# Why is it reasonable to consider deviation from null mean as test statistic?

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!

• 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.