Variance interval from a reported standard deviation

Question

A sample of n = 15 measurements from a normal population has sample standard deviation s = 5. Construct a 95% confidence interval for the population variance \sigma^{2}.

Accepted Answer

Setup. First square the standard deviation: s^{2} = 25. Then use \left[\frac{(n-1)s^{2}}{\chi^{2}{	ext{upper}}},\ \frac{(n-1)s^{2}}{\chi^{2}{	ext{lower}}}ight], with n - 1 = 14. The critical values are \chi^{2}{0.025,14} = 26.119 and \chi^{2}{0.975,14} = 5.629. Here (n-1)s^{2} = 14 	imes 25 = 350. Compute. 	ext{lower} = \frac{350}{26.119} = 13.40, \qquad 	ext{upper} = \frac{350}{5.629} = 62.18. So the interval for \sigma^{2} is [13.40,\ 62.18]. Interpret. With only 14 degrees of freedom the interval is dramatically wide: the upper bound is more than four times the lower bound, a remind…

Variance interval from a reported standard deviation

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