Alright so my lecture notes start off by showing that, by doing the usual likelihood ratio test and Neyman Pearson stuff, the most powerful test at level $\alpha$ for $H_0: \mu = \mu_0$ versus $H_A: \mu >\mu_0$ rejects for large values of $\overline{X}-\mu_0$, and the test that is most powerful against the alternative $H_A:\mu<\mu_0$ rejects for small values of $\overline{X}-\mu_{0}$. It says that since the most powerful test is not the same for every alternative, then we don't have a uniformly most powerful (UMP) test.

It goes on to suggest that in such a case we should use the Generalised Likelihood Ratio test. Doing it this way tells us to reject the null at level $\alpha$ if $|\overline{X}-\mu_{0}|>z_{\alpha/2}/\sqrt{n}$.

My confusion is the following. After doing it the first way, we might not have a UMP test but we still have a test nonetheless... What's the point in going on to do the Generalised Likelihood Ratio test if that's normally not even optimal? Assuming there is an explanation for this, if I had to pick one method at the start of doing this question, how should I know whether to choose the Generalised Likelihood Ratio test instead of the normal UMP stuff, without going through both methods -- is there a way to tell immediately which will be better?

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    $\begingroup$ The point is that for two-tailed test UMP doesn't exist. Therefore we don't have a single "best" test in such case. Usual restriction is to consider only unbiased tests, i.e. when power level is no less than significance, and choose the most powerful one of them $\endgroup$ Mar 12 '17 at 20:46
  • $\begingroup$ @ŁukaszGrad Thanks for the reply. I'm still confused because the first method still gives a valid way of doing it, no? You can rewrite $H_0:\mu = \mu_0$ vs $H_A: \mu \neq \mu_0$ as $H_0:\mu=\mu_0$ vs $H_A: \mu=\mu_{A}$ for $\mu_{A}>\mu_0$ or $\mu_{A}<\mu_0$ and then consider both those simple versus simple cases separately to get a test using the Neyman Pearson (albeit not a UMP test). Why choose the Generalised Likelihood way instead? $\endgroup$
    – Alex.F
    Mar 12 '17 at 21:02
  • $\begingroup$ U could test it separately, that is, test against $H_{A1}: \mu > \mu_0$ and then against $H_{A2}: \mu < \mu_0$. But there is no gain in doing so, since this is what two-tailed test is doing at once $\endgroup$ Mar 12 '17 at 21:15
  • $\begingroup$ @Alex.F keep following that thread. What if $\mu_A > \mu_0$? What if $\mu_A < \mu_0$? These two cases will give you different tests. Then remember you don't know which case is true. One test will have higher power for one situation, and the other will have better power for the other. But no one test will have UNIFORMLY the highest power. $\endgroup$
    – Taylor
    Mar 12 '17 at 21:30
  • $\begingroup$ @Taylor Ah I see now, many thanks! Just a small, slightly unrelated question: Why can $H_0:\mu \geq \mu_0$ vs $H_A: \mu <\mu_0$ be rewritten as $H_0:\mu=\mu_0$ vs $H_1:\mu=\mu_1$ where $\mu_1<\mu_0$? What about the greater than condition? $\endgroup$
    – Alex.F
    Mar 12 '17 at 22:02

I made a small R plot of power levels of each test. Red and blue are one-tailed and dotted black is two-tailed one. Dotted green is significance level $\alpha = 0.05$, $\mu_0 = 0$.

enter image description here

  • $\begingroup$ Ah brilliant, certainly helps to see it visually! Very kind of you. $\endgroup$
    – Alex.F
    Mar 12 '17 at 22:16

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