Question

9) The mean salary of people living in a certain city is $37,500 with a standard deviation of $2,103. A sample of n people will be selected at random from those living in the city. Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income. Round your answer up to the next largest whole number.

Answer 1

9. Here z value for 90% CI is 1.645 as P(-1.645<z<1.645)=0.90

Now Margin of Error is E=500

Now we will find n using formula of E

$E=z*\frac{\sigma }{\sqrt{n}}$

So $n=(\frac{z*\sigma}{E})^2=(\frac{1.645*2103}{500})^2=47.87=48$