CHAPTER
Hypothesis Testing with One Sample
7
7.1 INTRODUCTION TO HYPOTHESIS TESTING
7.1 Try It Yourself Solutions
1a. (1) The mean is not 74 months.
74µ≠
(2) The variance is less than or equal to 2.7.
22.7σ≤
(3) The proportion is more than 24%
2a. 0: 0.01; : 0.01
a
Hp Hp≤>
b. A type I error will occur if the actual proportion is less than or equal to 0.01, but you reject 0
H.
A type II error will occur if the actual proportion is greater than 0.01, but you fail to reject 0
H.
c. A type II error is more serious because you would be misleading the consumer, possibly causing
serious injury or death.
3a. (1) 0:H The mean life of a certain type of automobile battery is 74 months.
:
a
H The mean life of a certain type of automobile battery is not 74 months.
0: 74; : 74
a
HHµµ=≠
b. (1) Two-tailed (2) Right-tailed (3) Right-tailed
c. (1) (2) (3)
258 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
7.1 EXERCISE SOLUTIONS
1. The two types of hypotheses used in a hypothesis test are the null hypothesis and the alternative
hypothesis.
The alternative hypothesis is the complement of the null hypothesis.
4. No; Failing to reject the null hypothesis means that there is not enough evidence to reject it.
5. False. In a hypothesis test, you assume the null hypothesis is true.
9. False. A small P–value in a test will favor a rejection of the null hypothesis.
10. False. If you want to support a claim, write it as your alternative hypothesis.
11. 0: 645Hµ≤ (claim); : 645
a
Hµ> 12. 0: 128Hµ≥; : 128
a
Hµ< (claim)
21. Right-tailed 22. Left tailed 23. Two-tailed 24. Two-tailed
25. 750µ> 26. 3σ<
0: 750Hµ≤; : 750
a
Hµ> (claim) 0: 3H
σ≥
; : 3
a
H
σ<
(claim)
CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 259
27. 320σ≤ 28. 85µ≥
0: 320Hσ≤ (claim); : 320
a
Hσ> 0: 85Hµ≥ (claim); : 85
a
Hµ<
32. A type I error will occur if the actual mean flow rate of the hose is 16 gallons per minute, but you
reject 0: 16Hµ=.
A type II error will occur if the actual mean flow rate of the hose is not 16 gallons per minute, but
you fail to reject 0: 16Hµ=.
33. A type I error will occur if the actual standard deviation of the length of time to play a game is
34. A type I error will occur if the actual proportion of U.S. adults who own a video game system is
0.26, but you reject 0: 0.26Hp≤.
A type II error will occur if the actual proportion of U.S. adults who own a video game system is
not 0.26, but you fail to reject 0: 0.26Hp≤.
37. 0:H The proportion of homeowners who have a home security alarm is greater than or equal to
14%.
:
H The proportion of homeowners who have a home security alarm is less than 14%.
260 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
38. 0:H The mean time that the manufacturer’s clocks lose is less than or equal to 0.02 second per
day.
:
a
H The mean time that the manufacturer’s clocks lose is greater than 0.02 second per day.
0: 0.02;Hµ≤ : 0.02
a
Hµ>
Right-tailed because the alternative hypothesis contains >.
40. 0:H The proportion of lung cancer cases that are due to smoking is 87%.
:
a
H The proportion of lung cancer cases that are due to smoking is not87%.
0: 0.87;Hp= : 0.87
a
Hp≠
Two-tailed because the alternative hypothesis contains .
42. 0:H The mean tuition of the state’s universities is less than or equal to $25,000 per year.
:
a
H The mean tuition of the state’s universities is greater than $25,000 per year.
0: 25,000;Hµ≤ : 25,000
a
Hµ>
Right-tailed because the alternative hypothesis contains >.
44. (a) There is enough evidence to reject the claim that the standard deviation of the life of the lawn
mower is at most 2.8 years.
(b) There is not enough evidence to reject the claim that the standard deviation of the life of the
lawnmower is at most 2.8 years.
CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 261
48. (a) There is enough evidence to reject the sports drink maker’s claim that the mean calorie
content of its beverages is 72 calories per serving.
(b) There is not enough evidence to reject the sports drink maker’s claim that the mean calorie
content of its beverages is 72 calories per serving.
49. 0: 60;Hµ≥ : 60
a
Hµ< 50.
0: 21;Hµ= : 21
a
Hµ≠
a
a
53. If you decrease ,α you are decreasing the probability that you reject 0.H Therefore, you are
54. If 0,α= the null hypothesis cannot be rejected and the hypothesis test is useless.
55. Yes; If the P-value is less than 0.05,α= it is also less than 0.10.α=
58. (a) Fail to reject 0
H because the CI includes values less than 54.
(b) Fail to reject 0
H because the CI includes values less than 54.
(c) Reject 0
H because the CI is located to the right of 54.
262 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
7.2 HYPOTHESIS TESTING FOR THE MEAN (LARGE SAMPLES)
7.2 Try It Yourself Solutions
1a. (1) 0.0347 0.01Pα=>=
(2) 0.0347 0.05Pα=<=
b. (1) Fail to reject 0
H because 0.0347 > 0.01.
(2) Reject 0
H because 0.0347 < 0.05.
3a. Area to the right of z = 1.64 is 0.0505.
b. 2(area) 2(0.0505) 0.1010p== =
c. Fail to reject 0
H because 0.1010 0.10P=>.
5a. The claim is “one of your distributors reports an average of 150 sales per day.”
0: 150 (claim); : 150HH
α
µµ=≠
b. 0.01α=
CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 263
6a. 0.0440 0.01Pα=>= b. Fail to reject 0
H.
7a. 8a.
9a. The claim is “the mean work day of the company’s mechanical engineers is less than 8.5 hours.”
0: 8.5; : 8.5 (claim)
a
HHµµ≥<
b. 0.01α=
c. 02.33;z=− Rejection region: 2.33z<−
10a. 0.01α=
b. 02.575;z±=± Rejection regions: 2.575, 2.575zz<− >
c.
Fail to reject 0
H.
264 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
7.2 EXERCISE SOLUTIONS
1. In the z-test using rejection region(s), the test statistic is compared with critical values. The z-test
using a P–value compares the P–value with the level of significance .α
4.
P = 0.0606; Fail to reject 0
H because P = 0.0606 > 0.05.
CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 265
7.
P = 2(Area) = 2(0.0465) = 0.0930; Fail to reject 0
H because P = 0.0930 > 0.05.
9. b 10. d 11. c 12. a
13. (a) Fail to reject 0
H (P = 0.0461 > 0.01).
(b) Reject 0
H (P = 0.0461 < 0.05).
17. 1.645 18. 1.41 19. 1.88−
266 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
24. (a) Reject 0
H because z > 1.96.
25. 0: 40 (claim); : 40
a
HHµµ=≠
0
0.05 1.96zα=→=±
26. 0: 1745; : 1745 (claim)
a
HHµµ≤>
0
0.10 1.28zα=→=
1752 1745 1.222
38
44
x
zs
n
µ−−
== ≈
Fail to reject 0
H. There is not enough evidence at the 10% level of significance to support the
claim.
28. 0: 22,500 (claim); : 22,500
a
HHµµ≤>
0
0.01 2.33zα=→=
23,250 22,500 4.193
1200
45
x
zs
n
µ−−
== ≈
Reject 0
H. There is enough evidence at the 1% level of significance to reject the claim.
CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 267
(e) There is enough evidence at the 1% level of significance to support the claim that the mean
raw score for the school’s applicants is more than 30.
