Confidence interval; exponential distribution (normal or student approximation?)

Question

Let's say we have got a sample of size $n$ from an exponential distribution with an unknown mean $\lambda$.

We want to construct a confidence interval and so we can compare this: $$\frac{\bar{X}-\lambda}{\sqrt{\frac{S^2}{n}}}$$ to a student t-distribution with $n-1$ degrees of freedom.

However, as in the case of the exponential distribution, we know that $Var[X]=(E[X])^2$, so rather than introducing an estimator for variance, we can simply use one estimator, i.e: $$\frac{\bar{X}-\lambda}{\sqrt{\frac{\bar{X}^2}{n}}}.$$ And now the question is:

Do we compare this statistic with normal distribution or again with student t-distribution?

Answer 1

I do not recommend to use the Student approximation. Instead, it is better to observe that $X_1 + \dots + X_n \simeq \Gamma(n,\theta)$ with $\theta = 1/\lambda$. Therefore, $\frac{2}{\lambda}(X_1 + \dots + X_n) \simeq \Gamma(n,\frac12)=\chi^2_{2n}$. Hence you can construct the required interval through the quantiles of $\chi^2_{2n}$.