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.

CI 99%   =  ( - 0.0009 ;  0.0541 )

Step-by-step 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 )

answer: this is not a question

step-by-step explanation: this is a answer