Sample Proportion

Let there be $x$ successes out of $n$ Bernoulli Trials. The sample proportion is the fraction of samples which were successes, so

$$\hat p={x\over n}.$$ (1)

For large $n$, $\hat p$ has an approximately Normal Distribution. Let RE be the Relative Error and SE the Standard Error, then

$$\left\langle{p}\right\rangle{} = p$$ (2)

$$\mathop{\rm SE}(\hat p) \equiv \sigma(\hat p) = \sqrt{p(1-p)\over n}$$ (3)

$$\mathop{\rm RE}(\hat p) = \sqrt{2\hat p(1-\hat p)\over n}\mathop{\rm erf}\nolimits ^{-1}(\mathop{\rm CI}\nolimits ),$$ (4)

where CI is the Confidence Interval and $\mathop{\rm erf}\nolimits {x}$ is the Erf function. The number of tries needed to determine $p$ with Relative Error RE and Confidence Interval CI is

$$n={2[\mathop{\rm erf}\nolimits ^{-1}(\mathop{\rm CI}\nolimits )]^2\hat p(1-\hat p)\over (\mathop{\rm RE})^2}.$$ (5)