Question

A random sample of 31 charge sales showed a sample standard deviation of $50. a 90% confidence interval estimate of the population standard deviation is

Answer 1

The $100(1-\alpha)\%$ confidence interval of a standard deviation is given by:

$\sqrt{ \frac{(n-1)s^2}{\chi^2_{1- \frac{\alpha}{2} } }} \leq\sigma\leq\sqrt{ \frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2} } }}$

Given a sample size of 31 charge sales, the degree of freedom is 30 and for 90% confidence interval, 

$\chi^2_{1- \frac{\alpha}{2} }=43.773 \\ \\ \chi^2_{\frac{\alpha}{2} }=18.493$

Therefore, the 90% confidence interval for the standard deviation is given by

$\sqrt{ \frac{(31-1)50^2}{43.773 }} \leq\sigma\leq\sqrt{ \frac{(31-1)50^2}{18.493 }} \\ \\ \Rightarrow\sqrt{ \frac{(30)2500}{43.773 }} \leq\sigma\leq\sqrt{ \frac{(30)2500}{18.493 }} \\ \\ \Rightarrow\sqrt{ \frac{75000}{43.773 }} \leq\sigma\leq\sqrt{ \frac{75000}{18.493 }} \\ \\ \Rightarrow \sqrt{1713.38} \leq\sigma\leq \sqrt{4055.59} \\ \\ \Rightarrow41.4\leq\sigma\leq63.7$