In a scientific association, an electoral process for the president of the association will take place. There are 2 candidates, A and B. Every member (of the N in total) will vote either for A or for B. We would like to estimate the percentage of A candidate (let’s call it P) and for this, we use a sample of size $n (n<N)$.
What is the optimum number of n, so that the percentage of voters of A in the sample varies from the real percentage P of voters of A by less than $1%$ (in absolute value), with probability $95%$?
I have found a wikipedia link https://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval#Normal_approximation_interval but am not sure which formula to use, since there are no numbers for the population $(N)$ and also for $P$ and $(1-P)$.
As I am not at all familiar with statistics and this kind of stuff, I did a bit of research - the only (easy) thing I managed to find is that the value of $z$ for $95%$ is $1.96$.
So we would like $P(|\hat{p}-p| \le 0.01)$ to be $95%$. But we don't have any indication of $P$!
(I am not doing any homework - this is a "real life" problem to calculate a suggested "exit poll" size).
Thank you very much in advance!
Edit: I now found this equation: $n \geq \left(\dfrac{z_{\alpha/2} \sigma}{\delta}\right)^2$ and wonder if I can use it.
If you're assuming $\hat{P}\sim \mathcal{N}\Big(P,\frac{P(1-P)}{n}\Big)$ then $\hat{P}-P\sim \mathcal{N}\Big(0,\frac{P(1-P)}{n}\Big)$ whose variance is bounded above by $1 \over 4n$. This implies $$P\Big(|\hat{P}-P|<0.01\Big)\geq P\big(|X|<0.01\big)$$ where $X\sim \mathcal{N}\big(0,\frac{1}{4n}\big)$. The value of $n$ which forces the RHS to equal 0.95 is $$n=2500z_{0.025}\approx 4900$$