## 3 Results
### Trained Model Performance
**Gradient Boosting:**  
**Training score**: Gradient boosting also shows that the model is fitting very well the training data (Figure 9). **Cross-validation score**: The learning curve for Gradient boosting indicates an initial steep rise during training until 200 samples, then an almost steady learning curve in the region of 200 and 320 samples, and then a further incremental way from 320 to 600 samples, indicating that sample numbers of around 350 and 600 are suitable for this model (Figure 9). Gradient Boosting shows a similar learning curve with Random Forest with the cross-validation score almost meeting the training score but with a higher divergence. **Ridge Classifier:**  
**Training score**: Finally, Ridge Classifier shows inferior data fitting starting at 0.94 while decreasing around the 200 samples and then steadily increasing again until 600 samples were used (Figure 10). **Cross-validation score**: Despite the poor data fitting, the learning curve steadily increases from around 200 samples to the point of 600 samples, where the cross-validation score data tries to meet the training score (Figure 10). Figure 8: Learning curve of Random Forest Classifier after training. Since cross-validation alone is insufficient to confirm the superiority of our models, we used the paired \(t\)-test technique [43] for each of our models on the cross-validation results in order to determine if there were statistically significant differences. Following cross-validation, metrics (accuracy, precision, F1 and recall) from all models were compared. The result provided a t-statistic and \(p\)-value. For a statistically significant difference between our three models, the \(p\)-value should be less than 0.05. A greater t-statistic value indicates a more significant difference in model performance. A statistically significant difference is considered a t-statistic value larger than 2 or less than \(-2\). As seen in Figure 11, we performed paired \(t\)-tests on our models for all their metrics and evaluated the relative performance. The results were inherently paired since the models were assessed and trained on identical cross-validation splits, which is why the paired \(t\)-test was selected. **F1 Score:** No significant difference was found between Random Forest and Gradient Boosting with t = 1.633 and \(p\) = 0.178. Random Forest outperformed Ridge Classifier with  
Figure 11: Paired \(t\)-test statistic results across all models and metrics. Figure 10: Learning curve of Ridge Classifier after training. t = 5.302 and \(p\) = 0.006. Similarly, Gradient Boosting outperformed Ridge Classifier with t = 4.986 and \(p\) = 0.008. **Recall:** In this metric, the results suggest that all models performed similarly in terms of recall, with the comparison between Random Forest and Gradient Boosting being t = 1.000 and \(p\) = 0.374. Similarly, between Random Forest and Ridge Classifier the difference was t = 2.743 and \(p\) = 0.052. Finally, Between Gradient Boosting and Ridge Classifier the significant differences were negligible with t = 2.260 and \(p\) = 0.087. **Precision:** No statistically significant difference was shown between Random Forest and Gradient Boosting with t = 1.521 and \(p\) = 0.203. However, Random Forest outperformed Ridge Classifier significantly with t = 6.678 and \(p\) = 0.003. Gradient Boosting also outperformed Ridge Classifier with t = 7.115 and \(p\) = 0.002. **Accuracy:** Random Forest and Gradient Boosting comparison showed values of t-statistic = 1.633 and \(p\) = 0.178, indicating no statistical significance between these models. Comparing Random Forest and Ridge Classifier the difference was statistically significant with t = 5.222 and \(p\) = 0.006, indicating that Random Forest outperforms Ridge Classifier in accuracy. Similar results were found when comparing Gradient Boosting and Ridge Classifier with t = 4.961 and \(p\) = 0.008.