. We can conclude that the new bonus plan

increases the mean sales volume. 3. a. H0: µ = 32 Specified filling weight Ha: µ ≠ 32 Overfilling or underfilling exists b. There is no evidence that the production line is not operating properly. Allow the production

process to continue. c. Conclude µ ≠ 32 and that overfilling or underfilling exists. Shut down and adjust the production

line. 4. a. H0: µ ≥ 220 Ha: µ < 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 µ < 220. Consider implementing the new method based on the conclusion that it lowers

the mean cost per hour. 5. a. The Type I error is rejecting H0 when it is true. In this case, this error occurs if the researcher

concludes that the mean newspaper-reading time for individuals in management positions is greater than the national average of 8.6 minutes when in fact it is not.

b. The Type II error is accepting H0 when it is false. In this case, this error occurs if the researcher

concludes that the mean newspaper-reading time for individuals in management positions is less than or equal to the national average of 8.6 minutes when in fact it is greater than 8.6 minutes.

6. a. H0: µ ≤ 1 The label claim or assumption. Ha: µ > 1 b. Claiming µ > 1 when it is not. This is the error of rejecting the product’s claim when the claim is

true.

Hypothesis Testing

9 - 3

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

7. a. H0: µ ≤ 8000 Ha: µ > 8000 Research hypothesis to see if the plan increases average sales. b. Claiming µ > 8000 when the plan does not increase sales. A mistake could be implementing the

plan when it does not help. c. Concluding µ ≤ 8000 when the plan really would increase sales. This could lead to not

implementing a plan that would increase sales. 8. a. H0: µ ≥ 220 Ha: µ < 220 b. Claiming µ < 220 when the new method does not lower costs. A mistake could be implementing

the method when it does not help. c. Concluding µ ≥ 220 when the method really would lower costs. This could lead to not

implementing a method that would lower costs.

9. a. 0 19.4 20 2.12/ 2 / 50

xzn

µσ

− −= = = −

b. Area = .4830 p-value = .5000 - .4830 = .0170 c. p-value ≤ .05, reject H0 d. Reject H0 if z ≤ -1.645 -2.12 ≤ -1.645, reject H0

10. a. 0 26.4 25 1.48/ 6 / 40

xzn

µσ

− −= = =

b. Area = .4306 p-value = .5000 - .4306 = .0694 c. p-value > .01, do not reject H0 d. Reject H0 if z ≥ 2.33 1.48 < 2.33, do not reject H0

11. a. 0 14.15 15 2.00/ 3/ 50

xzn

µσ

− −= = = −

b. Area = .4772

Chapter 9

9 - 4

p-value = 2(.5000 - .4772) = .0456 c. p-value ≤ .05, reject H0 d. Reject H0 if z ≤ -1.96 or z ≥ 1.96 -2.00 ≤ -1.96, reject H0

12. a. 0 78.5 80 1.25/ 12 / 100

xzn

µσ

− −= = = −

p-value = .5000 - .3944 = .1056 p-value > .01, do not reject H0

b. 0 77 80 2.50/ 12 / 100

xzn

µσ

− −= = = −

p-value = .5000 - .4938 = .0062 p-value ≤ .01, reject H0

c. 0 75.5 80 3.75/ 12 / 100

xzn

µσ

− −= = = −

p-value ≈ 0 p-value ≤ .01, reject H0

d. 0 81 80 .83/ 12 / 100

xzn

µσ

− −= = =

Area to left of z = .83 p-value = .5000 + .2967 = .7967 p-value > .01, do not reject H0 13. Reject H0 if z ≥ 1.645

