Utilizing Supervised Learning in Learning Analytics
Case Study 4
Author
Shahin
Business Scenario: Predicting Student Performance
In this case study, you are an analyst at an online education platform. The management is interested in predicting student performance based on various factors to provide personalized support and improve the learning experience. Your task is to develop a supervised learning model to predict students’ final grades using simulated data.
Objective:
Your goal is to build a predictive model using supervised learning techniques in R. You will utilize simulated student data with features such as study hours, quiz scores, forum participation, and previous grades to predict the final grades.
Data Generation:
# Set a fixed random seed for reproducibilityset.seed(10923)# Number of students#TODO: set num_students to 500# Enter code below:num_students=500# Simulate study hours (ranging from 1 to 20 hours)study_hours <-sample(1:20, num_students, replace =TRUE)# Simulate quiz scores (ranging from 0 to 100)quiz_scores <-sample(0:100, num_students, replace =TRUE)# Simulate forum participation (ranging from 0 to 50 posts)forum_posts <-sample(0:50, num_students, replace =TRUE)# Simulate previous grades (ranging from 0 to 100)previous_grades <-sample(0:100, num_students, replace =TRUE)# Simulate final grades (ranging from 0 to 100)final_grades <-0.3* study_hours +0.4* quiz_scores +0.2* forum_posts +0.1* previous_grades +rnorm(num_students, mean =0, sd =5) +25# Create a data framestudent_data <-data.frame(StudyHours = study_hours, QuizScores = quiz_scores, ForumPosts = forum_posts, PreviousGrades = previous_grades, FinalGrades = final_grades)# View the first few rows of the generated datahead(student_data)
StudyHours QuizScores ForumPosts PreviousGrades
Min. : 1.00 Min. : 0.00 Min. : 0.00 Min. : 0.00
1st Qu.: 6.00 1st Qu.: 24.00 1st Qu.:12.00 1st Qu.: 23.00
Median :11.00 Median : 48.00 Median :24.00 Median : 51.00
Mean :10.67 Mean : 48.54 Mean :24.26 Mean : 50.05
3rd Qu.:16.00 3rd Qu.: 73.00 3rd Qu.:37.00 3rd Qu.: 75.00
Max. :20.00 Max. :100.00 Max. :50.00 Max. :100.00
FinalGrades
Min. :24.19
1st Qu.:47.15
Median :57.18
Mean :57.35
3rd Qu.:67.01
Max. :95.36
# finding the correlation between StudyHours and FinalGradescorrelation <-cor(student_data$StudyHours, student_data$FinalGrades)# Print the correlation coefficientcat("Correlation between StudyHours and FinalGrades: ", correlation, "\n")
Correlation between StudyHours and FinalGrades: 0.1521135
# Fit a linear regression model to predict FinalGrades using all variablesmodel <-lm(FinalGrades ~ StudyHours + QuizScores + ForumPosts + PreviousGrades, data = student_data)# Get the summary of the modelsummary(model)
Call:
lm(formula = FinalGrades ~ StudyHours + QuizScores + ForumPosts +
PreviousGrades, data = student_data)
Residuals:
Min 1Q Median 3Q Max
-13.3924 -3.4734 0.3027 3.0976 16.7901
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 25.076304 0.762633 32.88 < 2e-16 ***
StudyHours 0.298397 0.037113 8.04 6.66e-15 ***
QuizScores 0.404363 0.007692 52.57 < 2e-16 ***
ForumPosts 0.202482 0.015217 13.31 < 2e-16 ***
PreviousGrades 0.090967 0.007480 12.16 < 2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 4.93 on 495 degrees of freedom
Multiple R-squared: 0.8696, Adjusted R-squared: 0.8685
F-statistic: 825.1 on 4 and 495 DF, p-value: < 2.2e-16
Modeling
Use 80% of the data for training and 20% for testing to predict final grades. Compute the Mean Squared Error and model accuracy based on prediction interval.
# Todo:set.seed(10923) # Set seed for reproducibilitysample_index <-sample(1:nrow(student_data), 0.8*nrow(student_data))train_data <- student_data[sample_index, ]test_data <- student_data[-sample_index, ]test_data
# Building a Linear Regression model using the train data and assign it to an object # called model.model <-lm(FinalGrades ~ StudyHours + QuizScores + ForumPosts + PreviousGrades, data = train_data)# Making predictions on the test settest_predictions <-predict(model, newdata = test_data)# Compute mean squared errormse <-mean((test_data$FinalGrades - test_predictions)^2)# Compute R-squaredssr <-sum((test_predictions -mean(test_data$FinalGrades))^2)sst <-sum((test_data$FinalGrades -mean(test_data$FinalGrades))^2)r_squared <- ssr / sst# Print evaluation metricscat("Mean Squared Error (MSE):", mse, "\n")
Mean Squared Error (MSE): 22.34656
cat("R-squared (R^2):", r_squared, "\n")
R-squared (R^2): 0.8598451
Model Accuracy based on Prediction Interval
# Get the predictions and prediction intervalspred_int <-predict(model, newdata = test_data, interval ="prediction")# Extract lower and upper bounds of the prediction intervallower_bound <- pred_int[, "lwr"]upper_bound <- pred_int[, "upr"]# Actual values from the test dataactual_values <- test_data$FinalGrades# Check if the actual values fall within the prediction intervalcorrect_predictions <- actual_values >= lower_bound & actual_values <= upper_bound# Compute accuracyaccuracy <-sum(correct_predictions) /length(correct_predictions)# Print accuracycat("Model Accuracy using Prediction Interval:", accuracy, "\n")
Model Accuracy using Prediction Interval: 0.96
The accuracy is calculated as the proportion of correct predictions.