## 4 Experimental Results  
ANOVA multifactorial analysis of variance is used to identify which factors are relevant. In this sense, a normal probability plot is constructed to determine which factors are relevant, as shown in Figure 1 for the AIC index. Figure 1 and Table 3 show the CE effect as the most relevant. The AIC response is only affected by the CE effect, because the \(p\)-value is less than 0.05. It is necessary to verify that the model fulfills the normality, homoscedasticity, and independent conditions to validate the result reached through the ANOVA analysis. \begin{table}
\begin{tabular}{c c c} \hline \hline \multirow{2}{*}{**Factors**} & \multicolumn{2}{c}{**Levels**} \\ \cline{2-3}  & **Low** & **High** \\ \hline Training algorithm & Levenberg–Marquardt & Resilient Backpropagation \\ Number of hidden layers & 1 & 5 \\ Number of neurons in the & 1 & 10 \\ hidden layer & 60 & 120 \\ Data number & 10 & 30 \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Factors design. Table 3 shows the results reached through the ANOVA analysis for AIC. Because the Chi-square test has a \(p\)-value greater than 0.05, it is not possible to reject the null hypothesis; hence, the residuals fit into a normality plot. To verify the homoscedasticity condition, we used the Levene's test, as shown in Table 5.  
According to the results in Table 5, four factors fulfill the homoscedasticity condition because their \(p\)-values are greater than 0.05. The third and last condition is lag 1 autocorrelation, and the results are as follows: Durbin-Watson = 2.3533 (\(p\)-value = 0.5515), Lag 1 autocorrelation = \(-\)0.057753. The Durbin-Watson statistic is equal to 2 approximately; hence, residuals are independent. \begin{table}
\begin{tabular}{c c} \hline \hline
**Test** & \(p\)**-Value** \\ \hline Chi-square & 0.1319 \\ \hline \hline \end{tabular}
\end{table} 
Table 4: Normality test for AIC. Figure 1: Normal probability plot for Akaike information criterion (AIC). For MAPE analysis the same procedure is carried out. As is seen in Figure 2 and Table 6, the CE effect is the most relevant. The MAPE response is only affected by the CE effect, because the \(p\)-value is less than 0.05. It is necessary to verify that the model fulfills the normality, homoscedasticity, and independent conditions to validate the result reached through the ANOVA analysis for MAPE. \begin{table}
\begin{tabular}{c c c c c c} \hline \hline
**Source** & **Sum of Squares** & **Degrees of Freedom** & **Mean Square** & **F-Ratio** & \(p\)**-Value** \\ \hline A: Train\_Alg & 0.871079 & 1 & 0.871079 & 2.78 & 0.1096 \\ \hline B: Hidd\_Num & 0.0182764 & 1 & 0.0182764 & 0.06 & 0.8114 \\ \hline C: Neu\_Num & 0.0201734 & 1 & 0.0201734 & 0.06 & 0.8020 \\ \hline D: Data\_Num & 0.00900434 & 1 & 0.00900434 & 0.03 & 0.8669 \\ \hline E: Val\_Per & 0.331972 & 1 & 0.331972 & 1.06 & 0.3145 \\ \hline AB & 0.0507545 & 1 & 0.0507545 & 0.16 & 0.6912 \\ \hline AC & 0.00168203 & 1 & 0.00168203 & 0.01 & 0.9422 \\ \hline AD & 0.0519588 & 1 & 0.0519588 & 0.17 & 0.6877 \\ \hline AE & 0.446753 & 1 & 0.446753 & 1.43 & 0.2451 \\ \hline BC & 0.567389 & 1 & 0.567389 & 1.81 & 0.1921 \\ \hline BD & 0.0440187 & 1 & 0.0440187 & 0.14 & 0.7114 \\ \hline BE & 0.412097 & 1 & 0.412097 & 1.32 & 0.2637 \\ \hline CD & 0.048383 & 1 & 0.048383 & 0.15 & 0.6981 \\ \hline CE & 1.75907 & 1 & 1.75907 & 5.62 & 0.0270 \\ \hline DE & 0.00784171 & 1 & 0.00784171 & 0.03 & 0.8757 \\ \hline Error total & 6.89167 & 22 & 0.313258 & & \\ \hline Total (corrected error) & 11.5321 & 37 & & & \\ \hline \hline \end{tabular}
\end{table} 
Table 5: Homoscedasticity test for AIC. \begin{table}
\begin{tabular}{c c c} \hline \hline  & **Levene’s Test** & \(p\)**-Value** \\ \hline B & 1.38557 & 0.263 \\ C & 1.54509 & 0.22749 \\ D & 1.61809 & 0.2127 \\ E & 1.73115 & 0.1919 \\ \hline \hline \end{tabular}
\end{table} 
Table 5: Homoscedasticity test for AIC. Figure 2: Normal probability plot for Mean Absolute Percentage Error (MAPE). Chi-square is used to validate the normality condition.