a. 0 52.5 50 2.42/ 8 / 60

xzn

µσ

− −= = =

2.42 ≥ 1.645, reject H0

b. 0 51 50 .97/ 8 / 60

xzn

µσ

− −= = =

.97 < 1.645, do not reject H0

Hypothesis Testing

9 - 5

c. 0 51.8 50 1.74/ 8 / 60

xzn

µσ

− −= = =

1.74 ≥ 1.645, reject H0

14. a. 0 23 22 .87/ 10 / 75

xzn

µσ

− −= = =

p-value = 2(.5000 - .3078) = .3844 p-value > .01, do not reject H0

b. 0 25.1 22 2.68/ 10 / 75

xzn

µσ

− −= = =

p-value = 2(.5000 - .4963) = .0074 p-value ≤ .01, reject H0

c. 0 20 22 1.73/ 10 / 75

xzn

µσ

− −= = = −

p-value = 2(.5000 - .4582) = .0836 p-value > .01, do not reject H0 15. a. H0: µ ≥ 1056 Ha: µ < 1056

b. 0 910 1056 1.83/ 1600 / 400

xzn

µσ

− −= = = −

p-value = .5000 - .4664 = .0336 c. p-value ≤ .05, reject H0. Conclude the mean refund of “last minute” filers is less than $1056. d. Reject H0 if z ≤ -1.645 -1.83 ≤ -1.645, reject H0 16. a. H0: µ ≤ 895 Ha: µ > 895

b. 0 915 895 1.19/ 225 / 180

xzn

µσ

− −= = =

Area = .3830

Chapter 9

9 - 6

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 before

drawing a final conclusion. 17. a. H0: µ = 39.2 Ha: µ ≠ 39.2

b. 0 38.5 39.2 1.54/ 4.8 / 112

xzn

µσ

− −= = = −

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 has

changed. d. Reject H0 if z ≤ -1.96 or z ≥ 1.96 z = -1.54; cannot reject H0 18. a. H0: µ = 26,133 Ha: µ ≠ 26,133

b. 0 25, 457 26,133 2.09/ 7600 / 550

xzn

µσ

− −= = = −

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

c. p-value ≤ .05. Reject H0; Collier County has a mean annual wage different from the state mean annual wage.

19. H0: µ ≤ 14.32 Ha: µ > 14.32

0 14.68 14.32 2.15/ 1.45 / 75

xzn

µσ

− −= = =

Area = .4842 p-value = .5000 - .4842 = .0158 p-value ≤ .05, reject H0. Conclude that there has been an increase in the mean hourly wage of

production workers. 20. a. H0: µ ≥ 181,900

Hypothesis Testing

9 - 7

Ha: µ < 181,900

b. 166,400 181,900 2.93/ 33,500 / 40

xzn

µσ

− −= = = −

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: µ ≤ 15 Ha: µ > 15

b. 17 15 2.96/ 4 / 35

xznµ

σ− −

= = =

c. p-value = .5000 - .4985 = .0015 d. p-value ≤ .01; reject H0; the premium rate should be charged. 22. a. H0: µ = 8 Ha: µ ≠ 8

b. 0 8.5 8 1.71/ 3.2 / 120

xz

nµ

σ− −

= = =

p-value = 2(.5000 - .4564) = .0872

c. Do not reject H0. Cannot conclude that the population mean waiting time differs from 8 minutes. d. .025 ( / )x z nσ± 8.5 ± 1.96 (3.2 / 120) 8.5 ± .57 (7.93 to 9.07) Yes; µ = 8 is in the interval. Do not reject H0.

23. a. 0 14 12 2.31/ 4.32 / 25

xts n

µ− −= = =

b. Degrees of freedom = n – 1 = 24 Using t table, p-value is between .01 and .025. Actual p-value = .0147 c. p-value ≤ .05, reject H0.

Chapter 9

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d. With df = 24, t.05 = 1.711 Reject H0 if t ≥ 1.711 2.31 > 1.711, reject H0.

24. a. 0 17 18 1.54/ 4.5 / 48

xts n

µ− −= = = −

b. Degrees of freedom = n – 1 = 47

Area in lower tail is between .05 and .10

Using t table, p-value (two-tail) is between .10 and .20 Actual p-value = .1304

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

d. With df = 47, t.025 = 2.012 Reject H0 if t ≤ -2.012 or t ≥ 2.012 t = -1.54; do not reject H0

