A researcher is going to estimate the proportion of individu

A researcher is going to estimate the proportion of individuals that have health issues. She is going to sample approximately 800 people and will compute a 95% confidence interval. What is her margin of error likely to be?

.25

.035

.07

.01

Solution

Given a=1-0.95= 0.05, Z(0.025) = 1.96 (from standard normal table)

We use p=0.5 as estimated.

So margin of error is

Z*sqrt(p*(1-p)/n) = 1.96*sqrt(0.5*0.5/800)

=0.03464823

Answer: B. .035