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.

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:

$$\beta(\theta)=1-\Phi\left(c+\frac{\theta_0-\theta}{\sigma/\sqrt{n}}\right)$$

• Thanks. Makes sense. The missing fourth equal sign did put me off. – cbro Nov 30 '18 at 18:06

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.