Accuracy of price scanners at Wal-Mart. The National Institute for Standards and Technology (NIST) mandates that for every 100 items scanned through the electronic checkout scanner at a retail store, no more than 2 should have an inaccurate price. A recent study of the accuracy of checkout scanners at Wal-Mart stores in California was conducted (Tampa Tribune, Nov. 22, 2005). At each of 60 randomly selected Wal-Mart stores, 100 random items were scanned. The researchers found that 52 of the 60 stores had more than 2 items that were inaccurately priced.

Question

Accuracy of price scanners at Wal-Mart. The National Institute for Standards and Technology (NIST) mandates that for every 100 items scanned through the electronic checkout scanner at a retail store, no more than 2 should have an inaccurate price. A recent study of the accuracy of checkout scanners at Wal-Mart stores in California was conducted (Tampa Tribune, Nov. 22, 2005). At each of 60 randomly selected Wal-Mart stores, 100 random items were scanned. The researchers found that 52 of the 60 stores had more than 2 items that were inaccurately priced.

a. Give an estimate of p, the proportion of Wal-Mart stores in California that have more than 2 inaccurately priced items per 100 items scanned.

b. Construct a 95% confidence interval for p.

c. Give a practical interpretation of the interval, part b.

d. Suppose a Wal-Mart spokesperson claims that 99% of California Wal-Mart stores are in compliance with the NIST mandate on accuracy of price scanners. Comment on the believability of this claim.

e. Are the conditions required for a valid large-sample confidence interval for p satisfied in this application? If not, comment on the validity of the inference in part d.

Answer 1

a. To estimate the proportion p, we need to calculate the sample proportion. The sample proportion is the number of Wal-Mart stores with more than 2 inaccurately priced items divided by the total number of stores sampled. In this case, the sample proportion is 52/60 = 0.8667.

b. To construct a 95% confidence interval for p, we need to calculate the margin of error and use it to create the interval. The formula for the margin of error is:

Margin of error = Z * sqrt((p * (1 - p)) / n)

where Z is the z-score corresponding to the desired confidence level (in this case, 95%), p is the sample proportion, and n is the number of stores sampled.

For a 95% confidence level, the z-score is approximately 1.96. Plugging in the values:

Margin of error = 1.96 * sqrt((0.8667 * (1 - 0.8667)) / 60) ≈ 0.0937

The confidence interval is then given by:

Confidence interval = p ± Margin of error

Confidence interval = 0.8667 ± 0.0937

c. The practical interpretation of the confidence interval is that we are 95% confident that the true proportion of Wal-Mart stores in California that have more than 2 inaccurately priced items per 100 scanned items falls within the range 0.7730 to 0.9604.

d. To comment on the believability of the claim that 99% of California Wal-Mart stores are in compliance with the NIST mandate, we compare the claim to the confidence interval. Since the lower bound of the confidence interval is 0.7730 and the claim is 0.99, it is highly unlikely that the claim is true. The claim falls well outside the range of the confidence interval.

e. The conditions required for a valid large-sample confidence interval for p are not satisfied in this case. The conditions assume that the sample is a random sample from a large population with a finite population correction factor applied (if applicable). In this case, the sample consists of 60 stores, which may not be considered large, and we do not have information about the population size. Therefore, the inference made in part d based on a large-sample confidence interval may not be valid.