Let $X_1, X_2 ... X_n \text{ ~ } N(\mu, \sigma^{2})$ be identically distributed and consider the estimate $$ \sigma_* ^{2} = \frac{1}{n} \sum\limits_{i=1}^n (X_i - \mu)^2 $$ for the variance.
Find number $a$ and $b \in \mathbb{R} $ such that $ P(\sigma_*^2 < a\sigma ^2) =0.025$ and $ P(\sigma_*^2 > b\sigma ^2) =0.025$
This is all that was given in the question, and I'm not sure how to approach it, I was thinking of using the equation for $\sigma ^2$ but its the same (or I think it is) as $ \sigma_* ^{2}$. This question is in relation to hypothesis testing and so far I've looked at z and t tests, but I'm not sure how i can standardise these two values. Any help is much appreciated.
$\sigma_*^2$ is a random variable, being a function of the random variables $X_1, \ldots, X_n$. Its probability distribution is derived from the sampling distribution, and your goal is to determine what this distribution is. Note that $$X_i - \mu \sim \operatorname{Normal}(0, \sigma^2)$$ and $$\frac{X_i - \mu}{\sigma} \sim \operatorname{Normal}(0,1)$$ is a pivotal quantity. So $$\Pr[\sigma_*^2 < a\sigma^2] = \Pr\left[\frac{\sigma_*^2}{\sigma^2} < a\right],$$ and since $$\frac{\sigma_*^2}{\sigma^2} = \frac{1}{n} \sum_{i=1}^n \left(\frac{X_i - \mu}{\sigma}\right)^2$$ is the mean of $n$ IID standard normal random variables, $$n\frac{\sigma_*^2}{\sigma^2} \sim \chi^2_n.$$ How would you find the constant $a$ based on this information?