What proportion of secretaries of Fortune 500 companies has a personal computer at his or her workstation? You want to answer this question by conducting a random survey. How large a sample should you take if you want to be 95% confident of the results and you want the error of the confidence interval to be no more than .05? Assume no one has any idea of what the proportion actually is.


The formula to estimate the sample size required to estimate the proportion is

$$ n =p*(1-p)\bigg(\frac{z}{E}\bigg)^2 $$

where $p$ is the proportion of success, $z$ is the $Z_{\alpha/2}$ and $E$ is the margin of error.

Given that margin of error $E =0.05$. The confidence coefficient is $0.95$. Assume that the proportion is $p =0.5$.

The critical value of $Z$ is $Z_{\alpha/2} =Z_{0.025}= 1.96$.

Z-critical value
Z-critical value

The minimum sample size required to estimate the proportion is
$$ \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.5(1-0.5)\bigg(\frac{1.96}{0.05}\bigg)^2\\ &=384.16\\ &\approx 385. \end{aligned} $$

Thus, the sample of size $n=385$ will ensure that the $95$% confidence interval for the proportion will have a margin of error $0.05$.

Further Reading