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 111 of those records. You find that 80 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 3 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. 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

Solution :

Given that,

n = 111

x = 80

= x / n = 80 / 111 = 0.721

1 - = 1 - 0.721 = 0.279

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.96 * (((0.721 * 0.279) / 111)

= 0.083

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.721 - 0.083 < p < 0.721 + 0.083

0.638 < p < 0.804

(0.638,0.804)

Lower bound for confidence interval = 0.638

Upper bound for confidence interval = 0.804