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 119 of those records. You find that 89 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. Sample proportion =_______________

You decide to construct a 95% confidence interval for the proportion of all enrolling students at the university that complete their degree. You may find this standard normal table useful for the following questions. If you use your answer to part a) in the following calculations, use the rounded version.

b)Calculate the lower bound for the confidence interval. Give your answer as a decimal to 3 decimal places. Lower bound for confidence interval = _______________

c)Calculate the upper bound for the confidence interval. Give your answer as a decimal to 3 decimal places. Upper bound for confidence interval =________________

Answer 1

Answer:

a)0.75

b)0.67

c)0.826

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 interval $1-\alpha$, 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:

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 $n = 119, \pi = \frac{89}{119} = 0.7478$.

95% confidence interval

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7478 - 1.96\sqrt{\frac{0.7478*0.2522}{119}} = 0.67$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7478 + 1.96\sqrt{\frac{0.7478*0.2522}{119}} = 0.826$

So answering the questions:

a)Calculate the proportion of sampled students that complete their degree.

This is $\pi = 0.75$.

b)Calculate the lower bound for the confidence interval.

This is 0.67

c)Calculate the upper bound for the confidence interval.

This is 0.826.