Question

In a randomly selected group of 650 automobile deaths, 180 were alcohol related. Construct a 95 percent confidence interval for the true proportion of all automobile accidents caused by alcohol.

a. [.243, .311]
b. [.262, .292]
c. [.259, .294]
d. [.263, .291]

Answer 1

a. [.243, .311]

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 zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

In a randomly selected group of 650 automobile deaths, 180 were alcohol related. This means that
`n = 650, \pi = (180)/(650) = 0.277`

95% confidence level

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

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.277 - 1.96\sqrt{(0.277*0.723)/(650)} = 0.243`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.277 + 1.96\sqrt{(0.277*0.723)/(650)}{119}} = 0.311`

So the correct answer is:

a. [.243, .311]