According to Saukah et.al 1997:210, a teacher made-test is considered adequate if it has a reliability coefficient of 0.50. It showed that the test items were reliable as well
because the reliability coefficient of the test items was 0.56 which was higher than 0.50. From the result of the try out test analysis, there were three test items which
were indicated as not fair. There were 43 test items that were categorized as fair. Therefore, the test items still could be used for the vocabulary post test since the
number of the test items in the vocabulary post test was 40 items.
4.5 The Result of Primary Data
The primary data were gained from the students’ Vocabulary achievement in the form of the vocabulary post test scores.
4.5.1 The Results of Post Test The vocabulary post test was done to both groups of the experimental group
and the control group. The vocabulary test was administered on December 4
th
, 2014 after the groups were taught twice by using different treatments. The scores of the
vocabulary post test were used to investigate the significant difference between the experimental group and the control group. The vocabulary post test consisted of 40
test items in the form of multiple choices.
4.5.2 The Analysis of the Vocabulary Post Test Scores The results of the vocabulary post test were analyzed statistically by using the
independent sample t-test formula to know whether the mean difference between the experimental and the control groups was significant or not.
1. The mean score of the experimental group M
x
=
�� ��
=
3340 38
= 87.89 2. The mean score of the control group
M
y
=
�� ��
=
3822.5 46
= 83.09 3. The individual score deviation square of M
x
� x
2
= � X
2
− ∑ X
2
N
x
= 294837.5 −
3340
2
38 = 294837.5
− 11155600
38 = 294837.5
− 293568.45 = 1269.05
4. The individual score deviation square of M
y
� y
2
= � Y
2
− ∑ Y
2
N
y
= 319556.3 −
3822.5
2
46 = 319556.3
− 14611506.25
46 = 319556.3
− 317641.4 = 1914.9
5. The calculation of t-test of students’ reading comprehension achievement � =
�� − �� �� ∑�
2
+ ∑�
2
�
�
+ �
�
− 2� � 1
�
�
+ 1
�
�
�
� = 87.89
− 83.09 ��
1269.05 + 1914.9 38 + 46
− 2 � � 1
38 +
1 46�
� = 4.8
�� 3183.95
82 � [0.026 + 0.021]
� = 4.8
�[38.82][0.047]
� = 4.8
√1.82
� = 4.8
1.34 = 3.582
notes: t
: the value of t Mx
: the mean score of the experimental group My
: the mean score of the control group x
: individual score deviation of the experimental group y
: individual score deviation of the control group Nx
: the number of respondents in the experimental group Ny
: the number of respondents in the control group Arikunto, 2006:311
6. The calculation of the degree of freedom Df = N
x
+ N
y
– 2 = 38 + 46 – 2
= 82
From the calculation above, it was found that the value of t was 3.582. Then, it was consulted to the t-table of 5 of the significance level and Df=82. The value of
t-table of 5 with Df=81 was 1.989. The value of t-test was higher than the value of t-table 3.582 1.989. It means that there was a significant mean difference between
the experimental group and the control group.
4.6 The Hypothesis Verification
