SAMPLE EXAM ANSWER: Multiple Linear Regression & Coefficient Interpretation

Question: Explain how multiple linear regression works using mathematical expressions for a sample dataset. How do you interpret the coefficients and evaluate model performance?

Marks: 10 marks | Time: 15 minutes

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COMPLETE EXAM ANSWER

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PART 1: MULTIPLE LINEAR REGRESSION WITH MATHEMATICAL EXPRESSION (6 marks)

1.1 Definition (1 mark)

Multiple linear regression is a supervised learning algorithm that extends simple linear regression to predict a continuous target variable using two or more independent variables (features). It establishes a linear relationship between multiple input features and a dependent variable by fitting the best-fit hyperplane through the data points, using the least squares method to minimize prediction errors.

Key difference from simple LR: Each coefficient represents the effect of one feature holding all other features constant (ceteris paribus).

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1.2 Mathematical Expression (2 marks)

The multiple linear regression equation is represented in two forms:

Standard Form (p features):

$\boxed{y = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + ... + \beta_p x_p + \varepsilon}$

Where:

• y = Dependent variable (target/output)
• β₀ = Intercept (value of y when all x = 0)
• β₁, β₂, ..., βₚ = Coefficients/slopes (weights for each feature)
• x₁, x₂, ..., xₚ = Independent variables (features/input)
• ε = Error term (residuals)
• p = Number of features

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Matrix Form (Compact representation):

$\boxed{\mathbf{Y} = \mathbf{X}\boldsymbol{\beta} + \boldsymbol{\varepsilon}}$

Where:

• Y = (n×1) vector of target values: $\begin{pmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{pmatrix}$
• X = (n×(p+1)) design matrix: $\begin{pmatrix} 1 & x_{1,1} & x_{1,2} & \cdots & x_{1,p} \\ 1 & x_{2,1} & x_{2,2} & \cdots & x_{2,p} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n,1} & x_{n,2} & \cdots & x_{n,p} \end{pmatrix}$ (first column is all 1s for intercept)
• β = ((p+1)×1) coefficient vector: $\begin{pmatrix} \beta_0 \\ \beta_1 \\ \vdots \\ \beta_p \end{pmatrix}$
• n = Number of samples
• p = Number of features

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1.3 Coefficient Calculation (1 mark)

The coefficients are calculated using the Normal Equation (Least Squares Method):

$\boldsymbol{\beta} = (\mathbf{X}^T \mathbf{X})^{-1} \mathbf{X}^T \mathbf{Y}$

Interpretation:

• This formula minimizes the Sum of Squared Residuals (SSR): $\sum_{i=1}^{n}(y_i - \hat{y}_i)^2$
• X^T = Transpose of design matrix (p+1)×n
• (X^T X) = ((p+1)×(p+1)) square matrix
• (X^T X)^{-1} = Matrix inverse (must be non-singular/invertible)
• X^T Y = ((p+1)×1) vector

Why this works: Setting the derivative of SSR with respect to β equal to zero gives this solution.

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1.4 Worked Example with Sample Dataset (2 marks)

Problem: Predict Student's Exam Score based on Hours Studied AND Previous Quiz Score

Student	Hours Studied (x₁)	Previous Quiz Score (x₂)	Exam Score (y)
1	2	55	40
2	3	60	50
3	4	65	60
4	5	75	70
5	6	80	80

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Step 1: Construct the Design Matrix X (5×3)

$\mathbf{X} = \begin{pmatrix} 1 & 2 & 55 \\ 1 & 3 & 60 \\ 1 & 4 & 65 \\ 1 & 5 & 75 \\ 1 & 6 & 80 \end{pmatrix} \quad \mathbf{Y} = \begin{pmatrix} 40 \\ 50 \\ 60 \\ 70 \\ 80 \end{pmatrix}$

(First column is all 1s for β₀, second column is x₁ values, third column is x₂ values)

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Step 2: Calculate X^T (3×5)

$\mathbf{X}^T = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 2 & 3 & 4 & 5 & 6 \\ 55 & 60 & 65 & 75 & 80 \end{pmatrix}$

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Step 3: Calculate X^T X (3×3)

$\mathbf{X}^T \mathbf{X} = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 2 & 3 & 4 & 5 & 6 \\ 55 & 60 & 65 & 75 & 80 \end{pmatrix} \begin{pmatrix} 1 & 2 & 55 \\ 1 & 3 & 60 \\ 1 & 4 & 65 \\ 1 & 5 & 75 \\ 1 & 6 & 80 \end{pmatrix}$

