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{ 2008 05 11 }

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) --





Posted by lixiaoxu on Sunday, May 11, 2008, at 12:43 am. Filed under English. Tagged . Follow any responses to this post with its comments RSS feed. You can post a comment or trackback from your blog.

{ 2 } Comments

  1. lixiaoxu | May 11, 2008 at 1:33 am | Permalink

    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.

  2. liutom | August 18, 2008 at 4:37 am | Permalink

    厉害

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