Question

The university would like to conduct a study to estimate the true proportion of all university students who have student loans. According to the study, in a random sample of 215 university students, 86 have student loans.

(a) Construct a 95% confidence interval for estimating the true proportion of all university students who have student loans (2 marks)

(b) Provide an interpretation of the confidence interval in part (a). (1mark)

(c) Conduct an appropriate hypothesis test, at the 5% level of significance to test the claim that more than 30% of all university students have student loans.

1. Provide the hypothesis statement
2. Calculate the test statistic value
3. Determine the probability value

Answer 1

a)
sample proportion, = 0.4
sample size, n = 215
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.4 * (1 - 0.4)/215) = 0.0334

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.4 - 1.96 * 0.0334 , 0.4 + 1.96 * 0.0334)
CI = (0.3345 , 0.4655)

b)
We are 95% confident that the true proportion of students having loan is between 0.3345 and 0.4655

c)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.3
Alternative Hypothesis, Ha: p > 0.3

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.4 - 0.3)/sqrt(0.3*(1-0.3)/215)
z = 3.2

P-value Approach
P-value = 0.0007
As P-value < 0.05, reject the null hypothesis.