25. a. 0 44 45 1.15/ 5.2 / 36

xts n

µ− −= = = −

Degrees of freedom = n – 1 = 35

Using t table, p-value is between .10 and .20

Actual p-value = .1282 p-value > .01, do not reject H0

b. 0 43 45 2.61/ 4.6 / 36

xts n

µ− −= = = −


Actual p-value = .0066 p-value ≤ .01, reject H0

c. 0 46 45 1.20/ 5 / 36

xts n

µ− −= = =

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

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

Hypothesis Testing

9 - 9

p-value > .01, do not reject H0

26. a. 0 103 100 2.10/ 11.5 / 65

xts n

µ− −= = =


Using t table, 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

xts n

µ− −= = = −



c. 0 102 100 1.54/ 10.5 / 65

xts n

µ− −= = =


Actual p-value = .1295 p-value > .05, do not reject H0 27. a. H0: µ ≥ 238 Ha: µ < 238

b. 0 231 238 .88/ 80 / 100

xts n

µ− −= = = −



Actual p-value = .1918 c. p-value > .05; do not reject H0. Cannot conclude mean weekly benefit in Virginia is less than the

national mean.

Chapter 9

9 - 10

d. df = 99 t.05 = -1.66 Reject H0 if t ≤ -1.66 -.88 > -1.66; do not reject H0 28. a. H0: µ ≤ 3530 Ha: µ > 3530

b. 0 3740 3530 2.49/ 810 / 92

xts n

µ− −= = =



Actual p-value = .0074 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: µ = 5600 Ha: µ ≠ 5600

b. 0 5835 5600 2.26/ 520 / 25

xts n

µ− −= = =


Using t table, area in tail is between .01 and .025

Actual p-value = .0332 c. p-value ≤ .05; reject H0. The mean diamond price in New York City differs. d. df = 24 t.025 = 2.064 Reject H0 if t < -2.064 or t > 2.064 2.26 > 2.064; reject H0 30. a. H0: µ = 600 Ha: µ ≠ 600

b. 0 612 600 1.17/ 65 / 40

xts n

µ− −= = =

df = n - 1 = 39

Hypothesis Testing

9 - 11

Using t table, area in tail is between .10 and .20 p-value is between .20 and .40

Actual p-value = .2501 c. Withα = .10 or less, we cannot reject H0. We are unable to conclude there has been a change in the

mean CNN viewing audience. d. The sample mean of 612 thousand viewers is encouraging but not conclusive for the sample of 40

days. Recommend additional viewer audience data. A larger sample should help clarify the situation for CNN.

31. H0: µ ≤ 47.50 Ha: µ > 47.50

0 51 47.50 2.33/ 12 / 64

xts n

µ− −= = =

Degrees of freedom = n - 1 = 63

Using t table, p-value is between .01 and .025 Actual p-value = .0114

Reject H0; Atlanta customers are paying a higher mean water bill. 32. a. H0: µ = 10,192 Ha: µ ≠ 10,192

b. 0 9750 10,192 2.23/ 1400 / 50

xts n

µ− −= = = −


Using t table, area in tail is between .01 and .025

p-value is between .02 and .05 Actual p-value = .0302 c. p-value ≤ .05; reject H0. The population mean price at this dealership differs from the national mean

price $10,192. 33. a. H0: µ ≤ 280 Ha: µ > 280 b. 286.9 - 280 = 6.9 yards

c. 0 286.9 280 2.07/ 10 / 9

xts n

µ− −= = =

Chapter 9

9 - 12



Actual p-value = .0361 c. p-value ≤ .05; reject H0. The population mean distance for the new driver is greater than the USGA

approved driver. 34. a. H0: µ = 2 Ha: µ ≠ 2

b. 22 2.210

ixx

nΣ

= = =

c. ( )2

.5161

ix xs

nΣ −

= =−

d. 0 2.2 2 1.22/ .516 / 10

xts n

µ− −= = =

Degrees of freedom = n - 1 = 9


Actual p-value = .2518 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 pzp 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 H0 if z ≤ -1.96 or z ≥ 1.96 z = − 1.25; do not reject H0

Hypothesis Testing

9 - 13

36. a. 0

0 0

.68 .75 2.80(1 ) .75(1 .75)

300

p pzp p

n

− −= = = −

− −

p-value = .5000 - .4974 = .0026 p-value ≤ .05; reject H0

b. .72 .75 1.20.75(1 .75)

300

z −= = −

−

p-value = .5000 - .3849 = .1151 p-value > .05; do not reject H0

c. .70 .75 2.00.75(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 40 1.88(1 ) .40(1 .40)

425

p pzp p

n

− −= = =

− −

Area = .4699 p-value = .5000 - .4699 = .0301 c. p-value ≤ .05; reject H0. Conclude more than 40% receive more than 10 e-mail messages per day.

