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
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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
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
9 - 8
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− −= = = −
Using t table, p-value is between .005 and .01
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− −= = =
Degrees of freedom = n – 1 = 64
Using t table, area in tail is between .01 and .025 p-value (two tail) is between .02 and .05
Actual p-value = .0394 p-value ≤ .05, reject H0
b. 0 96.5 100 2.57/ 11/ 65
xts n
µ− −= = = −
Using t table, area in tail is between .005 and .01 p-value (two tail) is between .01 and .02
Actual p-value = .0127 p-value ≤ .05, reject H0
c. 0 102 100 1.54/ 10.5 / 65
xts n
µ− −= = =
Using t table, area in tail is between .05 and .10 p-value (two tail) is between .10 and .20
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− −= = = −
Degrees of freedom = n – 1 = 99
Using t table, p-value is between .10 and .20
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− −= = =
Degrees of freedom = n – 1 = 91
Using t table, p-value is between .005 and .01
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− −= = =
Degrees of freedom = n – 1 = 24
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− −= = = −
Degrees of freedom = n – 1 = 49
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
Degrees of freedom = n – 1 = 8
Using t table, p-value is between .025 and .05
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
Using t table, area in tail is between .10 and .20 p-value is between .20 and .40
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
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
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
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.
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
Using t table, area in tail is between .10 and .20 p-value is between .20 and .40
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