Question

A reputable polling organization in a certain country surveyed 106,600 ​adults, and 18​% of those polled reported that they smoked. Complete parts a and b below.

​a) Calculate the margin of error for the proportion of all adults who smoke with 90​% confidence.

ME=________

​(Round to four decimal places as​ needed.)

Answer 1

sample size		n                    =		106600.0
sample proportion     p̂		x/n=		0.1800
std error       =Se		=√(p*(1-p)/n) =		0.0012
for 90 % CI value of z=				1.645
margin of error E=z*std error                            =				0.0020

from above:

margin of error for the proportion =0.0020