The waiting times in minutes of a random sample of 20 people

The waiting times (in minutes) of a random sample of 20 people at a bank have a sample standard deviation or3.7 minutes. Construct a confidence interval for the population variance sigma^2 and the population standard deviation sigma. Use a 99% level of confidence. Assume the sample is from a normally distributed population. what is the confidence interval for the population variance sigma^2?

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 =0.01
^2 right = (1 - Confidence Level)/2 = (1 - 0.99)/2 = 0.01/2 = 0.005
^2 left = 1 - ^2 right = 1 - 0.005 = 0.995
the two critical values ^2 left, ^2 right at 19 df are 38.5823 , 6.844
S.D( S^2 )=3.7
Sample Size(n)=20
Confidence Interval = [ 19 * 13.69/38.5823 < ^2 < 19 * 13.69/6.844 ]
= [ 260.11/38.5823 < ^2 < 260.11/6.844 ]
Confidence Interval for ^2 = [ 6.7417 < ^2 < 38.0056 ] ~ [ 6.7 < ^2 < 38 ]