ISYE 6501 Final Exam Actual 2026/2027 – Complete Q&A with
Detailed Rationales | 100% Verified | Pass Guaranteed – A+
Graded
Section A: Classification Models — Logistic Regression, SVM, & K-NN
Q1: A logistic regression model is fit to predict customer churn with a single predictor:
monthly spend. The estimated coefficient for monthly spend is −0.04. Which
interpretation is correct?
A. For every $1 increase in monthly spend, the probability of churn decreases by 4
percentage points
B. For every $1 increase in monthly spend, the log-odds of churn decrease by 0.04
[CORRECT]
C. For every $1 increase in monthly spend, the odds of churn are multiplied by −0.04
D. For every $1 increase in monthly spend, the probability of churn decreases by a factor
of 0.04
Correct Answer: B
Rationale: In logistic regression, coefficients represent the change in log-odds (logit) per
unit change in the predictor, not direct probability changes. Option A incorrectly
interprets the coefficient as a direct probability change. Option C incorrectly suggests
negative odds multiplication. Option D misapplies the coefficient as a probability scaling
factor.
,Q2: An SVM with a linear kernel achieves 92% training accuracy but only 78% test
accuracy on a customer classification task. Which adjustment is most likely to improve
generalization?
A. Switch to an RBF kernel with a very small gamma value
B. Decrease the regularization parameter C to allow more misclassifications in training
[CORRECT]
C. Increase the regularization parameter C to enforce stricter margin constraints
D. Add polynomial features of degree 5 to the linear kernel
Correct Answer: B
Rationale: The 14-point gap between training and test accuracy indicates overfitting.
Decreasing C increases regularization, allowing more training misclassifications and
improving generalization. Option A with very small gamma would make the RBF kernel
too rigid. Option C would worsen overfitting by enforcing stricter margins. Option D
would increase model complexity and exacerbate overfitting.
Q3: In a K-NN classifier applied to a dataset with features measuring customer age
(18–85), annual income ($25,000–$250,000), and number of purchases (0–50), why is
feature scaling essential before fitting the model?
A. K-NN is a parametric model sensitive to multicollinearity
B. Features with larger numerical ranges would dominate the distance calculation,
biasing predictions [CORRECT]
C. Scaling is required to ensure the algorithm converges to a global optimum
,D. K-NN cannot handle continuous variables without standardization
Correct Answer: B
Rationale: K-NN uses distance metrics (typically Euclidean) where features with larger
scales (e.g., income in thousands) dominate those with smaller scales (e.g., number of
purchases). Option A is incorrect—K-NN is non-parametric. Option C is wrong—K-NN has
no optimization convergence. Option D is false—K-NN handles continuous variables
natively.
Q4: A confusion matrix from a medical diagnostic test shows: True Positives = 85, True
Negatives = 920, False Positives = 45, False Negatives = 15. What is the sensitivity
(recall) of this classifier?
A. 0.654
B. 0.850 [CORRECT]
C. 0.953
D. 0.984
Correct Answer: B
Rationale: Sensitivity = TP ÷ (TP + FN) = 85 ÷ (85 + 15) = 85 ÷ 100 = 0.850. Option A
incorrectly calculates 85 ÷ 130. Option C is specificity = 920 ÷ 965. Option D is precision
= 85 ÷ 130.
Q5: Which of the following is a key assumption of logistic regression that does NOT
apply to ordinary least squares linear regression?
A. Independence of observations
, B. Linearity between predictors and the response variable
C. The response variable follows a binomial distribution [CORRECT]
D. No perfect multicollinearity among predictors
Correct Answer: C
Rationale: Logistic regression assumes the response follows a binomial (Bernoulli)
distribution, whereas linear regression assumes normally distributed errors. Options A
and D are assumptions shared by both models. Option B is incorrect—logistic
regression assumes linearity in the log-odds, not between predictors and the raw
response.
Q6: An analyst is tuning an SVM for a high-dimensional text classification problem with
10,000 features and 5,000 observations. The classes are linearly separable in the
transformed feature space. Which kernel is most appropriate?
A. Linear kernel [CORRECT]
B. Polynomial kernel of degree 10
C. RBF kernel with gamma = 0.1
D. Sigmoid kernel
Correct Answer: A
Rationale: With 10,000 features (high-dimensional sparse data typical of text), a linear
kernel is computationally efficient and often performs as well as or better than
non-linear kernels due to the curse of dimensionality. Options B and C would be
computationally expensive and prone to overfitting in high dimensions. Option D
Detailed Rationales | 100% Verified | Pass Guaranteed – A+
Graded
Section A: Classification Models — Logistic Regression, SVM, & K-NN
Q1: A logistic regression model is fit to predict customer churn with a single predictor:
monthly spend. The estimated coefficient for monthly spend is −0.04. Which
interpretation is correct?
