# Confidence Interval of estimator for the exponential distribution

We have an exponential distribution $$f(x)=\frac{e^\frac{-x}{\theta}}{\theta}$$

We are told that $n=3$ and that the data are given as $x_1=1, x_2=2.5, x_3=5.5$

a.) Determine an approximate 95% confidence interval based on asymptotic normality of $\overline{x}$

b.) Determine an approximate 95% confidence interval based on the asymptotic distribution of the deviance $D(\theta)$. The solution does not have to be exact. Pay attention to the function g, defined as $$g(\theta)= \frac{3}{\theta}+\ln\theta -\ln3-1$$

My questions:

a.) I have to take the sample mean, and then I add up the $z_{ \frac{\alpha}{2} }\frac{\sigma}{\sqrt n}$ thing? With $z_{0.025}=1.96?$ What value do I use for $\sigma$?

b.) What deviance $D(\theta)$ mean? The question added a graph, on the x-axis I see theta from 0 to 14 and on the y-axis I see (a probability?) from 0.0 up to 1.0 Has this something to do with the power function of the test? I should have knowledge about Neyman-Pearson, likelihood ratio tests, power function of a test.

• Regarding the first question, one would think that $n=3$ is too small for an asymptotic distribution to hold. An exact confidence interval would be better here. But if one has to use the large sample approximation, then the relevant asymptotic distribution of $\overline X$ is given by $\sqrt{n}(\overline X-\theta)\stackrel{a}\sim N(0,\theta^2)$. Then use the pivot $\frac{\sqrt{n}(\overline X-\theta)}{\theta}\stackrel{a}\sim N(0,1)$. – StubbornAtom Apr 15 at 7:16