Question

A consumer agency wants to estimate the proportion of all drivers who wear seat belts while driving. A preliminary study of 600 people has shown that 436 of drivers wear seat belts while driving. How large should the sample size be so that the 99% confidence interval for the population proportion has a maximum error of .03?

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = x / n = 436 / 600 = 0.727

1 - $\hat p$ = 1 - 0.727 = 0.273

margin of error = E = 0.03

Z$\alpha$/2 = Z0.005  = 2.576

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

= (2.576 / 0.03)2 * 0.727 * 0.273

= 1463.34

sample size = n = 1464