Question

6. Determine the minimum sample size required to have a margin of error of 2% with 99% confidence. [4p

Answer 1

Solution :

Given that,

$\hat p$ = 0.5

1 - $\hat p$ = 1 - 0.5 = 0.5

margin of error = E = 2% = 0.02

At 99% confidence level the z is,

$\alpha$ = 1 - 99%

$\alpha$ = 1 - 0.99 = 0.01

$\alpha$/2 = 0.005

Z_$\alpha$/2 = 2.576

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

= (2.576 / 0.02)² * 0.5 * 0.5

=4147.36

Sample size = 4148