Answer Happy

Suppose that Williams Apparel is a national chain of clothing stores and is planning to print a number of copies of an expensive four- color sales catalog. Each catalog will include a coupon that gives a $70 discount on a purchase of $250 or more. Williams desires to send the catalog and coupon to only those customers who have at least a .50 probability of redeeming the coupon. A pilot study is conducted using 20 randomly selected previous customers who have a Williams credit card (Card-1) and 20 randomly selected previous customers who do not have a Williams credit card (Card=0). Each of the 40 randomly selected previous customers is sent a copy of the catalog and coupon (from a small preliminary print run), and at the end of the test period, the response variable Coupon is set equal to 1 for a particular customer if the customer redeemed the coupon and is set equal to 0 for the customer if the customer did not redeem the coupon. The values of the response variable Coupon are given for the 40 customers in Figure 5.11(c), along with the corresponding values of the predictor variable Card and of the predictor variable Purchases (last year's purchases, in hundreds of dollars).3 Figure 5.11(d) gives the JMP classification tree analyzing the data in Figure 5.11(c) using the default minimum split size of 5. Figures 5.12 and 5.13 give the leaf report and fit details for the classification tree, and Figure 5.14 gives the coupon redemption probability estimates and classifications for the 40 sampled customers the 40 samples customers. a. Interpret the confusion matrix in Figure 5.13. (Round your answers to the nearest whole number. Leave no cells blank. Enter "0" wherever required.) First Row Misclassification Rate Second Row Misclassification Rate % %

b-1. Using the classification tree or leaf report, find the coupon redemption probability estimates and classifications for customers 41 and 42. (Customer 41 used the card and 42 did not use the card. Round your answers to 4 decimal places.) Customer 41 Customer 42 Prob(Coupon=0) Prob(Coupon=1) b-2. Which customers will be sent a catalog with a coupon? Customer 41 O Customer 42

FIGURE 5.11 Classification Trees, Regression Trees, and Data for Exercises 5.2 and 5.3 (a) Mortgage default tree (b) Second heart attack tree No 0.10 1 1 1 0 0 0 0 0 1 1 1 0 1 0 (c) The coupon redemption data s CoupRedemp Coupon Purchases Card 0 10.52 0 0 14.89 0 15.75 1 29.27 48.15 0 1 0 1 1 0 0 1 0 0 0 1 1 1 1 0 1 1 0 0 Household income > $50,000? No On Job >5 yr? 0 0 51.40 29.89 21.23 17.55 46.61 28.01 15.97 50.10 51.62 27.48 4.95 43.25 35.04 28.94 53.67 13.24 51.54 42.05 29.71 39.24 58.38 23.75 23.89 50.94 36.37 34.40 38.01 55.87 21.46 40.75 46.52 34.93 31.97 42.28 26.28 0.05 0 1 0 1 0 Yes No 0 0 1 0 1 0 1 1 1 1 0 1 1 0 0 1 0 0 0 1 1 1 1 0 1 1 0 0 0 0 Yes Debt> $15,000? 0.01 Yes 0.04 Purchases 36.37 Card(1) Count G^2 LogWorth 20 20.016097 16007719 Level Rate Prob Count 0 0.2000 0.2167 4 16 1 0.8000 0.7833 Count 12 G^2 0 Level Rate Prob Count 0 0.0000 0.0397 1 10000 0.9603 0 1 1 1 (d) JMP output of a classification tree for the coupon redemption data (for Exercise 5.3) Count Systolic Blood Pressure <91? Yes .70 RSquare N 0.644 40 0 Count 1 All Rows Purchases<36.37 G2 LogWorth 40 55.051105 5.5715641 Level Rate G^2 8 11.090355 Level Rate Prob Count 00 0.5000 0.5019 4 12 1 0.5000 0.4981 4 Yes .02 Prob Count 0.5500 0.5500 22 0.4500 0.4500 18 Number of Splits 3 0 1 Count No Age ≤62.5? 0 1 Yes .50 Sinus Tachycardia? Purchases>=42.28 No Card(0) Count G2 LogWorth 20 13.003319 0.9797146 Level Rate Prob Count 0.9000 0.8833 18 0.1000 0.1167 2 G^2 5 6.7301167 Level Rate Prob Count 0.6000 0.5972 3 0.4000 0.4028 No Purchases<42.28 .11 Count G^2 15 0 Level Rate Level Rate Prob Count 0 10000 0.9740 15 0 0.0000 0.0260

FIGURE 5.13 The JMP Fit Detalls for the Coupon Redemption Tree Fit Detalls Measure Entropy RSquare Generalized RSquare Mean -Log p RMSE Mean Abs Dev Misclassification Rate N Actual Coupon Training Definition 0.6442 1-Loglike(model)/Loglike(0) 0.7866 (1-L(0)/L(model) (2/n)/(1-L0Y(2/n) Confusion Matrix Training 0 1 0.2448 0.2841 0.1818 0.1500 40 n Predicted Count 0 22 6 -Log(p[J]/n √(y)-PU])²/n lyl-PU]/n (p[J]#pMax)/n 1 0 12