Section 8-4
8-38
31
.
18
2
10
,
05
.
0
=
χ
49
.
27
2
15
,
025
.
0
=
χ
22
.
26
2
12
,
01
.
0
=
χ
93
.
46
2
25
,
005
.
0
=
χ
85
.
10
2
20
,
95
.
0
=
χ
01
.
7
2
18
,
99
.
0
=
χ
14
.
5
2
16
,
995
.
0
=
χ
8-39
99% lower confidence bound for
σ
2
For
α
= 0.01 and n = 15,
29.14
=
−
2
1
,
n
α
χ
=
2
14
,
01
.
0
χ
2
2
2
00003075
.
0
14
.
29
)
008
.
0
(
14
σ
σ
≤
≤
8-40
95% two sided confidence interval for
σ
8
.
4
10
=
=
s
n
02
.
19
2
9
,
025
.
0
2
1
,
2
/
=
=
−
χ
χ
α
n
and
70
.
2
2
9
,
975
.
0
2
1
,
2
/
1
=
=
−
−
χ
χ
α
n
76
.
8
30
.
3
80
.
76
90
.
10
70
.
2
)
8
.
4
(
9
02
.
19
)
8
.
4
(
9
2
2
2
2
<
<
≤
≤
≤
≤
σ
σ
σ
8-41
95% confidence interval for
σ
: given n = 51, s = 0.37
First find the confidence interval for
σ
2
:
For
α
= 0.05 and n = 51,
71.42 and
32.36
χ
α
/ ,
2
1
2
n
−
= χ
0 025 50
2
.
,
=
χ
α
1
2
1
2
−
−
=
/ ,n
χ
0 975 50
2
.
,
=
36
.
32
)
37
.
0
(
50
42
.
71
)
37
.
0
(
50
2
2
2
≤
≤
σ
0.096
≤
σ
2
≤
0.2115
Taking the square root of the endpoints of this interval we obtain,
0.31 <
σ
< 0.46
8-42
95% confidence interval for
σ
09
.
0
17
=
=
s
n
85
.
28
2
16
,
025
.
0
2
1
,
2
/
=
=
−
χ
χ
α
n
and
91
.
6
2
16
,
975
.
0
2
1
,
2
/
1
=
=
−
−
χ
χ
α
n
137
.
0
067
.
0
0188
.
0
0045
.
0
91
.
6
)
09
.
0
(
16
85
.
28
)
09
.
0
(
16
2
2
2
2
<
<
≤
≤
≤
≤
σ
σ
σ
8-14
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8-43
The data appear to be normally distributed based on examination of the normal probability plot
below.
Therefore, there is evidence to support that the mean temperature is normally distributed.

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- Spring '09
- KAILASHKAPUR
- Normal Distribution
-
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