# Aligning EMR Structure with Treatment Cycles: Enhancing Data Management and CDSS Functionality
## 6 Empirical Evaluation of the Cycle Approach's Utilization in EMRs
### Statistical Analysis  
To rigorously assess the impact of the cyclical approach on treatment outcomes, we conducted a statistical evaluation using analysis of variance (ANOVA) [61]. This method was chosen because it allows us to compare means across multiple treatment cycles and determine whether the observed differences are statistically significant. Unlike simple pairwise comparisons, which increase the risk of Type I errors [62], ANOVA efficiently tests for variations across all cycles simultaneously, ensuring a more reliable interpretation of the treatment effects. \begin{table}
\begin{tabular}{c c c c c c} \hline \hline
**Treatment Cycle** & \begin{tabular}{c} **Post\_Treatment** \\ **Glucose\_Mean** \\ \end{tabular} & \begin{tabular}{c} **Treatment\_Success** \\ **Rate\_Mean** \\ \end{tabular} & \begin{tabular}{c} **Treatment\_Success** \\ **Rate\_Std** \\ \end{tabular} & \
\begin{tabular}{c} **Treatment\_Success** \\ **Rate\_Min** \\ \end{tabular} &
\begin{tabular}{c} **Treatment\_Success** \\ **Rate\_Max** \\ \end{tabular} \\ \hline Overall & 8158.12 & 14.53 & 11.15 & 0 & 25 \\ \hline \hline \end{tabular}
\end{table} 
Table 3: Overall metrics (non-cyclical approach). According to the structure of our dataset (where each treatment cycle may involve a distinct treatment plan tailored to evolving clinical judgments), we considered each cycle an independent analysis unit. While some individual patients may have undergone multiple cycles, the treatment strategy in each cycle was often adjusted depending on patient-specific responses. Therefore, treatment cycles were not strictly repeated measures of a fixed protocol but potentially distinct interventions with unique clinical contexts. Given the independent nature of the treatment cycles in our study, ANOVA was selected as the most suitable statistical method, as it effectively evaluates the variance across cycles without assuming systematic correlations within patients. The frequent adjustments in treatment plans reinforce this choice. These adjustments minimize the dependency between treatment cycles. As a result, the assumptions of a mixed-effects model become less applicable. At the same time, ANOVA ensures the delivery of robust and clinically relevant insights into cycle-specific outcomes [63]. ANOVA determines whether the means of multiple groups differ significantly by comparing the variance within groups to the variance between groups. The test statistics (F-statistic) are calculated as follows:  
\[F=\frac{\text{Variance between groups}}{\text{Variance within groups}}\]  
ANOVA works by comparing two types of variances. **Variance between groups (treatment cycles):** This specifies how much the average values (means) of different treatment cycles differ from the overall mean of all cycles combined. If the treatment cycles are firmly different, their means should be spread apart. Conversely, if they are not, their means will be close together. **Variance within groups (individual patients in the same cycle):** This specifies how much individual patient outcomes vary within each treatment cycle. If a treatment cycle is consistent, individual patients will have similar results, and this variance will be low. Otherwise, much variation within the same treatment cycle might indicate external factors affecting treatment. The first ANOVA test examined post-treatment glucose levels across different cycles. The results are as follows:  
F-statistic: 64.6814, \(p\)-value: 0.0. It is interpreted that, since the \(p\)-value is zero, we reject the null hypothesis and conclude that glucose levels significantly differ across treatment cycles. This suggests that the treatment effectiveness is not uniform across cycles and that iterative interventions are critical in stabilizing glucose levels. Similarly, an ANOVA test was conducted to analyze the treatment success rates across cycles. The results obtained are as follows:  
F-statistic = 14.4841, \(p\)-value = 0.0.