Confidence interval for a variance using chi-square

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A sample of n = 20 measurements from a normal population has sample variance s^{2} = 16. Construct a 95% confidence interval for the population variance \sigma^{2}.

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Setup. For a normal population, (n-1)s^{2}/\sigma^{2} follows a chi-square distribution with n-1 degrees of freedom. Inverting gives \left[\frac{(n-1)s^{2}}{\chi^{2}{	ext{upper}}},\ \frac{(n-1)s^{2}}{\chi^{2}{	ext{lower}}}ight]. Here n - 1 = 19. The two critical values that leave 2.5% in each tail are \chi^{2}{0.025,19} = 32.852 (upper) and \chi^{2}{0.975,19} = 8.907 (lower). Note (n-1)s^{2} = 19 	imes 16 = 304. Compute. 	ext{lower} = \frac{304}{32.852} = 9.25, \qquad 	ext{upper} = \frac{304}{8.907} = 34.13. So the interval for \sigma^{2} is [9.25,\ 34.13]. Interpret. The interval is v…

Confidence interval for a variance using chi-square

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