Respuesta :
Answer:
Option E) 0.092
Step-by-step explanation:
Given that 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.
Sample size = 50
Of those students who regular access social networking websites from their college library=44
Sample proportion p =[tex]\frac{44}{50} =0.88[/tex]
Std error of p = [tex]\sqrt{\frac{pq}{n} } =\sqrt{\frac{0.88*0.12}{50}\\ =0.04596[/tex]
Margin of error for 95% level=1.96*0.04596
=0.09008
Approximately this equals option E) 0.092
