Question

A random sample of students were asked whether they were satisfied with their school's lunch menu. The resulting confidence interval for the proportion of students who are satisfied is (0.33,0.51).

What is the margin of error?

Answer 1

Solution :

Given that,

confidence interval = 0.33, 0.51

Point estimate = = (Lower confidence interval + Upper confidence interval ) / 2

Point estimate = = (0.33 + 0.51) / 2 = 0.42

Margin of error = E = Upper confidence interval -

Margin of error = E = 0.51 - 0.42

Margin of error = E = 0.09