Assignment 8
Caleb McNaull & Greg Roman
2024-04-18
Objective
The objective is to predict customer attrition from Credit Card Data using a classification tree.
Data Prep
-The data has 10127 observations.
-The data was split into training and validation datasets, with a 70/30 split.
The Tree
Two Interpretations from the Tree
- Customers with total transaction counts above 55 and transaction amounts of $5,423 or more are highly unlikely to leave
- Customers with total transaction counts below 55, revolving balances below $614, and changes in transaction count below 0.65% Q/Q are those most likely to leave
Variable Importance
This plot shows that total transaction count and total transaction amount are the two most important variables in predicting which customers may leave.
Model Accuracy
Correctly predicted a customer to leave 75.5% of the time.
Incorrectly predicted a customer to leave 3.6% of the time.
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## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
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## Total Observations in Table: 2990
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##
## | validation_tree$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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