Confidence interval for a standard deviation

Question

A sample of n = 10 measurements from a normal population has sample variance s^{2} = 12. Construct a 95% confidence interval for the population standard deviation \sigma.

Accepted Answer

Setup. First build the interval for the variance, then take square roots. Use \left[\frac{(n-1)s^{2}}{\chi^{2}{	ext{upper}}},\ \frac{(n-1)s^{2}}{\chi^{2}{	ext{lower}}}ight], with n - 1 = 9. The critical values are \chi^{2}{0.025,9} = 19.023 and \chi^{2}{0.975,9} = 2.700. Here (n-1)s^{2} = 9 	imes 12 = 108. Compute the variance interval. 	ext{lower} = \frac{108}{19.023} = 5.68, \qquad 	ext{upper} = \frac{108}{2.700} = 40.0. Take square roots for the standard deviation. [\sqrt{5.68},\ \sqrt{40.0}] = [2.38,\ 6.32]. Interpret. Because the square root is monotonic, the endpoints for \sigma …

Confidence interval for a standard deviation

Your answer

More Statistics questions