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12. Let
μ
be the true mean time to completion for this semester.
(a) Hypotheses:
H
0
:
μ
= 35
H
a
:
μ
6
= 35
Test Statistic:
t
=
X

μ
0
S/
√
n
=
38
.
2

35
4
.
3
/
√
20
≈
3
.
33
Rejection Region:
Twotailed from a
t
curve with
n

1 = 19 degrees
of freedom.
The rejection region for
α
= 0
.
05 is above 2
.
093 and below

2
.
093.
Conclusion: The test statistic is in the rejection region, therefore we reject
H
0
in favor of
H
a
at 0
.
05 level of signiFcance. The data suggests that the
true mean time to completion this semester is di±erent from past semesters.
(b)
X
±
t
α/
2
,n

1
S
√
n
38
.
2
±
(2
.
093)
4
.
3
√
20
(36
.
18756
,
40
.
21244)
(c) Yes, the conFdence interval supports the conclusion from (a) that
μ
6
= 35
because 35 is not in this interval of “plausible values” for
μ
.
13. Let
p
be the true proportion of adults who will get relief from the drug.
ˆ
p
=
70
/
100 = 0
.
70.
(a) Hypotheses:
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