August 2011
Intermediate to advanced
634 pages
21h 57m
English
Content preview from A First Course in Quality Engineering, 2nd Edition
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Statistics for Quality 119
Solution
First, we find the CI for σ
2
, and then obtain the CI for σ:
α/2 = 0.005 (n – 1) = 3 s
2
= 0.133 (calculated from the sample).
We have to calculate the limits:
3(0. 133) 3(0.133)
0.005,3
2
0.995,3
2
χ
,
χ
From the χ
2
tables: χ
2
0.005,3
= 12.838 χ
2
0.995,3
= 0.0717
99
3 0 133
12 838
3 0 133
0 0717
0 03
2
%
.
.
,
.
.
. CI for σ =
( )( ) ( )( )
=
11 5 56
99 031 5 56 0 176 2 36
, .
% . , . . , . .
[ ]
=
=
[ ]
CI for σ
Now, suppose we want a 95% CI for σ in the above example:
α χ χ
σ
2 = 0.025
CI for
0 025 3
2
0 975 3
2
2
9 348 0 216
95 0 042
. , . ,
. .
% .
= =
= ,, .
% . , . .
1 847
95 0 205 1 359
[ ]
=
[ ]
CI for σ
We see that the smaller t
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