Question

You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately 17%. You would like to be 99% confident that your estimate is within 1% of the true population proportion. How large of a sample size is required?

n =

Answer 1

Solution :

Given that,

$\hat p$ = 0.17

1 - $\hat p$ = 1 - 0.17 = 0.83

margin of error = E =1 % = 0.01

At 99% confidence level the z is,

$\alpha$ = 1 - 99%

$\alpha$ = 1 - 0.99 = 0.01

$\alpha$/2 = 0.005

Z_$\alpha$/2 = 2.58 ( Using z table ( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )

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

= (2.58 / 0.01)² * 0.17 * 0.83

=9392.18

Sample size = 9392