Using the traditional formula, a $95\%$ CI for $p_1 - p_2$ is to be constructed based on equal sample sizes from the two populations. For what value of $n ( = m)$ will the resulting interval have width at most $0.1$ irrespective of the results of the sampling?
The CI in question is: $\hat{p}_1 - \hat{p}_2 \pm z_{0.025} \sqrt{\frac{\hat{p_1}\hat{q_1}}{m} + \frac{\hat{p_2}\hat{q_2}}{n} } $
$\hat{p_1} = X/m$ and $\hat{p_2} = Y/n$ , where $X \in bin(m,p_1)$ and $Y \in bin(n,p_2)$, The CI given above is for the difference between population proportions!
My attempt is this:
$\Bigg \vert \hat{p}_1 - \hat{p}_2 + z_{0.025} \sqrt{\frac{\hat{p_1}\hat{q_1} +\hat{p_2}\hat{q_2} }{m} } - \Big ( \hat{p}_1 - \hat{p}_2 - z_{0.025} \sqrt{\frac{\hat{p_1}\hat{q_1} +\hat{p_2}\hat{q_2} }{m} + } \Big ) \Bigg\vert \leq 0.1 $
what i come to is:
$2 \cdot z_{0.025}\sqrt{\frac{\hat{p_1}\hat{q_1} +\hat{p_2}\hat{q_2} }{m} } \leq 0.1$
The problem is the $\hat{p}_1\hat{q}_1$ and $\hat{p}_2\hat{q}_2$, since they are: $(x/m) \cdot(1-x/m) $ and $(y/m) \cdot(1- y/m)$ i need the values of $x$ and $y$ unless i can cancel them somewhere in the calculations but i can't see how.