Question

In a study of the accuracy of​ fast-food drive-through​ orders, a restaurant had 4141 orders that were not accurate among 441 orders observed. Use the bootstrap method to construct a 90​% confidence interval estimate of the proportion of orders that are not accurate. Use the 20 accompanying bootstrap samples. How does the result compare to the 9090​% confidence interval 0.07 less than

Answer 1

p = 441/4141 = 0.11

sd = [ p*(1-p) ]^0.5 = [0.11*0.89]^0.5 = 0.31

z for 90% confidence interval = 1.645

confidence interval = [ p - z*sd/(n^0.5) , p + z*sd/(n^0.5) ]

90% confidence interval = [ 0.11 - 1.645*0.31/(4141^0.5) , 0.11 + 1.645*0.31/(4141^0.5) ]

90% confidence interval = [ 0.1021 , 0.1179 ]

How does the result compare to the 90% confidence interval 0.07 less than , this confidence interval is wider than the CI that we calculated : [0.1021 , 0.1179]

P.S. (please upvote if you find the answer satisfactory)