Question

Out of 400 adults selected randomly from Minneapolis-St. Paul area, 42 are found to be smoker. Find the margin of error to construct a 86% confidence interval of the true estimate of smoker in the area.

Question 3 options:

	0.0269
	0.0133
	0.0226
	0.1920
	0.0196

Answer 1

Given:

Number of sample, n = 400

Number of success, X = 42

Sample proportion, = X/n = 42/400 = 0.105

Significance level, = 1-0.86 = 0.14

At 86% confidence level, the critical value of Z is

Z/2 = Z0.14/2 = 1.48

Margin of error, E :

E = Z/2 × √ [ (1 - ) / n ]

= 1.48 × √[ 0.105 ( 1 - 0.105 ) / 400 ]

= 1.48 × 0.01532

= 0.0226

Therefore the margin of error is E = 0.0226

Answer - option C