Question

When 332 college students are randomly selected and surveyed, it is found that 113 own a car. find a 99% confidence interval for the true proportion of all college students who own a car?

Answer 1

Given:
n = 332, sample size
p = 113/332 = 0.3404, sample proportion
99% confidence interval

The confidence interval for the population is calculated from
$p \pm z^{*} \sqrt{ \frac{p(1-p)}{n} }$
where z* = 2.58 for the 99% confidence level (from tables)..

$2.58 \sqrt{ \frac{0.3404(1-0.3404)}{332} } =0.0671$
Therefore the 99% confidence interval is
(0.3404 - 0.0671, 0.3404 + 0.0671) = (0.2733, 0.4075)

Answer:
The 99% confidence interval is (0.273, 0.408) or (27%, 41%).
That is, between 27% and 41% of the students own cars.