As a marketing director for a local university, you want to advertise the mean salary of your graduates. As an ethical marketing person, you want to be 99% sure that the number you advertise is not more than \$500 from the true average salary. A quick, small sample reveals a standard deviation of \$6278. How large a sample should you get?


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 $1-\alpha=0.99$. Thus $\alpha = 0.01$.

The sample standard deviation is $\sigma = 6278$.

z-critical 0.01

The critical value of $Z$ is $z=Z_{\alpha/2} = 2.58$.

The minimum sample size required to estimate the mean is

$$ \begin{aligned} n &= \bigg(\frac{z* \sigma}{E}\bigg)^2\\ & = \bigg(\frac{2.58*6278}{500}\bigg)^2\\ & =1049.4023\\ &\approx 1050. \end{aligned} $$

Thus, the sample of size $n=1050$ will ensure that the $99$% confidence interval for the mean will have a margin of error $500$.

Further Reading