Question

A software developer wants to know how many new computer games people buy each year. Assume a previous study found the variance to be 2.25. He thinks the mean is 6.3 computer games per year. What is the minimum sample size required to ensure that the estimate has an error of at most 0.13 at the 80% level of confidence? Round your answer up to the next integer.

Answer 1

z value for 80% is 1.28 as P(-1.28<z<1.28)=0.80. Also it is given that Margin of Error is 0.13

Further variance is 2.25

So we will find n using formula of E

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

$n=(\frac{z*\sigma}{E})^2=(\frac{1.28*\sqrt{2.25}}{0.13})^2=218.13=219$