Regression evaluation (MSE, RMSE, R²) in ML Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this MSE calculation code?

Given the true values and predicted values below, what is the Mean Squared Error (MSE) computed by the code?

ML Python

import numpy as np
true = np.array([3, -0.5, 2, 7])
pred = np.array([2.5, 0.0, 2, 8])
mse = np.mean((true - pred) ** 2)
print(round(mse, 2))

A 0.38

B 0.50

C 0.75

D 1.25

💡 Hint

Calculate the squared differences, then average them.

MSE is the average of squared differences between true and predicted values. Here, squared errors are [0.25, 0.25, 0, 1], average is 0.375 rounded to 0.38.

❓ Predict Output

intermediate

What is the RMSE value printed by this code?

Calculate the Root Mean Squared Error (RMSE) from the true and predicted values below.

ML Python

import numpy as np
true = np.array([1, 2, 3, 4])
pred = np.array([1.1, 1.9, 3.2, 3.8])
rmse = np.sqrt(np.mean((true - pred) ** 2))
print(round(rmse, 3))

A 0.250

B 0.129

C 0.158

D 0.200

💡 Hint

RMSE is the square root of MSE.

Squared errors: [0.01, 0.01, 0.04, 0.04], mean = 0.025, sqrt(0.025) = 0.158.

❓ Model Choice

advanced

Which model has the highest R² score based on these predictions?

Given true values and predictions from two models, which model has the higher R² score?

ML Python

import numpy as np
from sklearn.metrics import r2_score
true = np.array([10, 20, 30, 40, 50])
pred_model1 = np.array([12, 18, 29, 41, 49])
pred_model2 = np.array([9, 21, 31, 39, 52])
r2_1 = r2_score(true, pred_model1)
r2_2 = r2_score(true, pred_model2)
print(round(r2_1, 3), round(r2_2, 3))

A Model 2 has higher R²

B Both models have the same R²

C Model 1 has higher R²

D Cannot determine without more data

💡 Hint

Check which prediction set fits true values better overall.

Model 2 predictions are closer overall to true values, resulting in a higher R² score.

❓ Metrics

advanced

What is the R² score for a model that always predicts the mean?

If a regression model always predicts the mean of the true values, what is its R² score?

A 1.0

B Cannot be determined

C -1.0

D 0.0

💡 Hint

R² compares model error to variance of true values.

Predicting the mean results in residual sum of squares equal to total sum of squares, so R² = 0.

🔧 Debug

expert

Why does this R² calculation code raise an error?

Identify the error in this R² score calculation code and what error it raises.

ML Python

import numpy as np
true = np.array([1, 2, 3])
pred = np.array([1, 2])
r2 = 1 - np.sum((true - pred) ** 2) / np.sum((true - np.mean(true)) ** 2)
print(r2)

A No error, prints R² value

B ValueError due to shape mismatch in subtraction

C ZeroDivisionError due to zero variance in true values

D TypeError because np.mean returns a list

💡 Hint

Check if true and pred arrays have the same length.

true and pred arrays have different lengths, so (true - pred) causes ValueError for shape mismatch.

Practice

1. What does the Mean Squared Error (MSE) measure in regression evaluation?

easy

A. The total number of data points

B. The average of the squared differences between predicted and actual values

C. The correlation between features

D. The percentage of correct predictions

Solution

1. Step 1: Understand MSE definition

   MSE calculates the average of squared differences between predicted and actual values to measure error size.

2. Step 2: Compare options to definition

   Only The average of the squared differences between predicted and actual values correctly describes MSE as average squared error; others describe unrelated concepts.

3. Final Answer:

   The average of the squared differences between predicted and actual values -> Option B

4. Quick Check:

   MSE = average squared error [OK]

Hint: MSE = average squared error between prediction and truth [OK]

Common Mistakes:

• Confusing MSE with accuracy
• Thinking MSE measures correlation
• Mixing MSE with count of data points

2. Which of the following Python code snippets correctly calculates the Root Mean Squared Error (RMSE) given true values y_true and predicted values y_pred?

easy

A. rmse = np.sum((y_true - y_pred)**2)

B. rmse = np.mean(np.abs(y_true - y_pred))

C. import numpy as np
rmse = np.sqrt(np.mean((y_true - y_pred)**2))

D. rmse = np.sqrt(np.sum(y_true - y_pred))

Solution

1. Step 1: Recall RMSE formula

   RMSE is the square root of the mean of squared differences between true and predicted values.

