Question

When 328 college students are randomly selected and surveyed, it is found that 122 own a car. find a 99% confidence interval for the true proportion of all college students who own a car?

Answer 1

Answer:

$0.303<p<0.441$

Step-by-step explanation:

We know that,

$\text{Proportion}=p=\dfrac{r}{n}=\dfrac{122}{328}=0.372$

where

r = number of successful trials = 122,

n = 328

p = proportion

From probability distribution we also know that,

$Z_{critical}$ for a 99% confidence level = 2.576

Marginal error will be,

$M.E=Z_{critical}\cdot \sqrt{\dfrac{p(1-p)}{n}$

Putting the values,

$M.E=2.576\cdot \sqrt{\dfrac{0.372(1-0.372)}{328}}=0.069$

So the interval will be,

$=p\pm M.E$

$=0.372\pm 0.069$

$=0.372+ 0.069,0.372- 0.069$

$=0.441,0.303$