Question

The journal article also reported that the study originally had 1,474 participants, but only 1,234 reported drinking any alcohol in the past year. (Only those who reported drinking in the past year were retained for the hangover symptom questions.) Use this information to find a 95% confidence interval for the proportion of all students who would report drinking any alcohol in the past year. (Round your answers to three decimal places.)

Answer 1

The 95% confidence interval for the proportion of all students who would report drinking any alcohol in the past year is (0.818, 0.856).

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:

`n = 1474, \pi = (1234)/(1474) = 0.8372`

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)} = 0.8372 - 1.96\sqrt{(0.8372*0.1628)/(1474)} = 0.818`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.8372 + 1.96\sqrt{(0.8372*0.1628)/(1474)} = 0.856`

The 95% confidence interval for the proportion of all students who would report drinking any alcohol in the past year is (0.818, 0.856).