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Let $X_{1}, \ldots, X_{n}$ be a sample from the distribution $\mathrm{N}(\mu, 1)$ and consider testing $H_{0}: \mu=\mu_{0}$ versus $H_{1}: \mu=\mu_{1}$ where $\mu_{0}<\mu_{1}$ are known numbers.
(a) For a given level $\alpha,$ find the most powerful test.
(b) Calculate the power
(c) For given $\mu_{0}, \mu_{1}, \alpha,$ determine the minimal number of observations needed to have power at least $\beta$ (i.e., to reject $H_{0}$ with probability at least $\beta$ when $H_{1}$ holds).

Using Neyman Pearson lemma I have found the test statistic to be $\bar X$ and reject $H_{0}$ if $\bar X > k$, where $k= z_\alpha / \sqrt{n} + \mu_0$ .

And calculated the power to be Power = P{Reject $H_0$} = P{$Z> z_\alpha - \sqrt{n} (\mu-\mu_0)$}

But for the Part (c) I don't know how to find the minimum number of observation for given condition.
I think I have to solve the equation $\beta = P_{\mu_1}\{\bar X > z_\alpha / \sqrt{n} + \mu_0 \}$ = P{$Z> z_\alpha - \sqrt{n} (\mu_1-\mu_0)$}
But I don't know how to solve this for $n$.
Please help me with this. Thnakyou.

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2 Answers 2

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As you found, UMP test is given by Neyman Pearson's Lemma with rejection region

$$\mathbb {P}[\overline{X}_n>k|\mu=\mu_0]=\alpha$$

Now $\overline{X}_n>k$ is your decision rule ($k$ now is fixed) and you can calculate the power ( usually indicated with $\gamma$ because $\beta$ is normally used for type II error)

$$\mathbb {P}[\overline{X}_n>k|\mu=\mu_1]=\gamma$$

Understood this, finally fix $\gamma$ and get $n$


Example


$\mu_0=5$

$\mu_1=6$

$\alpha=5\%$

$n=4$

The critical region is

$$(\overline {X}_4-5)2=1.64\rightarrow \overline {X}_4=5.8224$$

Thus your decision rule is

$$\overline {X}_4\geq 5.8224$$

and you can calculate the power

$$\gamma=\mathbb {P}[\overline {X}_4\geq 5.8224|\mu=6]=1-\Phi(-0.36)\approx 64\%$$

Now suppose you want a fixed power $\gamma \geq 90\%$, simply re-solve the same inequality in $n$

$$\mathbb {P}[\overline {X}_n\geq 5.82|\mu=6]\geq 0.90$$

Getting

$$(5.8224-6)\sqrt{n}\leq-1.2816$$

That is

$$n\geq\Bigg \lceil \Bigg(\frac{1.2816}{0.1776}\Bigg)^2\Bigg\rceil=53$$

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You can state it in terms of the CDF $\Phi(z)=P(Z \le z)$; I don't think there is a way to avoid it.

\begin{align} \beta &\le 1 - \Phi(z_\alpha - \sqrt{n}(\mu_1-\mu_0)) \\ \Phi^{-1}(1-\beta) &\ge z_\alpha - \sqrt{n} (\mu_1 - \mu_0) \\ \sqrt{n} &\ge \frac{z_\alpha - \Phi^{-1}(1-\beta)}{\mu_1 - \mu_0} \end{align}

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