Question

During the recent impeachment inquiry of the President, a Monmouth University poll of n= 1,161 adults found that 49% supported the impeachment inquiry. What is the Margin of Error for this poll for 95% confidence?

Answer 1

Solution:

Given in the question

No. Of sample n = 1161

P(supported impeachment inquiry) = 0.49

95% magrin of error can be calcualted as, here we will use one proportion Z test to calcualte margin of error

Margin of error = Zalpha/2*sqrt(p*(1-p)/n)

Here alpha = 0.05, alpha/2 = 0.95

Zalpha/2 can be found from Z table

Zalpha/2 = 1.96

So 95% margin of error can be calcualted as

Margin of error = 1.96*sqrt(0.49*(1-0.49)/1161) = 1.96*0.0147 = 0.0288

So margin of error for this poll is 0.0288 for 95% confidence.