The monthly incomes of 12 randomly selected individuals The

The monthly incomes of 12 randomly selected individuals...

The monthly incomes of 12 randomly selected individuals who have recently graduated with a bachelor\'s degree in economics have a sample standard deviation of $338. Construct a confidence interval for the population variance sigma^2 and the population standard deviation sigma. Use a 90% level of confidence. Assume the sample is from a normally distributed population. What is the confidence interval for the population variance sigma^2? What is the confidence interval for the population standard deviation sigma?

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 =0.1
^2 right = (1 - Confidence Level)/2 = (1 - 0.9)/2 = 0.1/2 = 0.05
^2 left = 1 - ^2 right = 1 - 0.05 = 0.95
the two critical values ^2 left, ^2 right at 11 df are 19.6751 , 4.575
S.D( S^2 )=338
Sample Size(n)=12
Confidence Interval = [ 11 * 114244/19.6751 < ^2 < 11 * 114244/4.575 ]
= [ 1256684/19.6751 < ^2 < 1256684/4.5748 ]
CI Population Variance = [ 63871.7973 < ^2 < 274697.0359 ] =[ 63872 < < ^2 < 274698 ]

CI Population SD = [ Sqrt(63872) < < Sqrt(274697.0359) ] =[ 253 < < ^2 < 525 ]