CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 277
27. (a) 0: $105 (claim); : $105
a
HHµµ=≠
(b) 110 105 2.63
8.50
20
x
ts
n
µ−−
== ≈
P-value = 2{Area to right of t = 2.63} = 2(0.00825) = 0.0165
(c) Fail to reject 0
H.
(d) There is not enough evidence at the 1% level of significance to reject the claim that the mean
daily meal cost for two adults traveling together on vacation in San Francisco is $105.
29. (a) 0: 32; : 32 (claim)
a
HHµµ≥<
(b) 30.167x≈ 4.004s≈
30.167 32 1.942
4.004
18
x
ts
n
µ−−
== ≈−
P-value = {Area to left of t = 1.942−} = 0.0344
(c) Reject 0
H.
(d) There is enough evidence at the 5% level of significance to support the claim that the mean
class size for full-time faculty is fewer than 32 students.
278 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
(c) Fail to reject 0
H.
(d) There is not enough evidence at the 1% level of significance to reject the claim that the mean
number of classroom hours per week for full-time faculty is 11.0.
32. Hypothesis test results:
0
: population mean
: 27
: 27
A
H
H
µ
µ
µ
=
≠
Mean Sample
Mean
Std. Err DF T-Stat P-value
µ 31.5 1.3567731 11 3.316693 0.0069
Reject 0
H.
There is enough evidence at the 1% level of significance to support the claim.
34. Hypothesis test results:
0
: population mean
: 2118
H
µ
µ
=
CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 279
35. 0: 5000; : 5000 (claim)
a
HHµµ≤>
H.
36. (a) Because 0.0748 > 0.01, fail to reject 0
H.
(b) Because 0.0748 < 0.10, reject 0
H.
(c) 5434 5000 1.96
625
8
x
ts
n
µ−−
=≈ ≈
37. Because σ is unknown, n < 30, and the gas mileage is normally distributed, use the t–distribution.
0: 23 (claim); : 23
a
HHµµ≥<
22 23 0.559
4
5
x
ts
n
µ−−
=≈ ≈−
P-value = {Area left of t = 0.559−} = 0.303
Fail to reject 0
H. There is not enough evidence at the 5% level of significance to reject the claim
that the mean gas mileage for the luxury sedan is at least 23 miles per gallon.
280 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
39. More likely; For degrees of freedom less than 30, the tails of the t–distribution curve are thicker
than those of a standard normal distribution curve. So, if you incorrectly used a standard normal
7.4 HYPOTHESIS TESTING FOR PROPORTIONS
7.4 Try It Yourself Solutions
1a.
()() ()()
125 0.25 31.25 5, 125 0.75 93.75 5np nq==>==>
b. The claim is “more than 25% of U.S. adults have used a cellular phone to access the internet.”
0: 0.25; : 0.25 (claim)
a
Hp Hp≤>
c. 0.05α=
l
f. Reject 0
H.
g. There is enough evidence at the 5% level of significance to support the claim that more than 25%
of U.S. adults have used a cellular phone to access the Internet.
e.
l
()()
0.36 0.30 2.07
0.30 0.70
250
pp
zpq
n
−−
== ≈
CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 281
7.4 EXERCISE SOLUTIONS
1. If 5np ≥ and 5,nq ≥ the normal distribution can be used.
4.
()( )
90 0.48 43.2 5
(90)(0.52) 46.8 5 use normal distribution
np
nq
==>
==>→
0: 0.48 (claim); : 0.48
a
Hp Hp≥<
01.405;z=− Rejection region: 1.405z<−
l
()()
0.40 0.48 1.52
0.48 0.52
90
pp
zpq
n
−−
== ≈−
Reject 0
H. There is enough evidence at the 8% level of significance to reject the claim.
l
6.
()( )
225 0.70 157.5 5
(225)(0.30) 67.5 5 use normal distribution
np
nq
==>
==>→
0: 0.70; : 0.70 (claim)
a
Hp Hp≤>
01.75;z= Rejection region: 1.75z>
l
282 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
8.
()( )
50 0.95 47.5 5
(50)(0.05) 2.5 5
np
nq
==>
==<
Cannot use normal distribution because 5nq <.
10. (a) 0: 0.40 (claim); : 0.40
a
Hp Hp≥<
(b)
02.05;z=− Rejection region: 2.05z<−
11. (a) 0: 0.50 (claim); : 0.50
a
Hp Hp≤>
(b)
02.33;z= Rejection region: 2.33z>
(c)
l
()()
0.58 0.50 1.96
0.50 0.50
pp
zpq
n
−−
== ≈
CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 283
13. (a) 0: 0.75; : 0.75 (claim)
a
Hp Hp≤>
(b)
01.28;z= Rejection region: 1.28z>
(c)
l
()()
0.82 0.75 1.98
0.75 0.25
150
pp
zpq
n
−−
== ≈
(d) Reject 0
H.
(e) There is enough evidence at the 10% level of significance to support the claim that more than
75% of females ages 20-29 are taller than 62 inches.
l
15. (a) 0: 0.35; : 0.35 (claim)
a
Hp Hp≥<
(b)
01.28;z=− Rejection region: 1.28z<−
(c)
l
()()
0.39 0.35 1.68
0.35 0.65
400
pp
zpq
n
−−
== ≈
(d) Fail to reject 0
H.
(e) There is not enough evidence at the 10% level of significance to support the claim that less
than 35% of U.S. households own a dog.
l
284 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
(e) There is not enough evidence at the 5% level of significance to reject the claim that 30% of
U.S. households own a cat.
