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
alternative hypothesis
Type I error
Type II error
one-tailed test
two-tailed test
p-value
level of significance
critical value
power curve
9-1
Chapter 9
Solutions:
1.
a.
H0: µ ≤ 600
Manager’s claim.
Ha: µ > 600
2.
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.
a.
H0: µ ≤ 14
Ha: µ > 14
3.
4.
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.
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.
a.
H0: µ ≥ 220
Ha: µ < 220
5.
Research hypothesis
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.
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.
H a: 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.
9-2
Hypothesis Testing
c.
7.
a.
Concluding µ ≤ 1 when it is not. In this case, we miss the fact that the product is not meeting its
label specification.
H0: µ ≤ 8000
Ha: µ > 8000
8.
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.
a.
H0: µ ≥ 220
Ha: µ < 220
9.
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.
a.
z=
b.
Area = .4830
x − µ0
σ/ n
=
19.4 − 20
2 / 50
= −2.12
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.
b.
z=
x − µ0
σ/ n
=
26.4 − 25
6 / 40
= 1.48
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.
b.
z=
x − µ0
σ/ n
=
14.15 − 15
3 / 50
= −2.00
Area = .4772
9-3
Chapter 9
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.
z=
x − µ0
σ/ n
=
78.5 − 80
12 / 100
= −1.25
p-value = .5000 - .3944 = .1056
p-value > .01, do not reject H0
b.
z=
x − µ0
σ/ n
=
77 − 80
12 / 100
= −2.50
p-value = .5000 - .4938 = .0062
p-value ≤ .01, reject H0
c.
z=
x − µ0
σ/ n
=
75.5 − 80
12 / 100
= −3.75
p-value ≈ 0
p-value ≤ .01, reject H0
d.
z=
x − µ0
σ/ n
=
81 − 80
12 / 100
= .83
Area to left of z = .83
p-value = .5000 + .2967 = .7967
p-value > .01, do not reject H0
Reject H0 if z ≥ 1.645
13.
a.
z=
x − µ0
σ/ n
=
52.5 − 50
8 / 60
= 2.42
2.42 ≥ 1.645, reject H0
b.
z=
x − µ0
σ/ n
=
51 − 50
8 / 60
= .97
.97 < 1.645, do not reject H0
9-4
Hypothesis Testing
c.
z=
x − µ0
σ/ n
=
51.8 − 50
= 1.74
8 / 60
1.74 ≥ 1.645, reject H0
14. a.
z=
x − µ0
σ/ n
=
23 − 22
10 / 75
= .87
p-value = 2(.5000 - .3078) = .3844
p-value > .01, do not reject H0
b.
z=
x − µ0
σ/ n
=
25.1 − 22
= 2.68
10 / 75
p-value = 2(.5000 - .4963) = .0074
p-value ≤ .01, reject H0
c.
z=
x − µ0
σ/ n
=
20 − 22
10 / 75
= −1.73
p-value = 2(.5000 - .4582) = .0836
p-value > .01, do not reject H0
15. a.
H0: µ ≥ 1056
Ha: µ < 1056
b.
z=
x − µ0
σ/ n
=
910 − 1056
1600 / 400
= −1.83
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.
z=
x − µ0
σ/ n
=
915 − 895
225 / 180
= 1.19
Area = .3830
9-5
Chapter 9
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.
z=
x − µ 0 38.5 − 39.2
=
= −1.54
σ / n 4.8 / 112
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.
z=
x − µ0
σ/ n
25, 457 − 26,133
=
7600 / 550
= −2.09
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
z=
x − µ0
σ/ n
=
14.68 − 14.32
1.45 / 75
= 2.15
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
9-6
Hypothesis Testing
Ha: µ < 181,900
x −µ
z=
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.
=
166, 400 − 181,900
b.
σ/ n
33,500 / 40
= −2.93
H0: µ ≤ 15
Ha: µ > 15
x −µ
z=
c.
p-value = .5000 - .4985 = .0015
d.
p-value ≤ .01; reject H0; the premium rate should be charged.
22. a.
σ/ n
=
17 − 15
b.
