∑ Y : the number of even items
N : the number of the respondents
From the calculation, it was found that the reliability coefficient of a half of test items was 0.97. In order to find the reliability coefficient of the whole test, the
following calculation was done.
= 2 0.97
1 + 0.97 =
1.94 1.97
= 0.98 notes:
r
11
: reliability coefficient for the whole test items r
xy :
reliability coefficient for the half of the test items According to Saukah et.al 1997:210, a teacher made-test is considered adequate if it
has the reliability coefficient of 0.50. Therefore, it could be said that the test used in this research reliable because the reliability coefficient of the test was 0.98 which was
higher than 0.50.
4.5 The Result of Primary Data
The primary data were gained by administering a vocabulary achievement test as a 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 April 6
th
, 2015 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 30
test items in the form of multiple choice.
The results of the vocabulary post test were analyzed statistically by using t- test formula to know whether the mean difference between the experimental and the
control groups was significant or not. The calculation was done as the following. 1. The mean score of the experimental group
M
x
= =
.
= 88.4 2. The mean score of the control group
M
y
= =
.
= 82.6 3. The individual score deviation square of M
x
x = X
X N
= 266750.6 3006.3
34 = 266750.6
9037839.69 34
= 266750.6 265818.8
= 931.8 4. The individual score deviation square of M
y
y = Y
Y N
= 226871.37 2725.8
33 = 226871.37
7429985.64 33
= 226871.37 225151.08
= 1720.29
5. The calculation of t-test of students’ vocabulary achievement =
+ +
2 1
+ 1
= 88.4
82.6 931.8 + 1720.29
34 + 33 2
1 34 +
1 33
= 5.8
2652.09 65
[0.029 + 0.030]
= 5.8
[40.8][0.056]
= 5.8
2.28
= 5.8
1.5 = 3.866
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 = 34 + 33 – 2
= 65 From the calculation above, it was found that the value of t was 3.866. Then,
it was consulted to the t-table with significant level of 5 in two tails. The value of t- table of 5 with Df=65 in two tails was
. Therefore the value of t-computation was higher than the value of the t-table 3.866 2.000.
4.6 The Hypothesis Verification
