Assignment 8
Libby Camardella
2024-04-04
Data Preparation
- Split the data (10127), into training and validation datasets (70%/30%, 7046/3081).
The Tree
Two Interpretations From the Tree
Customers likely to stay with the credit card company:
- Customers with 55 or more transaction counts
- Customers with less than 5423 total transaction amounts
Customers likely to leave the credit card company:
- Customers with less than 55 transaction counts
- Customers with less than 614 total revolving balance
- Customers with less than 0.65 change in transaction amount from Q4 to Q1
Variable Importance
Model Accuracy
Using the validation data, it is observed that the model is able to:
Predict correctly 75.5% of attrited customers will leave.
Predict correctly 3.6% of existing customers will leave.
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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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