Need help with a test hypothesis involving an exponential distribution You may ignore the questions $a$, $b$ and $c$ as they are relatively easy. I know how to test hypotheses for a sample data but question $d$ is confusing me. How do I know what type of test statistic I should be using for this and what quantile distribution I should be comparing this with? We've only really dealt with testing hypotheses for sample distributions. Thanks in advance for any help.

• It seems to me that this can be rephrased as: give a rule such that if $\theta=2$ then you reject the null hypothesis $\theta=2$ with probability 0.05. So this would amount to finding a symmetric interval centered around 2 whose complement has probability 0.05 under a Exp (2) distribution. This is easy enough to do...
– Ian
Sep 18 '16 at 4:16
• I found this interval to be -2log(0.05) to infinity. Is this correct? Although I'm not entirely sure what this interval means. Sep 22 '16 at 12:12
• I said a symmetric one: so you're looking for $a$ such that $\int_{2-a}^{2+a} \frac{1}{2} e^{-t/2} dt = 0.95$. So $e^{-(2-a)/2}-e^{-(2+a)/2}=0.95$ so $\sinh(a/2)=\frac{0.95e}{2}$, $a=2\sinh^{-1}(0.95e/2)$. This breaks down because actually this $a$ is greater than $2$; so you're really looking for $1-e^{-x/2}=0.95$ so $x=-2\log(0.05)$. Then the relevant interval is $[0,-2\log(0.05)]$.
– Ian
Sep 22 '16 at 12:20
• To clarify what this means: if you have the null hypothesis $H_0 : \theta=2$, you will accept the null hypothesis if your result is in $[0,-2\log(0.05)]$ and you will reject it otherwise, with the overall level of significance of the test being 0.95 (i.e. the probability of rejecting the null when it holds is at most $0.05$).
– Ian
Sep 22 '16 at 12:58
• Thank you a lot, I understand it now Sep 22 '16 at 14:04

If you draw a single observation attempting to test $H_0 : \theta=2$, then the standard test is the same as it usually is for the mean: you reject this null hypothesis with level of significance $1-\alpha$ if your result falls outside of an interval $[2-\delta,2+\delta]$ which has probability $1-\alpha$ when $\theta=2$. The probability of such an interval is $\int_{2-\delta}^{2+\delta} \frac{1}{2} e^{-t/2} dt=e^{-(2+\delta)/2}-e^{-(2-\delta)/2}=2e^{-1}\sinh(\delta/2)$ if $0 \leq \delta<2$ and $\int_0^{2+\delta} \frac{1}{2} e^{-t/2} dt = 1-e^{-(2+\delta)/2}$ if $\delta \geq 2$. If $\alpha=0.05$, you note that $2e^{-1}\sinh(1)<0.95$, so that you look at the second case. This gives the interval as $[0,-2\log(0.05)] \approx [0,6]$. You accept this null hypothesis given a result in this interval and reject it otherwise.