In a random sample of 50 undergraduate students at a college, it was found that 44 students regularly access social networking w
ebsites 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?
<span>sqrt(pq/n) </span><span>p and q generally represent the proportion of success to failure; and n is the sample size </span>lets define them first. it tells us that we have 50 students, and 44 are our success rate:
p = 44/50
<span>our q (our fail rate) is 0.88 </span><span>then p = .88, q must equal 1-0.88 </span><span>p = .88, q = .12, and n = 50 from this we can calcuate: sqrt(pq/n)=sqrt (0.88x0.12x50) </span> =2.29
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
