Solutions to Homework 11
Problem 9.25
For all parts of the problem
μ
0
= 45 and
n
= 36. This means we have
n

1 = 36

1 = 35
degrees of freedom for our
t
distribution.
a. The test statistic is
t
=
¯
x

μ
0
s/
√
n
=
44

45
5
.
2
/
√
36
=

1
.
15
.
The
p
value is between 0.10 and 0.15. Since this
p
value is greater than
α
= 0
.
01, we
do not reject
H
0
and conclude that it is possible the mean is greater than or equal to
45.
b. The test statistic is
t
=
¯
x

μ
0
s/
√
n
=
43

45
4
.
6
/
√
36
=

2
.
61
.
The
p
value is between 0.005 and 0.01. Since this
p
value is less than
α
= 0
.
01, we
reject
H
0
and conclude that the mean is less than 45.
c. The test statistic is
t
=
¯
x

μ
0
s/
√
n
=
46

45
5
/
√
36
= 1
.
2
.
The
p
value is between 0.90 and 0.85. Since this
p
value is greater than
α
= 0
.
01, we
do not reject
H
0
and conclude that it is possible the mean is greater than or equal to
45.
Problem 9.33
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Jager,AbigailL
 Statistics, Degrees Of Freedom, 30%, 15 degrees

Click to edit the document details