## 6 Experimentation
### Results and Statistical Tests  
The results were gathered from a questionnaire and were analyzed as follows: the null hypothesis (H0) held if the resolution times (RT) used by the technicians in Group A were statistically similar to the times used by technicians in Group B. If RT were less than for Group B, the alternative hypothesis (H1) could hold, on which grounds only two options would be considered: the resolution times were better for Group B than for Group A. We asked the technicians to record the resolution time for each incident on the time sheet (Table 2), which we used to calculate the mean and variance (Table 3). As a binomial distribution with only two possible values was used, the possibility of an improvement in the resolution times would be \(p=0.5\) for both options. The test statistic is the number of positive results for the comparison between the times of both groups, and neither the negative nor the equal results are considered. This is a one-tailed test, as H1 is only analyzed in one direction. We decided that the acceptable risk of test result error, that is, the significance value, would be 5%, meaning that the confidence level of the results is 95 %. Therefore, according to [PERSON]'s \(t\) distribution table for the above values (Table 4), the critical value for 10 tests would be 1.812. This means that the region of rejection for H0 could be verified if the resulting statistic is less than 1.812. According to [PERSON] [61], it would be recommendable in small tests like this to use two values. On this ground, we also considered an error margin of 0.10 for the analysis. In this case, according to Student's \(t\) distribution table (Table 4), the critical number would be 1.372. \begin{table}
\begin{tabular}{c c c c c} \hline \hline
0.10 & 0.05 & 0.025 & 0.01 & 0.05 & 0.0005 \\
1.372 & 1.812 & 2.228 & 2.764 & 3.169 & 4.587 \\ \hline \hline \end{tabular}
\end{table} 
Table 4: Student’s table entry for 10 tests with a significance level of 0.05 and 0.1 (df 10). \begin{table}
\begin{tabular}{c c c} \hline \hline
**Incident** & **Resolution Time (RT), Minutes** & **Technician (A/B)** \\ \hline \hline \multirow{2}{*}{1} & 120 & A \\  & 30 & B \\ \hline \multirow{2}{*}{2} & 105 & A \\  & 45 & B \\ \hline \multirow{2}{*}{3} & 180 & A \\  & 105 & B \\ \hline \multirow{2}{*}{4} & 45 & A \\  & 30 & B \\ \hline \multirow{2}{*}{5} & 80 & A \\  & 50 & B \\ \hline \multirow{2}{*}{6} & 50 & A \\  & 45 & B \\ \hline \multirow{2}{*}{7} & 60 & A \\  & 60 & B \\ \hline \multirow{2}{*}{8} & 55 & A \\  & 20 & B \\ \hline \multirow{2}{*}{9} & 20 & A \\  & 15 & B \\ \hline \multirow{2}{*}{10} & 55 & A \\  & 35 & B \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Resolution time by incident and technician. \begin{table}
\begin{tabular}{c c c c} \hline \hline
**Technicians** & **Resolution Time** & **Mean** & **Variance** \\ \hline A & 770 & 77 & 1849 \\ B & 465 & 46.5 & 530.22 \\ \hline \hline \end{tabular}
\end{table} 
Table 3: Resolution time mean and variance by technician groups. For the calculation of resolution time, mean and variance, the basic statistics have simply been obtained from the data collected in Table 2 for both groups (A and B). From the data shown in Table 3, we can conclude that the standard deviation for Group A (43) is greater than for Group B (23.026). Thanks to the use of Student's \(t\), it is intended to check whether the average difference in the resolution times obtained in Table 3 is statistically significant or not. For this purpose, a study was carried out with a significance level of 0.05 and 0.1, shown in Table 4. We used the data from Table 4 to calculate the value of the statistic (t) for comparison with the critical value, depending on the significance level (1.812 and 1.372 with a significance level of 0.05 and 0.10, respectively). As recommended by [PERSON] et al. [61], we analyzed the results taking into account two significance values.