Variance interval from 25 observations

Question

A sample of n = 25 measurements from a normal population has sample variance s^{2} = 9. Construct a 95% 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 = 24. The critical values are \chi^{2}{0.025,24} = 39.364 and \chi^{2}{0.975,24} = 12.401. Here (n-1)s^{2} = 24 	imes 9 = 216. Compute. 	ext{lower} = \frac{216}{39.364} = 5.49, \qquad 	ext{upper} = \frac{216}{12.401} = 17.42. So the interval for \sigma^{2} is [5.49,\ 17.42]. Interpret. With more degrees of freedom than a smaller sample, the chi-square distribution is less skewed and the interval is comparatively tighter, though still asymmetric a…

Variance interval from 25 observations

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