Question

When 258 college students are randomly selected and surveyed, it is found that 106 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car. 0.351 < p < 0.471 0.332 < p < 0.490 0.339 < p < 0.482 0.360 < p < 0.461

Answer 1

Answer: 0.332 < p < 0.490

Step-by-step explanation:

We know that the confidence interval for population proportion is given by :-

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where n= sample size

$\hat{p}$ = sample proportion

z* = critical z-value.

As per given , we have

n= 258

Sample proportion of college students who own a car = $\hat{p}=\dfrac{106}{258}\approx0.411$

Critical z-value for 99% confidence interval is 2.576. (By z-table)

Therefore , the  99% confidence interval for the true proportion(p) of all college students who own a car will be :$0.411\pm (2.576)\sqrt{\dfrac{0.411(1-0.411)}{258}}\\\\=0.411\pm (2.576)\sqrt{0.00093829}\\\\= 0.411\pm (2.576)(0.0306315197142)\\\\=0.411\pm 0.0789=(0.411-0.0789,\ 0.411+0.0789)\\\\=(0.3321,\ 0.4899)\approx(0.332,\ 0.490)$

Hence, a 99% confidence interval for the true proportion of all college students who own a car : 0.332 < p < 0.490