Question

A researcher wants to determine what percentage of the population skips breakfast. A study was done several years ago that found 39% of people do not eat breakfast. How large of a sample would the researcher need if he wants to be at a 95% confidence level as well as a margin of error of 2%?

Answer 1

Solution :

Given that,

$\hat p$ = 0.39

1 - $\hat p$ = 1 - 0.61

margin of error = E = 0.02

Z$\alpha$/2 = 1.96

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

= (1.96 / 0.02)2 * 0.39 * 0.61

= 2285

sample size = 2285