Homogeneity of the Pre-test Homogeneity of the Post-test

No Score x = X - � x² y= Y- � y² X Y 21 5 5 0.256 0.066 0.769 0.592 22 -5 5 -9.744 94.938 0.769 0.592 23 -10 10 -14.744 217.373 5.769 33.284 24 20 40 15.256 232.758 35.769 1279.438 25 15 10 10.256 105.194 5.769 33.284 26 15 15 10.256 105.194 10.769 115.976 27 5 5 0.256 0.066 0.769 0.592 28 10 5.256 27.630 -4.231 17.899 29 15 -5 10.256 105.194 -9.231 85.207 30 35 30.256 915.450 -4.231 17.899 31 -5 5 -9.744 94.938 0.769 0.592 32 5 5 0.256 0.066 0.769 0.592 33 5 5 0.256 0.066 0.769 0.592 34 -4.744 22.502 -4.231 17.899 35 -5 -5 -4.744 22.502 10.769 85.207 36 15 -4.744 22.502 -9.231 369.822 37 -5 -4.744 22.502 -19.231 0.592 38 -15 -4.744 22.502 0.769 4976.923 39 5 -4.744 4447.436 0.000 85.207 ∑ 185 165 0.000 22.502 10.769 369.822 Based on the data from Table 4.8, the statistical calculation was done in the following steps: a. Determine the Mean of Variable X with formula: � � = �� � � − = 185 39 � − = 4.74 b. Determine the Mean of Variable Y with formula: � � = �� � � − = 165 39 � − = 4.23 c. Determine the Standard Deviation Score of Variable X with formula: �� � = � �x 2 � 1 �� � = � 4447.436 39 �� � = √114.037 �� � = 10.678 d. Determine the Standard Deviation Score of Variable Y with formula: �� � = � �y 2 � 2 �� � = � 4976.923 39 �� � = √127.613 �� � = 11.296 e. Determine the Standard Error Mean of Variable X with formula: �� � � = �� � �� 1 − 1 �� � � = 10.678 √39 − 1 �� � � = 10.678 √38 �� � � = 10.678 6.16 �� � � = 1.732 f. Determine the Standard Error Mean of Variable Y with formula: �� � � = �� � �� 2 − 1 �� � � = 11.296 √39 − 1 �� � � = 11.296 √38 �� � � = 11.296 6.16 �� � � = 1.833 g. Determine the Standard Error of Difference with formula: �� � � − �� � � = ��� � � 2 + �� � � 2 = �1.732 2 + 1,832 2 = √3.0 + 3.359 = √6.359 = 2.522 h. Determine t with formula: � = � � − � � �� � � − �� � � � = 4.74 − 4.23 2.522 � = 0.51 2.522 � = 0.203 i. Determine the degree of freedom �� with formula: �� = � 1 + � 2 – 2 �� = 39 + 39– 2 �� = 78 − 2 �� = 76 From the result of statistical calculation above, it can be seen that the value of t or t test is 0.203 and the degree of freedom �� was 76. The value of t in the degree of freedom of 76 and at the degree of significance 1 or t table of df 75 with ɑ=0.01 is 2.376. Subsequently, after the manual calculation was done, the calculation was verified by employing independent samples t test in IBM SPSS Statistics 20. Here are the results: Table 4.9 Group Statistics Group N Mean Std. Deviation Std. Error Mean GainScore EXP 39 4.74 10.818 1.732 CONTROL 39 4.23 11.444 1.833