Answer:
The 99% confidence interval for the true proportion of all college students who own a car is (0.315, 0.463).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the z-score that has a p-value of
.
When 293 college students are randomly selected and surveyed, it is found that 114 own a car.
This means that ![n = 293, \pi = \frac{114}{293} = 0.389](https://tex.z-dn.net/?f=n%20%3D%20293%2C%20%5Cpi%20%3D%20%5Cfrac%7B114%7D%7B293%7D%20%3D%200.389)
99% confidence level
So
, z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.389 - 2.575\sqrt{\frac{0.389*0.621}{293}} = 0.315](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.389%20-%202.575%5Csqrt%7B%5Cfrac%7B0.389%2A0.621%7D%7B293%7D%7D%20%3D%200.315)
The upper limit of this interval is:
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.389 + 2.575\sqrt{\frac{0.389*0.621}{293}} = 0.463](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.389%20%2B%202.575%5Csqrt%7B%5Cfrac%7B0.389%2A0.621%7D%7B293%7D%7D%20%3D%200.463)
The 99% confidence interval for the true proportion of all college students who own a car is (0.315, 0.463).