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I did search a bit and couldn't find this specific one. So decided to ask. The question is the following.

We want to test the mean of a normal population. Suppose that $X_1,\dots,X_n\overset{i.i.d.}\sim N(\mu,1)$. Suppose that we are testing $H_0:\mu = \mu_0$ against the alternative $H_1:\mu\neq \mu_0$. We are looking for a rejection region of the form $|\overline{X}-\mu_0|>c$.

Therefore \begin{eqnarray*} P_{\mu_0}(|\overline{X}-\mu_0|>c) &=& \alpha\\ P_{\mu_0}(\sqrt{n}(|\overline{X}-\mu_0|)>c\sqrt{n}) &=& \alpha \end{eqnarray*} This implies that $c\sqrt{n} = z_{{\alpha}/{2}}$.

Thus we arrive at a test that calls for rejection of the null-hypothesis if $\sqrt{n}(|\overline{X}-\mu_0|) >z_{{\alpha}/{2}}$.

In this test, why is the p-value $P(|Z|>\text{TS})$, where $\text{TS} = \sqrt{n}(|\overline{X}-\mu_0|)$?

Ashok
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  • What about this is unclear? I have an answer ready to go (probably will post anyway) but am quite unsure that it addresses your particular area of confusion. – Dave Jan 24 '21 at 07:54
  • @Dave: Can you explain me how is the formula for p-value derived? In other words, why is the p-value $\text{Pr}(|Z|>\text{TS})$? – Ashok Jan 24 '21 at 07:58
  • @Dave: Have edited the question a bit. Hope it's clear now. – Ashok Jan 24 '21 at 08:06

1 Answers1

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This may look unfamiliar because the variance is $1$, so it can be omitted from the calculation, resulting in no denominator.

Dave
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