Chapter 9
Hypothesis Tests
Learning Objectives
1. Learn how to formulate and test hypotheses about a population mean and a population proportion.
2. Understand the types of errors possible when conducting a hypothesis test.
3. Understand the level of significance and be able to determine the probability of making a Type I
error 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. 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-tail test
Solutions:
1. a. H0:
600 Manager’s claim, the assumption being challenged
Ha:
> 600
2. a. H0:
14
Ha:
> 14 Research hypothesis
increases the mean sales volume.
3. a. H0:
= 32 Specified filling weight, the assumption being challenged
Ha:
32 Overfilling or underfilling exists
4. a. H0:
220
Ha:
< 220 Research hypothesis to see if mean cost is less than $220.
b. We are unable to conclude that the new method reduces costs.
5. a. Conclude that the population mean monthly cost of electricity in the Chicago neighborhood is
greater than $104 and hence higher than in the comparable neighborhood in Cincinnati.
is not.
6. a. H0:
1 The label claim or assumption.
Ha:
> 1
label specification.
7. a. H0:
8000
Ha:
> 8000 Research hypothesis to see if the plan increases average sales.
implementing a plan that would increase sales.
8. a. H0:
220
Ha:
< 220
implementing a method that would lower costs.
9. a.
019.4 20 2.12
/ 2/ 50
x
zn
−−
= = = −
b. p-value is the area in the lower tail
c. p-value .05, reject H0
-2.12 -1.645, reject H0
10. a.
026.4 25 1.48
/ 6/ 40
x
zn
−−
= = =
b. p-value is the area in the upper tail: P(z ≥ 1.48)
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.
014.15 15 2.00
/ 3/ 50
x
zn
−−
= = = −
b. Because z < 0, p-value is two times the lower tail area
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.
078.5 80 1.25
/ 12/ 100
x
zn
−−
= = = −
p-value is the lower-tail area
p-value > .01, do not reject H0
077 80 2.50
x
−−
Using Excel: p-value = NORM.S.DIST(-2.50,TRUE) = . 0062
p-value .01, reject H0
c.
075.5 80 3.75
/ 12/ 100
x
zn
−−
= = = −
p-value .01, reject H0
d.
Using Excel: p-value = NORM.S.DIST(.83,TRUE) = .7967
p-value > .01, do not reject H0
13. Reject H0 if z 1.645
a.
052.5 50 2.42
/ 8/ 60
x
zn
−−
= = =
1.74 1.645, reject H0
14. a.
023 22 .87
/ 10/ 75
x
zn
−−
= = =
Because z > 0, p-value is two times the upper tail area
p-value > .01, do not reject H0
b.
025.1 22 2.68
/ 10/ 75
x
zn
−−
= = =
Because z > 0, p-value is two times the upper tail area
p-value .01, reject H0
c.
020 22 1.73
/ 10/ 75
x
zn
−−
= = = −
Because z < 0, p-value is two times the lower tail area
15. a. H0:
 
Ha:
< 1056
b.
0910 1056 1.83
/ 1600/ 400
x
zn
−−
= = = −
Using Excel: p-value = NORM.S.DIST(-1.83,TRUE) = .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:
3173
Ha:
> 3173
b.
03325 3173 2.04
/ 1000 / 180
x
zn
−−
= = =
c. p-value < .05. Reject H0. The current population mean credit card balance for undergraduate
students has increased compared to the previous all-time high of $3173 reported in April 2009.
17. a. H0:
= 24.57
Ha:
24.57
b.
023.89 24.57 1.55
/ 2.4 / 30
x
zn
−−
= = = −
c. p-value > .05, do not reject H0. We cannot conclude that the population mean hourly wage for
manufacturing workers differs significantly from the population mean of $24.57 for the goods-
producing industries.
18. a. H0:
= 192
Ha:
192
b.
0182 192 2.23
/ 55/ 150
x
zn
−−
= = = −
19. H0:
≥ 12
Ha:
< 12
10 12
01.77
8 50
x
zn
−−
= = = −
claim of 12 minutes made by the taxpayer advocate.
20. a. H0:
838
Ha:
< 838
b.
0745 838 2.40
/ 300 60
x
zn
−−
= = = −
c. Lower tail p-value is area to left of the test statistic.
21. a. H0:
15
Ha:
> 15
b.
17 15 2.96
/ 4/ 35
x
zn
−−
= = =
22. a. H0:
= 8
Ha:
8
b.
8.4 8.0 1.37
/ 3.2 / 120
x
zn
−−
= = =
c. Do not reject H0. Cannot conclude that the population mean waiting time differs from 8 minutes.
23. a.
014 12 2.31
/ 4.32/ 25
x
tsn
−−
= = =
b. Degrees of freedom = n – 1 = 24
p-value is upper-tail area
24. a.
017 18 1.54
/ 4.5/ 48
x
tsn
−−
= = = −
b. Degrees of freedom = n – 1 = 47
Because t < 0, p-value is two times the lower tail area
Using t table: area in lower tail is between .05 and .10; therefore, p-value is between .10 and .20.
/ 5.2/ 36
Degrees of freedom = n – 1 = 35
p–value is lower-tail area
Using t table: p-value is between .10 and .20
/ 4.6/ 36
p-value is lower-tail area
Using t table: p–value is between .005 and .01
Using Excel: p-value = T.DIST(-2.61,35,TRUE) = .0066