Question

Construct a​ 95% confidence interval for the population standard deviation sigma of a random sample of 15 men who have a mean weight of 165.2 pounds with a standard deviation of 12.5 pounds. Assume the population is normally distributed.

Answer 1

From chi square distribution tables for n-1 = 14 degrees of freedom, we get:

$P(\chi^2_{14} < 26.12) = 0.975$

$P(\chi^2_{14} < 5.629) = 0.025$

Therefore the confidence interval for the population standard deviation here is computed as:

$\sqrt{\frac{(n-1)s^2}{26.12}} < \sigma < \sqrt{\frac{(n-1)s^2}{5.629}}$

$\sqrt{\frac{14*12.5^2}{26.12}} < \sigma < \sqrt{\frac{14*12.5^2}{5.629}}$

$9.15 < \sigma < 19.71$

This is the required 95% confidence interval for the population standard deviation here.