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.
1.50
a. To estimate p, the proportion of Wal-Mart stores in California that have more than 2 inaccurately priced items per 100 items scanned, we can use the observed data. Out of the 60 randomly selected Wal-Mart stores, we know that 52 stores had more than 2 inaccurately priced items. Therefore, our estimate for p is 52/60 = 0.8667.
b. To construct a 95% confidence interval for p, we can use the formula for a confidence interval for a proportion. The formula is:
CI = p̂ ± Z * √(p̂(1-p̂)/n)
where p̂ is the sample proportion, Z is the z-score corresponding to the desired level of confidence, and n is the sample size.
In this case, p̂ = 0.8667, n = 60, and the desired level of confidence is 95%, which corresponds to a z-score of 1.96.
Plugging in the values, we get:
CI = 0.8667 ± 1.96 * √(0.8667(1-0.8667)/60)
Calculating this expression, we find:
CI = 0.8667 ± 1.96 * 0.0489
Therefore, the 95% confidence interval for p is approximately 0.8667 ± 0.0959, or (0.7708, 0.9626).
c. The practical interpretation of the interval is that we can be 95% confident that the true proportion of Wal-Mart stores in California that have more than 2 inaccurately priced items per 100 items scanned falls within the interval (0.7708, 0.9626).
d. The claim made by the Wal-Mart spokesperson that 99% of California Wal-Mart stores are in compliance with the NIST mandate seems unlikely. According to our estimate from the study, 86.67% of the stores had more than 2 inaccurately priced items. Moreover, the lower bound of the confidence interval is 0.7708, which is less than 0.99. Hence, it is not believable that 99% of the stores are in compliance.
e. The conditions required for a valid large-sample confidence interval for p are typically that the sample size n is large enough, and both np and n(1-p) are greater than or equal to 10. In this case, the sample size is 60, and we estimated p̂ to be 0.8667. Therefore, np̂ = 60 * 0.8667 = 52 and n(1-p̂) = 60 * (1-0.8667) = 8, which are both greater than 10. Hence, the conditions for a valid large-sample confidence interval for p are satisfied. Therefore, the inference made in part d is valid, and we can conclude that the Wal-Mart spokesperson's claim is not believable based on the study's findings.