Question

Construct the indicated confidence interval for the difference in proportions. Assume that the samples are independent and that they have been randomly selected.A marketing survey involves product recognition in New York and California. Of 558 New Yorkers surveyed, 193 knew the product while 196 out of 614 Californians knew the product. Construct a 99% confidence interval for the difference in the proportions of New Yorkers and Californians who knew the product.

Answer 1

CI 99% = ( - 0.0009 ; 0.0541 )

Explanation:

Sample 1 New Yorkers

sample size n₁ = 558

x₁ = 193

p₁ = x₁/n₁ = 193/ 558 p₁ = 0.3458 q₁ = 1 - p₁ q₁ = 1 - 0.3458

q₁ = 0.6542

Sample 2 Californians

sample size n₂ = 614

x₂ = 196

p₂ = x₂/n₂ = 196 / 614 p₂ = 0.3192 q₂ = 1 - p₂ q₂ = 1 - 0.3192

q₂ = 0.6808

CI 99 % means significance level α = 1 αα% α = 0.01

α/2 = 0.005

In z-table we look for z score for 0.005 z (c) = 2.57

CI 99 % = [ ( p₁ - p₂ ) ± z(c) * √( p₁*q₁)/n₁ + ( p₂*q₂)/n₂

p₁ - p₂ = ( 0.3458 - 0.3192 ) = 0.0266

√( p₁*q₁)/n₁ + ( p₂*q₂)/n₂ =√ 0.3458*0.6542)/558 + 0.3192*0.6808)/614

√( p₁*q₁)/n₁ + ( p₂*q₂)/n₂ = √ 4.05*10⁻⁴ + 3.54 * 10⁻⁴

√( p₁*q₁)/n₁ + ( p₂*q₂)/n₂ = 10⁻² * √7.59 = 10⁻² * 2.75

Then:

CI 99 % = 0.0266 ± 2.75 * 10⁻²

CI 99% = 0.0266 ± 0.0275

CI 99% = ( - 0.0009 ; 0.0541 )