Question

the starting salaries of college instructors have a standard deviation of 2000$. how large a sample is needed if we wish to be 95% confident that our sample mean will be within 500$ of the true average salary of all college instructors?

Answer 1

For a sample size n, a 95% confidence interval would be:\[\mu\pm \Phi(0.975)\frac{\sigma}{\sqrt{n}}=\mu \pm 1.96\frac{2000}{\sqrt{n}}=\mu \pm \frac{3920}{\sqrt{n}}\]Now, if we just consider the right half of the confidence interval (which we can because the left half is identical) then you have:\[\mu +\frac{3920}{\sqrt{n}}=\mu +500 \implies \frac{3920}{\sqrt{n}}=500\implies \sqrt{n}=\frac{3920}{500}\implies n=61.4656\]So, as long as the sample size is 61 or less, we can be 95% sure the mean will lie within $500 of the average.