Answer:
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.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 300, \pi = \frac{47}{300} = 0.1567[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1567 - 1.96\sqrt{\frac{0.1567*0.8433}{300}} = 0.1156[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1567 + 1.96\sqrt{\frac{0.1567*0.8433}{300}} = 0.1978[/tex]
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.
