Question

When 293 college students are randomly selected and surveyed, it is found that 114 own a car. Construct a 99% confidence interval for the true proportion of all college students who own a car. Round your answer to 3 decimal places.

Answer 1

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 $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

When 293 college students are randomly selected and surveyed, it is found that 114 own a car.

This means that $n = 293, \pi = \frac{114}{293} = 0.389$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.389 - 2.575\sqrt{\frac{0.389*0.621}{293}} = 0.315$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.389 + 2.575\sqrt{\frac{0.389*0.621}{293}} = 0.463$

The 99% confidence interval for the true proportion of all college students who own a car is (0.315, 0.463).