Decision Tree Evaluation

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Questions and Answers

Calculate the individual entropy at each node and leaf based on the provided decision tree.

Root Node (100 Pass, 50 Fail): Entropy = $ - (100/150) \log_2(100/150) - (50/150) \log_2(50/150) \approx 0.918 $. Left Leaf (Age < 22) (95 Pass, 15 Fail): Entropy = $ - (95/110) \log_2(95/110) - (15/110) \log_2(15/110) \approx 0.576 $. Right Leaf (Age >= 22) (5 Pass, 35 Fail): Entropy = $ - (5/40) \log_2(5/40) - (35/40) \log_2(35/40) \approx 0.544 $.

Compute the weighted average entropy at each level (layer) of the tree (Layers 1 & 2) based on the provided decision tree.

Layer 1 (Root Node): Entropy = 0.918. Layer 2 (Leaves after split): Weighted Average Entropy = $ (110/150) \times \text{Entropy}{Left} + (40/150) \times \text{Entropy}{Right} = (110/150) \times 0.576 + (40/150) \times 0.544 \approx 0.567 $.

Calculate the Information Gain (IG) for the split between layers 1 and 2 based on the provided decision tree.

Information Gain (IG) = $ \text{Entropy}{Parent} - \text{WeightedAverageEntropy}{Children} = \text{Entropy}{Layer1} - \text{WeightedAverageEntropy}{Layer2} = 0.918 - 0.567 = 0.351 $.

Interpret the Information Gain (IG) value calculated in the previous step. Was it worthwhile to apply decision tree modeling to this dataset based on the split shown? Answer YES or NO, and explain why.

<p>YES. The Information Gain (IG) is 0.351. Since this value is positive, it indicates that splitting the data based on the 'age' attribute resulted in daughter nodes that are more pure (less random) than the parent node. Therefore, the split was worthwhile as it provided a reduction in uncertainty.</p>

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Complete the “Confusion Matrix” for Layer 2 (only) based on the given decision tree model in Question 1. Assume 'Pass' is the positive class and 'Fail' is the negative class.

<p>Based on the Layer 2 leaves: Left node (Age < 22) predicts Pass (95 Actual Pass, 15 Actual Fail). Right node (Age >= 22) predicts Fail (5 Actual Pass, 35 Actual Fail). The confusion matrix is:</p> <table> <thead> <tr> <th></th> <th>Predicted Pass</th> <th>Predicted Fail</th> </tr> </thead> <tbody> <tr> <td><strong>Actual Pass</strong></td> <td>95 (TP)</td> <td>5 (FN)</td> </tr> <tr> <td><strong>Actual Fail</strong></td> <td>15 (FP)</td> <td>35 (TN)</td> </tr> </tbody> </table>

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Calculate the following metrics using the completed “Confusion Matrix” for Layer 2 (only): Error Rate, Accuracy, Precision, Recall, F1 Score.

<p>Using TP=95, FP=15, TN=35, FN=5 (Total=150):</p> <ul> <li>Error Rate = (FP + FN) / Total = (15 + 5) / 150 = 20 / 150 $ \approx $ 0.133</li> <li>Accuracy = (TP + TN) / Total = (95 + 35) / 150 = 130 / 150 $ \approx $ 0.867</li> <li>Precision = TP / (TP + FP) = 95 / (95 + 15) = 95 / 110 $ \approx $ 0.864</li> <li>Recall = TP / (TP + FN) = 95 / (95 + 5) = 95 / 100 = 0.95</li> <li>F1 Score = 2 * (Precision * Recall) / (Precision + Recall) = 2 * (0.864 * 0.95) / (0.864 + 0.95) $ \approx $ 0.905</li> </ul>

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A small e-commerce company wants to understand what influences whether a customer will make a purchase. Based on the provided dataset snippet and scenario, choose which variable/attribute should be the “Label” or dependent variable. Also, identify which other variables/attributes should be selected as independent variables.

<p>Dependent Variable (Label): "Made Purchase". Independent Variables: "Age", "Gender", "Device Used", "Time on Website (min)", "Clicked Ad".</p>

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Flashcards

Decision Tree

A tree-like structure used for decision-making, where each internal node represents a test on an attribute, each branch represents the outcome of the test, and each leaf node represents a class label.

Entropy

A measure of the uncertainty or randomness in a set of data.