Question

A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Step 2 of 2: Suppose a sample of 498 floppy disks is drawn. Of these disks, 464 were not defective. Using the data, construct the 90% confidence interval for the population proportion of disks that are defective. Round your answers to three decimal places.

Answer 1

sample proportion, pcap = (498 - 464)/498= 0.068
sample size, n = 498
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.068 * (1 - 0.068)/498) = 0.0113

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.645

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.068 - 1.645 * 0.0113 , 0.068 + 1.645 * 0.0113)
CI = (0.049 , 0.087)