The pivotal qty for the normal approximation to the binomial is given by (X-np)/sqrt(npq) whereas for the normal approximation to the binomial for proportions, the numerator is as expected but the denominator sqrt(p1q1/n1+p2q2/n2). I would have expected we just add the variances together and so get sqrt(n1p1q1+n2p2q2). Thanks.
X ~ Bin (n, p) ~ N(np, npq) So: (X - np)/sqrt(npq) ~ N(0,1) But p hat = X/n ~ N(p, pq/n) So: (p hat - p)/sqrt(pq/n) ~ N(0,1)
Right. Its the distribution of p hat (and hence the variance of p hat) we are looking at not the distn of X. That makes sense. And var(p hat)=var(X/n)=1/(n^2)var(X)=npq/n^2=pq/n?
