In: Statistics and Probability
Use technology to construct the confidence intervals for the population variance sigma squared and the population standard deviation sigma. Assume the sample is taken from a normally distributed population. cequals0.98, ssquaredequals7.84, nequals28 The confidence interval for the population variance is
Solution :
Given that,
c = 0.98
s = 7.84
n = 28
At 98% confidence level the
is ,
= 1 - 98% = 1 - 0.98 = 0.02
/ 2 = 0.02 / 2 = 0.01
/2,df =
0.01,27 = 23.68
and
1-
/2,df =
0.99,27 = 6.57
Point estimate = s2 = 7.84
2L
=
2
/2,df
= 23.68
2R
=
21 -
/2,df = 6.57
The 98% confidence interval for
2 is,
(n - 1)s2 /
2
/2
<
2 < (n - 1)s2 /
21 -
/2
( 27*7.84) / 23.68 <
2 < ( 27 * 7.84) / 6.57
4.51 <
2 < 16.44
(4.51 , 16.44)