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 217 university students, 88 have student loans.

(a) Construct a 99% confidence interval for estimating the true proportion of all university students who have student loans

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

(c) Conduct an appropriate hypothesis test, at the 1% 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, = 0.4055
sample size, n = 217
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.4055 * (1 - 0.4055)/217) = 0.0333

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

Margin of Error, ME = zc * SE
ME = 2.58 * 0.0333
ME = 0.0859

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.4055 - 2.58 * 0.0333 , 0.4055 + 2.58 * 0.0333)
CI = (0.3196 , 0.4914)

b)
we are 99% confident that the true proportion of all university students who have student loans is between (0.3196 , 0.4914)

c)

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

Rejection Region
This is right tailed test, for α = 0.01
Critical value of z is 2.33.
Hence reject H0 if z > 2.33

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.4055 - 0.3)/sqrt(0.3*(1-0.3)/217)
z = 3.39

P-value Approach
P-value = 0.0003
As P-value < 0.01, reject the null hypothesis.