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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}.
\end{displaymath} (1)

For large $n$, $\hat p$ has an approximately Normal Distribution. Let RE be the Relative Error and SE the Standard Error, then
$\displaystyle \left\langle{p}\right\rangle{}$ $\textstyle =$ $\displaystyle p$ (2)
$\displaystyle \mathop{\rm SE}(\hat p)$ $\textstyle \equiv$ $\displaystyle \sigma(\hat p) = \sqrt{p(1-p)\over n}$ (3)
$\displaystyle \mathop{\rm RE}(\hat p)$ $\textstyle =$ $\displaystyle \sqrt{2\hat p(1-\hat p)\over n}\mathop{\rm erf}\nolimits ^{-1}(\mathop{\rm CI}\nolimits ),$ (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}.
\end{displaymath} (5)

© 1996-9 Eric W. Weisstein