4 / 35
= 2.96
H0: µ = 8
H a: µ ≠ 8
b.
z=
x − µ0
σ/ n
=
8.5 − 8
3.2 / 120
= 1.71
p-value = 2(.5000 - .4564) = .0872
c.
Do not reject H0. Cannot conclude that the population mean waiting time differs from 8 minutes.
d.
x ± z.025 (σ / n )
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.
b.
t=
x − µ0
s/ n
=
14 − 12
4.32 / 25
= 2.31
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.
9-7
Chapter 9
d.
With df = 24, t.05 = 1.711
Reject H0 if t ≥ 1.711
2.31 > 1.711, reject H0.
24. a.
b.
t=
x − µ0
s/ n
=
17 − 18
4.5 / 48
= −1.54
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.
t=
x − µ0
s/ n
=
44 − 45
5.2 / 36
= −1.15
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.
t=
x − µ0
s/ n
=
43 − 45
4.6 / 36
= −2.61
Using t table, p-value is between .005 and .01
Actual p-value = .0066
p-value ≤ .01, reject H0
c.
t=
x − µ0
s/ n
=
46 − 45
5 / 36
= 1.20
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
9-8
Hypothesis Testing
p-value > .01, do not reject H0
26. a.
t=
x − µ0
s/ n
=
103 − 100
11.5 / 65
= 2.10
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.
t=
x − µ0
s/ n
=
96.5 − 100
11/ 65
= −2.57
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.
t=
x − µ0
s/ n
=
102 − 100
10.5 / 65
= 1.54
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.
t=
x − µ0
s/ n
=
231 − 238
80 / 100
= −.88
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.
9-9
Chapter 9
d.
t.05 = -1.66
df = 99
Reject H0 if t ≤ -1.66
-.88 > -1.66; do not reject H0
28. a.
H0: µ ≤ 3530
Ha: µ > 3530
b.
t=
x − µ0
s/ n
=
3740 − 3530
810 / 92
= 2.49
Degrees of freedom = n – 1 = 91
Using t table, p-value is between .005 and .01
Actual p-value = .0074
c.
29. a.
p-value ≤ .01; reject H0. The mean attendance per game has increased. Anticipate a new all-time
high season attendance during the 2002 season.
H0: µ = 5600
Ha: µ ≠ 5600
b.
t=
x − µ0
s/ n
=
5835 − 5600
520 / 25
= 2.26
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.
t=
x − µ0
s/ n
=
612 − 600
65 / 40
= 1.17
df = n - 1 = 39
9 - 10
Hypothesis Testing
Using t table, area in tail is between .10 and .20
p-value is between .20 and .40
Actual p-value = .2501
c.
With α = .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.
H0: µ ≤ 47.50
31.
Ha: µ > 47.50
t=
x − µ0
s/ n
=
51 − 47.50
12 / 64
= 2.33
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.
t=
x − µ0
s/ n
=
9750 − 10,192
1400 / 50
= −2.23
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.
33. a.
p-value ≤ .05; reject H0. The population mean price at this dealership differs from the national mean
price $10,192.
H0: µ ≤ 280
Ha: µ > 280
b.
286.9 - 280 = 6.9 yards
c.
t=
x − µ0
s/ n
=
286.9 − 280
10 / 9
= 2.07
9 - 11
Chapter 9
Degrees of freedom = n – 1 = 8
Using t table, p-value is between .025 and .05
Actual p-value = .0361
c.
34. a.
p-value ≤ .05; reject H0. The population mean distance for the new driver is greater than the USGA
approved driver.
H0: µ = 2
H a: µ ≠ 2
b.
x=
c.
s=
d.
t=
Σxi 22
=
= 2.2
10
n
Σ ( xi − x )
2
n −1
x − µ0
s/ n
=
= .516
2.2 − 2
.516 / 10
= 1.22
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.
