Question

What sample size is needed to give a margin of error within ±2.5% in estimating a population proportion with 99% confidence? An initial small sample has p^=0.78.

Round the answer up to the nearest integer.

Answer 1

Solution:

Given that:

E= 0.025

= 0.78

1 - = 1 - 0.78= 0.22

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

sample size = (Z_/2 / E)² * * (1 - )     

   = (2.576 / 0.025)² * (0.78 * 0.22)

                   = 1821.92

sample size = 1822