Question

A researcher at a major clinic wishes to estimate the proportion of the adult population of the United States that has a certain health condition. How large of a sample is needed in order to be 95% confident that the sample proportion will not differ from the true proportion by more than 3%?

Group of answer choices

1113

1068

752

1344

1535

Answer 1

Solution :

Given that,

$\hat p$= 0.5

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

margin of error = E = 3% = 0.03

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 / 0.03)2 * 0.5 * 0.5

= 1068

Sample size =1068