A car dealer wants to estimate the price its competitors charge for similar cars. The dealer wants to be 95% sure that the maximum margin of error is no larger than \$500. A pilot sample showed a standard deviation of \$3215. How large a sample does the dealer need to take?


The formula to estimate the sample size required to estimate the population mean is

$$ n =\bigg(\frac{z* \sigma}{E}\bigg)^2 $$

where $\sigma$ is the population stabdard deviation, $z$ is the $Z_{\alpha/2}$ and $E$ is the margin of error.

Given that the margin of error $E =500$. The confidence coefficient is $1-\alpha=0.95$. Thus $\alpha = 0.05$.

The population standard deviation is $\sigma = 3215$.

Z-critical value
Z-critical value

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

The minimum sample size required to estimate the mean is

$$ \begin{aligned} n &= \bigg(\frac{z* \sigma}{E}\bigg)^2\\ & = \bigg(\frac{1.96*3215}{500}\bigg)^2\\ & =158.8306\\ &\approx 159. \end{aligned} $$

Thus, the sample of size $n=159$ will ensure that the $95$% confidence interval for the mean will have a margin of error $500$.

Further Reading