Suppose that I have a finite population of A's and B's, with properties:
Population size: $n$
There are $n_1$ A's and $n - n_1$ B's (so that $p = \frac{n_1}{n}$, $q = \frac{n - n_1}{n}$)
I'm taking a sample of size r without replacement. What I need to demonstrate (if it can be demonstrated, I'm not sure) is that: if p > 0,5 then the probability that sample proportion $\bar{p}$ > p is greater than the probability that $\bar{p}$ < p.
Put another way: If p > 0,5, then Prob($\bar{p} \ge p$) > Prob($\bar{p} \le p$)
Plainly, that if deviations from the expected value occur, then they are more likely to occur in the direction of the more frequent type of member.
As an additional premise I have that $pr \in \mathbb{Z}$.
So, instead of using a cumulative hypergeometric distribution, I thought it would be simpler (avoiding the sums) to reformulate the problem in this manner:
$$\frac{\binom{n_1}{pr} \binom{n - pr}{r - pr}}{n \choose r} > \frac{\binom{n - n_1}{qr} \binom{n - qr}{r - qr}}{n \choose r}$$
Is this ok? What it says -on the left for example- is basically "from the population, take a subset of pr A's (so you know now that the sample will contain at least the same proportion of A's than the population), and then take the rest of the sample from the remainder of the population as you please".
If this is ok, then how do I go on from there? I managed to reduce what I need to prove to:
$$\frac{(pn)!}{(pn -pr)!} (n - pr)! > \frac{(n - pn)!}{(n - r - (pn -pr))!} (n - (r - pr))!$$
(since the denominators cancel and, after applying the binomial coeeficient formula, (r - pr)! = (qr)! and (r - qr)! = pr! can also be cancelled).
Is what i've got so far all right? where to go from here? Thank you
EDIT: I've been trying the conjecture for various values here: http://stattrek.com/online-calculator/hypergeometric.aspx and the result seems to hold. Furthermore, there seem to be some interesting properties, such as, the greater p is, the greater the difference between $Prob(\hat{p} \ge p)$ and $Prob(\hat{p} \le p)$.
It seems to hold for sampling with replacement as well, which can be tried here: http://onlinestatbook.com/simulations/CLT/clt.html
If we forget about the fact that $pr \in {\mathbb Z}$ for a moment, take $p = 0.51$ and $n = 2$. Then the only way that ${\hat{p}} \geq p$ is if you get 2 successes out of 2 trials, which has probability approximately $1/4$. So it is more likely that ${\hat{p}} \leq p$. Your condition that $pr \in {\mathbb Z}$ makes things more interesting. I think your conjecture might be true in that case but I'm not sure how to show it yet. I'll update if I do.