Data Description Research Finding

- Class = 1 + 3.3 log n = 1 + 3.3 log 32 = 1 + 3.3 1.50 = 5.95 = 6 - Interval = = = 8 Table 4.3 Frequency of Post-test of Experimental class Interval score F Percentage 44 —51 1 3 52 —59 3 9 60 —67 8 25 68 —75 10 31 76 —83 8 25 84 —91 1 3 92 —99 1 3 N = 32 Note: P = x 100 P = Percent F = Frequency N = Number of Students The Table 4.3 and 4.4 showed about the number of frequency of pretest and posttest score from the students of experiment class. The Table 4.3 above showed that most students’ gained score was in interval 31—37. It means that there were about 11 34 students who had score from interval 31 —37 for pretest score. Whereas in Table 4.4, students mostly got score in interval 68 —75 or there were about 10 students 31 who had gained score in interval 68 —75. Table 4.4 Students’ Score of VIII.A Control Class Students Pretest Y 1 Posttest Y 2 Gained Score 1. 32 64 32 2. 56 72 16 3. 40 64 24 4. 32 48 16 5. 40 52 12 6. 32 48 16 7. 32 60 28 8. 52 62 10 9. 64 72 8 10. 44 52 8 11. 28 52 24 12. 20 36 16 13. 32 36 4 14. 48 52 4 15. 52 60 8 16. 32 36 4 17. 52 72 20 18. 56 72 16 19. 54 60 6 20. 56 60 4 21. 20 40 20 22. 56 68 12 23. 68 76 8 24. 28 56 28 25. 44 52 16 26. 48 54 6 27. 48 52 4 Students Pretest Y 1 Posttest Y 2 Gained Score 28. 56 72 16 29. 56 68 12 30. 52 60 8 31. 68 72 4 32. 56 72 16 N = 32 ∑Y 1 = 147432 = 46.06 ∑Y 2 = 187232 = 58.50 ∑Y =426 42632 =13.31 It can be seen from the table above that the st udents’ average pretest score was 40.06 and the average posttest score was 58.50 from the class VIII.A as the control class. Whereas the average gained score among pretest and posttest score was 13.31. a. Statistical Data of Pretest of Control Class - Range: H – L 68 – 20 = 48 - Class = 1 + 3.3 log n = 1 + 3.3 log 32 = 1 + 3.3 1.50 = 5.95 = 6 - Interval = = = 8 Table 4.5 Frequency of Pre-test of Control class Interval score of Pre-test F Percentage 20 —27 2 6 28 —35 8 25 36 —43 2 6 44 —51 5 15 52 —59 12 37 60 —67 1 3 68 —75 2 6 N= 32 b. Statistical Data of Posttest of Control Class - Range: H – L 76 – 36 = 40 - Class = 1 + 3.3 log n = 1 + 3.3 log 32 = 1 + 3.3 1.50 = 5.95 = 6 - Interval = R = 40 = 6.66666 = 7 C 6 Table 4.6 Frequency of Post-test of Control class Interval score of Pre-test F Percentage 36 —42 4 12 43 —49 2 6 50 —56 7 21 57 —63 7 21 64 —70 5 15 71 —77 7 21 N= 32 The Table 4.5 and 4.6 showed about the number of frequency of pretest and posttest score from the students of control class. The Table 4.5 above showed that most students’ gained score for pretest is in interval 52—59. It means that there are about 12 or 37 students who had score from interval 52 —59. Whereas in Table 4.6, from posttest score there were some students gained the same frequency of posttest score that was 21 in each interval of 50 —56, 57—63, and 71 —77. The average pretest and posttest score of experimental score were 40.06 and 66.62 with the gained score average was 24.56. The average score of control class was 46.06 for pretest and 58.50 for posttest score with the gained score average was 13.31. It can be seen that there was a differences among students’ score of experiment and control class. The average posttest score of control class was 58.50 which were lower than the experiment class 66.62.

2. Data Analyzing

a. Normality Test

Before go further to the final calculation of the research, normality test of the data have been done on this research see appendix 6. As it has been stated by Budi Susetyo, this kind of test is necessary to be used to see whether the sample that is used in this research has been normally distributed, from the normally distributed population, or not. 1 The Lyllifors test was used in this research. The data needed to be transformed into the basic value as the function of the distribution will be found later in the table. The maximum dispute T from the calculation must be in the absolute value +. The final result can be seen by comparing the value of T maks gained with T table , with the criteria of test hypothesis as follows: H i is rejected if T T table H o is accepted if T ≤ T table 1 Budi Susetyo, StatistikaUntukAnalisis Data Penelitian, Bandung: PT. Refika Aditama, 2010, p. 144 – 149. Hypothesis: H o : Data of X is normally distributed. H i : Data of X is not normally distributed. Based on the Table of Lyllifors test, the T table T 0.0532 with the degree significance of 0.05 was in value 0.156. From the table calculation of using Lyllifors test formula, the result of pretest score from the experiment class had T maks T table = 0.0965 0.156 see appendix of Lyllifors Table, which means that H o was accepted and data was normally distributed. For the posttest score of experiment class showed the result of normality test with T maks T table = 0.0818 0.156, means that H o was accepted as well and the data were normally distributed. Meanwhile, for the control class pretest and posttest normality test result showed T maks T table = 0.1399 0.156 for pretest and T maks T table = 0.0989 0.156, hence for both result it can be concluded that H o was accepted and both data were normally distributed.

b. t-test Calculation

Finally, the last step in analyzing the data, the statistic calculation of the t- test formula with the degree of significance 5 and 1 is used as follows: Table 4.7 Result Calculation of Post-test of Experiment Class Interval score F Xi x ʹ fx ʹ fx ʹ² 44 —51 1 47.5 3 3 9 52 —59 3 55.5 2 6 12 60 —67 8 63.5 1 8 8 68 —75 10 71.5 76 —83 8 79.5 -1 -8 8 84 —91 1 87.5 -2 -2 4 92 —98 1 95.5 -3 -3 9 N = 32 ∑fxʹ=4 ∑fxʹ² =50 a. Determining Mean of Variable X, with formula: M 1 = M ʹ + i ʹ = 71.5 + 8 = 71.5 + 8 0.13 = 71.5 + 0.96 = 72.46 b. Determining Standard of Deviation score of Variable X, with formula: SD 1 = i √ ∑ ฀ ∑ ฀ = 8 √ = 8 √ = 8 √ = 8 1.24 = 9.92 c. Determining Standard Error of Different Mean of Variable X, with formula: SE M 1 = √ = √ = = 1.78 Table 4.8 Result Calculation of Post-test of Control Class d. Determining Mean of Variable Y, with formula: M 2 = M ʹ + i ʹ = 60 + 7 = 60 + 7 0.12 = 60 + 0.84 = 60.84 e. Determining Standard of Deviation Score of Variable Y, with formula: SD 2 = i √ ∑ ʹ ∑ ʹ = 7 √ = 7 √ = 7 √ = 7 1.61 = 11.27 Interval Score F Yi y ʹ Fy ʹ fy ʹ ² 36 —42 4 39 3 12 36 43 —49 2 46 2 4 8 50 —56 7 53 1 7 7 57 —63 7 60 64 —70 5 67 -1 -5 5 71 —77 7 74 -2 -14 28 N= 32 ∑fyʹ =4 ∑fyʹ ²=84