Predict Customer Attrition using the Credit Card data set , using the classification algorithm from the rpart package.
Split the data (10127 observations), into training and validation datasets (70%/30% , 7137/2990)
Using the validation data, it is observed that the model is able to:
Predict correctly 27% of the cheaters Misclassify faithful people as cheaters 8.6% of the time
##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 2990
##
##
## | validation_tree$Attrition_Flag
## validation_tree$Attrition_Flag_predicted | Attrited Customer | Existing Customer | Row Total |
## -----------------------------------------|-------------------|-------------------|-------------------|
## Attrited Customer | 364 | 90 | 454 |
## | 1155.572 | 222.084 | |
## | 0.802 | 0.198 | 0.152 |
## | 0.755 | 0.036 | |
## | 0.122 | 0.030 | |
## -----------------------------------------|-------------------|-------------------|-------------------|
## Existing Customer | 118 | 2418 | 2536 |
## | 206.873 | 39.758 | |
## | 0.047 | 0.953 | 0.848 |
## | 0.245 | 0.964 | |
## | 0.039 | 0.809 | |
## -----------------------------------------|-------------------|-------------------|-------------------|
## Column Total | 482 | 2508 | 2990 |
## | 0.161 | 0.839 | |
## -----------------------------------------|-------------------|-------------------|-------------------|
##
##
## Statistics for All Table Factors
##
##
## Pearson's Chi-squared test
## ------------------------------------------------------------
## Chi^2 = 1624.287 d.f. = 1 p = 0
##
## Pearson's Chi-squared test with Yates' continuity correction
## ------------------------------------------------------------
## Chi^2 = 1618.706 d.f. = 1 p = 0
##
##