Example 9.10 Sales Response to Coupons for Discounts on Appliances Confidence Interval for the Difference Between Proportions

| 9.2 | 9.3 | 9.4 | 9.5 | 9.6 | 9.7 | 9.8 | 9.9 | 9.11 | 9.12 | 9.13 | 9.14 | Objective To illustrate how to find a confidence interval for the difference between proportions of customers purchasing appliances with and without 5% discount coupons.

| 9.2 | 9.3 | 9.4 | 9.5 | 9.6 | 9.7 | 9.8 | 9.9 | 9.11 | 9.12 | 9.13 | 9.14 | Background Information n An appliance store is about to run a big sale. n It selects 300 of its best customers and randomly divides them into two sets of 150 customers each. n It then mails a notice of the sale to all 300 customers but includes a coupon for an extra 5% off the sale price to the second set of customers only. n As the sale progresses, the store keeps track of which of these customers purchase appliances.

| 9.2 | 9.3 | 9.4 | 9.5 | 9.6 | 9.7 | 9.8 | 9.9 | 9.11 | 9.12 | 9.13 | 9.14 | COUPONS.XLS n The data from the sale are recorded in this file. n What can the store’s manager conclude about the effectiveness of the coupons?

| 9.2 | 9.3 | 9.4 | 9.5 | 9.6 | 9.7 | 9.8 | 9.9 | 9.11 | 9.12 | 9.13 | 9.14 | Solution n The data has been arranged in a “contingency” table. n Of the 150 customers who received coupons, 55 purchased an appliance. n Of the 150 who did not receive coupons, only 35 purchased an appliance. n These translate to the sample proportions and , calculated in cells B8 and B9 with the formula =B4/D4 and =B5/D5. By subtraction these lead directly to the sample difference between proportions, , in cell B11.

| 9.2 | 9.3 | 9.4 | 9.5 | 9.6 | 9.7 | 9.8 | 9.9 | 9.11 | 9.12 | 9.13 | 9.14 | Solution -- continued n The standard error of this difference is calculated in cell B12 with the formula =SQRT(B8*(1-B8)/D4+B9*(1-B9)/D5 n The z-multiple for he confidence interval is calculated in cell B15 with the formula =NORMSINV(B14+(1-B14/2) n Finally, the limits of the confidence interval for the difference are calculated in cells B18 and B19 with the formulas =B11-B15*B12 =B11+B15*B12

| 9.2 | 9.3 | 9.4 | 9.5 | 9.6 | 9.7 | 9.8 | 9.9 | 9.11 | 9.12 | 9.13 | 9.14 | Conclusions n Because the confidence limits are both positive, we can conclude that the effect of coupons is almost surely to increase the proportion of buyers. Our best guess is that the increase is about 13% (from 23.3% to 36.7%) n How can the store manager interpret this mean difference? He can use it to estimate the extra business he will receive by including coupons as opposed to not including them. n The confidence interval implies that for every 100 customers, the coupons will probably induce an extra 3 to 23 customers to purchase an appliance who otherwise would not have made a purchase.

| 9.2 | 9.3 | 9.4 | 9.5 | 9.6 | 9.7 | 9.8 | 9.9 | 9.11 | 9.12 | 9.13 | 9.14 | Conclusions -- continued n However, the difference between proportions does not directly indicate the difference in profit from including coupons. n This is because the customers with coupons pay 5% less than the customers without them. n For every 100 customers who receive a mailing with no coupon, the store can expect to make $ in profit. For every 100 that receive coupons the expected profit is $ n Thus the store makes less money including coupons.