c. The Result of Post Test.
For the need of the research, the writer had done the trustworthiness of the test by using discriminating power and difficulty
item for pretest, posttest 1, and posttest 2. According to the data, there was found no item would be dropped. It could be seen in the appendix.
Furthermore, the following table illustrated t he data on students’
achievement ’ score of pretest, post test 1, and post test 2
Table: 4.8 The Result of Post Test
Students’ Number
Pretest Cycle 1 Post Test Cycle 2 Post Test
1 70
70 80
2 60
70 90
3 60
75 95
4 70
80 90
5 50
50 60
6 60
65 95
7 40
70 90
8 40
45 60
9 40
45 85
10 20
50 75
11 60
65 75
12 45
65 90
13 35
40 75
14 65
70 85
15 65
70 95
16 85
85 100
17 70
80 100
18 25
75 95
19 65
70 75
20 70
80 100
21 50
75 85
22 60
60 90
23 75
75 80
24 85
85 100
25 50
60 75
26 45
45 50
27 55
60 80
28 70
80 95
29 70
75 95
30 70
85 100
31 60
75 95
Mean: � =
� �
57.58 67.58
85.64
: The students who passed KKM 75
To know the students’ improvement score from pretest to posttest in each cycle, the writer used some steps. The steps are calculating the
students’ mean score of the test, calculating the class percentage, and calculating the students’ improvement score from pretest to posttest1 and 2
into percentage. To analyze the data of pre test, the first step is to get the mean
score of the class. The following is the calculation: X =
�
X = �
= 57.58 From the calculation above, it was known that the mean score of
the class in pretest is 57.58 . In other words, the students’ achievement
score of passive voice present continuous tense before implementing Classroom Action Research CAR is 57.58
The next step is to know the percentage of students’ score who passed the KKM 75. The writer computes as follows:
P =
� �
X 100 P = X100
P = 9.68 From the compu
tation above the students’ score percentage in the pretest was 9.7. It means that the students who pass the KKM are 3
students and the other 28 students were below the KKM 75. Furthermore, in the cycle 1
after getting students’ score in the posttest 1, the writer analyzed the data in order to compare the result
between pretest and posttest 1. There are three steps to know the comparing result of pretest and posttest1. Those are calculating the
students’ mean score of the class, calculating the students’ improvement into percentage, and calculating the class percentage.
The first step was calculating the mean score of posttest 1. It was calculated as follows:
� =
�
� =
� = 67.58
The calculation above shown that the stud ents’ mean score of
posttest 1 was 67.58. It was shown that there were some improvements score from the pretest mean score. It could be seen from the pretest mean
score 57.58 to the mean score of posttest 1 67.58. In other word there was an improvement about 10 67.58
– 57.58. Next, the percentage of students’ improvement score could be
explained from the following computation:
P =
−
X 100
P =
. − .
.
X 100 P =
.
X 100 P = 17.37
Based on the result above, the percentage of students’ improvement score from the pretest to the posttest 1 was 17.37. It means
that the score in the cycle 1 improved about 17.37 from the pretest score. After that, the writer would like to know the percentage of students
who passed the KKM. It used the calculation as following: P =
� �
X 100 P = X100
P = 38.71 According to the calculation, the class percentage which passed the
KKM in the posttest 1 was 38.71. In the other word, there were 12 students who derived the KKM and the other 19 students were out of
target. The class percentage of posttest 1 saw some improvements of the class percentage in the pretest 9.68. It could
be concluded the students’ improvement which derived the KKM is 29.03 38.71
– 9.68. In cycle 2, the writer used the same steps to get the mean score of
the class, to get the percentage of the students’ improvement score, and to know the class percentage which derived the KKM 75.
Firstly, to get the mean score of the class, the writer used calculation as follows:
� =
�
� =
�
� = 85.64
From that calculation, the mean score of posttest 2 is 85.64. It means
that there are some students’ improvements scores from the mean score of posttest 1 67.58.
Next, to get the percentage of students’ improvement scores, the following is the calculation:
P =
−
X 100 P =
. − .
.
X 100 P =
. .
X 100 P = 48.73
According to that calculation, It could be said that the posttest 2 improves 48.73 from the pretest and improves 31.36 from the posttest
1 48.73 – 17.37.
Then, to know the percentage of students who passed the KKM could be explained in the following computation:
P =
� �
X 100 P = X100
P = 90.32 From that calculation, the class percentage is 90.32. It means that
in the cycle 2 there were 28 students who pass the KKM and 3 students were below the KKM. Furthermore, the class percentage of posttest 2 saw
the improvements 80.64 90.32 – 9.68 from the percentage of
pretest or 51.61 90.32 – 38.71 from the percentage of posttest 1.
B. Data Interpretation
1. The Interpretation of Interview
In this part the writer would like to interpret the data results from the interview before the action and the interview after implementing the action.
According to the pre interview result with the English teacher of SMAI Al –
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