I've been trying to solve the following problem:
Let $X_1,\ldots,X_n$ be a sample of a random variable $X\sim\text{Exp}(\lambda)$, i.e., the exponential distribution with parameter $\lambda>0$. Consider the null hypothesis $H_0: \lambda=\lambda_0$ and the alternative hypothesis $H_1: \lambda\neq\lambda_0$. Construct a confidence interval for the likelihood-ratio test with significance level $\alpha$.
I know that $\sum\limits_{i=1}^nX_i\sim\text{Gamma}(n,\lambda)$, so $2\lambda n\mathbf{\overline X}\sim\text{Gamma}(n,\frac{1}{2})\sim\chi^2_{2n}$, and normally I'd use that to construct the confidence interval picking $c_1$ and $c_2$ quantiles of the $\chi^2_{2n}$ distribution such that $P(c_1\le2\lambda n\mathbf{\overline X}\le c_2)=1-\alpha$ and then isolate $\lambda$ (usually I take $c_1$ and $c_2$ such that there is a probability of $\frac{\alpha}{2}$ on both sides of the opposite inequality).
However, now I must use the likelihood-ratio test first, and then construct the confidence interval, and that's where I'm stumped.
I obtained the likelihood-ratio test statistic: $\lambda(\mathbf{x})=\lambda_0^n\,\mathbf{\overline x}^ne^n\exp(-n\mathbf{\overline x}\lambda_0)$, so the test would be
$\begin{cases}\text{Reject }H_0\text{ if }\lambda(\mathbf{x})\le c\\\text{Not reject }H_0\text{ otherwise}\end{cases}$
with $c$ such that $P_{\lambda_0}\big(\lambda(\mathbf X)\le c\big)=\alpha$, where the subindex indicates we calculate that probability assuming $\lambda=\lambda_0$.
I can isolate a nicer part of the likelihood-ratio test in the inequality $\lambda(\mathbf x)\le c\Rightarrow\lambda_0\,\mathbf{\overline x}\exp(-\lambda_0\mathbf{\overline x})\le\dfrac{\sqrt[n] c}{e}$, so naming $k=\dfrac{\sqrt[n] c}{e}$ we get $P_{\lambda_0}\big(\lambda(\mathbf X)\le c\big)=P_{\lambda_0}\Big(\lambda_0\,\mathbf{\overline X}\exp(-\lambda_0\,\mathbf{\overline X})\le k\Big)$. This is the point of the problem where I need help, since I have no idea how to proceed.
Any ideas? Thanks in advance!