Question

National statistics show that 23% of men smoke and 18.5% of women smoke. A random sample of 175 men indicated that 45 were smokers, and a random sample of 155 women surveyed indicated that 42 smoked. Construct a 98% confidence interval for the true difference in proportions of male and female smokers.

Answer 1

98% C.I. = [-0.1270, 0.0994]

Explanation:

Here

p1= proportion of men smokers

p1= 45/175= 0.2571

p2= proportion of women smokers

p2= 42/155= 0.2709

1) Choose the significance level ∝= 0.02

2) The formula for confidence interval is

(p1^- p2^)- z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))

and

(p1^- p2^) z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))

3) The z value is 2.05

4) Calculations:

(p1^- p2^)- z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))

Putting the values

= 0.2571-0.2709)(-2.05) sqrt ( 0.2571(1-0.2571)/175+ 0.2709(1-0.2709)/155)

=-0.1270 ( using calculator)

And

(p1^- p2^)z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))

Putting the values

= 0.2571-0.2709)(2.05) sqrt ( 0.2571(1-0.2571)/175+ 0.2709(1-0.2709)/155)

=0.0994 ( using calculator)

and Putting ± 2.05 in the confidence interval formula we get limits (-0.1270, 0.0994)