## 4 Result Analysis
### Hypothesis Testing
#### 4.4.1 First Hypothesis  
The formulated H10 null hypothesis states that:  
\(\mu\) (Time taken by T1) = \(\mu\) (Time Taken by T2)  
\(\mu\) (Valid Requirements gathered by T1) = \(\mu\) (valid requirements gathered by T2)  
For the testing the first part of the null hypothesis H10, we used a t-test on our data using the IBM SPSS statistical tool. We take a 95% confidence interval, therefore, statistical significance is set to \(\alpha\) = 0.05. As a result of our t-test, we generated the following results. Table 10 shows general information such as the number of participants in each team, mean time for both teams, standard deviation, and standard error mean for our data set.  
Table 11 shows upper and lower bounds on the 95% confidence interval, Mean difference, standard-error difference, and other important significant levels of dependent variable efficiency of both teams T1 and T2. As shown in Table 11, the significance value for the SNS-based proposed approach is 0.001, compared to the traditional RE approach carried out by team T2. The significant value was less than \(\alpha\) = 0.05, which means that team T1 has shown more efficiency of eliciting user requirements than team T2.  
Figure 12 also shows that team T2 has taken more time than team T1 to conduct RE activities for smartwatch domain. The boxplot shows that team T1 has a median 89 min and team T2 has a median 120 min. This means that the median of T1 is less than the median of T2.  
It is also shown in Table 10 that the mean time for team T1 was less than the mean time of team T2. Therefore, we can describe it as:  
\(\mu\) (Time taken by team T1) = 89 min  
\(\mu\) (Time Taken by team T2) = 120 min  
Hence,  
\(\mu\) (Time taken by T1) \(<\mu\) (Time Taken by T2)  
Thus, we reject the part \(a\) of the null hypothesis H10 because it states that \(\mu\) (Time taken by T1) = \(\mu\) (Time Taken by T2).  
For testing part b of the null hypothesis H10, we refer to Table 8. As shown in Table 8, the total functional requirements collected by team T1 were 175 and total non-functional requirements collected by team T1 were 52 requirements. Functional requirements gathered by team T2 were 63 and non-functional requirements were 11. So,  
Total requirements collected by team T1= 227  
\begin{table}
\begin{tabular}{c c c c c c} \hline \hline
**Team** & **N** & **Mean** & **Std.Deviation** & **Std.Error Mean** \\ \hline \multirow{2}{*}{**Time**} & T1 & 15 & 82.2667 & 18.87730 & 4.87410 \\  & T2 & 15 & 114.2000 & 25.64928 & 6.62262 \\ \hline \hline \end{tabular}
\end{table}
Table 10: Group Statistics: Mean Time for two Teams.  
Total requirements collected by team T2= 74  
Hence, it clearly tells that valid requirements gathered by team T1 were more than the valid requirements collected by team T2. Therefore, it can be stated as: \(\mu\) (Valid Requirements gathered by T1) \(>\mu\) (valid requirements gathered by T2). Therefore, based on this evidence, we reject the part \(b\) of the null hypothesis H10 which states that \(\mu\) (Valid Requirements gathered by T1) = \(\mu\) (valid requirements gathered by T2).  
We have also seen that part \(a\) and part \(b\) of the null hypothesis H10 are rejected. Hence, the null hypothesis H10 is rejected and the alternate hypothesis H1 is accepted.