2
$\begingroup$

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

enter image description here

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 enter image description here

Since when finding if it is the uniformly most powerful test
enter image description here

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, $\sum X_i>=4$

| cite | improve this question | |
$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – jbowman Nov 28 '15 at 20:30
  • $\begingroup$ @Jbowman that looks to me like it would be a good answer $\endgroup$ – Glen_b Nov 29 '15 at 0:21
  • $\begingroup$ @jbowman Thank you. So at $\alpha=0.05$ is the uniformly most powerful test $\sum X_i>=4 $ or $\sum X_i>=3$ $\endgroup$ – sam_rox Nov 29 '15 at 7:21
2
$\begingroup$

(Converted and expanded from a comment):

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.