Amy works for a company that produces plastic lids and cups for restaurants. She wants to estimate what proportion of lids produced in a day have a defect that makes them not fit the cups. She takes a random sample of 250 250250 lids from the more than 10 , 000 total lids produced that day and finds 15 lids in the sample with this defect. Based on this sample, calculate the 99% confidence interval for the proportion of lids produced that day with this defect?

99% confidence interval for the proportion of lids produced that day with this defect somewhere between ( .0213 ,  .0987) or between to

Step-by-step explanation:

Given -

Sample size ( n) = 250

Sample proportion = = .06

confidence interval = 1 - .99 = .01

= = 2.58

99% confidence interval for the proportion of lids produced that day with this defect =

            =  

             =

             =

              = ( .06 + .0387 ) ,  ( .06 - .0387 )

              =   ( .0987 , .0213 )

(0.021,0.099)

Step-by-step explanation:

khan