Show that a $100(1-\gamma)\%$ percent confidence interval for $\sigma^2$ is: $\left[ \dfrac{n \widehat{\sigma^2} }{\chi^2_{\gamma/2}(n-2)} , \dfrac{n \widehat{\sigma^2} }{\chi^2_{1-\gamma/2}(n-2)} \right]$
Any help would be appreciated. Normally I have some idea where to start and I state a direction in my question but this one has me wondering where to even start.
Thank you.
You have $$ \frac{n\widehat{\sigma^2}}{\sigma^2} \sim \chi^2(n-2), $$ so $$ \Pr\Big(\chi^2_{1-\gamma/2} (n-2) \le \frac{n\widehat{\sigma^2}}{\sigma^2} \le \chi^2_{\gamma/2} (n-1) \Big) = \gamma $$ Now forget for a moment about probability and look at the inequalities: $$ \chi^2_{1-\gamma/2} (n-2) \le \frac{n\widehat{\sigma^2}}{\sigma^2} \le \chi^2_{\gamma/2} (n-2). $$ Solve that system for $\sigma^2$ and you've got it. (Actually, I've seen some students get confused by trying to work with both of these two inequalities at the same time. Working on them separately may avoid that.)
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