I am having some problems calculating the power in this exercise.
This is what i've tried. So since i know that $\alpha$ is 0.05. I thereby know that the upper boundary is given by the formula 0.05 has a Z-score = 1.64485
$$ 1.64485 =\frac{\bar{x} - 4}{50/\sqrt{9}} \Rightarrow \bar{x} = 31.414166666666667` % note I already here knows that something is wrong$$
Then i calculate $\beta$ by $P(z \ge \frac{31.41417 - 4.5}{50/\sqrt{9}}) = 0.0531694$ and thereby the power beeing $1- 0.0531694 = 0.946831 $ This doesn't match what the "cheat sheet" => it says 1.. So what am I doing wrong?
The hypotheses are:
$H_0: \mu=4$
$H_A: \mu=4.5$
Calculating the critical value. Above it we do not accept $H_0$
We are only interested at the upper bound of the interval. If we standardized the random variable, we get $$\Phi \left( \frac{\overline x_c-\mu}{\frac{\sigma}{\sqrt{n}}}\right)=\Phi \left(\frac{\overline x_c-4}{\frac{2}{\sqrt{50}}}\right)=0.95$$
$\Phi(\cdot)$ ist the function of the standard normal distribution.
Note, that the standard deviation of $\overline x $ is $\large{\sigma_{\overline x}=\frac{\sigma}{\sqrt{n}}}$
$\frac{\overline x_c-4}{\frac{2}{\sqrt{50}}}=\Phi (0.95)=1.645$
If you solve the equation, you will get $\overline x_c=4.047$.
Calculating the probability of rejecting $H_A$, given $H_A$ is true
Now we assume, that $\overline x$ is $\mathcal N\left( 4.5,\left( \frac{2}{\sqrt{50}} \right) ^2 \right)$ distributed.
$\beta$ is the probability, that we accept $H_0$, given that $H_A$ is true.
Thus we calculate $\beta=\Phi \left(\frac{4.047-4.5}{\frac{2}{\sqrt{50}}}\right)$
After you have calculated $\beta$, you can calculate $1-\beta$.
I have added a sketch.