Critical value for Simple vs. Simple Normal Test Let $X \sim N(\mu,1)$ and $\mu_0$, $\mu_1$ $\in \mathbb{R}$ with $\mu_0<\mu_1$.
We have the following hypothes $H_0:\mu=\mu_0$ and $H_1:\mu=\mu_1$.
I found with the LR test that for small and big values of $x-\mu_0$ we should reject $H_0$ for $H_1$. But now I'm asked to make "small " and "big" quantitative. 
The significance level is given to be $\alpha=0.05$,  and we can choose $\mu_0=0$ and $\mu_1=1$. How to proceed?
 A: For this one-sided alternative you would reject for $\bar X > \mu_0.$
To find the 'critical value' $k$ for $\alpha = 0.05$ so that you reject for $\bar X \ge k$ and
fail to reject otherwise, you need to find $k$ such that $P(\bar X \ge k\,|\, \mu_0).$ 
For example, if $\mu_0 = 10$ and $n = 16$ then you want the 95th quantile of $\mathsf{Norm}(\mu = 10, \sigma = 1/4).$ According to R statistical
software (where qnorm is the normal 'quantile' or 'inverse CDF' function),
the answer is $k = 10.41.$
> qnorm(.95, 10, 1/4)
[1] 10.41121

If also $\mu_1 = 11,$ then the probability of Type II Error (failing to
reject when $H_1$ is true) can be found as $P(\bar X < k\,|\,\ mu_1) = 0.009.$
> pnorm(10.41, 11, 1/4)
[1] 0.009137468

The figure below shows the null distribution of $\bar X$ in blue, the
alternative distribution of $\bar X$ in orange, and the critical value $k$
as a vertical dashed line.

I suppose this question arises in a theoretical setting, and that you
are expected to answer in terms of the standard normal CDF $\Phi.$ I will
leave it to you to generalize from my specific numerical example.
