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

Table of contents
- Solution g

Solution

Under Neyman allocation, sample sizes are proportional to the arm standard deviations. Here

\frac{n _{1}}{n _{0}} = \frac{p _{1} ( 1 - p _{1} )}{p _{0} ( 1 - p _{0} )} = \frac{0.5}{0.3} = \frac{5}{3} .

For total size

n_{m a x}

, the minimized asymptotic variance is

Var (p_{1} - p_{0}) = \frac{( 0.25 + 0.09 ) ^{2}}{n _{m a x}} = \frac{0.64}{n _{m a x}} .

Equal allocation gives

\frac{0.25}{n _{m a x} /2} + \frac{0.09}{n _{m a x} /2} = \frac{0.68}{n _{m a x}} .

The Neyman allocation therefore reduces the large-sample variance by

0.04/ n_{m a x}

, about

5.9%

of the equal-allocation variance.

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