ISYE 6501 Final Quiz Winter 2026/2027 – Complete Q&A with
Detailed Rationales | 100% Verified | Pass Guaranteed – A+
Graded
Section A: Classification & Regression Models - Logistic Regression,
Linear Regression, K-NN, SVM (8 Questions)
Q1: A data analyst fits a logistic regression model to predict customer churn (1 = churn,
0 = no churn). The coefficient for "MonthlyCharges" is 0.08. Which interpretation is
correct?
A. A one-unit increase in MonthlyCharges increases the probability of churn by 8%
B. A one-unit increase in MonthlyCharges multiplies the odds of churn by e^0.08 ≈ 1.083
C. A one-unit increase in MonthlyCharges multiplies the odds of churn by e^0.08 ≈ 1.083
[CORRECT]
D. A one-unit increase in MonthlyCharges increases the log-odds of churn by 0.08
percentage points
Correct Answer: C
Rationale: In logistic regression, coefficients represent log-odds changes. A coefficient
of 0.08 means a one-unit increase in MonthlyCharges increases the log-odds by 0.08,
which corresponds to multiplying the odds by e^0.08 ≈ 1.083 (an 8.3% increase in odds,
not probability). Option A confuses odds with probability. Option D incorrectly adds
"percentage points" to log-odds.
,Q2: A support vector machine with an RBF kernel is trained on a dataset. The model
achieves 99% training accuracy but only 72% test accuracy. Which parameter
adjustment is most likely to improve generalization?
A. Increase gamma
B. Decrease gamma [CORRECT]
C. Decrease gamma [CORRECT]
D. Increase C
Correct Answer: C
Rationale: High training accuracy with low test accuracy indicates overfitting. For RBF
SVM, high gamma causes overfitting by giving each training sample excessive influence
(small region of influence). Decreasing gamma increases the kernel's reach, smoothing
the decision boundary and improving generalization. Option A would worsen overfitting.
Option D (increasing C) reduces regularization, also worsening overfitting.
Q3: In a K-Nearest Neighbors classifier, the analyst observes that training accuracy is
95% but validation accuracy is 62%. Which adjustment is most appropriate?
A. Increase K
B. Increase K [CORRECT]
C. Add more features without scaling
D. Decrease K
Correct Answer: B
, Rationale: The large gap between training and validation accuracy indicates overfitting.
In K-NN, small K leads to overfitting (high variance, low bias) because the model
memorizes local noise. Increasing K smooths the decision boundary by averaging over
more neighbors, reducing variance and improving generalization. Option D would
worsen overfitting. Option C would exacerbate the curse of dimensionality.
Q4: A multiple linear regression model has R² = 0.85 and Adjusted R² = 0.78. A second
model with an additional predictor has R² = 0.87 and Adjusted R² = 0.75. Which
statement is correct?
A. The second model is better because it has higher R²
B. The first model is preferred because its Adjusted R² is higher [CORRECT]
C. The first model is preferred because its Adjusted R² is higher [CORRECT]
D. Both models are equivalent because R² differences are small
Correct Answer: C
Rationale: Adjusted R² penalizes model complexity by accounting for the number of
predictors. A decrease in Adjusted R² when adding a variable indicates the new
predictor does not contribute enough explanatory power to justify its inclusion. The first
model has higher Adjusted R² (0.78 > 0.75) and is preferred. Option A ignores the
penalty for unnecessary complexity.
Q5: A confusion matrix for a binary classifier shows: TP = 80, FP = 20, FN = 30, TN = 70.
What is the F1-score?
A. 0.727
Detailed Rationales | 100% Verified | Pass Guaranteed – A+
Graded
Section A: Classification & Regression Models - Logistic Regression,
Linear Regression, K-NN, SVM (8 Questions)
Q1: A data analyst fits a logistic regression model to predict customer churn (1 = churn,
0 = no churn). The coefficient for "MonthlyCharges" is 0.08. Which interpretation is
correct?
A. A one-unit increase in MonthlyCharges increases the probability of churn by 8%
B. A one-unit increase in MonthlyCharges multiplies the odds of churn by e^0.08 ≈ 1.083
C. A one-unit increase in MonthlyCharges multiplies the odds of churn by e^0.08 ≈ 1.083
[CORRECT]
D. A one-unit increase in MonthlyCharges increases the log-odds of churn by 0.08
percentage points
Correct Answer: C
Rationale: In logistic regression, coefficients represent log-odds changes. A coefficient
of 0.08 means a one-unit increase in MonthlyCharges increases the log-odds by 0.08,
which corresponds to multiplying the odds by e^0.08 ≈ 1.083 (an 8.3% increase in odds,
not probability). Option A confuses odds with probability. Option D incorrectly adds
"percentage points" to log-odds.
,Q2: A support vector machine with an RBF kernel is trained on a dataset. The model
achieves 99% training accuracy but only 72% test accuracy. Which parameter
adjustment is most likely to improve generalization?
A. Increase gamma
B. Decrease gamma [CORRECT]
C. Decrease gamma [CORRECT]
D. Increase C
Correct Answer: C
Rationale: High training accuracy with low test accuracy indicates overfitting. For RBF
SVM, high gamma causes overfitting by giving each training sample excessive influence
(small region of influence). Decreasing gamma increases the kernel's reach, smoothing
the decision boundary and improving generalization. Option A would worsen overfitting.
Option D (increasing C) reduces regularization, also worsening overfitting.
Q3: In a K-Nearest Neighbors classifier, the analyst observes that training accuracy is
95% but validation accuracy is 62%. Which adjustment is most appropriate?
A. Increase K
B. Increase K [CORRECT]
C. Add more features without scaling
D. Decrease K
Correct Answer: B
, Rationale: The large gap between training and validation accuracy indicates overfitting.
In K-NN, small K leads to overfitting (high variance, low bias) because the model
memorizes local noise. Increasing K smooths the decision boundary by averaging over
more neighbors, reducing variance and improving generalization. Option D would
worsen overfitting. Option C would exacerbate the curse of dimensionality.
Q4: A multiple linear regression model has R² = 0.85 and Adjusted R² = 0.78. A second
model with an additional predictor has R² = 0.87 and Adjusted R² = 0.75. Which
statement is correct?
A. The second model is better because it has higher R²
B. The first model is preferred because its Adjusted R² is higher [CORRECT]
C. The first model is preferred because its Adjusted R² is higher [CORRECT]
D. Both models are equivalent because R² differences are small
Correct Answer: C
Rationale: Adjusted R² penalizes model complexity by accounting for the number of
predictors. A decrease in Adjusted R² when adding a variable indicates the new
predictor does not contribute enough explanatory power to justify its inclusion. The first
model has higher Adjusted R² (0.78 > 0.75) and is preferred. Option A ignores the
penalty for unnecessary complexity.
Q5: A confusion matrix for a binary classifier shows: TP = 80, FP = 20, FN = 30, TN = 70.
What is the F1-score?
A. 0.727
