In a sample of 200 bottles at a factory, 38 had major defects. The sampling method had a margin of error of 0.02. What is the in
terval estimate for the proportion that would have defects in the form (lower limit, upper limit) ? a. (0.17, 0.21) b. (0.19, 0.21) c. (0.16, 0.20) d. (0.15, 0.19)
The confidence interval is always centered at the mean of the sample you take. In this case, the average value of the sample is 38/200 = 0.19. The only interval of length 0.04 centered at 0.19 is (0.17; 0.21). Note that the intervals should have length 0.04 in order for the margin of error to be 0.02.
