In a random sample of 50 undergraduate students at a college, it was found that 44 students regularly access social networking websites from their college library. what is the margin of error for the true proportion of all undergraduates who access social networking sites from their college library?

Based on the given data the margin of error for the sample population is 6.15. This is computed based on the formula of margin of error for a sample proportion which states that the margin of error (z) is  = squareroot of ((p(p-1))) divided by n, where p is the sample and n is the population. with that formula z = sqrt ((44x43)/50), z = 6.15

sqrt(pq/n)
p and q generally represent the proportion of success to failure; and n is the sample size
lets define them first.
it tells us that we have 50 students, and 44 are our success rate: p = 44/50 our q (our fail rate) is 0.88
then p = .88, q must equal 1-0.88
p = .88, q = .12, and n = 50
from this we can calcuate:
sqrt(pq/n)=sqrt (0.88x0.12x50)
               =2.29