Question

Suppose you are interested in estimating the proportion of Florida State University students who make use of the university's mental health services. You take a sample of 300 students, and find that 47 made use of the university's mental health services. Which of the following is a 95% confidence interval for the true proportion of Florida State University students who make use of the university's mental health services (rounded to two decimal places)? a. 16.84 b. 12.20 c. 10.21 d. 45.57

Answer 1

The 95% confidence interval for the true proportion of Florida State University students who make use of the university's mental health services is (0.12, 0.20). Option b.

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 = 300, \pi = (47)/(300) = 0.1567`

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)} = 0.1567 - 1.96\sqrt{(0.1567*0.8433)/(300)} = 0.1156`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.1567 + 1.96\sqrt{(0.1567*0.8433)/(300)} = 0.1978`

Rounded to two decimal places: (0.12, 0.20)

The 95% confidence interval for the true proportion of Florida State University students who make use of the university's mental health services is (0.12, 0.20). Option b.