Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 387 drivers and find that 298 claim to always buckle up. Construct a 84% confidence interval for the population proportion that claim to always buckle up.

Answer 1

Answer:

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

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

They randomly survey 387 drivers and find that 298 claim to always buckle up.

This means that $n = 387, \pi = \frac{298}{387} = 0.77$

84% confidence level

So $\alpha = 0.16$, z is the value of Z that has a p-value of $1 - \frac{0.16}{2} = 0.92$, so $Z = 1.405$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.74$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.8$

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).