Question

When 339 college students are randomly selected and surveyed, it is found that 111 own a car. How do you construct a 95% confidence interval for the percentage of all college students who own a car?

Assume that necessary conditions and assumptions are met.

Answer 1

First, look up the z-value for 95% (2.5% in each tail) ...

Explanation:

z-value `=1.96` [from Standard Normal table]

Find the mean and standard error of the sample proportion ...

sample mean #= 111/339 ~~ 0.3274

standard error `=sqrt((pq)/n)=sqrt([0.3274xx(1-0.3274)]/339]~~0.0255`

Confidence Interval `="mean"+-"z-value"xx"standard error"`

CI `=0.3274+-(1.96)(0.0255)=(0.277, 0.377)`

hope that helped