ymochelows

2022-09-14

Exponential Confidence Interval
It is known that, for large n:
$\sqrt{n}\left(\lambda \overline{x}-1\right)\sim \mathrm{N}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\left(0,1\right)$
With this approximation, show that the 95% confidence interval for $\lambda$ is:
$\frac{\sqrt{n}-1.96}{\sqrt{n}\overline{x}},\phantom{\rule{1em}{0ex}}\frac{\sqrt{n}+1.96}{\sqrt{n}\overline{x}}$
I think I need to manipulate the formula for confidence intervals for exponential distributions but I'm not sure where to start (simplified version since large n?)

Step 1
About 95% of the probability distribution of a standard normal distribution lies in the interval (-1.96, 1.96).
Step 2
So all you need to do is manipulate
$P\left(-1.96<\sqrt{n}\left(\lambda \overline{x}-1\right)<1.96\right)=0.95$
to something like
$P\left(\frac{\sqrt{n}-1.96}{\sqrt{n}\overline{x}}<\lambda <\frac{\sqrt{n}+1.96}{\sqrt{n}\overline{x}}\right)=0.95$
remembering that $\overline{x}$ is a random variable while $\lambda$ is not.

Do you have a similar question?