We wish to estimate the proportion of people who get more in

We wish to estimate the proportion of people who get more information from television than from newspapers. If we wish to find 90% interval with an error of .05, how large a sample will we need?

Solution

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.1 is = 1.64
Samle Proportion = 0.5
ME = 0.05
n = ( 1.64 / 0.05 )^2 * 0.5*0.5
= 268.96 ~ 269