Find the maximum value of \(\sqrt{p(1-p)}\) as \(p\) varies between \(0\) and \(1\).
Suppose that a proportion \(p\) of the population is female. In order to estimate \(p\) we pick a sample of \(n\) people at random and find the proportion of them who are female. Find the value of \(n\) which ensures that the chance of our estimate of \(p\) being more than \(0.01\) in error is less than 1\%.
Discuss how the required value of \(n\) would be affected if
(a) \(p\) were the proportion of people in the population who are left-handed;
(b) \(p\) were the proportion of people in the population who are millionaires.
Solution:
\(\,\) \begin{align*}
&& \sqrt{p(1-p)} &= \sqrt{p-p^2} \\
&&&= \sqrt{\tfrac14-(\tfrac12-p)^2} \\
&&&\leq \sqrt{\tfrac14} = \tfrac12
\end{align*} Therefore the maximum is \(\tfrac12\) when \(p=\frac12\)
Notice that our estimate \(\hat{p}\) will (for large \(n\)) be follow a normal distribution \(N(p, pq/n)\) by either the normal approximation to the binomial or central limit theorem.
We would like \(0.01 > \mathbb{P}\left ( |\hat{p}-p| < 0.01 \right)\) or in other words
\begin{align*}
&& 0.01 &> \mathbb{P}\left ( |\hat{p}-p| > 0.01 \right) \\
&&&=\mathbb{P}\left ( |\sqrt{\frac{pq}{n}}Z+p-p| > 0.01 \right) \\
&&&= \mathbb{P} \left (|Z|>\frac{0.01\sqrt{n}}{\sqrt{pq}}\right)
\end{align*} therefore we need \(\frac{0.01\sqrt{n}}{\sqrt{pq}}> 2.58 \Rightarrow n > 258^2 pq \approx 2^{14} \approx 16\,000\), where we are using \(pq = \frac14\) as the worst case possibility and \(258 \approx 256 = 2^8\)
If we were looking at when we are looking at left handed people (maybe ~\(10\%\), we would be looking at \(pq = \frac{9}{100}\) so we need a smaller sample). If we are looking at millionaires (an even smaller again percentage), we would need an even smaller sample. This is surprising since you would expect you would need a larger sample to accurately gauge smaller proportions. However, this surprise can be resolved by considering that this is an absolute error. For smaller values the relative error is larger, but the absolute error is smaller.