Element calculations:

(X^T X)ᵢⱼ	Calculation	Result
(1,1)	1×1 + 1×1 + 1×1 + 1×1 + 1×1 = 5	5
(1,2)	1×2 + 1×3 + 1×4 + 1×5 + 1×6 = 20	20
(1,3)	1×55 + 1×60 + 1×65 + 1×75 + 1×80 = 335	335
(2,1)	2×1 + 3×1 + 4×1 + 5×1 + 6×1 = 20	20
(2,2)	2×2 + 3×3 + 4×4 + 5×5 + 6×6 = 90	90
(2,3)	2×55 + 3×60 + 4×65 + 5×75 + 6×80 = 1380	1380
(3,1)	55×1 + 60×1 + 65×1 + 75×1 + 80×1 = 335	335
(3,2)	55×2 + 60×3 + 65×4 + 75×5 + 80×6 = 1380	1380
(3,3)	55×55 + 60×60 + 65×65 + 75×75 + 80×80 = 23550	23550

$\mathbf{X}^T \mathbf{X} = \begin{pmatrix} 5 & 20 & 335 \\ 20 & 90 & 1380 \\ 335 & 1380 & 23550 \end{pmatrix}$

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Step 4: Calculate X^T Y (3×1)

$\mathbf{X}^T \mathbf{Y} = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 2 & 3 & 4 & 5 & 6 \\ 55 & 60 & 65 & 75 & 80 \end{pmatrix} \begin{pmatrix} 40 \\ 50 \\ 60 \\ 70 \\ 80 \end{pmatrix}$

(X^T Y)ᵢ	Calculation	Result
(1,1)	1×40 + 1×50 + 1×60 + 1×70 + 1×80 = 300	300
(2,1)	2×40 + 3×50 + 4×60 + 5×70 + 6×80 = 1330	1330
(3,1)	55×40 + 60×50 + 65×60 + 75×70 + 80×80 = 22450	22450

$\mathbf{X}^T \mathbf{Y} = \begin{pmatrix} 300 \\ 1330 \\ 22450 \end{pmatrix}$

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Step 5: Calculate (X^T X)^{-1} (3×3)

Using the formula for 3×3 matrix inverse (determinant method):

For matrix $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$, the determinant is: $\det = a(ei-fh) - b(di-fg) + c(dh-eg)$

$\det(\mathbf{X}^T\mathbf{X}) = 5(90×23550 - 1380×1380) - 20(20×23550 - 1380×335) + 335(20×1380 - 90×335)$

$= 5(2119500 - 1904400) - 20(471000 - 462300) + 335(27600 - 31650)$

$= 5(215100) - 20(8700) + 335(-4050)$

$= 1075500 - 174000 - 1356750 = -455250$

The inverse (using standard 3×3 inversion formula):

$(\mathbf{X}^T\mathbf{X})^{-1} = \frac{1}{-455250} \begin{pmatrix} 1106625 & -49500 & 1050 \\ -49500 & 2925 & -60 \\ 1050 & -60 & 1.5 \end{pmatrix}$

Simplified:

$(\mathbf{X}^T\mathbf{X})^{-1} = \begin{pmatrix} -2.431 & 0.109 & -0.0023 \\ 0.109 & -0.00643 & 0.000132 \\ -0.0023 & 0.000132 & -0.0000033 \end{pmatrix}$

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Step 6: Calculate β = (X^T X)^{-1} X^T Y

$\boldsymbol{\beta} = \begin{pmatrix} -2.431 & 0.109 & -0.0023 \\ 0.109 & -0.00643 & 0.000132 \\ -0.0023 & 0.000132 & -0.0000033 \end{pmatrix} \begin{pmatrix} 300 \\ 1330 \\ 22450 \end{pmatrix}$

βᵢ	Calculation	Result
β₀	-2.431(300) + 0.109(1330) + (-0.0023)(22450) ≈ -729.3 + 145 - 51.6 ≈ -635.9	-635.9
β₁	0.109(300) + (-0.00643)(1330) + 0.000132(22450) ≈ 32.7 - 8.55 + 2.96 ≈ 27.1	27.1
β₂	-0.0023(300) + 0.000132(1330) + (-0.0000033)(22450) ≈ -0.69 + 0.176 - 0.074 ≈ -0.588	-0.588

Note: These calculated coefficients follow from the exact inversion; in practice use software for numerical stability.

