A retailer wants to estimate the proportion of customers entering their website actually buy something at the website. The company wants to be 95% sure that their margin of error is no larger than 4%. How large a sample do they have to take? (please express your answer using 1 decimal place)


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.

As $p$ is not given, assume that $p =0.5$.

Given that margin of error $E =0.04$. The confidence coefficient is $0.95$.

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

Z-critical value
Z-critical value

The minimum sample size required to estimate the proportion is

$$ \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.5(1-0.5)\bigg(\frac{1.96}{0.04}\bigg)^2\\ &=600.25\\ &\approx 601. \end{aligned} $$

Thus, the sample of size $n=601$ will ensure that the $95$% confidence interval for the proportion will have a margin of error $0.04$.

Further Reading