H : The condition of experiment class is not different from controlled
class H
1
: The sample of experiment class is differentfrom controlled class. Based on the criteria, it can be concluded that H
is accepted. It means that the sample in experiment class and controlled clss were homogenous.
Moreover, the writer also used SPSS to calculated the homogeneity of the data. The result that the writer got can be seen on the table below:
Table 4.9 Test of Homogeneity of Variance
Test of Homogeneity of Variances
pretest Levene Statistic
df1 df2
Sig. ,130
1 62
,720
Based on the table above, it can be seen that the result of homogeneity test Lavene Statistic score shows 0.130 with the significant 0.720.
Therefore, the significant score is higher than 0.05 0.720 0.05. it means that the sample in experiment class and controlled class were homogenous.
Table 4.10 ANOVA Test
ANOVA
pretest Sum of Squares
df Mean Square
F Sig.
Between Groups 58,141
1 58,141 ,973
,328 Within Groups
3704,594 62
59,752 Total
3762,734 63
Based on the table above, F score from the result of calculation is 0.973 with the significant score 0.328. the writer found H
is accepted from
the comparison between F
o
F
observation
and F
table
which shows F
table
1.99 is higher than F
o
0.973. Therefore, the writer concluded that the data is homogenous.
C. Test of Hypothesis T-test
In this part, the researcher calculated the data to test hypohesis that whether there is significant different between stuents‟ vocabulary
understanding using crossword puzzle in experiment class and students‟
vocabulary understanding without crossword puzzle in controlled class. the researcher calculated the data using T-test formula. Two classes were
compared, the experimental class was X variable and the controlled class was Y variable. The next table is staistical calculation of the gain score both
experimental class using crossword puzzle and controlled class without crossword puzzle.
Table 4.11 The Statistical Calculaion of the Gain Score of Both
the Controlled and the Experimental Class
Students X
Y X=Mx-
X Y=My-
Y X
2
Y
2
1
17
7 -3,59
-6,38 289,00
289,00 2
26 29
26,00 29,00
676,00 676,00
3 33
9 11,00
9,00 1089,00
1089,00 4
17 24
17,00 24,00
289,00 289,00
5 12
29 12,00
29,00 144,00
144,00 6
10 26
10,00 26,00
100,00 100,00
7 26
3 26,00
3,00 676,00
676,00 8
34 6
34,00 6,00
1156,00 1156,00
9 9
9 9,00
9,00 81,00
81,00 10
30 6
30,00 6,00
900.00 900,00
11 22
20 22,00
20,00 484,00
484,00 12
20 14
20,00 14,00
400,00 400,00
13 11
17 11,00
17,00 121,00
121,00
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
