One-sided hypothesis test (error rates) Can someone please help me understand how the expressions after the equals (marked in red) are arrived at? I don't quite understand the very last one where the c.d.f. $\phi$ comes into the picture. My guess for the first expression is that the term added to $c$ is always going to be negative and hence its added in order to get to the next expression involving a c.d.f. 
 A: They are calculating the power function $\beta(\theta)$ of the test.
Note that $\beta(\theta)$, a function of $\theta$, is the probability of rejecting the null hypothesis $H_0$ under $\theta$.
Formally, for testing $H_0:\theta\in\Theta_0$ vs $H_1:\theta\in\Theta_1$, where $\Theta_0\subset\Theta$ and $\Theta_1\subset \Theta-\Theta_0$, we define the power function of the test as 
$$\beta(\theta)=P_{\theta}(\text{reject }H_0)\quad\forall\,\theta\in\Theta$$
Distribution of the sample mean $\overline X_n$ is given by
$$\overline X_n\sim N(\theta,\sigma^2/n)$$
So they standardize $\overline X_n$ after the second equal sign:
\begin{align}
\beta(\theta)&=P_{\theta}\left(\overline X_n>c\sigma/\sqrt{n}+\theta_0\right)
\\&=P_{\theta}\left(\frac{\overline X_n-\theta}{\sigma/\sqrt{n}}>c+\frac{\theta_0-\theta}{\sigma/\sqrt{n}}\right)
\end{align}
Hence the cdf $\Phi$ comes into play after a missing fourth equal sign:
$$\beta(\theta)=1-\Phi\left(c+\frac{\theta_0-\theta}{\sigma/\sqrt{n}}\right)$$
A: For the first $=$, in the numerator of the term on the LHS of the equality, add and subtract $\theta$, rearrange terms to get $\bar{X}_n-\theta-(\theta_0-\theta)$ on the numerator and then move the second part to the RHS. For the second $=$, the LHS of the inequality inside the $P()$ is just a standardized $\bar{X}_n$, so the probability it exceeds the RHS is the probability a standard normal ($Z$) exceeds the RHS. Note that there should be another $=$ sign just before the $1-\Phi()$ term at the end.