35. a.
b.
p-value > .05; do not reject H0. No reason to change from the 2 hours for cost estimating purposes.
z=
p − p0
p0 (1 − p0 )
n
=
.175 − .20
.20(1 − .20)
400
= −1.25
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
9 - 12
Hypothesis Testing
36. a.
z=
p − p0
p0 (1 − p0 )
n
=
.68 − .75
.75(1 − .75)
300
= −2.80
p-value = .5000 - .4974 = .0026
p-value ≤ .05; reject H0
b.
z=
.72 − .75
.75(1 − .75)
300
= −1.20
p-value = .5000 - .3849 = .1151
p-value > .05; do not reject H0
c.
z=
.70 − .75
.75(1 − .75)
300
= −2.00
p-value = .5000 - .4772 = .0228
p-value ≤ .05; reject H0
d.
z=
.77 − .75
.75(1 − .75)
300
= .80
p-value = .5000 + .2881 = .7881
p-value > .05; do not reject H0
37. a.
H0: p ≤ .40
Ha: p > .40
b.
p=
z=
189
= .45
425
p − p0
p0 (1 − p0 )
n
=
.45 − 40
.40(1 − .40)
425
= 1.88
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.
9 - 13
Chapter 9
38. a.
H0: p = .64
Ha: p ≠ .64
b.
p=
z=
52
= .52
100
p − p0
=
p0 (1 − p0 )
n
.52 − .64
.64(1 − .64)
100
= −2.50
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.
252
= .72
350
Wisconsin p =
z=
p − p0
p0 (1 − p0 )
n
=
.72 − .70
.70(1 − .70)
350
= .82
Area = .2939
p-value = 2(.5000 - .2939) = .4122
Cannot reject H0.
California p =
z=
189
= .63
300
.63 − .70
.70(1 − .70)
300
= −2.65
Area = .4960
p-value = 2(.5000 - .4960) = .0080
Reject H0. California has a different (lower) percentage of adults who do not exercise regularly.
9 - 14
Hypothesis Testing
40. a.
b.
p=
414
= .2702 (27%)
1532
H0: p ≤ .22
Ha: p > .22
z=
p − p0
p0 (1 − p0 )
n
=
.2702 − .22
.22(1 − .22)
1532
= 4.75
p-value ≈ 0
Reject H0; There has been a significant increase in the intent to watch the TV programs.
c.
41. a.
These studies help companies and advertising firms evaluate the impact and benefit of commercials.
H0: p ≥ .75
Ha: p < .75
b.
z=
p − p0
p0 (1 − p0 )
n
=
.72 − .75
.75(1 − .75)
300
= −1.20
Area = .3849
p-value = .5000 - .3849 = .1151
c.
42.
p-value > .05; do not reject H0. The executive's claim cannot be rejected.
H0: p ≤ .24
Ha: p > .24
p=
z=
93
= .31
300
p − p0
p0 (1 − p0 )
n
=
.31 − .24
.24(1 − .24)
300
= 2.84
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.
9 - 15
Chapter 9
43. a.
H0: p = .48
Ha: p ≠ .48
b.
p=
z=
360
= .45
800
p − p0
p0 (1 − p0 )
n
=
.45 − .48
.48(1 − .48)
800
= −1.70
Area = .4554
p-value = 2(.5000 - .4554) = .0892
c.
44. a.
p-value > .05; do not reject H0. There is no reason to conclude the proportion has changed.
H0: p ≤ .50
Ha: p > .50
b.
p=
z=
285
= .57
500
p − p0
p0 (1 − p0 )
n
=
.57 − .50
.50(1 − .50)
500
= 3.13
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.
p=
z=
205
= .41
500
p − p0
p0 (1 − p0 )
n
=
.41 − .44
.44(1 − .44)
500
= −1.35
9 - 16
Hypothesis Testing
Area = .4115
p-value = 2(.5000 - .4115) = .1770
p-value > .05; do not reject H0. No change.
c.
p=
245
= .49
500
.49 − .44
z=
.44(1 − .44)
500
= 2.25
Area = .4878
p-value = 2(.5000 - .4878) = .0244
p-value ≤ .05; reject H0. There has been a change: an increase in repeat customers.
σx =
46.
σ
n
=
5
120
= .46
c
Ha : µ < 10
H0: µ ≥ 10
.05
x
10
c = 10 - 1.645 (5 / 120 ) = 9.25
Reject H0 if x ≤ 9.25
a.
When µ = 9,
z=
9.25 − 9
5 / 120
= .55
Prob (H0) = (.5000 - .2088) = .2912
b.
Type II error
9 - 17
Chapter 9
c.
