Consider a random sample $X_1,X_2,..,X_n$ from a variable with density function $$f_X(x)=2\lambda\pi xe^{-\lambda\pi x^2} \ \ \ \ \ \ \ x>0$$ A useful estimator of $\lambda$ is $$\hat{\lambda}=\frac{n}{\pi\sum_{i=1}^{n} X^2_1}$$ Derive a 95% confidence interval for $\lambda$ $\big($a hint is provided suggesting to consider the distribution of $\frac{\lambda}{\hat{\lambda}}\big).$
Taking the advice of the hint, I attempted to find the distribution of $\frac{\lambda}{\hat{\lambda}}.$ $$\frac{\lambda}{\hat{\lambda}}=\frac{\lambda}{\frac{n}{\pi\sum_{i=1}^{n} X^2_1}}=\frac{\pi\lambda}{n}\sum_{i=1}^{n} X^2_i$$ It can be shown that $$\frac{\pi}{n}\sum_{i=1}^{n} X^2_i\sim\text{Gamma}\Big(n,\frac{1}{n\lambda}\Big)$$ and hence $$\lambda\Bigg(\frac{\pi}{n}\sum_{i=1}^{n} X^2_i\Bigg)\sim\text{Gamma}\Big(n,\frac{1}{n}\Big)$$ But how does this help find a confidence interval? The only formula I really know for confidence intervals for similar types of questions is $\hat{\lambda}\pm z_{0.975}\text {se}(\hat{\lambda})$ where $Z\sim N(0,1)$
Edit, using the pivotal method: $$\mathbb{P}\Big(g_{\frac{\alpha}{2}}\leq \frac{\lambda}{\hat{\lambda}}\leq g_{1-\frac{\alpha}{2}}\Big)=0.95$$ $$\mathbb{P}\Big(\hat{\lambda}g_{0.025}\leq\lambda\leq\hat{\lambda}g_{0.975}\Big)=0.95$$ Hence a 95% confidence interval is $$\Big(\hat{\lambda}g_{0.025},\hat{\lambda}g_{0.975}\Big)$$ where $g_{\frac{\alpha}{2}}$ is the $\frac{\alpha}{2}$%tile of the $\text{Gamma}(n,\frac{1}{n})$ distribution.
You can take the corresponding quantiles for the Gamma distribution and proceed similarly as if the distribution was normal - let $Z$ be Gamma distributed. Then you have:
$$\mathbb{P}(Z\in(q_{\frac{\alpha}{2}}, q_{1-{\frac{\alpha}{2}}}))=1-\alpha$$
where $q_{\frac{\alpha}{2}}, q_{1-{\frac{\alpha}{2}}}$ are the corresponding quantiles