Chapter 9

9 - 14

38. a. H0: p = .64 Ha: p ≠ .64

b. 52 .52100

p = =

0

0 0

.52 .64 2.50(1 ) .64(1 .64)

100

p pzp 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. Wisconsin 252 .72350

p = =

0

0 0

.72 .70 .82(1 ) .70(1 .70)

350

p pzp p

n

− −= = =

− −

Area = .2939 p-value = 2(.5000 - .2939) = .4122 Cannot reject H0.

California 189 .63300

p = =

.63 .70 2.65.70(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.

Hypothesis Testing

9 - 15

40. a. 414 .27021532

p = = (27%)

b. H0: p ≤ .22 Ha: p > .22

0

0 0

.2702 .22 4.75(1 ) .22(1 .22)

1532

p pzp 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 .75 1.20(1 ) .75(1 .75)

300

p pzp 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 ≤ .24 Ha: p > .24

93 .31300

p = =

0

0 0

.31 .24 2.84(1 ) .24(1 .24)

300

p pzp p

n

− −= = =

− −

Area = .4977 p-value = .5000 - .4977 = .0023 p-value ≤ .05; reject H0. In 2003, an estimated 31% of people who moved selected to be convenient

to work as their primary reason. This is an increase compared to 1990.

Chapter 9

9 - 16

43. a. H0: p = .48 Ha: p ≠ .48

b. 360 .45800

p = =

0

0 0

.45 .48 1.70(1 ) .48(1 .48)

800

p pzp 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 .50 3.13(1 ) .50(1 .50)

500

p pzp p

n

− −= = =

− −

p-value ≈ 0 c. p-value ≤ .01; reject H0. Conclude Burger King fries are preferred by more than .50 of the

population. d. Yes; statistical evidence shows Burger King fries are preferred. The give-away was a good way to

get 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 pzp p

n

− −= = = −

− −

Hypothesis Testing

9 - 17

Area = .4115 p-value = 2(.5000 - .4115) = .1770 p-value > .05; do not reject H0. No change.

c. 245 .49500

p = =

.49 .44 2.25.44(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 .46120x n

σσ = = =

x

10

.05

c

H0: µ ≥ 10Ha: µ < 10

c = 10 - 1.645 (5 / 120 ) = 9.25 Reject H0 if x ≤ 9.25 a. When µ = 9,

9.25 9 .555 / 120

z −= =

Prob (H0) = (.5000 - .2088) = .2912 b. Type II error

Chapter 9

9 - 18

c. When µ = 8,

9.25 8 2.745 / 120

z −= =

β = (.5000 - .4969) = .0031 47. Reject H0 if z ≤ -1.96 or if z ≥ 1.96

10 .71200x n

σσ = = =

xxc1 c220

Ha: µ ≠ 20 Ha: µ ≠ 20H0: µ = 20

.025 .025

c1 = 20 - 1.96 (10 / 200 ) = 18.61 c2 = 20 + 1.96 (10 / 200 ) = 21.39 a. µ = 18

18.61 18 .8610 / 200

z −= =

β = .5000 - .3051 = .1949 b. µ = 22.5

21.39 22.5 1.5710 / 200

z −= = −

β = .5000 - .4418 = .0582

Hypothesis Testing

9 - 19

c. µ = 21

21.39 21 .5510 / 200

z −= =

β = .5000 +.2088 = .7088 48. a. H0: µ ≤ 15 Ha: µ > 15 Concluding µ ≤ 15 when this is not true. Fowle would not charge the premium rate even though

the rate should be charged. b. Reject H0 if z ≥ 2.33

0 15 2.33/ 4 / 35

x xzn

µσ

− −= = =

Solve for x = 16.58 Decision Rule: Accept H0 if x < 16.58 Reject H0 if x ≥ 16.58 For µ = 17,

