90% chi-square interval for a variance

Question

A sample of n = 30 measurements from a normal population has sample variance s^{2} = 25. Construct a 90% confidence interval for the population variance \sigma^{2}.

Accepted Answer

Setup. Use the chi-square interval \left[\frac{(n-1)s^{2}}{\chi^{2}{	ext{upper}}},\ \frac{(n-1)s^{2}}{\chi^{2}{	ext{lower}}}ight], with n - 1 = 29. For 90% confidence each tail holds 5%, so the critical values are \chi^{2}{0.05,29} = 42.557 (upper) and \chi^{2}{0.95,29} = 17.708 (lower). Here (n-1)s^{2} = 29 	imes 25 = 725. Compute. 	ext{lower} = \frac{725}{42.557} = 17.04, \qquad 	ext{upper} = \frac{725}{17.708} = 40.94. So the interval for \sigma^{2} is [17.04,\ 40.94]. Interpret. Two things narrow this interval relative to the smaller-sample cases: the larger sample of 30, and the l…

90% chi-square interval for a variance

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