Predict attrition using the CreditCardData data from the
AERpackage, using the classification algorithm from the
rpart package. ### Data Prep Classified observations by
attrition and existing customer Split the data (10127 observations) into
training and validation datasets (70%/30%)
Staying with the card is most likely when total transation count > 55 and whentotal transaction count >= 5243.
Attrition is least likely when: total transaction count < 55, total revolving balance < $614, total count change from Q4 -> Q1 < 0.65, and the relationship count < 3.
It makes sense that the higher the higher the total transaction
count, the more likely attrition is.
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## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 2990
##
##
## | validation_tree$Attrition_Flag
## validation_tree$Attrition_Flag | Attrited Customer | Existing Customer | Row Total |
## -------------------------------|-------------------|-------------------|-------------------|
## Attrited Customer | 482 | 0 | 482 |
## | 2103.700 | 404.300 | |
## | 1.000 | 0.000 | 0.161 |
## | 1.000 | 0.000 | |
## | 0.161 | 0.000 | |
## -------------------------------|-------------------|-------------------|-------------------|
## Existing Customer | 0 | 2508 | 2508 |
## | 404.300 | 77.700 | |
## | 0.000 | 1.000 | 0.839 |
## | 0.000 | 1.000 | |
## | 0.000 | 0.839 | |
## -------------------------------|-------------------|-------------------|-------------------|
## Column Total | 482 | 2508 | 2990 |
## | 0.161 | 0.839 | |
## -------------------------------|-------------------|-------------------|-------------------|
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## Statistics for All Table Factors
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## Pearson's Chi-squared test
## ------------------------------------------------------------
## Chi^2 = 2990 d.f. = 1 p = 0
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## Pearson's Chi-squared test with Yates' continuity correction
## ------------------------------------------------------------
## Chi^2 = 2982.609 d.f. = 1 p = 0
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