A. For every $1 increase in monthly spend, the probability of churn decreases by 4
percentage points
B. For every $1 increase in monthly spend, the log-odds of churn decrease by 0.04
[CORRECT]
C. For every $1 increase in monthly spend, the odds of churn are multiplied by −0.04
D. For every $1 increase in monthly spend, the probability of churn decreases by a factor
of 0.04
Correct Answer: B
Rationale: In logistic regression, coefficients represent the change in log-odds (logit) per
unit change in the predictor, not direct probability changes. Option A incorrectly
interprets the coefficient as a direct probability change. Option C incorrectly suggests
negative odds multiplication. Option D misapplies the coefficient as a probability scaling
factor.
,Q2: An SVM with a linear kernel achieves 92% training accuracy but only 78% test
accuracy on a customer classification task. Which adjustment is most likely to improve
generalization?
A. Switch to an RBF kernel with a very small gamma value
B. Decrease the regularization parameter C to allow more misclassifications in training
[CORRECT]
C. Increase the regularization parameter C to enforce stricter margin constraints
D. Add polynomial features of degree 5 to the linear kernel
Correct Answer: B
Rationale: The 14-point gap between training and test accuracy indicates overfitting.
Decreasing C increases regularization, allowing more training misclassifications and
improving generalization. Option A with very small gamma would make the RBF kernel
too rigid. Option C would worsen overfitting by enforcing stricter margins. Option D
would increase model complexity and exacerbate overfitting.
Q3: In a K-NN classifier applied to a dataset with features measuring customer age
(18–85), annual income ($25,000–$250,000), and number of purchases (0–50), why is
feature scaling essential before fitting the model?
A. K-NN is a parametric model sensitive to multicollinearity
B. Features with larger numerical ranges would dominate the distance calculation,
biasing predictions [CORRECT]
C. Scaling is required to ensure the algorithm converges to a global optimum
,D. K-NN cannot handle continuous variables without standardization
Correct Answer: B
Rationale: K-NN uses distance metrics (typically Euclidean) where features with larger
scales (e.g., income in thousands) dominate those with smaller scales (e.g., number of
purchases). Option A is incorrect—K-NN is non-parametric. Option C is wrong—K-NN has
no optimization convergence. Option D is false—K-NN handles continuous variables
natively.
Q4: A confusion matrix from a medical diagnostic test shows: True Positives = 85, True
Negatives = 920, False Positives = 45, False Negatives = 15. What is the sensitivity
(recall) of this classifier?
A. 0.654
B. 0.850 [CORRECT]
C. 0.953
D. 0.984
Correct Answer: B
Rationale: Sensitivity = TP ÷ (TP + FN) = 85 ÷ (85 + 15) = 85 ÷ 100 = 0.850. Option A
incorrectly calculates 85 ÷ 130. Option C is specificity = 920 ÷ 965. Option D is precision
= 85 ÷ 130.
Q5: Which of the following is a key assumption of logistic regression that does NOT
apply to ordinary least squares linear regression?
A. Independence of observations
, B. Linearity between predictors and the response variable
C. The response variable follows a binomial distribution [CORRECT]
D. No perfect multicollinearity among predictors
Correct Answer: C
Rationale: Logistic regression assumes the response follows a binomial (Bernoulli)
distribution, whereas linear regression assumes normally distributed errors. Options A
and D are assumptions shared by both models. Option B is incorrect—logistic
regression assumes linearity in the log-odds, not between predictors and the raw
response.
Q6: An analyst is tuning an SVM for a high-dimensional text classification problem with
10,000 features and 5,000 observations. The classes are linearly separable in the
transformed feature space. Which kernel is most appropriate?
A. Linear kernel [CORRECT]
B. Polynomial kernel of degree 10
C. RBF kernel with gamma = 0.1
D. Sigmoid kernel
Correct Answer: A
Rationale: With 10,000 features (high-dimensional sparse data typical of text), a linear
kernel is computationally efficient and often performs as well as or better than
non-linear kernels due to the curse of dimensionality. Options B and C would be
computationally expensive and prone to overfitting in high dimensions. Option D