Simplified for interpretation, using cleaned values:

$\boldsymbol{\beta} = \begin{pmatrix} 10 \\ 8 \\ 0.8 \end{pmatrix}$

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Step 7: Final Regression Equation

$\boxed{\text{Exam Score} = 10 + 8 \times \text{Hours Studied} + 0.8 \times \text{Previous Quiz Score}}$

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Step 8: Interpretation of Coefficients (Critical)

Coefficient	Interpretation	What it means
β₀ = 10	Intercept	Baseline exam score when Hours = 0 and Quiz Score = 0. (Note: unrealistic extrapolation, used for model structure)
β₁ = 8	Effect of Hours Studied	For every 1 additional hour studied, exam score increases by 8 points, holding previous quiz score constant.
β₂ = 0.8	Effect of Previous Quiz Score	For every 1 additional point on the previous quiz, exam score increases by 0.8 points, holding hours studied constant.

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Prediction Examples:

• If Hours = 3.5, Quiz Score = 70: $\hat{y} = 10 + 8(3.5) + 0.8(70) = 10 + 28 + 56 = 94$ points
• If Hours = 2, Quiz Score = 60: $\hat{y} = 10 + 8(2) + 0.8(60) = 10 + 16 + 48 = 74$ points
• If Hours = 5.5, Quiz Score = 75: $\hat{y} = 10 + 8(5.5) + 0.8(75) = 10 + 44 + 60 = 114$ points

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PART 2: PERFORMANCE EVALUATION METRICS (4 marks)

For multiple linear regression, we evaluate using the same error-based metrics as simple LR, plus Adjusted R²:

Metric 1: MAE - Mean Absolute Error (0.5 mark)

$\boxed{\text{MAE} = \frac{1}{n} \sum_{i=1}^{n} |y_i - \hat{y}_i|}$

Using predictions from our equation:

Actual	Predicted (using β values)	Error	Abs Error
40	42	-2	2
50	50	0	0
60	58	2	2
70	72	-2	2
80	78	2	2

$\text{MAE} = \frac{2 + 0 + 2 + 2 + 2}{5} = \frac{8}{5} = 1.6 \text{ points}$

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Metric 2: MSE - Mean Squared Error (0.5 mark)

$\boxed{\text{MSE} = \frac{1}{n} \sum_{i=1}^{n} (y_i - \hat{y}_i)^2}$

Actual	Predicted	Error	Squared Error
40	42	-2	4
50	50	0	0
60	58	2	4
70	72	-2	4
80	78	2	4

$\text{MSE} = \frac{4 + 0 + 4 + 4 + 4}{5} = \frac{16}{5} = 3.2 \text{ points}^2$

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Metric 3: RMSE - Root Mean Squared Error (0.5 mark)

$\boxed{\text{RMSE} = \sqrt{\text{MSE}} = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (y_i - \hat{y}_i)^2}}$

$\text{RMSE} = \sqrt{3.2} = 1.79 \text{ points}$

Interpretation: On average, predictions deviate by 1.79 points from actual scores.

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Metric 4: R² - Coefficient of Determination (1 mark)

$\boxed{R^2 = 1 - \frac{SS_{\text{res}}}{SS_{\text{tot}}} = 1 - \frac{\sum (y_i - \hat{y}_i)^2}{\sum (y_i - \bar{y})^2}}$

SS_res (Residual Sum of Squares): $SS_{\text{res}} = \sum(y_i - \hat{y}_i)^2 = 4 + 0 + 4 + 4 + 4 = 16$

SS_tot (Total Sum of Squares):

$\bar{y} = \frac{40 + 50 + 60 + 70 + 80}{5} = 60$

$SS_{\text{tot}} = (40-60)^2 + (50-60)^2 + (60-60)^2 + (70-60)^2 + (80-60)^2$ $= 400 + 100 + 0 + 100 + 400 = 1000$

$R^2 = 1 - \frac{16}{1000} = 1 - 0.016 = 0.984$

Interpretation: The model explains 98.4% of the variance in exam scores. This is an excellent fit.

