mont4e_sm_ch08_sec04 - Section 8-4 8-38 02.05,10 = 18.31...

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

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.