Question

of voters who will vote for Donald Trump in the next presidential election. How large does the sample need to be to estimate p with a 53. Suppose you want to estimate the proportion margin of error of 0.04 with 95% confidence? ооооо

Answer 1

Solution :

Given that,

$\hat p$= 0.5

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

margin of error = E = 0.04

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z$\alpha$/2 = Z0.025 = 1.96 ( Using z table )

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

= (1.96 / 0.04)2 * 0.5 * 0.5

= 600.25

Sample size =601