Typically, statistical power calculations seem to come from nowhere, although they contain elements of notation that are somewhat familiar. This is a compact derivation of the power calculation, in the simple case of one sample.

By the central limit theorem,

\begin{align*} \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} \sim Z\ \big|\ \text{H}_0 \\ \frac{\bar{x} - \mu_A}{\sigma/\sqrt{n}} \sim Z\ \big|\ \text{H}_A \end{align*}

where \(Z\) is distributed standard normal. Rearranging:

\begin{align*} \bar{x} &\sim \frac{\sigma}{\sqrt{n}}Z + \mu_0\ \big|\ \text{H}_0 \\ \bar{x} &\sim \frac{\sigma}{\sqrt{n}}Z + \mu_A\ \big|\ \text{H}_A \end{align*}

Without loss of generality, let \(\mu_A > \mu_0\). The situation looks like this:

We want to find the \(x\)-value where the yellow line is. Let's call it \(Z_c\). We want \(Z_c\) to satisfy the two properties:

\begin{align*} \mathbb{P}(\bar{x} < Z_c \big|\ \text{H}_0) &= 1 - \alpha\\ \mathbb{P}(\bar{x} < Z_c \big|\ \text{H}_A) &= \beta \end{align*}

These are met respectively when the following two equations hold:

\begin{align*} Z_c &= \mu_0 + \frac{\sigma}{\sqrt{n}}Z_{1 - \alpha} \\ Z_c &= \mu_A + \frac{\sigma}{\sqrt{n}}Z_{\beta} = \mu_A - \frac{\sigma}{\sqrt{n}}Z_{1 - \beta} \end{align*}

Because both are equal to \(Z_c\), we can drop it and set the two right-hand side expressions as equal:

\begin{align*} \mu_0 + \frac{\sigma}{\sqrt{n}}Z_{1 - \alpha} = \mu_A - \frac{\sigma}{\sqrt{n}}Z_{1 - \beta} \end{align*}

Solving for \(n\), we have:

\begin{align*} \mu_A - \mu_0 &= \frac{\sigma}{\sqrt{n}}\big[Z_{1 - \alpha} + Z_{1 - \beta}\big] \\ \Rightarrow\quad n &= \left[\frac{\sigma(Z_{1 - \alpha} + Z_{1 - \beta})}{\mu_A - \mu_0}\right]^2 \\ \end{align*}

The different flavors of power calculation come from:

• two-sided tests (just substitute \(Z_{1-\alpha/2}\) in place of \(Z_{1-\alpha}\))
• hypothesis testing using a more complicated statistic, like \(\bar{x}_1-\bar{x}_2\)
• using a different finite-sample approximation for the distribution of \(\bar{x}\).