A researcher wants to construct a 98 confidence interval for

A researcher wants to construct a 98% confidence interval for the proportion of elementary school students in Seward County who receive free or reduced-price school lunches. A state-wide survey indicates that the proportion is 0.45. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.08?

Solution

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.02 is = 2.33
Samle Proportion = 0.45
ME = 0.08
n = ( 2.33 / 0.08 )^2 * 0.45*0.55
= 209.9457 ~ 210