Given that $X_1...X_{50}$ are IID with $N(\mu,\sigma^2)$ (both parameters are unknown), $x_1.....x_{50}$ are realizations of the random variable, with $\bar{x} = 0.2, s^2 = 0.73$.
I am asked to find the exact 95% distribution CI for $\sigma^2$, given that ${\chi}^2_{49,0.975}$ = 31.6 and ${\chi}^2_{49,0.0.025}$ = 70.2.
I actually don't know how to start, except that by the definition, ${\chi}^2 = \frac{(n-1)s^2}{\sigma^2}$.
I suspect I am going to have to take the difference between 70.2 and 31.6, since 0.975 - 0.025 = 0.95, which is the confidence interval, but I am not too sure.
The confidence interval for $\sigma^2$ is: $$\chi^2_{(1-\alpha/2)}\le \frac{(n-1)s^2}{\sigma^2}\le \chi^2_{\alpha/2} \iff \\ \frac{(n-1)s^2}{\chi^2_{\alpha/2}}\le \sigma^2\le \frac{(n-1)s^2}{\chi^2_{(1-\alpha/2)}} \Rightarrow \\ \frac{(50-1)s^2}{\chi^2_{0.025;49}}\le \sigma^2\le \frac{(50-1)s^2}{\chi^2_{0.975;49}}.$$ Can you finish?