Objective
Predict Customer Attrition using the CreditCardData, using the
classification algorithim from the rpart package.
Data Prep
Created a new binary variable, Attrition_Flag by classifying
observations that are predicted to stay = 0 and observations that are
predicted to leave = 1.
Split the data (10127 observations), into training and validation
datasets (70%/30% , 7046/3081).
Tree
Two Interpretations From the Tree
1. Most Likely Attired Customers:
- Total_Revolving_Bal < 610
- Total_Ct_Chng_Q4_Q1 < 0.65
2. Customers to most likely Stay:
- Total_Trans_Amt >= 5365
Variable Importance
Total_Trans_Ct is the best predictor for Attrition, while
Dependent_count is the worst.
Model Accuracy
This model predicts Attrited customers 75.5% of the time
This model misses its attrited customers 3.6% of the time
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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$Attrition_Flag_predicted
## validation_tree$Attrition_Flag | Attrited Customer | Existing Customer | Row Total |
## -------------------------------|-------------------|-------------------|-------------------|
## Attrited Customer | 364 | 118 | 482 |
## | 1155.572 | 206.873 | |
## | 0.755 | 0.245 | 0.161 |
## | 0.802 | 0.047 | |
## | 0.122 | 0.039 | |
## -------------------------------|-------------------|-------------------|-------------------|
## Existing Customer | 90 | 2418 | 2508 |
## | 222.084 | 39.758 | |
## | 0.036 | 0.964 | 0.839 |
## | 0.198 | 0.953 | |
## | 0.030 | 0.809 | |
## -------------------------------|-------------------|-------------------|-------------------|
## Column Total | 454 | 2536 | 2990 |
## | 0.152 | 0.848 | |
## -------------------------------|-------------------|-------------------|-------------------|
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## Statistics for All Table Factors
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## Pearson's Chi-squared test
## ------------------------------------------------------------
## Chi^2 = 1624.287 d.f. = 1 p = 0
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## Pearson's Chi-squared test with Yates' continuity correction
## ------------------------------------------------------------
## Chi^2 = 1618.706 d.f. = 1 p = 0
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