Question

The annual earnings of 12 randomly selected computer software engineers have a sample standard deviation of

$ 3626 Assume the sample is from a normally distributed population. Construct a confidence interval for the population variance

sigma squared σ2 and the population standard deviation sigma σ. Use a 95%

level of confidence. Interpret the results.

What is the confidence interval for the population variance

sigma squared σ2​?

What is the confidence interval for the population standard deviation

sigma σ​?

Please show all steps

Answer 1

a)

Chi-square critical values at 0.05 level with 11 df = $\chi^{2}$ L = 3.816 , $\chi^{2}$ U = 21.920

95% confidence interval for $\sigma^{2}$ is

(n-1) S² /  $\chi^{2}$U <  $\sigma^{2}$ < (n-1) S² /  $\chi^{2}$L

11 * 3626² / 21.920 < $\sigma^{2}$ < 11 * 3626² / 3.816

6597930 < $\sigma^{2}$ < 37900062

We are 95% confident that population variance is between 6597930 and 37900062

b)

95% confidence interval for $\sigma$ is

Sqrt [ (n-1) S² /  $\chi^{2}$U ] <  $\sigma$ < sqrt [ (n-1) S² /  $\chi^{2}$L ]

sqrt [11 * 3626² / 21.920 ] < $\sigma$ < sqrt [11 * 3626² / 3.816]  

2569 < $\sigma$ < 6156

We are 95% confident that population variance is between 2569 and 6156