Question

1. Customers who participate in a store’s loyalty card program save money on their purchases but allow the store to keep track of their shopping habits and potentially sell the data to third parties. A Pew Internet and American life Project Survey revealed that out of a survey of 250 US adults, 100 agreed to participate in the store loyalty program.

a. What is the point estimate of the population proportion of all US adults who would agree to participate in a store loyalty card program?

b. Form a 90% confidence interval around the estimate in part a.

c. Provide a practical interpretation of the confidence interval in part b. Your answer should begin with, “We are 90% confident…

Answer 1

a) point estimate=Sample Proportion ,    p̂ = x/n = 100/250 = 0.4000

b)

z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.030984          
margin of error , E = Z*SE =    1.960   *   0.03098   =   0.0510
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.40000   -   0.05096   =   0.34904
Interval Upper Limit = p̂ + E =   0.40000   +   0.05096   =   0.45096
                  
90%   confidence interval is (   0.3490   < p <    0.4510   )

c)

We are 95% confident that true proportion lies within confidence interval