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) --

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

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.

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