please help me!!!

Question

A random sample of 15 colleges in Kansas showed that the annual salary has a sample variance of s2=83.2. Find a 95% confidence interval for the population variance.

Answer 1

In general, the sample variance follows a chi-squared distribution with n-1 degrees of freedom.

For a 95% confidence interval, we want σ² to fall within the middle 95% of the chi-squared distribution.

Population Variance Confident Interval Formula: [ (n - 1)s²] / B < σ² < [ (n - 1)s²] / A

B and A are the lower and upper bounds of our chi-square confidence interval 95% . B is at the 2.5 percentile while A is at the 97.5 percentile. We can look these up in a chi-square table or in a calculator.

https://www.medcalc.org/manual/chi-square-table.php

We have 15-1=14 degrees of freedom.

B is the 2.5 percentile value of the chi-squared distribution for 14 degrees of freedom. B=26.119.

A is the 97.5 percentile value of the chi-squared distribution for 14 degrees of freedom. A=5.629.

Using the Population Variance Confident Interval Formula, we have [(15-1)83.2] / 26.119 < σ² < [(15-1)83.2] / 5.629.

Simplify the numbers to find the lower and upper bounds of the confidence interval for σ².