Question

In a random sample of 500 people aged 20-24, 22% were smokers. In a random sample of 450 people aged 25-29, 14% were smokers. Construct a 95% confidence interval for the difference between the proportions of 20-24 year olds and 25-29 year olds who are smokers. Also, find the margin of error. What is the Parameter of interest? What is the underlying Distribution?

Answer 1

For Sample 1: n1 = 500, p̂1 = 0.22

For Sample 2: n2 = 450, p̂2 = 0.14

95% Confidence interval for the difference:

At α = 0.05, two tailed critical value, z_c = NORM.S.INV(0.05/2) = 1.960

Lower Bound = (p̂1 - p̂2) - z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.22 - 0.14) - 1.96*√[(0.22*0.78/500) + (0.14*0.86/450)] = 0.0316

Upper Bound = (p̂1 - p̂2) + z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.22 - 0.14) + 1.96*√[(0.22*0.78/500) + (0.14*0.86/450)] = 0.1284

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Margin of error , E = z*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ])

= 1.96*√[(0.22*0.78/500) + (0.14*0.86/450)] = 0.0484