Question

Insurance companies are interested in knowing the population percentage of drivers who always buckle up before riding in a car. When designing a study to determine this population proportion, what is the minimum number of drivers you would need to survey to be 95% confident that the population proportion is estimated to within 0.04

Answer 1

The minimum number of drivers you would need to survey is 601.

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{(\pi(1-\pi))/(n)}`

In which

z is the z-score that has a p-value of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a p-value of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

What is the minimum number of drivers you would need to survey to be 95% confident that the population proportion is estimated to within 0.04?

The number is n for which M = 0.04.

We don't have an estimate for the proportion, so we use
`\pi = 0.5`. Then

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.04 = 1.96\sqrt{(0.5*0.5)/(n)}`

`0.04√(n) = 1.96*0.5`

`√(n) = (1.96*0.5)/(0.04)`

`(√(n))^2 = ((1.96*0.5)/(0.04))^2`

`n = 600.25`

Rounding up:

The minimum number of drivers you would need to survey is 601.