Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 391 drivers and find that 319 claim to always buckle up. Construct a 85% confidence interval for the population proportion that claim to always buckle up.

Answer 1

Answer:

The 85% confidence interval for the population proportion that claim to always buckle up is (0.7877, 0.8441).

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

They randomly survey 391 drivers and find that 319 claim to always buckle up.

This means that $n = 391, \pi = \frac{319}{391} = 0.8159$

85% confidence level

So $\alpha = 0.15$, z is the value of Z that has a pvalue of $1 - \frac{0.15}{2} = 0.925$, so $Z = 1.44$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8159 - 1.44\sqrt{\frac{0.8159*0.1841}{391}} = 0.7877$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8159 + 1.44\sqrt{\frac{0.8159*0.1841}{391}} = 0.8441$

The 85% confidence interval for the population proportion that claim to always buckle up is (0.7877, 0.8441).