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?
Solution :
Given that,
Point estimate = sample proportion = = x / n = 436 / 600 = 0.727
1 - = 1 - 0.727 = 0.273
margin of error = E = 0.03
Z/2 = Z0.005 = 2.576
sample size = n = (Z / 2 / E )2 * * (1 - )
= (2.576 / 0.03)2 * 0.727 * 0.273
= 1463.34
sample size = n = 1464
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