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

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

(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)

p̂ = X / n = 86/215 = 0.4
p̂ ± Z(α/2) √( (p * q) / n)
0.4 ± Z(0.05/2) √( (0.4 * 0.6) / 215)
Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.4 - Z(0.05) √( (0.4 * 0.6) / 215) = 0.335
upper Limit = 0.4 + Z(0.05) √( (0.4 * 0.6) / 215) = 0.465
95% Confidence interval is ( 0.335 , 0.465 )

b)

Interpretation - We are 95% confident that the true proportion of all university students who have student loans

is between 0.335 and 0.465

c)

H0: p = 0.30

Ha: p > 0.30

Test statistics

z = ( - p) / sqrt [ p ( 1 - p) / n ]

= ( 0.4 - 0.3) / sqrt( 0.3 * ( 1- 0.3) / 215 ]

= 3.20

p-value = P(Z > z )

= P(Z > 3.20)

= 0.0007

Since p-value < 0.05 level, reject H0.