Question

A large department store routinely audits the accuracy of its price scanning system by taking a random sample of products off its shelves and scanning them. Each type of product the store stocks is called a "line item" in the inventory. Every line items corresponds to exactly one UPC code, and different line items have distinct UPC codes. However, the store typically stocks many units of each line item. The store designs the sample to be a simple random sample of UPC codes of items they stock---that is, line items are sampled without replacement, with equal probability of selecting each line item, regardless of how many units of each line item are in stock. Each product in the sample will be scanned, and the price the scanner gives will be compared to the correct price for the product. We assume that the auditors have a way to determine the correct price of the product with perfect accuracy. Each product in the sample will then be classified as correctly priced, underpriced, or overpriced. Technically, both overpricing and underpricing are illegal, but law enforcement agents tend to be more concerned with overcharges than with undercharges. As a first step, the store seeks to estimate the percentage of pricing errors. The internal auditors take a simple random sample of 22 line items from their inventory of 5,500 line items. The percentage of overpriced items in the sample is 0%, and the percentage of underpriced items in the sample is 9.1%. A conservative 72% confidence interval for the percentage of line items in the store that are priced incorrectly would extend from (Q21) (low) to (Q22) (high). The store also wants to estimate the average net price error (overcharges - undercharges) to customers. Because the net error to customers depends on the mix of items that customers buy (the items in customers' shopping baskets), the store management decides that an appropriate measure is the average net price error per shopping basket. Estimating this net error requires sampling customers' shopping baskets. The store samples shopping baskets by pulling customers aside at random, and having them check out in special lines in which the price of every product in their baskets is scanned and also verified manually. Suppose that this results essentially in a simple random sample of shopping baskets. In the course of a month, 170 shopping baskets are pulled aside; there were 34,000 shopping baskets of goods sold during that month. The mean net price error per shopping basket in the sample was -10.48 cents/basket (a net undercharge), with a sample standard deviation of 2.75 cents per basket. An approximate 95% confidence interval for the net price error per shopping basket of goods sold during the month would extend from (Q23) cents (low) to (Q24) cents (high).

Answer 1

Q21 8.68% and Q22 9.52%:

The 72% confidence interval for the line items in the store that are priced incorrectly would extend from 8.68% to 9.52%

Q23:-10.91 cents and Q24: -5.05 cents.

An approximate 95% confidence interval for the net price error per shopping basket of goods sold during the month would extend from -10.91 cents to -5.05 cents.

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Here we have n=5500 and the percentage of underpriced and overpriced items is 9.1% or

P=0.091

The 72% confidence interval for the line items in the store that are priced incorrectly is given by

$CI=\hat{p}\mp Z_{1-\alpha/2}*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$. The value for the 72% confidence level is 1.0803. Applying the values in the formula, we have

20-Iz

CI = 0.091 F 1.0803 * 0.091(1 – 0.0901) 5500

CI = 0.091 F 1.0803 * 0.091(0.909) 5500

$CI=0.091\mp 1.0803*\sqrt{\frac{0.0827}{5500}}$

$CI=0.091\mp 1.0803*\sqrt{0.00001503}$

$CI=0.091\mp 1.0803*0.0039$

$CI=0.091\mp 0.0042$

$CI=(0.0868,0.0952)$

We are given that n=170 and the mean net price error $\bar{x}=-10.48$ and the sample sd or SE is $se=2.75$ .

95% confidence interval is

$CI=\bar{x}\mp t_{1-\alpha /2,n-1}*se$

We get the value of $t_{0.975,169}=1.9741$ and plugging other values, we get

$CI=-10.48\mp 1.9741*2.75$

$CI=-10.48\mp 5.4288$

$CI=(-10.9088,-5.0512)$