2026-03-25
The dataset I will be using is the Student Exam Performance Dataset Analysis.
The dataset lists 19 different variables that affects a students’ exam score. There are socio-economic factors, lifestyle factors, and academic factors.
The question I will be answering is can we accurately predict a student’s exam score based on academic factors?
Uploading the dataset and selecting the academic variables that we will be using. When uploading I will also establish what counts as missing values because I had an issue with finding them while cleaning the data.
StudentPerformanceFactors =
read.csv("C:/Users/sophi/Downloads/StudentPerformanceFactors.csv",
na.strings = c("NA", "", "N/A"))
df = subset(StudentPerformanceFactors, select = c(Hours_Studied,
Attendance,Previous_Scores, Motivation_Level,
Tutoring_Sessions, Teacher_Quality, Exam_Score))
| Hours_Studied | Attendance | Previous_Scores | Motivation_Level | Tutoring_Sessions | Teacher_Quality | Exam_Score |
|---|---|---|---|---|---|---|
| 23 | 84 | 73 | Low | 0 | Medium | 67 |
| 19 | 64 | 59 | Low | 2 | Medium | 61 |
| 24 | 98 | 91 | Medium | 2 | Medium | 74 |
| 29 | 89 | 98 | Medium | 1 | Medium | 71 |
| 19 | 92 | 65 | Medium | 3 | High | 70 |
| 19 | 88 | 89 | Medium | 3 | Medium | 71 |
We need to check if the data needs cleaning or not.
anyNA(df)
[1] TRUE
colSums(is.na(df))
Hours_Studied Attendance Previous_Scores Motivation_Level
0 0 0 0
Tutoring_Sessions Teacher_Quality Exam_Score
0 78 0
sum(is.na(df$Teacher_Quality))/length(df$Teacher_Quality) * 100
[1] 1.180566
Now we that have identified which column has missing values, we will deal with the missing data by removing the rows with missing data. We also will remove duplicate rows.
cleaned_data = drop_na(df, Teacher_Quality)
cleaned_data = cleaned_data %>%
distinct()
colSums(is.na(cleaned_data))
Hours_Studied Attendance Previous_Scores Motivation_Level
0 0 0 0
Tutoring_Sessions Teacher_Quality Exam_Score
0 0 0
This is the summary of the new, cleaned dataset:
Hours_Studied Attendance Previous_Scores Motivation_Level Min. : 1.00 Min. : 60 Min. : 50.00 Length:6526 1st Qu.:16.00 1st Qu.: 70 1st Qu.: 63.00 Class :character Median :20.00 Median : 80 Median : 75.00 Mode :character Mean :19.98 Mean : 80 Mean : 75.05 3rd Qu.:24.00 3rd Qu.: 90 3rd Qu.: 88.00 Max. :44.00 Max. :100 Max. :100.00 Tutoring_Sessions Teacher_Quality Exam_Score Min. :0.000 Length:6526 Min. : 55.00 1st Qu.:1.000 Class :character 1st Qu.: 65.00 Median :1.000 Mode :character Median : 67.00 Mean :1.494 Mean : 67.24 3rd Qu.:2.000 3rd Qu.: 69.00 Max. :8.000 Max. :101.00
Now that we cleaned the data we will do exploratory data analysis. This histogram shows us the distribution of the exam scores. Most students scored below a 70. 
We can see that there is a small positive slope for this variable, meaning it should have significance in the model. 
Similarly to the last slide, this scatter plot displays a small positive slope as well. 
From this plot, we know that we must change the categorical variables to numerical so they can be used in the linear model. 
The conclusion from the last slide also applies to this categorical variable. 
We see little to no slope in this plot, so the previous scores seem to have little to no effect on the exam scores. 
The boxplot shows that there is a slight improvement in score and number of tutoring sessions. There also appears to be less outliers the more tutoring sessions but that may also be a result of less students attending tutoring sessions. 
Before we create the model, we need to change the categorical variables into numerical variables so it can be used in the linear regression prediction model.
cleaned_data$Motivation_Level <-
as.numeric(factor(cleaned_data$Motivation_Level,
levels = c("Low", "Medium", "High")))
cleaned_data$Teacher_Quality <-
as.numeric(factor(cleaned_data$Teacher_Quality,
levels = c("Low", "Medium", "High")))
sapply(cleaned_data, class)
Hours_Studied Attendance Previous_Scores Motivation_Level
"integer" "integer" "integer" "numeric"
Tutoring_Sessions Teacher_Quality Exam_Score
"integer" "numeric" "integer"
To determine which data we will train and test, we split the data. Then we create new variables to store and train and test the data.
df_split <- cleaned_data %>% initial_split(strata = Exam_Score) df_split
<Training/Testing/Total> <4892/1634/6526>
df_train = training(df_split) df_test = testing(df_split)
Then, we will create the linear model.
lm_fit = linear_reg() %>% fit(Exam_Score ~., data = cleaned_data) tidy(lm_fit)
# A tibble: 7 × 5 term estimate std.error statistic p.value <chr> <dbl> <dbl> <dbl> <dbl> 1 (Intercept) 39.1 0.318 123. 0 2 Hours_Studied 0.293 0.00504 58.1 0 3 Attendance 0.198 0.00261 75.8 0 4 Previous_Scores 0.0482 0.00209 23.0 9.69e-113 5 Motivation_Level 0.530 0.0433 12.2 5.38e- 34 6 Tutoring_Sessions 0.494 0.0245 20.2 7.91e- 88 7 Teacher_Quality 0.525 0.0503 10.4 2.42e- 25
We then create predictions with the model and find the results from both the training and testing data.
lm_fit %>% predict(new_data = df_train) results_train = lm_fit %>% predict(new_data = df_train) %>% mutate(truth = df_train$Exam_Score) results_test = lm_fit %>% predict(new_data = df_test) %>% mutate(truth = df_test$Exam_Score)
Now, we find the root mean square error which measures the difference between the predicted values from the model and the actual observed values from the data.
We will get the root mean square error from both the training data and the testing data as shown in the next slide. We expect to get a value that is close or similar in the estimate section for both the training and testing results.
results_train %>% rmse(truth = truth, estimate = .pred)
# A tibble: 1 × 3 .metric .estimator .estimate <chr> <chr> <dbl> 1 rmse standard 2.34
results_test %>% rmse(truth = truth, estimate = .pred)
# A tibble: 1 × 3 .metric .estimator .estimate <chr> <chr> <dbl> 1 rmse standard 2.68
Plotting the training and testing results 
Based on the plots in the previous slide, the linear model seems to accurately predict students’ exam scores based on the academic variables provided from the data. The RMSE from the training and testing data seem to be around 2.5, meaning that the model’s predicted results deviate from the actual value by 2.5%.
There are also quite a few outliers that we saw in the plots, however majority of the data seems to fit the linear regression line well, so we can say that academic variables can accurately predict a student’s exam score.