A recent survey reported that in a sample of 400 students wh

A recent survey reported that in a sample of 400 students who attend two-year colleges, 108 work at least 20 hours a week. In a sample of 200 students attending private universities, 24 students work at least 20 hours a week. What is the 95% confidence interval for the difference of the two proportions?

Solution

Confidence Interval for Diffrence of Proportion
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
Probability Succuses( X1 )=108
No.Of Observed (n1)=400
P1= X1/n1=0.27
Proportion 2
Probability Succuses(X2)=24
No.Of Observed (n2)=200
P2= X2/n2=0.12
C.I = (0.27-0.12) ±Z a/2 * Sqrt( (0.27*0.73/400) + (0.12*0.88/200) )
=(0.27-0.12) ± 1.96* Sqrt(0.001)
=0.15-0.063,0.15+0.063
=[0.087,0.213]