For equal arm size $n$, write $v^*=p_1(1-p_1)+p_0(1-p_0)$ at the clinically relevant alternative. A one-sided level-$\alpha$ <Wald test> rejects when $Z>z_{1-\alpha}$, and its approximate <statistical power> is
$$
1-\Phi\!\left(z_{1-\alpha}-\frac{\delta^*\sqrt n}{\sqrt{v^*}}\right).
$$
Equating this to $1-\beta$ gives the per-arm <sample size>
$$
n=\frac{v^*\bigl(z_{1-\alpha}+z_{1-\beta}\bigr)^2}{(\delta^*)^2},
$$
rounded up. If the design uses a null-based critical standard error $v_0$ but an alternative standard error $v^*$, the corresponding more general formula is
$$
n=\frac{\bigl(z_{1-\alpha}\sqrt{v_0}+z_{1-\beta}\sqrt{v^*}\bigr)^2}{(\delta^*)^2}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 207 / 2 / b / Solution

{title2=$1-\beta$}
{wiki}

The statistical power of a test at an alternative parameter value is the probability that the test rejects its null hypothesis at that value.


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