Question

It is known that the population standard deviation of an aptitude test score is 72. How large a sample must be taken in order to be 90% confident that the margin of error will not exceed 15?

Answer 1

For 90% CI, z value is 1.645 as P(-1.645<z<1.645)=0.90

Further Margin of Error is E=15 and standard deviation is 72

So we will find n using formula of E

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

So $n=(\frac{z*\sigma}{E})^2=(\frac{1.645*72}{15})^2=62.35=63$