Kim wants to determine a 95 percent confidence interval for the true proportion of high school students in the area who attend their home basketball games. How large of a sample must she have to get a margin of error less than 0.03?

Answer 1

Assuming the worst case of the point estimate of 50% (i.e. 50% participation), and
n=size of sample

Margin of error
= z_1-α * standard error of worst case point estimate
= z_95% * sqrt(0.5(1-0.5)/n)
=1.96 * 0.5/sqrt(n)

Equate to given margin of error of 0.03, then
0.03=1.96*0.5/sqrt(n)
Solve for n
n=(1.96*0.5/0.03)²
=32.67^2
=1067.11 (round up)
=1068