# Unexpectedly, the theoretically best reject-region of T-test is bounded.

$f_{t}\left(x;\mu,df\right)\equiv C\left(df\right)\left(1+\frac{\left(x-\mu\right)^{2}}{df}\right)^{-\frac{df+1}{2}}$

$\lambda\left(x;\mu_{0},\mu_{1},df\right)\equiv\frac{f_{t}\left(x;\mu_{1},df\right)}{f_{t}\left(x;\mu_{0},df\right)}=\left(\frac{v+\left(x-\mu_{1}\right)^{2}}{v+\left(x-\mu_{0}\right)^{2}}\right)^{-\frac{df+1}{2}}{\longrightarrow\atop x\rightarrow\infty}1$

For NHST $H_{0}:T\sim t_{df}$ vs $H_{1}:T-1\sim t_{df}$, theoretically, $p\left(t\right)=\int_{\left\{ x:\lambda\left(x\right)\ge\lambda\left(t\right)\right\} }f_{t}\left(x,\mu_{0},df\right)dx$ is s.t. $\lim_{t\rightarrow\infty}p\left(t\right)=\frac{1}{2}$ , rather than zero. Nevertheless, pratically a large t, rejecting both $H_{0}$ and $H_{1}$, should not be counted as any evidence to retain or reject $H_{0}$.

To verify the shape of $\lambda\left(x\right)$ --

## 2 thoughts on “Unexpectedly, the theoretically best reject-region of T-test is bounded.”

1. lixiaoxu says:

The complication dwells in that the alternative hypothesis is single. If the alternative are multiple, that is, $$\mu>0$$, the theoretically best reject-region is still $$T > criterion$$.

Multiple alternatives usually involve the bias problem. The best reject-region for one single alternative may be inappropriate (biased) for the other single alternative. The appropriate RD set for multiple alternatives is a strict subset of the appropriate RD set for a single alternative.