26. a.
0103 100 2.10
/ 11.5/ 65
x
tsn
−−
= = =
Degrees of freedom = n – 1 = 64
b.
096.5 100 2.57
/ 11/ 65
x
tsn
−−
= = = −
Because t < 0, p-value is two times the lower tail area
c.
0102 100 1.54
/ 10.5/ 65
x
tsn
−−
= = =
Because t > 0, p-value is two times the upper tail area
27. a. H0:
13.04
Ha:
< 13.04
b.
012.75 13.04 1.45
/ 2 / 100
x
tsn
−−
= = = −
Degrees of freedom = n – 1 = 99
c. p-value > .05; do not reject H0. We cannot conclude that the cost of a restaurant meal is significantly
cheaper than a comparable meal fixed at home.
28. a. H0:
9
Ha:
< 9
b.
07.27 9 2.50
/ 6.38/ 85
x
tsn
−−
= = = −
29. a. H0:
= 90,000
Ha:
90,000
b.
085,272 90,000.00 2.14
/ 11,039.23/ 25
x
tsn
−−
= = = −
d. df = 24 t.025 = 2.064
Reject H0 if t < -2.064 or t > 2.064
–2.14 < -2.064; reject H0. The conclusion is the same as in part (c).
30. a. H0:
= 6.4
Ha:
6.4
c. Most researchers would choose
.10
=
or less. If you chose
= .10 or less, you cannot reject H0.
You are unable to conclude that the population mean number of hours married men with children in
your area spend in child care differs from the mean reported by Time.
31. H0:
423
Ha:
> 423
0460.4 423.0 2.20
/ 101.9 / 36
x
tsn
−−
= = =
32. a. H0:
= 10,192
Ha:
10,192
b.
09750 10,192 2.23
/ 1400/ 50
x
tsn
−−
= = = −
Degrees of freedom = n – 1 = 49
33. a. H0:
1503
Ha:
< 1503
c.
01440 1503 1.91
/ 165/ 25
x
tsn
−−
= = = −
Degrees of freedom = n – 1 = 24
p–value is lower-tail area
Ha:
2
b.
22 2.2
10
i
x
xn
= = =
c.
( )
2
.516
1
i
xx
sn
−
==
−
d.
02.2 2 1.22
/ .516/ 10
x
tsn
−−
= = =
35. a.
0
00
.175 .20 1.25
(1 ) .20(1 .20)
400
pp
zpp
n
−−
= = = −
−−
b. Because z < 0, p-value is two times the lower tail area
Using normal table with z = -1.25: p-value = 2(.1056) = .2112
Using Excel: p-value = 2*NORM.S.DIST(-1.25,TRUE) = .2113
d. z.025 = 1.96
Reject H0 if z -1.96 or z 1.96
p-value is upper-tail area
100
0
00
.52 .64 2.50
(1 ) .64(1 .64)
100
pp
zpp
n
−−
= = = −
−−
Because z < 0, p-value is two times the lower tail area
Using normal table with z = -2.50: p-value = 2(.0062) = .0124
b. Proportion of stay at home residents in Arkansas
74 .6167
120
p==
0
00
.6167 .577 .88
(1 ) .577(1 .577)
120
pp
zpp
n
−−
= = =
−−
Because z > 0, p-value is two times the upper tail area
Using normal table with z = .88: p-value = 2(.1894) = .3788
.494 .577 2.25
.577(1 .577)
180
−
Because z < 0, p-value is two times the lower tail area
Using the normal table with z = -2.25: p-value = 2(.0122) = .0244
Using Excel: p-value = 2*NORM.S.DIST(-.98,TRUE) = . 0245
Reject H0. Conclude that the proportion of stay at home residents in Virginia does differ significantly
from the overall proportion of .577.
c. p-value
.01
=
; reject H0. We conclude that there has been a statistically significant decline in
the proportion of American families owning stocks or stock funds over the ten–year period.
42. a.
p
= 12/80 = .15
.15 1.96 (.0399)
.15 .0782 or .0718 to .2218
c. We can conduct a hypothesis test concerning whether the return rate for the Houston store is equal to
.06 at an α = .05 level of significance using the 95% confidence interval in part (b). Since the
confidence interval does not include .06, we conclude that the return rate for the Houston store is
92 .6133
150
0
00
.6133 .50 2.78
(1 ) (.50)(.50)
150
pp
zpp
n
−−
= = =
−
p-value is the area in the upper tail at z = 2.78
Using normal table with z = 2.78: p-value = 1 – .9973 = .0027
Using Excel: p-value = 1-NORM.S.DIST(2.78,TRUE) = .0027