## 4 Discussion
### Correlation Analysis: 4 in. vs. 6 in. Cylinders (Curing Method #C1)
\begin{table}
\begin{tabular}{c c c c c} \hline \hline  & **Linear Model** & **Nonlinear Model** & **Modified Linear Model** & **Evaluation** \\ \hline Model & y = \(1.14\times 10^{-9}+1.23\times 10^{-6}(x_{1})+0.001(x_{2})\) & & Better in \\ equation & y = \(228.0869+0.9325\)x & \(-4.63\times 10^{-7}(x_{3})+7.27\times 10^{-11}(x_{4})-4.03\times\) & y = \(0.9836\)x & simplicity and \\  & & \(10^{-15}(x_{3})\) & & easiness of use \\ \hline C.I. & 95\% & 95\% & 95\% & Same \\ \hline Correlation type & Pearson’s & Pearson’s & Modified Pearson & – \\ \hline Data points & 186 & 186 & 186 & Same \\ \hline Correlations & 0.9870 & 0.9870 & 0.987 & Same \\ \hline R & 0.9870 & 0.9840 & 0.9847 & A higher value is preferred \\ \hline R-squared & 0.9750 & 0.9670 & 0.9696 & A higher value is preferred \\ \hline RMSE & 251.2410 & 286.5370 & 276.4787 & A smaller value is preferred \\ \hline MAE & 196.9860 & 234.5930 & 216.3038 & A smaller value is preferred \\ \hline MAPE & 0.065 & 0.1100 & 0.0734 & A smaller value is preferred \\ \hline \hline \end{tabular}
\end{table} 
Table 6: Comparison of linear vs. nonlinear correlation models of the compressive strengths of 4 in. and 6 in. cylinders cured using Method #C1. \begin{table}
\begin{tabular}{c c} \hline \hline \# & **Explanation** \\ \hline (c) & Based on the PCA in Table 3, the variable curing days can affect the correlation between the compressive strengths of different cylinders. Hence, further analysis was performed after only one day of curing duration. The figure shows the result when setting the y-intercept to zero. The result model equation is y = \(1.0497\)x. The R-squared value = \(0.9889\). The model equation is accurate in describing the correlation of the observations. \\ \hline (d) & It shows further analysis of the linear correlation: (6 in. compressive strength = coefficient \(\times\) 4 in. compressive strength) using summer data. The y-intercept is zero. The result model equation is y = \(0.9603\)x. The R-squared value = \(0.9910\). The model prediction is accurate. \\ \hline (e) & It is a further analysis of the linear correlation using fall data. The y-intercept is zero. The result model equation is y = \(0.9947\)x. The R-squared value is \(0.9958\). The model equation is accurate. \\ \hline (f) & result model equation is y = \(0.9799\)x. The R-squared value is \(0.9962\). The results indicate that the model equation is accurate in describing the correlation of the observations. \\ \hline (g) & The result model equation is y = \(0.985\)x. The R-squared value is \(0.9956\). The results indicate that the model equation is accurate in describing the correlation of the observations. \\ \hline (h) & It is the fitted line plot for the nonlinear correlation. See Table 6 for details. \begin{tabular}{c c c c c} \hline \hline  & **Linear Model** & **Nonlinear Model** & **Modified Linear Model** & **Evaluation** \\ \hline Model & y = \(1.14\times 10^{-9}+1.23\times 10^{-6}(x_{1})+0.001(x_{2})\) & & Better in \\ equation & y = \(228.0869+0.9325\)x & \(-4.63\times 10^{-7}(x_{3})+7.27\times 10^{-11}(x_{4})-4.03\times\) & y = \(0.9836\)x & simplicity and \\  & & \(10^{-15}(x_{3})\) & & easiness of use \\ \hline C.I. & 95\% & 95\% & 95\% & Same \\ \hline Correlation type & Pearson’s & Pearson’s & Modified Pearson & – \\ \hline Data points & 186 & 186 & 186 & Same \\ \hline Correlations & 0.9870 & 0.9870 & 0.987 & Same \\ \hline R & 0.9870 & 0.9840 & 0.9847 & A higher value \\  & & & & is preferred \\ \hline R-squared & 0.9750 & 0.9670 & 0.9696 & A higher value \\  & & & & is preferred \\ \hline RMSE & 251.2410 & 286.5370 & 276.4787 & A smaller value \\  & & & & is preferred \\ \hline MAE & 196.9860 & 234.5930 & 216.3038 & A smaller value \\  & & & & is preferred \\ \hline MAPE & 0.065 & 0.1100 & 0.0734 & A smaller value is preferred \\ \hline \to check whether designed experiments (in which the runs are performed) are randomized. The residuals in both plots fluctuate in a random pattern around the center line of zero.