Question

The state education commission wants to estimate the percentage of tenth grade students that have reading skills at or below the eighth grade level. In an earlier study, the population percentage was estimated to be 22%. How large a sample would be required in order to estimate the percentage of tenth graders reading at or below the eighth grade level at the 95% confidence level with an error of at most 2%? (Round your answer up to the next integer) * O 583 O 1649 O 1699 O 1908

Answer 1

Solution,

Given that,

$\hat p$ = 0.22

1 - $\hat p$ = 1 - 0.22 = 0.78

margin of error = E = 0.02

At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ = 1 - 0.95 =0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z0.025  = 1.96

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

= (1.96 / 0.02 )2 * 0.22 * 0.78

= 1648.04

sample size = n = 1649