Finding minimum value of observation for a given power in hypothesis testing. 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.
 A: 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
$$n\geq\Bigg \lceil \Bigg(\frac{1.2816}{0.1776}\Bigg)^2\Bigg\rceil=53$$
A: 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}
