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I have the following question for my homework:

Suppose X~exp($\theta$).

We want to test $H_0: \theta=1 vs. H_a:\theta=2$, based on a sample of size 2 - ${X_1,X_2}.$

a. Obtain the most powerful test (MPT) at a significance level $\alpha=.01$

What I am confused about is how to go about next steps.

Since I have $H_0: \theta=1 vs. H_a:\theta=2$, should I just plug those in for $\lambda$ to get my likelihood function to be:

$\frac{L_1}{L_0}=\frac{\Pi2e^{-2x}}{\Pi e^{-x}}$

Then I end up with:

$\frac{L_1}{L_0}=\frac{2e^{-n2\Sigma x_i}}{e^{-n\Sigma x_i}}$

$\frac{L_1}{L_0}=\frac{2e^{-2\Sigma x_i}}{e^{-\Sigma x_i}}$

$\frac{L_1}{L_0}=2e^{-\Sigma x_i}$

I am not sure where to go from here or if I am doing this right.

Could someone please let me know whenever they have the chance?

Thanks!

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This should have an answer. The question is about the application of the NP lemma on the exponential distribution. Although the OP has made some minor arithmetic mistakes, it is true that the test will eventually be based on the sufficient statistic

$$Y = \sum_{i=1}^n X_i $$

The exponential distribution has a monotone likelihood ratio, so that was to be expected. The rejection region will be

$$Y \leq c$$

and from the additivity property of the Gamma distribution, the critical value corresponding to a $0.01$ test may be obtained, under $H_0$. This constant turns out to equal $6.63$ for $n=2$.

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  • $\begingroup$ Shouldn't the rejection region be $Y>c$? $\endgroup$ – StubbornAtom Dec 3 '19 at 14:58

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