Question

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 $ 328. Construct a confidence interval for the population variance sigma squared and the population standard deviation 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 squared​? left parenthesis nothing comma nothing right parenthesis ​(Round to the nearest integer as​ needed.) What is the confidence interval for the population standard deviation sigma​? left parenthesis nothing comma nothing right parenthesis ​(Round to the nearest integer as​ needed.)

Answer 1

Solution :

Given that,

s = 328

Point estimate = s² = 107584

n = 12

Degrees of freedom = df = n - 1 = 12 - 1 = 11

At 95% confidence level the ² value is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

1 - / 2 = 1 - 0.025 = 0.975

²_L = ²_/2,df = 21.92

²_R = ²_{1 - /2,df} = 3.816

The 95% confidence interval for ² is,

(n - 1)s² / ²_/2 < ² < (n - 1)s² / ²_{1 - /2}

11 * 107584 / 21.92 < ² < 11 * 107584 / 3.816

53988  < ² < 310146

The 95% confidence interval for ² is : (53988 , 310146)

The 95% confidence interval for is,

232 < < 557