# Which test statistic is better for testing population mean?

Consider the following situation: a stochastic variable $X: (\Omega, \mathcal{F}) \to (\mathbb{R}, \mathcal{R})$ is known to be normally distributed with some mean $\mu$ and some variance $\sigma^2$. We do not know $\sigma^2$, but the null hypothesis ($H_0$) is that $\mu = \mu_0$ for some known constant $\mu_0$. We want to test this against the alternative hypothesis ($H_1$) that $\mu \neq \mu_0$.

Given a sample $(X_1, \ldots, X_n)$ with sample mean $\overline{X}_n$, my statistics notes claim that a good test statistic would be $$T_1 = \sqrt{n}\frac{\overline{X}_n - \mu_0}{\sqrt{S_n^2}} \sim_{H_0} t_{n-1}$$ where $$S_n^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i - \overline{X}_n)^2$$ is the sample variance and $t_{n-1}$ is the $t$-distribution with $n-1$ degrees of freedom. My choice for a test statistic would intuitively be $$T_2 = \sqrt{n}\frac{\overline{X}_n - \mu_0}{\sqrt{R}} \sim_{H_0} t_{n}$$ where $$R = \frac{1}{n}\sum_{i=1}^n (X_i - \mu_0)^2 .$$

Did I make I mistake in determining that $T_2$ has $t_n$-distribution? In case I didn't, I would think that $T_2$ would be a better statistic, since we do not use the estimate $\overline{X}_n$ for $\mu_0$ in $R$.

## 1 Answer

The statistic $T_2$ does not have a $t_n$ distribution. The issue is that the numerator $\overline{X_n}-\mu_0$ is not independent of the denominator $\sqrt{R}$, unlike the situation with $T_1$. More importantly, using $T_2$ will not lead to a more powerful test; the Neyman-Pearson lemma can be used to show that $T_1$ provides a uniformly most powerful test, so there is no use trying to improve on it. (Well, technically a uniformly most powerful test only exists in the case of a one-sided alternative).

Intuitively, the reason the $T_2$ statistic is not better is because, as the true mean $\mu$ deviates further from the hypothesized value $\mu_0$ (holding $\sigma$ constant), the expected value of $R$ becomes larger (unlike $S_n^2$), preventing $T_2$ from becoming large.