18. Answers will vary. Sample answer: The company should continue the use of giveaways because
there is not enough evidence to say that less than 52% of adults are more likely to buy a product
when there are free samples.
19. 0: 0.35; : 0.35 (claim)
a
Hp Hp≥<
01.28:z=− Rejection region: 1.28z<−
()( )
()()()
156 400 0.35 1.68
400 0.35 0.65
xnp
znpq
−
−
== ≈
Reject 0
H. The results are the same.
7.5 HYPOTHESIS TESTING FOR VARIANCE
AND STANDARD DEVIATION
7.5 Try It Yourself Solutions
1a. 17, 0.01df α==
b. 2
033.409χ=
CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 285
4a. The claim is “the variance of the amount of sports drink in a 12-ounce bottle is no more than
0.40.”
22
0: 0.40 (claim); : 0.40
a
HHσσ≤>
5a. The claim is “the standard deviation of the lengths of response times is less than 3.7 minutes.”
0: 3.7; : 3.7 (claim)
a
HHσσ≥<
b. 0.05α= and d.f. = 18n−=
c. 2
6a. The claim is “the variance of the weight losses is 25.5.”
22
0: 25.5 (claim); : 25.5
a
HHσσ=≠
b. 0.10α= and d.f. = 112n−=
c. 22
5.226 and 21.026;
χχ== Rejection region: 22
21.026, 5.226χχ><
7.5 EXERCISE SOLUTIONS
1. Specify the level of significance α. Determine the degrees of freedom. Determine the critical
values using the 2-distributionχ. For a right-tailed test, use the value that corresponds to d.f. and
α. For a left-tailed test, use the value that corresponds to d.f. and 1α−. For a two-tailed test, use
286 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
3. The requirement of a normal distribution is more important when testing a standard deviation
than when testing a mean. If the population is not normal, the results of the 2-testχ can be
misleading because the 2-testχ is not as robust as the tests for the population mean.
11. (a) Fail to reject 0
H because 26.251χ<.
(b) Fail to reject 0
H because 26.251χ<.
(c) Fail to reject 0
H because 26.251χ<.
(d) Reject 0
H because 26.251χ>.
13. (a) Fail to reject 0
H because 2
8.547 22.307χ<< .
(b) Reject 0
H because 222.307χ>.
(c) Reject 0
H because 28.547χ<.
(d) Fail to reject 0
H because 2
8.547 22.307χ<< .
14. (a) Fail to reject 0
H because 210.645χ<.
(b) Fail to reject 0
H because 210.645χ<.
(c) Fail to reject 0
H because 210.645χ<.
(d) Reject 0
H because 210.645χ>.
CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 287
16. 22
0: 8.5 (claim); : 8.5
a
HHσσ≥<
2
012.338;χ= Rejection region: 212.338χ<
() ()( )
2
2
2
1227.45
19.28
8.5
ns
χσ
−
== ≈
Fail to reject 0
H. There is not enough evidence at the 5% level of significance to reject the claim.
18. 0: 40; : 40 (claim)
a
HHσσ≥<
2
03.053;χ= Rejection region: 23.053χ<
() ()( )
()
2
2
2
22
11140.8
11.444
40
ns
χσ
−
== ≈
Fail to reject 0
H. There is not enough evidence at the 1% level of significance to support the
claim.
20. (a) 22
0: 1.0 (claim); : 1.0
a
HHσσ=≠
(b)
22
12.401, 39.364;
LR
χχ== Rejection regions: 22
12.401, 39.364χχ<>
(c)
() ()( )
2
2
2
1241.65
39.6
1.0
ns
χσ
−
== =
(d) Reject 0
H.
(e) There is enough evidence at the 5% level of significance to reject the claim that the variance
of the gas mileages of its hybrid vehicles is 1.0.
288 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
(e) There is not enough evidence at the 10% level of significance to support the claim that the
standard deviation for eighth graders on the examination is less than 36 points.
23. (a) 0: 25 (claim); : 25
a
HHσσ≤>
(b)
2
036.741;χ= Rejection region: 236.741χ>
(c)
() ()()
()
2
2
2
22
12731
41.515
25
ns
χσ
−
== ≈
(d) Reject 0
H.
(e) There is enough evidence at the 10% level of significance to reject the claim that the standard
deviation of the number of fatalities per year from tornadoes is no more than 25.
25. (a) 0: $3500; : $3500 (claim)
a
HHσσ≥<
(b)
2
018.114;χ= Rejection region: 218.114χ<
(c)
() ()( )
()
2
2
2
22
1 27 4100 37.051
3500
ns
χσ
−
== ≈
(d) Fail to reject 0
H.
(e) There is not enough evidence at the 10% level of significance to support the claim that the
standard deviation of the total charges for patients involved in a crash in which the vehicle
struck a construction baracade is less than $3500.
CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 289
(d) Fail to reject 0
H.
(e) There is not enough evidence at the 1% level of significance to reject the claim that the
standard deviation of the room rates of hotels in the city is no more than $30.
28. (a) 0: $10,600 (claim); : $10,600
a
HHσσ≥<
(b)
2
011.651;χ= Rejection region: 211.651χ<
(c) 8284.89s≈
() ()( )
()
2
2
2
22
1 19 8284.89 11.607
10,600
ns
χσ
−
== ≈
(d) Reject 0
H.
(e) There is enough evidence at the 10% level of significance to reject the claim that the standard
deviation of the annual salaries of commodity buyers is at least $10,600.
30. Hypothesis test results:
2
: variance of variable
σ