Question

A random sample of 16 men have a mean height of 67.5 inches and a standard deviation of 2.22.2 inches. Construct a​ 99% confidence interval for the population standard​ deviation, sigmaσ. Assume the sample is from a normally distributed population.

Answer 1

Answer: $\approx1.49$

Step-by-step explanation:

We know that the confidence interval for population standard deviation is given by :-

$\sqrt{\dfrac{(n-1)s^2}{\chi^2_{\alpha/2}}}$

, where n= sample size

s = sample standard deviation.

$\chi^2_{\alpha/2}$  and $\chi^2_{1-\alpha/2}$= Chi-square critical value for degree of freedom (n-1) and significance level ($\alpha$).

Given : $\alpha=1-0.99=0.01$

n= 16

Critical ch-square values for degree of freedom 15 and $\alpha=0.01$will be :

$\chi_{\alpha/2}=\chi_{0.005}=32.80132$

$\chi_{1-\alpha/2}=\chi_{0.995}=4.6009$

Then , the required 99% confidence interval for the population standard​ deviation will be :

$\sqrt{\dfrac{(15)(2.2)^2}{32.80132}}$

$\sqrt{2.21332556129}$

$1.48772496157$

$\approx1.49$

Hence, the a​ 99% confidence interval for the population standard​ deviation :