Question

The Department of Transportion (DOT) is attempting to determine the proportion of drivers who require all passengers in the car to wear their seatbelt before putting the vehicle in drive. A survey of 114 drivers is performed and 62 people say they will not drive until all passengers in the vehicle are buckled up. To report their finding they want to create a 95% confidence interval. What would be the margin error for this confidence interval?
1) 0.0043
2) 0.0914
3) 0.0086
4) 0.0765
5) 0.0466

Answer 1

Answer:

2) 0.0914

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

The margin of error is of:

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

A survey of 114 drivers is performed and 62 people say they will not drive until all passengers in the vehicle are buckled up.

This means that $n = 114, \pi = \frac{62}{114} = 0.5439$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

What would be the margin error for this confidence interval?

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

$M = 1.96\sqrt{\frac{0.5439*0.4561}{114}}$

$M = 0.0914$

Thus the margin of error is 0.0914, option 2).