16.58 17 .624 / 35

z −= = −

β = .5000 -.2324 = .2676 c. For µ = 18,

16.58 18 2.104 / 35

z −= = −

β = .5000 -.4821 = .0179 49. a. H0: µ ≥ 25 Ha: µ < 25 Reject H0 if z ≤ -2.05

0 25 2.05/ 3 / 30

x xzn

µσ

− −= = = −

Solve for x = 23.88 Decision Rule:

Chapter 9

9 - 20

Accept H0 if x > 23.88 Reject H0 if x ≤ 23.88 b. For µ = 23,

23.88 23 1.613/ 30

z −= =

β = .5000 -.4463 = .0537 c. For µ = 24,

23.88 24 .223/ 30

z −= = −

β = .5000 +.0871 = .5871 d. The Type II error cannot be made in this case. Note that when µ = 25.5, H0 is true. The Type II

error can only be made when H0 is false. 50. a. Accepting H0 and concluding the mean average age was 28 years when it was not. b. Reject H0 if z ≤ -1.96 or if z ≥ 1.96

0 28/ 6 / 100

x xzn

µσ

− −= =

Solving for x , we find at z = -1.96, x = 26.82 at z = +1.96, x = 29.18 Decision Rule: Accept H0 if 26.82 < x < 29.18 Reject H0 if x ≤ 26.82 or if x ≥ 29.18 At µ = 26,

26.82 26 1.376 / 100

z −= =

β = .5000 +.4147 = .0853 At µ = 27,

26.82 27 .306 / 100

z −= = −

Hypothesis Testing

9 - 21

β = .5000 +.1179 = .6179 At µ = 29,

29.18 29 .306 / 100

z −= =

β = .5000 +.1179 = .6179 At µ = 30,

29.18 30 1.376 / 100

z −= = −

β = .5000 -.4147 = .0853

c. Power = 1 - β at µ = 26, Power = 1 - .0853 = .9147 When µ = 26, there is a .9147 probability that the test will correctly reject the null hypothesis that

µ = 28. 51. a. Accepting H0 and letting the process continue to run when actually over - filling or under - filling

exists. b. Decision Rule: Reject H0 if z ≤ -1.96 or if z ≥ 1.96 indicates Accept H0 if 15.71 < x < 16.29 Reject H0 if x ≤ 15.71 or if x ≥ 16.29 For µ = 16.5

16.29 16.5 1.44.8 / 30

z −= = −

β = .5000 -.4251 = .0749

Chapter 9

9 - 22

x

16.516.29

c

β

c. Power = 1 - .0749 = .9251 d. The power curve shows the probability of rejecting H0 for various possible values of µ. In

particular, it shows the probability of stopping and adjusting the machine under a variety of underfilling 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 Values of u

Power

52. 0 .01415 2.33 16.3250

c zn

σµ= + = + =

At µ = 17 16.32 17 1.20

4 / 50z −

= = −

β = .5000 - .3849 = .1151

At µ = 18 16.32 18 2.97

4 / 50z −

= = −

β = .5000 - .4985 = .0015 Increasing the sample size reduces the probability of making a Type II error.

Hypothesis Testing

9 - 23

53. a. Accept µ ≤ 100 when it is false. b. Critical value for test:

0 .0575100 1.645 119.5140

c zn

σµ= + = + =

At µ = 120 119.51 120 .04

75 / 40z −

= = −

β = .5000 - .0160 = .4840

c. At µ = 13 119.51 130 .88

75 / 40z −

= = −

β = .5000 - .3106 = .1894 d. Critical value for test:

