I am trying to solve the following problem for exercise purposes:
Let $ X_1, ..., X_n $ be independent and identically distributed random variables with the probability density function:
$$ f(x\mid \theta) = \frac{x^3e^{-\frac{x}{\theta}}}{6\theta^4} $$ where $x > 0$ and $\theta > 0$.
Assuming that we know that quantiles from the chi-square distribution can be computed exactly, please construct an exact $(1-\alpha)100\%$ confidence interval for $\theta$.
I have computed the Maximum Likelihood Estimator of $\theta$ to be $\hat{\theta} = \frac{\bar{X}}{4}$.
I also observed that based on the pdf $f(x\mid \theta)$, $X_i \sim \operatorname{Gamma}(4, \frac{1}{\theta})$ where 4 is the shape and $\frac{1}{\theta}$ is the $\textbf{rate}$ of the gamma distribution. Therefore, $\frac{2}{\theta}X_i \sim \chi^2_8$.
I am stuck at this step. Could someone please advise me on how I could take it from here?