2
2
2
1
2
1
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21
(
21
nn
xx
σσ
µµ
+
−−−
Practice Homework #2 Solutions
Pg. 352: 8.57 H0: µ=0.100 HA: µ>0.100 α=0.01
97.2
8
011.0
100.0210.0
71 =
−
=
−
=
=−=
n
s
x
tndf
µ
The rejection region is >2.998, so we fail to reject the null hypothesis. We cannot conclusively say it exceeds the
limits.
Pg. 357: 8.62 H
0: p=0.75 HA: p<0.75 α=0.05
a) Since 0.69 is less than 0.75 it does seem to contradict the null hypothesis.
b) 39.1
100
)75.01(75.0
75.069.0
)1(
ˆ−=
−
−
=
−
−
=
n
pp
pp
z
The rejection region is <-1.645, so we fail to reject the null hypothesis. We find that while it looks like p<0.75, we
do not have enough evidence to decide this conclusively.
c) We need to find the area less than z=-1.39. The table gives that the area between 0 and 1.39 is 0.4177, so the p-
value is 0.823 (draw the picture!). There is an 8.23% chance we would observe this much evidence against the
null hypothesis, even if it were true. This does not meet the 5% standard we set.
Pg. 370: 8.88
a) We must assume that the population is normally distributed (check it using a q-q plot).
b) H0: σ2=1 HA: σ2>1 α=0.05
04.29
1
84.4)17(
2
2
)1(
2
61 =
−
=
−
=
=−=
σ
χ
sn
ndf
The rejection region is >12.5916, so we reject the null hypothesis and conclude the variance is greater than one.
c) H0: σ2=1 HA: σ2≠1 α=0.05
04.29
1
84.4)17(
2
2
)1(
2
61 =
−
=
−
=
=−=
σ
χ
sn
ndf
The rejection region is <1.237347 or >14.4494, so we reject the null hypothesis and conclude the variance is not
equal to one.
Pg. 392: 9.10
a) The output shows a p-value of 0.1114 so we would not reject the null hypothesis even if the a-level was 0.10.
(Note SAS always gives the p-value for ≠, so we would have needed to draw the picture and figure out the p-value
manually if the alternate hypothesis was < or > ).
To check the numbers we first see that since we were testing H0:(µ1-µ2)=0 vs. HA:(µ1-µ2)≠0 we need the test
statistic that compares two means:
