# Hybrid LSTM Model to Predict the Level of Air Pollution in Montenegro
## 4 Results and Discussion
### Statistical Analysis  
A comparative statistical analysis was conducted on the calculated MAE values obtained from the training and testing of three models: the LSTM model, the best LSTM model without bagging, and the ensemble model with bagging. The summary statistics for the calculated MAE values obtained from the training and testing of the models are presented in Table 8. \begin{table}
\begin{tabular}{c c} \hline \hline
**Performance Metric Difference** & **Value** \\ \hline MSE Difference & \(-371.7075\) \\ MAE Difference & \(-9.3180\) \\ MAPE Difference & \(-0.1773\) \\ \(R^{2}\) Difference & \(0.5348\) \\ \hline \hline \end{tabular}
\end{table} 
Table 7: SVM model and ensemble model comparison. Figure 11: Actual and predicted values for the SVM model on a subset of the test dataset. Figure 12: Actual values, prediction of the SVM model, and prediction of the ensemble model on a subset of the test dataset. In order to examine whether datasets are following Gaussian distribution the following hypotheses were tested: the null hypothesis (\(H_{0}\)) assuming that the data is normally distributed and the alternative hypothesis (\(H_{a}\)) assuming that the data is not normally distributed. The Shapiro-Wilk normality test was performed for the calculated MAE values obtained from the training and testing of the models. The test statistics (\(W\)) were low, below 1, and the \(p\)-values were less than 0.0001, indicating that the observed data significantly deviated from a normal distribution, as is shown in Table 9. A probability-probability (P-P) plot was generated to illustrate that the calculated MAE values obtained from the training and testing of the three models did not adhere to a normal distribution. As can be seen in Figure 13, the points deviated significantly from the equality line, strongly suggesting that the examined datasets did not follow a normal distribution. Checking the normality (Shapiro-Wilk test) of the MAE distribution helps in selecting the appropriate statistical tests and in ensuring the validity of the conclusions drawn from these tests. Assuming normality is often a prerequisite for certain parametric tests like ANOVA or t-tests. If the MAE data is normally distributed, it implies that the underlying errors are evenly distributed around the mean error and that the ANOVA test would be appropriate for further statistical analysis. If the MAE data is not normally distributed, tests that do not assume a specific distribution of data (non-parametric tests like the Kruskal-Wallis test) are more appropriate. We further performed the non-parametric Kruskal-Wallis test [27] with a significance level (\(\alpha\)) set at 0.05 and 2 degrees of freedom. This test is employed to compare \(k\) samples when dealing with non-Gaussian distributions. The null hypothesis tested whether the samples came from the same population. For the calculated MAE values obtained from the training of the models, the calculated \(K\) statistic was 109.245, while the critical value was 5.991. The calculated two-tailed \(p\)-value was found to be less than 0.0001, indicating significant differences in the performance of the models under consideration. Consequently, the null hypothesis was rejected with a confidence level exceeding 99.99%. In the case of calculated MAE values obtained from the testing of the models, the calculated \(K\) statistic was 117.872, while the critical value was 5.991. The calculated two-tailed \(p\)-value was found to be less than 0.0001, indicating that the null hypothesis had to be rejected.