Normality Test The Analysis of Data

The Liliefors shows that the significant degree of 0.05 in L0.0525 = 0.173 H1 : L 0.173 H0 : L ≤ 0.173 In the table 4.4, the L max value is 0.135. Therefore, H is accepted because the result shows that L max is lower than L table 0.135 0.173. It means that the data in experiment class post-test is normally distributed. b. The Normality Test of Control Class Table 4.5 Calculation of Pre-test Normality in Control Class No. X f fX fX 2 p=fn ∑P z = Xi- Xs ф T=ф-∑p 1 43 1 43 1849 0.04 0.04 -2.22548 0.0102 0.0298 2 48 1 48 2304 0.04 0.08 -1.49341 0.0681 0.0119 3 50 2 100 5000 0.08 0.16 -1.20059 0.1151 0.0449 4 51 1 51 2601 0.04 0.2 -1.05417 0.1469 0.0531 5 52 3 156 8112 0.12 0.32 -0.90776 0.1841 0.1359 6 57 1 57 3249 0.04 0.36 -0.1757 0.4325 0.0725 7 59 2 118 6962 0.08 0.44 0.11713 0.5438 0.1038 8 60 7 420 25200 0.28 0.72 0.263543 0.6026 0.1174 9 62 1 62 3844 0.04 0.76 0.556369 0.7088 0.0512 10 64 2 128 8192 0.08 0.84 0.849195 0.7995 0.0405 11 65 2 130 8450 0.08 0.92 0.995608 0.8389 0.0811 12 70 1 70 4900 0.04 0.96 1.727672 0.9573 0.0027 13 72 1 72 5184 0.04 1 2.020498 0.9783 0.0217 Total ∑X= ∑ F= ∑ fX= ∑f X 2 = 753 25 1455 85847 S= 6.83 S 2 = 46.6 rata2x= 58.2 Lmax= 0.1359 Ltable= 0.173 ∑ [ ∑ ] [ ] √ The Liliefors shows that the significant degree of 0.05 in L0.0525 = 0.173 H1 : L 0.173 H0 : L ≤ 0.173 In the table 4.5, the L max value is 0.1359. Therefore, H is accepted because the result shows that L max is lower than L table 0.1359 0.173. It means that the data in control class pre-test is normally distributed. Table 4.6 Calculation of Post-test Normality in Control Class No. X f fX fX 2 p=fn ∑P z = Xi- Xs ф T=ф-∑p 1 50 1 50 2500 0.04 0.04 -2.46757 0.0069 0.0331 2 55 1 55 3025 0.04 0.08 -1.71342 0.0436 0.0364 3 57 2 114 6498 0.08 0.16 -1.41176 0.0793 0.0807 4 59 1 59 3481 0.04 0.2 -1.11011 0.1335 0.0665 5 61 1 61 3721 0.04 0.24 -0.80845 0.2119 0.0281 6 64 2 128 8192 0.08 0.32 -0.35596 0.3632 0.0432 7 65 1 65 4225 0.04 0.36 -0.20513 0.4207 0.0607 8 66 2 132 8712 0.08 0.44 -0.0543 0.4801 0.0401 9 67 1 67 4489 0.04 0.48 0.096531 0.5359 0.0559 10 68 3 204 13872 0.12 0.6 0.24736 0.5948 0.0052 11 70 4 280 19600 0.16 0.76 0.54902 0.7054 0.0546 12 72 2 144 10368 0.08 0.84 0.850679 0.8023 0.0377 13 73 1 73 5329 0.04 0.88 1.001508 0.8413 0.0387 14 75 1 75 5625 0.04 0.92 1.303167 0.9032 0.0168 15 76 2 152 11552 0.08 1 1.453997 0.9265 0.0735 Total ∑X= ∑ F= ∑ fX= ∑f X 2 = 978 25 1659 111189 S= 6.63 S 2 = 43.91 rata2x= 66.36 Lmax= 0.0807 Ttable= 0.173 ∑ [ ∑ ] [ ] √ The Liliefors shows that the significant degree of 0.05 in L0.0525 = 0.173 H1 : L 0.173 H0 : L ≤ 0.173 In the table 4.6, the L max value is 0.0807. Therefore, H is accepted because the result shows that L max is lower than L table 0.0807 0.173. It means that the data in control class post-test is normally distributed.

2. Homogeneity Test

In order to know the homogeneity of the data, the researcher did the homogeneity test. To do the homogeneity test, he analyzed the score of pre-test and post-test collected in both experimental and control classes using Fisher-test. Hypothesis: - H : F F t - H 1 : F F t Notes: - H : The experimental class is homogenous to the control class - H 1 : The experimental class is not homogenous to the control class In addition, the F value is calculated with the following formula: = Here are the results of the calculation of F-test both in terms of pre- test and post-test. First, the homogeneity test of pre-test data is presented as follows: Because the values of are already obtained from the calculations provided in Table 4.3 and Table 4.5 i.e. in this case, , the calculation of F-test for pre-test data can be directly conducted. The F-test calculation is as follows: = 1.2618 ≈ 1.26 With =n-1=25-1=24, and =n-1=25-1=24, the F t value at 95 level of significance α=0.05 obtained is 1.98 see Appendix. Due to the fact that F F t 0.05, 24, 24 = 1.26 1.98, H is accepted. Therefore, it can be concluded that the pretest data of experimental class and control class is considered homogenous. Second, the homogeneity test of the post data is presented as follows: Based on Table 4.4 and Table 4.6, the values for and obtained respectively are 83.42 and 43.91. The calculation of F-test for the post data is as follows: = 1.8998 ≈ 1.90 Likewise, with =n-1=25-1=24, and =n-1= 25-1=24, the F t value at 95 level of significance α=0.05 obtained is 1.98 see Appendix. Due to the fact that F F t 0.05, 24, 24 = 1.90 1.98, H is accepted. Therefore, it can be interpreted that the post-test data of the experimental class and control class is considered homogenous as well.

3. Hypothesis Test

After knowing that the data is normally distributed and homogenous, then he did the hypothesis test. The hypothesis test is used to see whether there is a significant difference between experiment and control class. The writer used t-test to do the hypothesis test. The formula of t-test is as follows Note:  M 1 : mean of variable X, the formula is: ∑  M 2 : mean of variable Y, the formula is: ∑  SE M1 : standard error mean of variable X, the formula is: √  SE M2 : standard error mean of variable Y, the formula is: √  SD 1 : standard of deviation score of variable X, the formula is: √ ∑  SD 2 : standard of deviation score of variable Y, the formula is: √ ∑ That is the main formula of t-test to do the hypothesis test. Before using the formula, the researcher also used some calculation procedure, such as: a. Determining mean of variable X ∑ b. Determining mean of variable Y ∑ c. Determining standard of deviation score of variable X √ ∑ √ √ √