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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?

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  • $\begingroup$ Is $\mu=\lambda$? $\endgroup$ – Mankind Mar 6 '16 at 18:26
  • $\begingroup$ ah, sorry, yes! $\endgroup$ – marco11 Mar 6 '16 at 18:27
  • $\begingroup$ The t interval assumes normality, and exponential is far from normal. Also, the sample mean is relatively slow to converge to normal by the CLT. So it is best to use the method of @zhoraster's Answer, which (in effect) uses the exact distribution of the sample mean. $\endgroup$ – BruceET Mar 6 '16 at 23:19
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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 $\chi^2_{2n}$.

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  • $\begingroup$ Ahh, then the confidence interval is has got one end only, right? I mean: $\lambda>\frac{2(X_1+...+X_n)}{\Phi^{-1}_{\chi^2_{2n}}(1-\alpha)}$ $\endgroup$ – marco11 Mar 6 '16 at 19:35
  • $\begingroup$ @marco11, not necessarily. You can use $\alpha/2$ and $1-\alpha/2$ quantiles for a two-sided interval. $\endgroup$ – zhoraster Mar 7 '16 at 7:10

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