Given:
n = 332, sample size
p = 113/332 = 0.3404, sample proportion
99% confidence interval
The confidence interval for the population is calculated from
[tex]p \pm z^{*} \sqrt{ \frac{p(1-p)}{n} } [/tex]
where z* = 2.58 for the 99% confidence level (from tables)..
[tex]2.58 \sqrt{ \frac{0.3404(1-0.3404)}{332} } =0.0671[/tex]
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.
