Question

You have been given the task of finding out what proportion of students that enroll in a local university actually complete their degree. You have access to first year enrolment records and you decide to randomly sample 115 of those records. You find that 85 of those sampled went on to complete their degree.
a)Calculate the proportion of sampled students that complete their degree. Give your answer as a decimal to 2 decimal places

Calculate lower bound and upper bound at 95% confidence interval. Give answer decimal to 3 places.

Answer 1

Answer:

The proportion of sampled students that complete their degree is 0.74.

The lower bound for the 95% confidence interval is 0.659 and the upper bound is 0.819.

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 115, \pi = \frac{85}{115} = 0.739$

Rounded to two decimal places, the proportion of sampled students that complete their degree is 0.74.

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.739 - 1.96\sqrt{\frac{0.739*0.261}{115}} = 0.659$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.739 + 1.96\sqrt{\frac{0.739*0.261}{115}} = 0.819$

The lower bound for the 95% confidence interval is 0.659 and the upper bound is 0.819.