Question

The FTC conducted a study investigating the accuracy of bar-code scanners. It was concluded that these computer scanners, mainly used by grocery stores, ring up the wrong price about 5% of the time. In most instances, the error was in favor of the shopper, according to the FTC. Assume 13 shoppers are randomly chosen and their bills, as indicted by the scanner, are checked against the correct bill computed by conventional means. Suppose that 200 items were scanned, 21 of the items were charged incorrectly by the scanner.

a. Construct a 95% confidence interval for the proportion of items that were rung up incorrectly by the scanner.

b. does it appear that the local grocery store has a larger error rate than 5%

Answer 1

sample proportion, pcap = 21/200 = 0.105
sample size, n = 200
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.105 * (1 - 0.105)/200) = 0.0217

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.105 - 1.96 * 0.0217 , 0.105 + 1.96 * 0.0217)
CI = (0.0625 , 0.1475)

b)
Yes, local grocery store has a larger error rate becuause 0.05 is less than the lower bound of the above CI