When µ = 8,
z=
9.25 − 8
5 / 120
= 2.74
β = (.5000 - .4969) = .0031
Reject H0 if z ≤ -1.96 or if z ≥ 1.96
47.
σx =
σ
10
=
n
200
= .71
Ha : µ ≠ 20
H0: µ = 20
Ha : µ ≠ 20
.025
.025
x
c1
20
c1 = 20 - 1.96 (10 / 200 ) = 18.61
c2 = 20 + 1.96 (10 / 200 ) = 21.39
a.
µ = 18
z=
18.61 − 18
10 / 200
= .86
β = .5000 - .3051 = .1949
b.
µ = 22.5
z=
21.39 − 22.5
10 / 200
= −1.57
β = .5000 - .4418 = .0582
9 - 18
c2
x
Hypothesis Testing
c.
µ = 21
z=
21.39 − 21
10 / 200
= .55
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
z=
x − µ0
σ/ n
=
x − 15
4 / 35
= 2.33
Solve for x = 16.58
Decision Rule:
Accept H0 if x < 16.58
Reject H0 if x ≥ 16.58
For µ = 17,
z=
16.58 − 17
4 / 35
= −.62
β = .5000 -.2324 = .2676
c.
For µ = 18,
z=
16.58 − 18
4 / 35
= −2.10
β = .5000 -.4821 = .0179
49. a.
H0: µ ≥ 25
Ha: µ < 25
Reject H0 if z ≤ -2.05
z=
x − µ0
x − 25
=
= −2.05
σ / n 3 / 30
Solve for x = 23.88
Decision Rule:
9 - 19
Chapter 9
Accept H0 if x > 23.88
Reject H0 if x ≤ 23.88
b.
For µ = 23,
z=
23.88 − 23
3 / 30
= 1.61
β = .5000 -.4463 = .0537
c.
For µ = 24,
z=
23.88 − 24
= −.22
3 / 30
β = .5000 +.0871 = .5871
d.
50. a.
b.
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.
Accepting H0 and concluding the mean average age was 28 years when it was not.
Reject H0 if z ≤ -1.96 or if z ≥ 1.96
z=
x − µ0
σ/ n
=
x − 28
6 / 100
Solving for x , we find
at
at
z = -1.96,
z = +1.96,
x = 26.82
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,
z=
26.82 − 26
6 / 100
= 1.37
β = .5000 +.4147 = .0853
At µ = 27,
z=
26.82 − 27
6 / 100
= −.30
9 - 20
Hypothesis Testing
β = .5000 +.1179 = .6179
At µ = 29,
z=
29.18 − 29
6 / 100
= .30
β = .5000 +.1179 = .6179
At µ = 30,
z=
29.18 − 30
6 / 100
= −1.37
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.
b.
Accepting H0 and letting the process continue to run when actually over - filling or under - filling
exists.
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
z=
16.29 − 16.5
.8 / 30
= −1.44
β = .5000 -.4251 = .0749
9 - 21
Chapter 9
c
β
x
16.29
16.5
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
1.00
.75
.50
Power
.25
.00
15.6
15.8
16.0 16.2
16.4
Possible Values of u
52.
c = µ0 + z.01
σ
= 15 + 2.33
n
At µ = 17 z =
16.32 − 17
4 / 50
4
50
= 16.32
= −1.20
β = .5000 - .3849 = .1151
At µ = 18 z =
16.32 − 18
4 / 50
β = .5000 - .4985 = .0015
= −2.97
Increasing the sample size reduces the probability of making a Type II error.
9 - 22
Hypothesis Testing
53. a.
b.
Accept µ ≤ 100 when it is false.
Critical value for test:
c = µ0 + z.05
σ
= 100 + 1.645
n
119.51 − 120
At µ = 120 z =
75 / 40
75
40
= 119.51
= −.04
β = .5000 - .0160 = .4840
c.
At µ = 13 z =
119.51 − 130
75 / 40
= −.88
β = .5000 - .3106 = .1894
d.
Critical value for test:
c = µ0 + z.05
σ
n
= 100 + 1.645
At µ = 120 z =
113.79 − 120
75 / 80
75
80
= 113.79
= −.74
β = .5000 - .2704 = .2296
At µ = 130 z =
113.79 − 130
75 / 80
= −1.93
β = .5000 - .4732 = .0268
Increasing the sample size from 40 to 80 reduces the probability of making a Type II error.
