I came across this question:
Given a sample of 50 values all coming from an exponential distribution where $f(x)=\theta x^{-x/\theta}$, find the MLE for theta. Then, find the 95% confidence interval for $\theta$.
My question is can I assume theta is normally distributed as n>30 so the CLT applies and then how do I find the expected value and variance of theta to then get the standard error of the interval
Thanks!
Assuming that you are dealing with the following density
$$f_X(x)=\frac{1}{\theta}e^{-\frac{x}{\theta}}$$
with mean $\mu=\theta$ and variance $\sigma^2=\theta^2$, $x\geq 0$
...easy find that $\hat{\theta}_{ML}=\overline{X}_{50}$ and remembering that the sample mean has the following asympthotic distribution
$$\overline{X}_{50}\dot{\sim} N(\theta;\frac{\theta^2}{50})$$
Thus to find you 95% approx CI you have only to solve in $\theta$ the following double inequality
$$ \bbox[5px,border:2px solid red] { -1.96<\frac{\overline{X}_{50}-\theta}{\overline{X}_{50}}\sqrt{50}<1.96 \qquad (1) } $$
... having estimated Sample mean's Standard deviation with $\frac{\overline{X}_{50}}{\sqrt{50}}$
On the contrary, if you are dealing with $f_X(x)=\theta e^{-\theta x}$ follow the same brainstorming modifying the forumlas I showed, accordingly
$$ \bbox[5px,border:2px solid red] { -1.96<\frac{\overline{X}_{50}-\frac{1}{\theta}}{\overline{X}_{50}}\sqrt{50}<1.96 \qquad (2) } $$
Personal opinion: even if the sample is large enough, I think it is easier to find the exact Confidence interval, using the $\chi^2$ distribution instead of having a Gaussian approx