Question

At one store, 33.4% of the 600 customers surveyed said checkout lines are too long. Find the margin of error for 95% confidence for the proportion of customers that said checkout lines are too long.

Answer 1

solution:

the given information as follows:

sample size = n = 600

proportion of customer who said checkout lines are too long = $\hat p$ = 33.4% = 0.334

confidence level = 95% = 0.95

so, $\alpha$ = 1 - 0.95 = 0.05

critical value of z from z table = $z_c=z_{1-0.05/2}=1.96$

margin of error:

$Me=z_c*\sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96*\sqrt{\frac{0.334(1-0.334)}{600}}=0.0377$

so , margin of error for 95% confidence interval = 0.0377