The maximal margin of error when constructing a 99 CI for a

The maximal margin of error when constructing a 99% CI for a normal population mean was 15 from sample size 500. How many observations are necessary to make the maximal margin of error 5 at the same confidence level. The answer is 4500, but I\'m lost on how to get it. Any help would be great. Thank you!

Solution

As for margins of error,

z = E*sqrt(n)/sigma

Then, as they are both 99% confidence intervals, then

E1 sqrt(n1) / sigma = E2 sqrt(n2) / sigma

E1 sqrt(n1) = E2 sqrt(n2)

15*sqrt(500) = 5*sqrt(n2)

3*sqrt(500) = sqrt(n2)

9*500 = n2

n2 = 4500 [ANSWER]