Section 7.4: Confidence Interval for Variance
and Standard Deviation of a Normal Population
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Suppose X1, · · · , Xn are iid observations of a normal sample, say N (µ, σ 2). Then,
(n − 1)S 2
2
∼
χ
n−1 .
σ2
Thus, we have
(n − 1)S 2
2
≤
≤
χ
α/2,n−1) = 1 − α
σ2
which gives the 1 − α level confidence interval for
σ 2 as
P (χ2
1−α/2,n−1
(n − 1)s2 (n − 1)s2
[ 2
,
].
χα/2,n−1 χ2
1−α/2,n−1
and for σ as
v
v
u
u
u (n − 1)s2 u (n − 1)s2
[u
,u
].
t 2
t 2
χα/2,n−1
χ1−α/2,n−1
2
First example of Section 7.4: example 7.15 on
textbook. In this example, the data collected 17
observations of breakdown voltage. We have n =
17 and s2 = 137324.3. Thus for 95% confidence
interval for σ 2, we need
χ2
0.975,16 = 6.908
and
χ2
0.025,16 = 28.845.
Thus, the 95% confidence interval for σ 2 is
16s2 16s2
[
,
] = [76172.3, 318064.4]
28.845 6.908
and the 95% confidence interval for σ is
√
√
[ 76172.3, 318064.4] = [276.0, 564.0].
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