Question

Out of 600 adults selected randomly from Minneapolis-St. Paul area, 90 are found to be smoker. Find the margin of error to construct a 82% confidence interval of the true estimate of smoker in the area.

Question 11 options:

	0.1920
	0.0267
	0.0211
	0.0133
	0.0196

Answer 1

Solution :

Given that,

n = 600

x = 90

Point estimate = sample proportion = = x / n = 90/600=0.15

1 -   = 1-0.15 =0.85

At 82% confidence level the z is ,

Z/2 = Z0.09 = 1.34 ( Using z table )

Margin of error = E = Z / 2    * ((( * (1 - )) / n)

= 1.34 (((0.15*0.85) /600 )

E = 0.0196