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)