Question

What sample size is needed to give a margin of error within plus-or-minus 2.5% in estimating a population proportion with 90% confidence? An initial small sample has p ^ =equals 0.81. Round your answer up to the nearest integer. The absolute tolerance is +/-2

Answer 1

Solution :

Given that,

$\hat p$ = 0.81

1 - $\hat p$ = 1 - 0.81 = 0.19

margin of error = E = 0.025

At 90% confidence level the z is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.10

$\alpha$ / 2 = 0.10 / 2 = 0.05

Z_$\alpha$/2 = Z_0.05 = 1.645

Sample size = n = (Z_$\alpha$/2 / E)² * $\hat p$ * (1 - $\hat p$ )

= (1.645 /0.025)² * 0.81* 0.19

=666

Sample size = 666