( zα + z β ) 2 σ 2
54.
n=
55.
n=
56.
At µ0 = 3,
( µ0 − µ a ) 2
( zα + z β ) 2 σ 2
( µ0 − µ a ) 2
=
(1.645 + 1.28) 2 (5) 2
= 214
(10 − 9) 2
=
(1.96 + 1.645) 2 (10) 2
= 325
(20 − 22) 2
α = .01.
z.01 = 2.33
At µa = 2.9375, β = .10.
z.10 = 1.28
σ = .18
n=
( zα + z β ) 2 σ 2
( µ0 − µ a ) 2
=
(2.33 + 1.28) 2 (.18) 2
= 108.09 Use 109
(3 − 2.9375) 2
9 - 23
Chapter 9
57.
At µ0 = 400,
α = .02.
z.02 = 2.05
At µa = 385,
β = .10.
z.10 = 1.28
σ = 30
n=
58.
( zα + z β ) 2 σ 2
( µ0 − µ a )
2
=
(2.05 + 1.28) 2 (30) 2
= 44.4 Use 45
(400 − 385) 2
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
n=
59.
( zα / 2 + z β ) 2 σ 2
(µ0 − µa )
=
2
(1.96 + 1.04) 2 (6) 2
= 324
(28 − 29) 2
At µ0 = 25,
α = .02.
z.02 = 2.05
At µa = 24,
β = .20.
z.20 = .84
σ =3
n=
60. a.
( zα + z β ) 2 σ 2
( µ0 − µ a )
2
=
(2.05 + .84) 2 (3) 2
= 75.2 Use 76
(25 − 24) 2
H0: µ = 16
Ha: µ ≠ 16
b.
z=
x − µ0
σ/ n
=
16.32 − 16
.8 / 30
= 2.19
Area = .4857
p-value = 2(.5000 - .4857) = .0286
p-value ≤ .05; reject H0. Readjust production line.
c.
z=
x − µ0
σ/ n
=
15.82 − 16
.8 / 30
= −1.23
Area = .3907
p-value = 2(.5000 - .3907) = .2186
p-value > .05; do not reject H0. Continue the production line.
9 - 24
Hypothesis Testing
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.
b.
H0: µ = 900
Ha: µ ≠ 900
x ± z.025
σ
n
935 ± 1.96
180
200
935 ± 25
(910 to 960)
c.
Reject H0 because µ = 900 is not in the interval.
d.
z=
x − µ0
935 − 900
=
= 2.75
σ / n 180 / 200
p-value = 2(.5000 - .4970) = .0060
62. a.
b.
H0: µ ≤ 45,250
Ha: µ > 45,250
z=
x − µ 0 47, 000 − 45, 250
=
= 2.71
6300 / 95
σ/ n
p-value = .5000 - .4966 = .0034
p-value ≤ .01; reject H0. Conclude New York City has higher mean salary.
63. a.
b.
H0: µ ≤ 37,000
Ha: µ > 37,000
t=
x − µ0
s/ n
=
38, 000 − 37, 000
5200 / 48
= 1.47
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.
9 - 25
Chapter 9
H0: µ = 6000
64.
Ha: µ ≠ 6000
t=
x − µ0
s/ n
=
5812 − 6000
1140 / 32
= −.93
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.
t=
x − µ0
s/ n
=
29.6 − 30
1.8 / 50
= −1.57
Degrees of freedom = 50 – 1 = 49
Using t table, p-value is between .05 and .10
Actual p-value = .0613
c.
66.
p-value > .01; do not reject H0. Cannot conclude miles per gallon is less than 30.
H0: µ ≤ 125,000
Ha: µ > 125,000
t=
x − µ0
s/ n
=
130, 000 − 125, 000
12,500 / 32
= 2.26
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
H a: µ ≠ 3
9 - 26
Hypothesis Testing
b.
x=
c.
s=
d.
t=
Σxi
= 2.8
n
Σ ( xi − x )
2
n −1
x − µ0
s/ n
=
= .70
2.8 − 3
.70 / 10
= −.90
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.
