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 study of the accuracy of checkout scanners at
Walmart stores in California was conducted. At each of 60
randomly selected Walmart 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 Walmart 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 Walmart spokesperson claims that 99% of
California Walmart 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.
f. Determine the number of Walmart stores that must
be sampled in order to estimate the true proportion to
within .05 with 90% confidence using the large-sample
method.
a. To estimate the proportion p, we need to find the number of Walmart stores in California that have more than 2 inaccurately priced items per 100 items scanned. In this case, out of the 60 randomly selected Walmart stores, 52 had more than 2 inaccurately priced items. Therefore, the estimate of p is 52/60 = 0.8667, or approximately 0.87.
b. To construct a confidence interval for p, we can use the formula:
p̂ ± Z * sqrt(p̂(1-p̂)/n)
Where p̂ is the sample proportion, Z is the Z-score corresponding to the desired confidence level, and n is the sample size.
Since we want a 95% confidence interval, the Z-score for a 95% confidence level is approximately 1.96. Given that p̂ = 0.87 and n = 60, we can plug these values into the formula:
0.87 ± 1.96 * sqrt(0.87(1-0.87)/60)
Calculating this expression, we get the confidence interval for p: (0.79, 0.95)
c. The practical interpretation of this confidence interval is that we are 95% confident that the true proportion of Walmart stores in California with more than 2 inaccurately priced items per 100 scanned items lies between 0.79 and 0.95.
d. The spokesperson's claim is that 99% of California Walmart stores are in compliance with the NIST mandate. However, our confidence interval from part b does not include 0.99. This suggests that the spokesperson's claim is not believable and does not align with the findings of the study.
e. The conditions required for a valid large-sample confidence interval for p are:
1. Random sample: The study states that the Walmart stores were randomly selected, so this condition is satisfied.
2. Independence: The accuracy of one store's checkout scanner should not be dependent on another store. If the stores were chosen independently, this condition is satisfied.
3. Large sample size: The sample size is 60, which is considered large enough for the large-sample confidence interval.
Therefore, the conditions for a valid large-sample confidence interval for p are satisfied, and the inference in part d is valid.
f. To determine the number of Walmart stores that must be sampled in order to estimate the true proportion to within 0.05 with 90% confidence using the large-sample method, we can use the formula for the sample size:
n = (Z^2 * p̂(1-p̂)) / E^2
Where Z is the Z-score corresponding to the desired confidence level, p̂ is the estimate of the proportion, and E is the desired margin of error.
For a 90% confidence level, the Z-score is approximately 1.645, and the desired margin of error is 0.05.
Plugging in these values along with the estimate of p̂ = 0.87 into the formula, we get:
n = (1.645^2 * 0.87(1-0.87)) / 0.05^2
Simplifying this expression, we find that the required sample size is approximately 157. Therefore, at least 157 Walmart stores need to be sampled in order to estimate the true proportion to within 0.05 with 90% confidence using the large-sample method.