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.

Answer 1

The 95 percent confidence interval for the true proportion of all automobile accidents caused by alcohol is (0.2425, 0.3113).

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, p = (180)/(650) = 0.2769`

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)} = 0.2769 - 1.96\sqrt{(0.2769*0.7231)/(650)} = 0.2425`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.2769 + 1.96\sqrt{(0.2769*0.7231)/(650)} = 0.3113`

The 95 percent confidence interval for the true proportion of all automobile accidents caused by alcohol is (0.2425, 0.3113).