The monthly incomes of 13 randomly selected individuals who

The monthly incomes of 13 randomly selected individuals who have recently graduated with a bachelor\'s degree in economics have a sample standard deviation of $339. Construct a confidence interval for the population variance sigma^2 and the population standard donation sigma. Use a 95% level of confidence. Assume the sample is from a normally distributed population. What is the confidence interval for the population variance sigma^2?

Solution

Confidence Interval
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 12 df are 23.3367 , 4.404
S.D( S^2 )=339
Sample Size(n)=13
Confidence Interval = [ 12 * 114921/23.3367 < ^2 < 12 * 114921/4.404 ]
= [ 1379052/23.3367 < ^2 < 1379052/4.4038 ]
Confidence Interval for ^2 = [ 59093.7022 < ^2 < 313150.461 ] ~ [ 59094 < ^2 < 313150 ]