Homogeneity Test Analysis of Data

14 11 9 11,00 9,00 121,00 121,00 15 10 6 8,00 6,00 64,00 64,00 16 18 8 18,00 8,00 324,00 324,00 17 17 13 17,00 13,00 289,00 269,00 18 29 12 29,00 12,00 841,00 144,00 19 28 14 28,00 14,00 784,00 196,00 20 16 18 16,00 1,00 256,00 324,00 21 23 22 23,00 22,00 529,00 484,00 22 17 7 17,00 7,00 289,00 49,00 23 37 17 37,00 17,00 1369,00 289,00 24 23 20 23,00 20,00 529,00 400,00 25 22 4 22,00 4,00 484,00 16,00 26 12 13 12,00 13,00 144,00 169,00 27 20 19 20,00 19,00 400,00 361,00 28 37 13 37,00 13,00 1369,00 169,00 29 13 12 13,00 12,00 169,00 144,00 30 20 17 20,00 17,00 400,00 289,00 31 17 17,00 0,00 289,00 0,00 32 24 5 24,00 5,00 576,00 25,00 Total 679 428 657,78 414,63 16189,00 7511,64 The table above described the result calculation of the gained score of the experimental class X and the controlled class Y. Based on the table above, it can be concluded that the total score of the experimantal class X is 679 and the controlled class Y is 428. From the table aove, the writer calculated them based on the steps of t- test formula, as follows: 1. Determining Mean of Variable X M x = = = 21.21 2. Determining Mean of Variable Y M y = = = 13.38 3. Determining Standar od Deviation Score of Variable X SD x = √ = √ = √ = 22.49 4. Determining Standard of Deviation Score of Variable Y SD y = √ = √ = √ = 15.3 5. Determining Standard Error of Mean of Variable X SEM x = √ = √ = √ = = 4.04 6. Determining Standard Error of Mean of Variable Y SEM x = √ = √ = √ = = 2.74 7. SEM x – My = √ = √ = √ = √ = 3.68 Determining t t = = = 2.13 Dtermining t table in significance level 5 with df degree of freedom Df = N1+N2 – 2 = 32+32 – 2 = 62 The value of df 62 at the degree of significance 5 or t table is 1.670. It means Ha is accepted because T 2.13 is higher than t table 1.999. 8. The Testing of Hypotheses The statistical hypotheses of this research can be seen as: Ho : There is no significant difference between students‟ vocabulary using crossword puzzle and without crossword puzzle. Ha : There is significant difference between students‟ vocabulary using crossword puzzle and without crossword puzzle. And then the criteria used as follows: 1. If t-test t o t-table t t in significant degree of 0.05, Ho null hypothesis is rejected. 2. If t-test t o t-table t t in significant degree of 0.05, Ho null hypothesis is accepted. In addition, the writer found To is 2.13 while t table is 1.999. After comparing between To and t table the writer found To is higher than t table. Therefore, it can be concluded that Ho is rejected. It means there is significance score gain between students‟ vocabulary using crossword puzzle and without crossword puzzle.

D. Interpretation of the Data and Discussion

In the description of the data which was taken from 32 students of experimental class, the writer could explain briefly about the data got from the students before they were analyzed. The description of the experimental class has the mean of pre-test 58.71 before using crossword puzzle. It means the mean score is bad because it is lower than the standard minimum KKM. After giving 4 times treatments for experimental class using crossword puzzle, the writer got the mean of post-test 74.56. So, the writer got the mean of the gain score 21.03. It is good because the mean score has reached KKM. The smallest score in the pre-test was 50 and the highest score was 76. The data showed in the post-test that the smallest score was 70 and the highest score was 94. It can be summarized that the lowest and the highest scores in post-test were higher than pre-test. Meanwhile, from the description of score in controlled class which was the writer got the mean of pre-test 61.40. It means the mean score is bad. In this class, the writer did not use crossword puzzle, but the writer only asked the students to memorize the vocabulary. After giving 4 times treatment without crossword puzzle, the writer got the mean of post-test 71.25. it is bad because the mean score is still lower than KKM. The writer got the mean of