With $X_i$ i.i.d. $N(\theta,1)$ and $$H_0: \theta=0 vs. H_1: \theta=1$$
I got that the test statistic is $\bar{X_n} > \frac{q_{\alpha}}{\sqrt{n}}+\theta$, where $\theta$ is either 0 for $H_0$ or 1 for $H_1$.
The type 1 test statistic is given with $\alpha = 0.05$.
To compute the power, I know that I have to compute the probability that I reject $H_0$ when $H_1$ is true (using the alternative distribution $N(1,1)$). So I need to get $$P_\theta(\bar{X_n}>\frac{q_{\alpha}}{\sqrt{n}}+\theta) = P_\theta(\bar{X_n}>\frac{1.64485}{\sqrt{n}}+1)$$
But from here on I am stuck on how to proceed, I suppose I need the CDF in the form $1-CDF(c)$ where $\theta=1$?
Can someone point me to the solution?
To calculate the power, first of all you have to set the decision rule. To calculate it $n$ is needed. Let's suppose just for an example that $n=4$
Then the decision rule is the following:
We reject $H_0$ iff
$z>1.64 \rightarrow \bar{X}_n \sqrt{n}>1.64 \rightarrow \bar{X}_n >\frac{1.64 }{2}=0.82 $
Now you can calculate the power
$$\mathbb{P}[\bar{X}_n>0.82|\theta=1]=\mathbb{P}[Z>(0.82-1)\sqrt{4}]\approx 0.64$$
Note that: the higher is $\theta$ under $H_1$the higher is the power