Question

Aresearcher wants to make a 99% confidence interval for a population proportion. A preliminary sample produced the sample proportion of 0.645. The sample size that would limit the margin of error to be within 0.025 of the population proportion is: i

Answer 1

Solution :

Given that,

$\hat p$ = 0.645

1 - $\hat p$ = 1 - 0.645 = 0.355

margin of error = E =0.025

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 - $\hat p$ )

= (2.58 / 0.025)² * 0.645 * 0.355

=2439

Sample size = 2439