Use technology to construct the confidence intervals for me

Use technology to construct the confidence intervals for me population variance sigma^2 and the population standard deviation sigma. Assume the sample is taken from 3. nermally diStrlOUIed population. c = 0.95, s = 32, n = 17 The confidence interval for the population variance is .

Solution

CI = (n-1) S^2 / ^2 right < ^2 < (n-1) S^2 / ^2 left
Where,
S = Standard Deviation
^2 right = (1 - Confidence Level)/2
^2 left = 1 - ^2 right
n = Sample Size

Since aplha =0.05
^2 right = (1 - Confidence Level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
^2 left = 1 - ^2 right = 1 - 0.025 = 0.975
the two critical values ^2 left, ^2 right at 16 df are 28.8454 , 6.908
S.D( S^2 )=32
Sample Size(n)=17
Confidence Interval = [ 16 * 1024/28.8454 < ^2 < 16 * 1024/6.908 ]
= [ 16384/28.8454 < ^2 < 16384/6.9077 ]
Confidence Interval for ^2 = [ 567.9935 < ^2 < 2371.8459 ] = [ 567.99 < ^2 < 2371.85 ]