Standard error of difference of estimated population proportions

Question

Can someone please explain this to me? I'm a little unclear with what the question is asking. I also don't know how they got this answer in the book.

Answer 1

You are estimating $p_1 - p_2$ by $\hat p_1 - \hat p_2 = X_1/n_1 - X_2/n_2,$ where you have observed $X_1$ successes from one population, and independently $X_2$ successes from the other. The 'standard error' of an estimator is simply its standard deviation.

First, $X_1 \sim Binom(n_1, p_1),\,$ so $Var(X_1) = n_1p_1(1-p_1).$ Then $$Var(\hat p_1) = Var(X_1/n_1) = \frac{1}{n_1^2}Var(X_1) = p_1(1-p_1)/n_1.$$ Similarly, for $Var(\hat p_2).$ Then, using independence, find $Var(\hat p_1 - \hat p_2)$ and, finally, $SD(\hat p_1 - \hat p_2)$ as required.

To find the answer given, I guess you are expected to use known values $\hat p_1$ and $\hat p_2$ to estimate $p_1$ and $p_2,$ respectively. I see that @Henry has done that in his (essentially contemporaneous) Answer (+1).

Answer 2

Possibly they think the standard error of an observed sample proportion $p$ can be estimated using $\sqrt{\dfrac{\hat p(1-\hat p)}{n-1}}$ so the standard error of the difference in sample proportions could be estimated with $\sqrt{\dfrac{\hat p_1(1-\hat p_1)+\hat p_2(1-\hat p_2)}{n-1}}$ and thus $\sqrt{\dfrac{0.15 \times 0.85 +0.12 \times 0.88}{n-1}}=\sqrt{\dfrac{0.2331}{n-1}}$ here. Remember that the variance of the difference between two independent random variables is the sum of their variances

So if you want this to be less than $0.05$ then you need $\sqrt{n-1}\gt \dfrac{ \sqrt{0.2331}}{0.05}$ i.e. ${n}\gt 1+\dfrac{{0.2331}}{0.05^2} = 94.24$ implying $n = 95$ is the smallest integer solution

This seems to reproduce their final result, even though it uses $\sqrt{n-1}$ rather than $\sqrt{n}$, so I wonder whether in fact they took a different approach: using $\sqrt{\tfrac{\hat p(1-\hat p)}{n}}$ instead would have reduced the final result by $1$