7.1
a.
z
α/
2
= 2
.
81 suggests that
α/
2 = 1

Φ(2
.
81) = 0
.
0025. Thus
α
= 0
.
005
and the corresponding confidence level is 99
.
5%.
b.
z
α/
2
= 1
.
44 suggests that
α/
2 = 1

Φ(2
.
81) = 0
.
075. Thus
α
= 0
.
15 and
the corresponding confidence level is 85%.
c. 99
.
7% implies an
α
value of 0
.
003 and consequently
z
α/
2=0
.
0015
= 2
.
96.
d. 75% implies an
α
value of 0
.
25 and consequently
z
α/
2=0
.
125
= 1
.
15.
7.4
a. 58
.
3
±
1
.
96
3
√
25
= [57
.
1
,
59
.
5]
b. 58
.
3
±
1
.
96
3
√
100
= [57
.
7
,
58
.
9]
c. 58
.
3
±
2
.
58
3
√
100
= [57
.
5
,
59
.
1]
d. 82% confidence implies an
α
value of 0
.
18 and consequently
z
α/
2=0
.
09
= 1
.
34
and the interval based on this would be 58
.
3
±
1
.
34
3
√
100
= [57
.
9
,
58
.
7].
e.
n
=
h
2
×
2
.
58
×
3
1
i
2
= 239
.
62
≈
240
7.13
a. ¯
x
±
1
.
96
0
.
163
√
69
= [0
.
990
,
1
.
066]
7.24 8
.
17
±
1
.
96
1
.
42
√
56
= [7
.
798
,
8
.
542]. No assumptions are made on the distribution
of percentage elongation.
7.34
a. A 95% lower confidence bound is 8
.
48

1
.
771
0
.
79
√
14
= 8
.
11. With 95% confi
dence, the true mean proportional limit stress of all such joints lies in the
interval [8
.
11
,
∞
). If this interval is calculated for infinitely many samples,
95% of these intervals will contain the true mean. We must assume that the
sample observations were coming from a normally distributed population.
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 Fall '08
 Staff
 Null hypothesis, Statistical hypothesis testing, H0, A. Ha

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