31. (a) 0: 28.5 (claim); : 28.5
a
HHµµ=≠
(b) 27.8 28.5 1.71
4.1
100
x
zs
n
µ−−
== ≈−
32. (a) 0: 24.2 (claim); : 24.2
a
HHµµ=≠
(b) 23.5 24.2 2.40
3.2
120
x
zs
n
µ−−
== ≈−
33. (a) 0: 15 (claim); : 15
a
HHµµ=≠
(b) 14.834x≈ 4.288s≈
14.384 15 0.22
4.288
32
x
zs
n
µ−−
== ≈−
268 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
34. (a) 0: 66,200; : 66,200 (claim)
a
HHµµ≤>
(b) 66,592.4x≈ 2384.376s≈
35. (a) 0: 40 (claim); : 40
a
HHµµ=≠
(b) 00
2.575, 2.575;zz−=− =
Rejection regions: 2.575, 2.575zz<− >
36. (a) 0: 874 (claim); : 874
a
HHµµ=≠
(b) 00
1.96, 1.96;zz−=− =
Rejection regions: 1.96, 1.96zz<− >
(c) 905 874 1.98
125
64
x
zs
n
µ−−
== ≈
(d) Reject 0
H.
(e) There is enough evidence at the 5% level of significance to reject the claim that the mean
monthly residential electricity consumption in your town is 874 kilowatt-hours.
38. (a) 0: 920 (claim); : 920HH
α
µµ≤>
(b) 01.28;z= Rejection region: 1.28z>
CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 269
(c) 925 920 1.84
18
44
x
zs
n
µ−−
== ≈
(d) Reject 0
H.
(e) There is enough evidence at the 10% level of significance to reject the claim that the mean
sodium content in one of their breakfast sandwiches is no more than 920 milligrams.
40. (a) 0: 10,000 (claim); : 10,000
a
HHµµ≥<
(b) 01.34;z=− Rejection region: 1.34z<−
(c) 9580.9x≈ 1772.4s≈
9580.9 10,000 1.38
1722.4
32
x
zs
n
µ−−
=≈ ≈−
(d) Reject 0
H.
(e) There is enough evidence at the 9% level of significance to reject the claim that the mean life
of fluorescent lamps is at least 10,000 hours.
42. (a) 0: 60; : 60 (claim)
a
HHµµ≥<
(b) 02.33;z=− Rejection region: 2.33z<−
(c) 49x= 21.51s≈
49 60 3.62
21.51
50
x
zs
n
µ−−
=≈ ≈−
270 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
(d) Reject 0
H.
(e) There is enough evidence at the 1% level of significance to support the claim that the mean
time it takes an employee to evacuate a building during a fire drill is less than 60 seconds.
43. Hypothesis test results:
: population mean
µ
44. Hypothesis test results:
0
: population mean
: 495
: 495
a
H
H
µ
µ
µ
=
>
45. Hypothesis test results:
0
: population mean
: 1210
H
µ
µ
=
46. Hypothesis test results:
0
: population mean
: 28750
: 28750
a
H
H
µ
µ
µ
=
≠
CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 271
47. 02.33;z=− Rejection region: 2.33z<−
48. 01.645;z= Rejection region: 1.645z>
22,200 22,000 1.548
775
36
x
zs
n
µ−−
== ≈
Fail to reject 0
H because the standardized test statistic 1.548z= is not in the rejection region
(1.645)z>.
49. (a) 0
0.02 2.05;zα=→=− Rejection region: 2.05z<−
Stays as fail to reject 0
H because the standardized test statistic 1.86z=− is not in the
rejection region (2.05)z<− .
Stays as fail to reject 0
H because the standardized test statistic 2.15z=− is not in the
rejection region (2.33)z<− .
(d) Same rejection region as Exercise 47: 2.33z<−
125,270 127,400 2.40
6275
50
x
zs
n
µ−−
== ≈−
50. (a) 0
0.06 1.555;zα=→= Rejection region: 1.555z>
Stays as fail to reject 0
H because the standardized test statistic 1.548z= is not in the
rejection region (1.555)z>.
272 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
(c) Same rejection region as Exercise 48: 1.645z>
22,200 22,000 1.63
775
40
x
zs
n
µ−−
== ≈
7.3 HYPOTHESIS TESTING FOR THE MEAN (SMALL SAMPLES)
7.3 Try It Yourself Solutions
1a. 13 b. 02.650t=−
4a. The claim is “the mean cost of insuring a 2008 Honda CR-V is less than $1200.”
