“…9, the collected data indicates a normal distribution. Thus, we also used a Paired T-Test by calculating: the difference of number of problems between both approaches, d = { 13, 14, 28, 15,11,14,9,17,14,8,15,17,11,12,8,8,15,7,12,16,16,8,17,14,8,11 }; the standard deviation of this difference, S d = 4.463183; the number of degrees of freedom, F = N − 1 = 26 − 1 = 25, where N is the number of subjects; t 0 = 4.463183; and t 0.05,25 = 1.708141. Since t 0 > t 0.05,25 , it is possible to reject the null hypothesis with a two sided test at 0.05 level.…”