Normality of Experimental Class

Table 4.8 Calculation of Normality Pre-test in Controlled Class X F X 2 fX fX 2 p=fn ∑p Z ᶲ T 30 1 900 30 900 0.02 0.02 -2.33 0.0099 0.010 35 1 1225 35 1225 0.02 0.04 -.1.94 0.0262 0.013 40 2 1600 80 3200 0.05 0.09 -1.55 0.0606 0.029 45 3 2025 135 6075 0.08 0.17 -1.16 0.1230 0.047 50 4 2500 200 10000 0.11 0.28 -0.77 0.2206 0.059 55 1 3025 55 3025 0.02 0.30 -0.38 0.3520 0.052 60 4 3600 240 14400 0.11 0.41 0.01 0.4960 0.086 65 10 4225 650 42250 0.28 0.49 0.40 0.6179 0.127 70 3 4900 210 14700 0.08 0.77 0.79 0.7852 0.015 75 4 5625 300 22500 0.11 0.88 1.18 0.8810 0.001 80 2 6400 160 12800 0.05 0.93 1.57 0.9418 0.011 ∑ 35 2095 131075 ̅ = ∑ � ∑ � = 2095 35 = 59.85 S 2 = ∑ � - ∑ � 2 = 131075 35 - 2095 35 2 S 2 = 3745 – 3582.02 = 162.98 S = √162.98 = 12.76 In the significant degree of 0.05, the value in the table Lillyfors showed T 0.05x35 = 0.149 H 1 = T 0.149 H = T ≤ 0.149 The result showed that T max T table 0.127 0.149. The conclusion is H accepted, it means that the data is normally distributed. 2 Normality of Post-test Hypothesis: H : data is normally distributed H 1 : data is not normally distributed Table 4.9 Calculation of Normality Post-test in Controlled Class X F X 2 fX fX 2 p=fn ∑p Z ᶲ T 30 1 900 30 900 0.02 0.02 -2.18 0.0146 0.005 35 2 1225 70 2450 0.05 0.07 -1.85 0.0322 0.037 45 1 2025 45 2025 0.02 0.09 -1.19 0.1170 0.027 50 2 2500 100 5000 0.05 0.14 -0.86 0.1949 0.054 55 4 3025 220 12100 0.11 0.25 -0.53 0.2981 0.048 60 7 3600 420 25200 0.20 0.45 -0.19 0.4247 0.025 65 3 4225 195 12675 0.08 0.53 0.13 0.5517 0.021 70 3 4900 210 14700 0.08 0.61 0.46 0.6722 0.067 75 1 5625 75 5625 0.02 0.63 0.79 0.7518 0.121 80 4 6400 320 25600 0.11 0.74 1.12 0.8686 0.128 85 4 7225 340 28900 0.11 0.85 1.45 0.9265 0.076 90 2 8100 180 16200 0.05 0.90 1.79 0.9633 0.063 ∑ 35 2205 146875 ̅ = ∑ � ∑ � = 2205 35 = 63 S 2 = ∑ � - ∑ � 2 = 146875 35 - 2205 35 2 S 2 = 4196.42 – 3969 = 227.42 S = √227.42 = 15.08 In the significant degree of 0.05, the value in the table Lillyfors showed T 0.05x35 = 0.149 H 1 = T 0.149 H = T ≤ 0.149 The result showed that T max T table 0.128 0.149. The conclusion is H accepted, it means that the data is normally distributed.

d. Homogeneity

Based on the calculation of normality, the writer got the result that all data in pre-test and post-test of both experimental class and controlled class have been distributed normally. The next step calculation was finding the homogeneity of the data. The purpose of this calculation was to see whether the datasample in both classes was homogenous or heterogeneous. Hypothesis: H : the condition of experimental class is not different from controlled class. H 1 : the condition of experimental class is different from controlled class. The criteria of the test: α = 0.05 H : � 1−� −1 F � 1 2 � 1−1 2−1 H 1 : F ≥ � 1 2 � 1, 2 The formula used to can be seen as follows: F = ℎ ℎ� ℎ � �� � ℎ � � �� � Or F = 1 2 2 2 The calculation can be seen as follows: F = = 198.94 114.64 = 1.73 N1-1 = 35-1=34 N2-1 = 35-1=34 � 0.05 34,34 = 1.84 It can be seen that F F 12 α n1-1n2-1 1.73 1.84. Based on the criteria, it can be concluded that H is accepted and the sample in experimental class and controlled class were homogenous.

3. Hypothesis Testing

To test the hypothesis whether there is significant different to classes, the writer calculated the data. The experimental class was X and the controlled class was Y. The procedures of calculated were as follows: