Prove the validity "four percent margin of error" method A sufficient number of voters are polled to determine the percentage in favor of a certain candidate. Assuming that an unknown proportion p of the voters favor him and they act independently of one another, how many should be polled to predict the value of p within 4.5% with 95% confidence?
 A: We give a brief correct answer. In a few hours, if you indicate curiosity, we can give a somwhat extended answer.
The number of votes for our candidate, in our sample, has nearly binomial distribution. (Under our assumptions, it has binomial distribution if we sample with replacement. However, one samples without replacement, so the distribution is hypergeometric. If the total population is large, and $n$ is very much smaller than that, it makes no practical difference.)
The sample mean $\bar{X}$ (total number of favourable, divided by $n$) has nearly normal distribution if $n$ is reasonably large. The mean of $\bar{X}$ is $p$, and the standard deviation is $\sqrt{\frac{p(1-p)}{n}}$.
We do not know $p$. So we take the pessimistic view, and find the largest possible value of $\sqrt{p(1-p)}$. In the interval $0\le x\le 1$, the function $\sqrt{x(1-x)}$ reaches its maximim at $x=1/2$. That maximum value is $1/2$. In fact, the curve $y-\sqrt{x(1-x)}$ is quite flat near $x=1/2$, so $\sqrt{x(1-x)}$ is close to $1/2$ even if $p$ is at some distance from $1/2$, like $0.4$. 
So we have 
$$\Pr\left(-\frac{1.96}{2\sqrt{n}} \le \bar{X}-p \le \frac{1.96}{2\sqrt{n}}\right)\ge 0.95.$$
A: I figured out the problem myself. Thought I'd post the answer here in case I forgot the procedure.
Each voter follows the Bernoullian distribution with probability $p$, so the total votes $S$ with $n$ voter should follow the normal distribution $N(np,\sqrt{np(1-p)})$. The 95% confidence interval for normal distribution is between $\pm 1.96\sigma$, which gives me
$$|{S-np}|< 1.96\sqrt{npq}$$
$$|\frac{S}{n}-p|<1.96\sqrt{\frac{pq}{n}}$$
And from the result $S$, I should assume that the supporter rate to be $p'=\frac{S}{n}$. To estimate $p$ within 4.5%, this inequation should hold:
$$|p'-p|<4.5\%$$
which is
$$|\frac{S}{np}-1|<\frac{4.5\%}{p}$$
so
$$1.96\sqrt{\frac{q}{np}}<\frac{4.5\%}{p}$$
with no prior estimate of p, we shall assume that $pq=p(1-p)$ takes its maximum $1/4$
This gives $$n>474.3$$
So to achieve the so-called "four percent margin of error", a minimum of 475 voters shall be surveyed.