68. a.
p-value > .05; do not reject H0. There is no evidence to conclude a difference compared to prior
year.
H0: p ≤ .50
Ha: p > .50
b.
p=
c.
z=
64
= .64
100
p − p0
p0 (1 − p0 )
n
=
.64 − .50
.50(1 − .50)
100
= 2.80
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.
p=
c.
z=
355
= .6502
546
p − p0
p0 (1 − p0 )
n
=
.6502 − .6667
.6667(1 − .6667)
546
= −.82
p-value = 2(.5000 - .2939) = .4122
9 - 27
Chapter 9
70. a.
p-value > .05; do not reject H0; Cannot conclude that the population proportion differs from 2/3.
H0: p ≤ .50
Ha: p > .50
b.
p=
c.
z=
67
= .6381 (64%)
105
p − p0
p0 (1 − p0 )
n
=
.6381 − .50
.50(1 − .50)
105
= 2.83
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.
b.
p=
330
= .825
400
H0: p = .78
Ha: p ≠ .78
z=
p − p0
p0 (1 − p0 )
n
=
.825 − .78
.78(1 − .78)
400
= 2.17
p-value = 2(.5000 - .4850) = .03
c.
72.
p-value ≤ .05; reject H0. The on-time arrival record has changed. It is improving.
H0: p ≥ .90
Ha: p < .90
p=
z=
49
= .8448
58
p − p0
p0 (1 − p0 )
n
=
.8448 − .90
.90(1 − .90)
58
= −1.40
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
9 - 28
Hypothesis Testing
b.
p=
c.
z=
44
= .352
125
p − p0
p0 (1 − p0 )
n
=
.352 − .47
.47(1 − .47)
125
= −2.64
p-value = .5000 - .4959 = .0041
d.
74. a.
p-value ≤ .01; reject H0. The proportion of foods containing pesticides has declined.
H0: µ ≤ 72
Ha: µ > 72
Reject H0 if z ≥ 1.645
z=
x − µ0
σ/ n
=
x − 72
20 / 30
= 1.645
Solve for x = 78
Decision Rule:
Accept H0 if x < 78
Reject H0 if x ≥ 78
b.
For µ = 80
z=
78 − 80
20 / 30
= −.55
β = .5000 -.2088 = .2912
c.
For µ = 75,
z=
78 − 75
20 / 30
= .82
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
9 - 29
Chapter 9
1.0
.8
P
o
w
e
r
.6
.4
.2
72
76
78
80
74
Possible Values of µ
Ho False
82
84
H0: µ ≥ 15,000
75.
Ha: µ < 15,000
At µ0 = 15,000, α = .02.
z.02 = 2.05
At µa = 14,000, β = .05.
z.10 = 1.645
n=
( zα + z β ) 2 σ 2
( µ0 − µ a ) 2
=
(2.05 + 1.645) 2 (4, 000) 2
= 218.5 Use 219
(15, 000 − 14, 000) 2
H0: µ = 120
76.
Ha: µ ≠ 120
At µ0 = 120,
α = .05. With a two - tailed test, zα / 2 = z.025 = 1.96
At µa = 117,
β = .02.
n=
b.
( zα / 2 + z β ) 2 σ 2
(µ0 − µa )2
=
z.02 = 2.05
(1.96 + 2.05) 2 (5) 2
= 44.7 Use 45
(120 − 117) 2
Example calculation for µ = 118.
Reject H0 if z ≤ -1.96 or if z ≥ 1.96
z=
x − µ 0 x − 120
=
σ / n 5 / 45
Solve for x .
At z = -1.96, x = 118.54
At z = +1.96, x = 121.46
9 - 30
Hypothesis Testing
Decision Rule:
Accept H0 if 118.54 < x < 121.46
Reject H0 if x ≤ 118.54 or if x ≥ 121.46
For µ = 118,
z=
118.54 − 118
5 / 45
= .72
β = .5000 +.2642 = .2358
Other Results:
If µ is
117
118
119
121
122
123
z
2.07
.72
-.62
+.62
+.72
-2.07
9 - 31
β
.0192
.2358
.7291
.7291
.2358
.0192