# Uniformly most powerful test in poisson

This is example in Hogg and Craig's book page 254

I have shown that the best critical region is $\sum X_i>=3$.

I don't understand why, after finding $\sum X_i>=3$ is the best critical region they go onto do

Since when finding if it is the uniformly most powerful test

is equivalent to $\sum X_i>=3$, and this is the critical region for testing for each simple hypothesis in $H_1$ and therefore from here can't I say that this is a uniformly most powerful test.

Have they done it like this because since Poisson distribution is discrete we can't find critical value so that Pr(type 1 error) is exactly equal to significance level. But in this case isn't the critical value selected so that size=largest possible values less than significance level. So in that case shouldn't the best critical region be, $\sum X_i>=4$

• Your supposition is correct; they have done this because the Poisson is discrete, so we can't find a critical value such that P(type 1 error) is exactly equal to the significance level, except for the significance levels equal to P(type 1 error) at some critical value. Note, however, that in the second part, they introduce a pre-specified $\alpha = 0.05$. Since the test with $Y \geq 3$ has P(type 1 error) = 0.08, it won't do any more. Instead, you have to go up to $Y \geq 4$ to get P(type 1 error) $\leq$ 0.05. You will note that in the second line of your second quoted block, they use 4. – jbowman Nov 28 '15 at 20:30
• @Jbowman that looks to me like it would be a good answer – Glen_b Nov 29 '15 at 0:21
• @jbowman Thank you. So at $\alpha=0.05$ is the uniformly most powerful test $\sum X_i>=4$ or $\sum X_i>=3$ – sam_rox Nov 29 '15 at 7:21

As you suspected, they have done this because the Poisson is discrete, so, except for critical values such that the probability of a type I error is equal to the significance level, we won't be able to match the target $\alpha$ exactly. In this situation, we try to find a region with the largest probability of type I error less than $\alpha$, in this case, $< 0.05$. Since $\sum X_i \geq 3$ as a region has a probability of 0.08, it won't do; we need a smaller region. $\sum X_i \geq 4$ has a probability of 0.019, which is less than 0.05, so it works.