Question

How large of the sample size is required, if we wish to estimate proportion p to within 0.03 with 95 % confidence?

Please answer in detail & explain the steps!

Answer 1

Solution :

Given that,

= 1 - = 0.5

margin of error = E = 0.03

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.96 / 0.03)² * 0.5 * 0.5

= 1067

sample size = 1067