2. Step 2: Match code to formula

   import numpy as np rmse = np.sqrt(np.mean((y_true - y_pred)**2)) correctly computes squared differences, averages them, then takes square root. Others miss steps or use wrong operations.

3. Final Answer:

   import numpy as np rmse = np.sqrt(np.mean((y_true - y_pred)**2)) -> Option C

4. Quick Check:

   RMSE = sqrt(mean squared error) [OK]

Hint: RMSE = sqrt of mean squared error, use np.sqrt(np.mean(...)) [OK]

Common Mistakes:

• Using sum instead of mean
• Taking sqrt before mean
• Using absolute difference instead of squared

3. Given the true values y_true = [3, -0.5, 2, 7] and predicted values y_pred = [2.5, 0.0, 2, 8], what is the R² score?

medium

A. 1.0

B. 0.865

C. 0.75

D. 0.948

Solution

1. Step 1: Calculate R² score formula

   R² = 1 - (sum of squared residuals / total sum of squares). Calculate residuals and totals:

2. Step 2: Compute sums

   Residuals squared: (3-2.5)^2 + (-0.5-0)^2 + (2-2)^2 + (7-8)^2 = 0.25 + 0.25 + 0 + 1 = 1.5
   Total sum squares: sum((y_true - mean(y_true))^2) with mean= (3-0.5+2+7)/4=2.875
   Totals: (3-2.875)^2 + (-0.5-2.875)^2 + (2-2.875)^2 + (7-2.875)^2 = 0.015625 + 11.390625 + 0.765625 + 17.015625 = 29.1875

3. Step 3: Calculate R²

   R² = 1 - (1.5 / 29.1875) ≈ 1 - 0.0514 = 0.9486

4. Final Answer:

   0.948 -> Option D

5. Quick Check:

   R² ≈ 0.95 means good fit [OK]

Hint: R² = 1 - (SS_res / SS_tot), calculate sums carefully [OK]

Common Mistakes:

• Using sum of residuals instead of squared residuals
• Forgetting to subtract mean in total sum squares
• Mixing up numerator and denominator

4. You have this Python code to compute MSE but it gives wrong results:

def mse(y_true, y_pred):
    return sum(y_true - y_pred)**2 / len(y_true)

What is the error?

medium

A. The subtraction is done element-wise but sum is outside, causing wrong squaring

B. The function should use absolute values instead of squares

C. The length should be squared, not the sum

D. The function is missing import statements

Solution

1. Step 1: Analyze the code operation order

   The code does sum(y_true - y_pred) first, then squares the total sum, which is incorrect for MSE.

2. Step 2: Correct MSE calculation

   MSE requires squaring each difference first, then averaging. The code should square element-wise before summing.

3. Final Answer:

   The subtraction is done element-wise but sum is outside, causing wrong squaring -> Option A

4. Quick Check:

   Square differences before sum for MSE [OK]

Hint: Square each difference before summing for MSE [OK]

Common Mistakes:

• Squaring sum instead of sum of squares
• Using absolute difference instead of squared
• Ignoring element-wise operations

5. You trained two regression models on different datasets. Model A has MSE=4 and R²=0.8, Model B has MSE=2 and R²=0.6. Which model is better and why?

hard

A. Model B is better because it has lower MSE, indicating smaller average error

B. Model A is better because it has higher R², showing better fit to data

C. Both models are equally good because MSE and R² contradict

D. Cannot decide without RMSE values

Solution

1. Step 1: Understand MSE and R² meanings

   MSE measures average squared error (lower is better). R² measures proportion of variance explained (higher is better).

2. Step 2: Compare models on error and fit

   Model B has lower MSE (2 < 4), so predictions are closer on average. Model A has higher R² but larger error, meaning it fits variance better but with bigger errors.

3. Step 3: Prioritize error for prediction quality

   Lower error (MSE) usually means better prediction accuracy, so Model B is better despite lower R².

4. Final Answer:

   Model B is better because it has lower MSE, indicating smaller average error -> Option A

5. Quick Check:

   Lower MSE means better prediction accuracy [OK]

Hint: Lower MSE beats higher R² for prediction accuracy [OK]

Common Mistakes:

• Choosing model with higher R² ignoring error size
• Thinking MSE and R² always agree
• Ignoring what metric matters for task