2
$\begingroup$

In a testing task, suppose we have already chosen a test statistic $T(X)$, and know its distribution under null hypothesis.

Let $\mu$ be the mean of the null distribution of $T(X)$. Why is it reasonable to use $|T(X) - \mu| \geq c$ for some $c$ as the testing procedure to reject the null hypothesis?

I think that that rejecting null if and only if $|T(X) - \mu| \geq c$ requires the null distribution of $T(X)$

  • to have its mass centered around its mean $\mu$. If the null distribution of $T(X)$ has none of its mass in an interval around its mean $\mu$, and most of its mass at a certain distance away from its mean $\mu$, it seems not reasonable to reject null if and only if $|T(X) - \mu| \geq c$, doesn't it?

  • to be symmetric around its mean $\mu$. If the null distribution of $T(X)$ isn't symmetric around its mean $\mu$, will $T(X) \in (a, b)$ perhaps be better than $|T(X) - \mu| \geq c$ as rejection region?

Thanks and regards!

$\endgroup$

1 Answer 1

2
$\begingroup$

When selecting your rejection region (say at 5%), following your reasoning (I'm deliberately abusing it here in the worst sense), you could pick any region of the support where 5% of the data is. You could even pick the 5% centered around its mean: this is just as "unlikely" as the 5% near the edges.

The point you are missing is that you are trying to find a rejection region that matches your alternative hypothesis. It doesn't actually matter so much that the distribution is centered around the mean, but for many statistics, values "away from" the mean are the ones that occur more often if the alternative is true.

The same goes for the symmetric part: only if values away from the mean in both directions are consider as equally strongly pointing to the alternative, does a symmetric CI makes the most sense, but even then, other arguments may come up to validate other choices (which may indeed depend on the shape of the distribution).

$\endgroup$
1
  • $\begingroup$ Thanks,Nick! I understand from your reply that the alternative hypothesis is what matters to my question. But I don't quite understand the following example, in Fisher's exact test for independence, the count $X_{1,1}$ in the top-left cell has the hypergeometric distribution under independence null hypothesis. I was wondering if it is reasonable to use $|X_{1,1} - \mu|$ as the test statistic, where $\mu$ is the mean of the hypergeometric distribution. See my question here. $\endgroup$
    – Tim
    Commented Apr 26, 2013 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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