Assignment 8
Pasquale Ferrara
2024-04-04
Objective
Predict attrition rate using the CreditCardData data from the AER package, using the classification algorithm from the rpart package.
Data Prep
- Created a new binary variable attrition_flag
- Split the data (10127 observations), into training and validation datasets (70%/30% , 7089/3038)
The Tree
Two Interpretations from the Tree
- The most likely existing customer:
Has a total transaction count greater than 55 and a total transaction amount less than 5,365
- The most likely attrited customer:
Has a total transaction count less than 55 with a total revolving balance less than 614 and a total count change in the 4Q to Q1 less than 0.65
Variable Importance
The total transaction count is the greatest indicator for whether someone is an existing customer.
Model Accuracy
Using the validation data, it is observed that the model is able to:
- Predict correctly 83.9% of the existing customers
- Misclassify existing customers as an attrited customer 4.7% of the time
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## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
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##
## Total Observations in Table: 2990
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##
## | 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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