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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. enter image description here

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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:

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

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

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