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

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 3 minutes ago

asked 6 minutes ago

asked 7 minutes ago

asked 10 minutes ago

asked 10 minutes ago

asked 16 minutes ago

asked 30 minutes ago

asked 40 minutes ago

asked 40 minutes ago

asked 43 minutes ago

asked 58 minutes ago

asked 59 minutes ago