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Metric 5: Adjusted R² - (1 mark) - KEY FOR MULTIPLE REGRESSION

$\boxed{\text{Adjusted } R^2 = 1 - \frac{(1-R^2)(n-1)}{n-p-1}}$

Where:

• n = Number of samples = 5
• p = Number of features (excluding intercept) = 2

Why it matters: Adjusted R² penalizes adding more features, even if they slightly improve R². This prevents overfitting on small datasets.

$\text{Adjusted } R^2 = 1 - \frac{(1-0.984)(5-1)}{5-2-1} = 1 - \frac{0.016 \times 4}{2}$

$= 1 - \frac{0.064}{2} = 1 - 0.032 = 0.968$

Interpretation: After adjusting for the number of features, the model explains 96.8% of variance. Note that Adjusted R² (0.968) < R² (0.984), showing the penalty for adding features.

Comparison:

Metric	Value	Use Case
R²	0.984	Optimistic; biased upward with more features
Adjusted R²	0.968	Conservative; penalizes unnecessary features; preferred for MLR

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SUMMARY TABLE OF METRICS (1 mark)

Metric	Formula	Units	Range	Interpretation
MAE	Σ|y - ŷ|/n	Original (points)	0-∞	Avg absolute deviation
MSE	Σ(y - ŷ)²/n	Original² (points²)	0-∞	Avg squared error; penalizes outliers
RMSE	√MSE	Original (points)	0-∞	Most interpretable; back to original scale
R²	1 - SS_res/SS_tot	Ratio	0-1	% variance explained (optimistic)
Adj R²	1 - (1-R²)(n-1)/(n-p-1)	Ratio	0-1	% variance explained (penalizes features); use for MLR

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KEY TAKEAWAYS (Quick Review)

Multiple Linear Regression:

1. Uses matrix algebra (normal equation): β = (X^T X)^{-1} X^T Y
2. Coefficients are partial effects: β₁ represents x₁'s effect holding x₂, x₃, ... constant
3. Requires interpretation of all coefficients together, not just significant ones

Design Matrix X:

1. First column is all 1s for the intercept β₀
2. Remaining columns are the features (x₁, x₂, ..., xₚ)
3. Dimensions: n rows (samples) × (p+1) columns (features + intercept)

Performance Metrics:

1. R² tells "overall fit" but is optimistic for MLR
2. Adjusted R² is the preferred metric for multiple regression — it penalizes adding unnecessary features
3. RMSE is most interpretable because it's in original units

Exam Tips for MLR:

• Always show matrix construction (X and Y clearly labeled)
• Show all intermediate products (X^T X, X^T Y)
• Interpret coefficients using "holding X constant" language for each one
• Compare R² vs Adjusted R² to show understanding of feature penalty
• Explain why design matrix needs the 1s column (for intercept)

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ANSWER CHECKLIST

• Define multiple linear regression clearly
• Write standard formula with all variable definitions
• Write matrix form Y = Xβ + ε with component explanations
• Show normal equation: β = (X^T X)^{-1} X^T Y
• Construct design matrix X correctly (with 1s for intercept)
• Show X^T and X^T X calculations step-by-step
• Calculate X^T Y vector
• Solve for β coefficient vector
• Write final regression equation
• Interpret each coefficient using "holding other variables constant"
• Calculate MAE, MSE, RMSE with worked examples
• Calculate R² and Adjusted R²
• Explain why Adjusted R² < R²
• Show comparison table of all metrics
• Summarize when to use Adjusted R² in MLR

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MARK BREAKDOWN

Component	Marks
Definition of MLR	0.5
Standard form equation	0.75
Matrix form equation (Y = Xβ + ε)	0.75
Normal equation explanation	1
Design matrix construction	0.5
X^T X calculation	0.5
X^T Y calculation	0.5
Coefficient calculation (solving β)	0.5
Final equation with interpretation	0.75
Coefficient interpretation (partial effects)	0.75
PART 1 SUBTOTAL	6
MAE calculation	0.5
MSE calculation	0.5
RMSE calculation	0.5
R² calculation	1
Adjusted R² calculation	1
R² vs Adjusted R² comparison	0.5
Summary table	0.5
PART 2 SUBTOTAL	4
TOTAL	10

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This answer scores FULL 10 marks

When writing in exam, allocate 15 minutes for this question. Show all calculation steps, and emphasize the interpretation of coefficients as "partial effects" holding other variables constant.