0 .0575100 1.645 113.7980

c zn

σµ= + = + =

At µ = 120 113.79 120 .74

75 / 80z −

= = −

β = .5000 - .2704 = .2296

At µ = 130 113.79 130 1.93

75 / 80z −

= = −

β = .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 20

( ) (1.645 1.28) (5) 214( ) (10 9)a

z zn α β σ

µ µ+ +

= = =− −

55. 2 2 2 2

2 20

( ) (1.96 1.645) (10) 325( ) (20 22)a

z zn α β σ

µ µ+ +

= = =− −

56. At µ0 = 3, α = .01. z.01 = 2.33 At µa = 2.9375, β = .10. z.10 = 1.28 σ = .18

2 2 2 2

2 20

( ) (2.33 1.28) (.18) 108.09( ) (3 2.9375)a

z zn α β σ

µ µ+ +

= = =− −

Use 109

Chapter 9

9 - 24

57. At µ0 = 400, α = .02. z.02 = 2.05 At µa = 385, β = .10. z.10 = 1.28 σ = 30

2 2 2 2

2 20

( ) (2.05 1.28) (30) 44.4( ) (400 385)a

z zn α β σ

µ µ+ +

= = =− −

Use 45

58. At µ0 = 28, α = .05. Note however for this two - tailed test, zα / 2 = z.025 = 1.96 At µa = 29, β = .15. z.15 = 1.04 σ = 6

2 2 2 2

/ 22 2

0

( ) (1.96 1.04) (6) 324( ) (28 29)a

z zn α β σ

µ µ+ +

= = =− −

59. At µ0 = 25, α = .02. z.02 = 2.05 At µa = 24, β = .20. z.20 = .84 σ = 3

2 2 2 2

2 20

( ) (2.05 .84) (3) 75.2( ) (25 24)a

z zn α β σ

µ µ+ +

= = =− −

Use 76

60. a. H0: µ = 16 Ha: µ ≠ 16

b. 0 16.32 16 2.19/ .8 / 30

xzn

µσ

− −= = =

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

xzn

µσ

− −= = = −

Area = .3907 p-value = 2(.5000 - .3907) = .2186 p-value > .05; do not reject H0. Continue the production line.

Hypothesis Testing

9 - 25

d. Reject H0 if z ≤ -1.96 or z ≥ 1.96 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: µ = 900 Ha: µ ≠ 900

b. .025x zn

σ±

180935 1.96200

±

935 ± 25 (910 to 960) c. Reject H0 because µ = 900 is not in the interval.

d. 0 935 900 2.75/ 180 / 200

xzn

µσ

− −= = =

p-value = 2(.5000 - .4970) = .0060 62. a. H0: µ ≤ 45,250 Ha: µ > 45,250

b. 0 47,000 45, 250 2.71/ 6300 / 95

xzn

µσ

− −= = =

p-value = .5000 - .4966 = .0034 p-value ≤ .01; reject H0. Conclude New York City has higher mean salary. 63. a. H0: µ ≤ 37,000 Ha: µ > 37,000

b. 0 38,000 37,000 1.47/ 5200 / 48

xts n

µ− −= = =

Degrees of freedom = n – 1 = 47 Using t table, p-value is between .05 and .10 Actual p-value = .0747 c. p-value > .05, do not reject H0. Cannot conclude mean greater than $37,000. A larger sample is

desirable.

Chapter 9

9 - 26

64. H0: µ = 6000 Ha: µ ≠ 6000

0 5812 6000 .93/ 1140 / 32

xts n

µ− −= = = −

Degrees of freedom = n – 1 = 31 Using t table, area in tail is between .10 and .20 p-value is between .20 and .40 Actual p-value = .3581 Do not reject H0. There is no evidence to conclude that the mean number of freshman applications

has changed. 65. a. H0: µ ≥ 30 Ha: µ < 30

b. 0 29.6 30 1.57/ 1.8 / 50

xts n

µ− −= = = −

Degrees of freedom = 50 – 1 = 49 Using t table, p-value is between .05 and .10 Actual p-value = .0613 c. p-value > .01; do not reject H0. Cannot conclude miles per gallon is less than 30. 66. H0: µ ≤ 125,000 Ha: µ > 125,000

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

xts n

µ− −= = =

Degrees of freedom = 32 – 1 = 31 Using t table, p-value is between .01 and .025 Actual p-value = .0154 p-value ≤ .05; reject H0. Conclude that the mean cost is greater than $125,000 per lot. 67. a. H0: µ = 3 Ha: µ ≠ 3