0: $1200; : $1200 (claim)
a
HHµµ≥<
b. 0.10α= and d.f. = 16n−=
c. 01.440;t=− Rejection region: 1.440t<−
5a. The claim is “the mean conductivity of the river is 1890 milligrams per liter.”
0: 1890 (claim); : 1890
a
HHµµ=≠
CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 273
d. 2500 1890 3.798
700
19
x
zs
n
µ−−
== ≈
e. Reject 0
H.
f. There is enough evidence at the 1% level of significance to reject the claim that the mean
conductivity of the river is 1890 milligrams per liter.
7.3 EXERCISE SOLUTIONS
1. Identify the level of significance α and the degrees of freedom, d.f. = 1n−. Find the critical
value(s) using the t–distribution table in the row with 1n− d.f. If the hypothesis test is:
(1) left-tailed, use the “One Tail, α” column with a negative sign.
(2) right-tailed, use the “One Tail, α” column with a positive sign.
(3) two-tailed, use the “Two Tail, α,” column with a negative and a positive sign.
7. 02.056t=± 8.
01.721t=±
9. (a) Fail to reject 0
H because 2.086t>− .
(b) Fail to reject 0
H because 2.086t>− .
(c) Fail to reject 0
H because 2.086t>− .
(d) Reject 0
H because 2.086t<− .
274 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
11. (a) Fail to reject 0
H because 2.602 2.602t−<< .
12. (a) Fail to reject 0
H because 1.725 1.725t−<<.
13. 0: 15 (claim); : 15
a
HHµµ=≠
0.01α= and d.f. = 15n−=
04.032t=±
13.9 15 0.834
3.23
6
x
ts
n
µ−−
== ≈−
Fail to reject 0
H. There is not enough evidence at the 1% level of significance to reject the claim.
Reject 0
H. There is enough evidence at the 5% level of significance to support the claim.
15. 0: 8000 (claim); : 8000
a
HHµµ≥<
0.01α= and d.f. = 124n−=
02.492t=−
7700 8000 3.333
450
25
x
ts
n
µ−−
== ≈−
Reject 0
H. There is enough evidence at the 1% level of significance to reject the claim.
CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 275
18. (a) 0: 7 (claim); : 7
a
HHµµ≤>
(b)
01.372;t
=
Rejection region: 1.372t>
(c) 8.7 7 2.09
2.7
11
x
ts
n
µ−−
== ≈
(d) Reject 0
H.
(e) There is enough evidence at the 10% level of significance to reject the claim that the mean
wait time for callers during a recent tax filing season was at most 7 minutes.
20. (a) 0: 30 (claim); : 30
a
HHµµ≥<
(b)
02.567;t=− Rejection region: 2.567t<−
(c) 28.5 30 3.74
1.7
18
x
ts
n
µ−−
== ≈−
(d) Reject 0
H.
(e) There is enough evidence at the 1% level of significance to reject the claim that the mean
battery life of their MP3 player is at least 30 hours.
276 CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE
(e) There is enough evidence at the 10% level of significance to support the claim that the mean
amount of waste recycled by adults in the United States is more than 1 pound per person per
day.
23. (a) 0: $26,000 (claim); : $26,000
a
HHµµ=≠
(b)
00
2.262, 2.262;tt−=− = Rejection region: 2.262, 2.262tt<− >
(c) 25,852.2,x≈ $3197.1s≈
25,852.2 26,000 0.15
3197.1
10
x
ts
n
µ−−
=≈ ≈−
(d) Fail to reject 0
H.
(e) There is not enough evidence at the 5% level of significance to reject the claim that the mean
salary for full-time male workers over age 25 without a high school diploma is $26,000.
25. (a) 0: 45; : 45 (claim)
a
HHµµ≤>
(b) 48 45 2.78
5.4
25
x
ts
n
µ−−
== ≈
P-value = {Area to right of t = 2.78 } = 0.0052
(c) Reject 0
H.
(d) There is enough evidence at the 10% level of significance to support the claim that the mean
speed of the vehicles is greater than 45 miles per hour.