Based on the data above, the writer gave bold numerical score to students who passed KKM 70,
it is showed there were five students who passed KKM in pretest, twenty one students in posttest I and twenty nine students in posttest II.
The writer also concluded the lowest score in pretest was 35, in posttest I was 50 and posttest II was 60.
To know the result among pretest, posttest I and posttest II, the writer calculated the student mean of the scores, calculated the class percentage and also
calculated the percentage of the achievement score from pretest, posttest I and posttest II.
20 55
60 65
21
75 80
80
22 70
75 85
23 55
75 80
24 50
70 75
25 60
75 90
26 55
65 70
27 45
65
70
28 60
75 85
29 55
75 70
30 50
65 75
31 65
75 90
32 65
65 75
33 55
70 75
34 50
65 80
� =
�
Mean: 55.29
69.26 75.15
The mean score of the pretest was computed such following:
_ ∑x
X = ──
n _
1880 X =
─── 34
_ X = 55.29
Based on that computation, it is showed the mean score of the class in pretest before implementation the action is 55.29. Then to know the percentage of
students’ score who passed the criterion of minimum completeness, the writer used the formula:
F P =
── X 100 N
5 P =
── X 100 34
P = 14.70
From the calculation above, it is known the students’ percentage score is
14.70 . It means there are five students, who passed the KKM of English Lesson
and students who got score below the target of KKM are twenty nine students. Next, after scoring the pretest before implementation, the writer
calculated the result of posttest 1. It was to know the improvement from the pretest to posttest 1 result. However to measure that improvement, it was needed
to know the mean score of the class by using the formula as:
_ ∑x
X = ──
n _
2355 X =
─── 34
_ X = 69.26
From that calculation, the students’ mean score of posttest I in cycle 1 is 69.26. It proves that there are some improvements from the pretest mean score. It
could be seen from the pretest mean score 55.29 to the mean score of posttest 1 69.26. It improves 13.97 69.26
– 55.29. The second step is to get the percentage of students’ improvement score
from pretest to posttest 1. The writer computes by using as follows:
y1 - y P =
─── X 100 y
69.26 – 55.29
P = ───────── X 100
55.29 13.97
P = ──── X 100
55.29 P = 25.26
Based on that computation, the percentage of the students’ improvement score from pretest to posttest 1 is 25.26. It shows that the score in the cycle 1
has improved 25.26 from the pretest score.
The third step is to know the percentage of students who pass the KKM. The calculation by using as follow:
