Under <Neyman allocation>, sample sizes are proportional to the arm standard deviations. Here
$$
\frac{n_1}{n_0}
=\frac{\sqrt{p_1(1-p_1)}}{\sqrt{p_0(1-p_0)}}
=\frac{0.5}{0.3}=\frac53.
$$
For total size $n_{\max}$, the minimized <asymptotic variance> is
$$
\operatorname{Var}(\widehat p_1-\widehat p_0)
=\frac{\bigl(\sqrt{0.25}+\sqrt{0.09}\bigr)^2}{n_{\max}}
=\frac{0.64}{n_{\max}}.
$$
Equal allocation gives
$$
\frac{0.25}{n_{\max}/2}+\frac{0.09}{n_{\max}/2}
=\frac{0.68}{n_{\max}}.
$$
The Neyman allocation therefore reduces the large-sample variance by $0.04/n_{\max}$, about $5.9\%$ of the equal-allocation variance.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-207/2/g/solution