Hypothesis Testing

9 - 27

b. 2.8ixx

nΣ

= =

c. ( )2

.701

ix xs

nΣ −

= =−

d. 0 2.8 3 .90/ .70 / 10

xts n

µ− −= = = −

Degrees of freedom = 10 - 1 = 9


Actual p-value = .3902 e. p-value > .05; do not reject H0. There is no evidence to conclude a difference compared to prior

year. 68. a. H0: p ≤ .50 Ha: p > .50

b. 64 .64100

p = =

c. 0

0 0

.64 .50 2.80(1 ) .50(1 .50)

100

p pzp 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 pzp p

n

− −= = = −

− −

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

Chapter 9

9 - 28

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

b. 67 .6381105

p = = (64%)

c. 0

0 0

.6381 .50 2.83(1 ) .50(1 .50)

105

p pzp p

n

− −= = =

− −

p-value = .5000 - .4977 = .0023 p-value ≤ .01; reject H0. Conclude that the four 10-hour day schedules is preferred by more than

50% of the office workers.

71. a. 330 .825400

p = =

b. H0: p = .78 Ha: p ≠ .78

0

0 0

.825 .78 2.17(1 ) .78(1 .78)

400

p pzp 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 .844858

p = =

0

0 0

.8448 .90 1.40(1 ) .90(1 .90)

58

p pzp 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

Hypothesis Testing

9 - 29

b. 44 .352125

p = =

c. 0

0 0

.352 .47 2.64(1 ) .47(1 .47)

125

p pzp p

n

− −= = = −

− −

p-value = .5000 - .4959 = .0041 d. p-value ≤ .01; reject H0. The proportion of foods containing pesticides has declined. 74. a. H0: µ ≤ 72 Ha: µ > 72 Reject H0 if z ≥ 1.645

0 72 1.645/ 20 / 30

x xzn

µσ

− −= = =

Solve for x = 78 Decision Rule: Accept H0 if x < 78 Reject H0 if x ≥ 78 b. For µ = 80

78 80 .5520 / 30

z −= = −

β = .5000 -.2088 = .2912 c. For µ = 75,

78 75 .8220 / 30

z −= =

β = .5000 +.2939 = .7939 d. For µ = 70, H0 is true. In this case the Type II error cannot be made. e. Power = 1 - β

Chapter 9

9 - 30

.2

.4

.6

.8

1.0

72 74 76 78 80 82 84

Ho FalsePossible Values of µ

Power

75. H0: µ ≥ 15,000 Ha: µ < 15,000 At µ0 = 15,000, α = .02. z.02 = 2.05 At µa = 14,000, β = .05. z.10 = 1.645

2 2 2 2

2 20

( ) (2.05 1.645) (4,000) 218.5( ) (15,000 14,000)a

z zn α β σ

µ µ+ +

= = =− −

Use 219

76. H0: µ = 120 Ha: µ ≠ 120 At µ0 = 120, α = .05. With a two - tailed test, zα / 2 = z.025 = 1.96 At µa = 117, β = .02. z.02 = 2.05

2 2 2 2

/ 22 2

0

( ) (1.96 2.05) (5) 44.7( ) (120 117)a

z zn α β σ

µ µ+ +

= = =− −

Use 45

b. Example calculation for µ = 118. Reject H0 if z ≤ -1.96 or if z ≥ 1.96

0 120/ 5 / 45

x xzn

µσ

− −= =

Solve for x . At z = -1.96, x = 118.54 At z = +1.96, x = 121.46

Hypothesis Testing

9 - 31

Decision Rule: Accept H0 if 118.54 < x < 121.46 Reject H0 if x ≤ 118.54 or if x ≥ 121.46 For µ = 118,

118.54 118 .725 / 45

z −= =

β = .5000 +.2642 = .2358 Other Results:

If µ is z β 117 2.07 .0192 118 .72 .2358 119 -.62 .7291 121 +.62 .7291 122 +.72 .2358 123 -2.07 .0192

ch9 solution

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