Question

A used car dealership surveyed 50 customers who bought a card from them two years ago. 78% of customers said that they would recommend this dealership to their friends. What is the 90% confidence interval for the percent of all former buyers who would recommend this dealership to their friends

Answer 1

Answer:

The 90% confidence interval for the percent of all former buyers who would recommend this dealership to their friends is (68.36%, 87.64%).

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 $1 - \frac{\alpha}{2}$.

A used car dealership surveyed 50 customers who bought a card from them two years ago. 78% of customers said that they would recommend this dealership to their friends.

This means that $n = 50, \pi = 0.78$.

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.78 - 1.645\sqrt{\frac{0.78*0.22}{50}} = 0.6836$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.78 + 1.645\sqrt{\frac{0.78*0.22}{50}} = 0.8764$

As a percent:

0.6836*100% = 68.36%

0.8764*100% = 87.64%

The 90% confidence interval for the percent of all former buyers who would recommend this dealership to their friends is (68.36%, 87.64%).