Hey, paying 100fg for a good answer to this problem, or if someone just wants to suggest where to start that would be appreciated as well.

Basically: We have a large population of N people, where it is so large N can be said to be infinite. This population consists of two equally sized groups. An opinion survey is to be conducted, where the answers can be "yes" or "no". We are trying to determine the proportion of "yes" in the population. A sample of size n is drawn, and the proportion of "yes" amongst them is given as an estimate. If in the entire population the proportion is p = (p1 + p2) / 2, where p1, p2 are the proportions of the two groups, exactly how much of a reduction in variance is obtained if a sample size of n/2 from each group (rather than selecting n from the entire population)?

Nvm, think I understand the question

This post was edited by zackill4 on Sep 24 2013 12:18pm

Unfortunately I have nothing to add at the moment, but I was confused by this:

Why would we need to estimate the number of "yes" responses in the sample group? Shouldn't we know that for certain? Isn't that the purpose of creating a sample?
Are am I just being dense today?

This post was edited by Azrad on Sep 24 2013 12:36pm

estimate for the population I assume.

I'm getting an odd solution for this btw, so I'm rechecking my algebra lol

Edit: goddamn sunglasses emote, supposed to say group B

This post was edited by zackill4 on Sep 24 2013 12:46pm

heh put it in code tags

This post was edited by Azrad on Sep 24 2013 12:45pm

Thanks lol

Let X count the number of yes in the size n sample.
X follows a binomial law with parameters n and p :

V(X) = np(1-p)

Let X1 count the number of yes in the first size n/2 sample, let X2 count the number of yes in the second size n/2 sample.
X1 and X2 both follow a binomial law with parameters n/2 and respectively p1 and p2 :

V(X1) = (n/2).p1(1-p1)
V(X2) = (n/2).p2(1-p2)

Now let Y = X1 + X2.

You're intersted in knowing the reduction V(Y) / V(X).

V(Y) = V(X1) + V(X2) + 2.cov (X1,X2)
but since X1 and X2 are independant :
V(Y) = V(X1) + V(X2)

Hence :

V(Y) = (n/2).[p1(1-p1)+p2(1-p2)]

And finally :

V(Y) / V(X) = [p1(1-p1) + p2(1-p2)] / [2p(1-p)]

Looks like we're in agreement then. It ain't gonna be a pretty solution though :(. You can get a 2p term in the numerator but you can't get rid of the p1 and p2 pretty sure.

Haven't forgotten about this problem, just don't have enough time to go through the work and check everything right now (which is why I haven't given fg yet). Thanks for the help though, guys. If it all works out, I'll give you both (zackill4 and feanur) 100fg though. That will most likely be midday tomorrow (the 25th).