An education software company did a survey of parents of the

An education software company did a survey of parents of their children. A random sample of 855 parents was asked if they were satisfied with the quality of education their children recieve. 432 of them indicated they were satisfied. Find the margin of error for a 95% confidence interval for the population proportion of all parents who are satisfied with the quality of education for their children.

A).0148

B).0264

C).0335

D).0411

Solution

Margin of Error = Z a/2 Sqrt(p*(1-p)/n))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=432
Sample Size(n)=855
Sample proportion =0.50526
Margin of Error = Z a/2 * ( Sqrt ( (0.5053*0.4947) /855) )
= 1.96* Sqrt(0.00029)
=0.03351