Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 401 drivers and find that 294 claim to always buckle up. Construct a 90% confidence interval for the population proportion that claim to always buckle up. Use interval notation, for example, [1,5].

Answer 1

Answer:

[0.6969, 0.7695]

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 401 drivers and find that 294 claim to always buckle up.

This means that $n = 401, \pi = \frac{294}{401} = 0.7332. 90% confidence level So [tex]\alpha = 0.1$, 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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7332 - 1.645\sqrt{\frac{0.7332*0.2668}{401}} = 0.6969$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7332 + 1.645\sqrt{\frac{0.7332*0.2668}{401}} = 0.7695$

The 90% confidence interval for the population proportion that claim to always buckle up is [0.6969, 0.7695]