Question

1. Some proportion of undergraduate students at Penn State are able to land full-time job offers by the time they graduate. A survey of 400 students from the Smeal College of Business showed that 268 students had job offers by their last semester. The College of Engineering surveyed 550 students and 330 of them had job offers by their last semester. (Note: this is not actual data)
   1. Construct a 95% confidence interval for the difference between the proportions of students from Smeal and engineering that have job offers before graduation.
   2. What can you say about this difference in proportions between the two colleges?

Answer 1

SolutionA:

Given n1=400 n2=550

x1=268 x2=330

p1^=x1/n1=268/400=0.67

p2^=x2/n2=330/550=0.6

z crit for 95%=1.96

95% confidence interval for the difference between the proportions of students from Smeal and engineering college

=(p1^-p2^) $\pm$ zsqrt(p1^(1-p1^)/n1+p2^*(1-p2^/n2)

=(0.67-0.6)+-1.96*sqrt(0.67*(1-0.67)/400+0.6*(1-0.6)/550)

=0.07+-0.0616

=0.07-0.0616,0.07+0.0616

=0.0084,0.1316

0.0084<p1-p2<0.131

we are 95% confident that true difference between the proportions of students from Smeal and engineering college lies in between 0.0084 and 0.131

Solutionb:

the 95% confidence does not contain 0

Reject H0

there is sufficient statistical evidence at 5% level of signficance to conclude that there exists a difference in proportion f students who got offers from Smeal and engineering college and it lies in between 0.84% to 13.1%