Question

Long answer question [7 marks] 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 (1 Marks) 2. Calculate the test statistic value (2 Marks) 3. Determine the probability value (1 Marks)

Answer 1

n = 215 x = 86 2 )s c=0.gs » 2=1-C = 0.06 > = 0.025 2012 = 20.02s - 1.96 margin of Emor (E) E = Za 2. V - " ν P CI- η. - 1.96 x locations 21s = 1096X Full -1.96 X V0.00111628 = 1.96 X 0.0334) E = 0.0655

lower limit i E -0.4-0.0655 = 0.3345 Upper limit - PTE = 0.4to.o65s = 0.4655 95% confidence interval is (0.3345, 0.4655) b) If we select 100 samples, 95% of the times the true proportion of all university students who have student loans falls in the interval (0.3345, 004655). 5) L=0.05 1. Ho P = 0.3 Hii p > 0.3 2. ² - Po Zo = Poxlrpo) in 0.4- 0.3 0.3 40 7 215 0.0312

Zo = 3.203) Zo = 3. 20 3. Hi contain > sign, so this is right tail test. :P-value = Plzzzo) = P(233.20) = I-PLZ <3.20) 3.20 = ) - 0.9993 from standard Normal probability table s row no. a 3.2 1 I column no=o.o0 J p-value = 0.0007 Q = 0.05 p-value is less than a & Reject Ho. Conclusion: There is sufficient evidence to conclude more than 30% of all university students sufficient & tudent loans. that have