Answer:
Option c:
[tex]0.25 \pm 1.645\sqrt{\frac{0.25*(1-0.25)}{90}}[/tex]
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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
You have access to first year enrolment records and you decide to randomly sample 119 of those records. You find that 89 of those sampled went on to complete their degree. This means that [tex]n = 119, \pi = \frac{89}{119} = 0.7478[/tex].
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
Sample of 90 people with a sample proportion of 0.25
This means that [tex]n = 90, p = 0.25[/tex].
Confidence interval:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.25 \pm 1.645\sqrt{\frac{0.25*(1-0.25)}{90}}[/tex]
Which is option c.
