pre- test up to students’ average score in cycle 1 and cycle 2. In analyzing that is
phase, the writer uses the formula:
4
P : percentage of students’ improvement y : pre-test result
y1 : post-test 1
P : percentage of students’ improvement y : pre-test result
y2 : post-test 2.
II. FINDING RESEACRH
A. Data Description
The writer put the result in the result of the data including the pre-test, post- test 1, and post-test 2 into a tables as following:
Table 4.2 The Students’ Vocabulary Score of Pre-test, Post-test 1 and Post-test 2
STUDENTS’ NUMBER
PRE-TEST POST-TEST
POST-TEST 2
1 40
50 70
2 30
55 80
3 35
55 60
4 30
55 75
5 55
75 85
6 25
45 55
7 45
65 80
4
David E. Meltzer, The Relationship between Mathematics Preparation and Conceptual Learning Gain in Physics: A Possible Hidden Variable in Diagnostic Pretest Scoces, Iowa:
Departement of Physics and Astronomy, 2008, p. 3.
y - y1 P =
─── X 100 y
y2 - y P =
─── X 100 y
8 30
55 75
9 25
55 60
10 25
55 60
11 35
40 50
12 35
70 80
13 35
45 50
14 30
35 45
15 30
55 75
16 30
55 75
17 60
70 80
18 30
55 75
19 35
50 55
20 35
50 60
21 40
55 65
22 45
50 60
23 40
75 80
24 30
55 85
25 60
75 85
26 30
55 65
27 35
75 85
28 50
70 75
29 35
55 75
30 45
80 85
31 30
55 80
32 45
55 65
33 45
70 85
34 30
70 85
35 60
65 70
36 30
55 70
Mean :
_
∑x X = ──
N
37.36 58.47
71.11
: The Students who passed the KKM 60
B. The Analysis the Data
In analyzing the data of both pre-test and post-test in each cycle, the writer compared the test result between pre-test and post-test. Before the writer carrying
out the classroom action research CAR, the writer gave students the pre-test, and the mean score of students in the pre-test as followings:
_ ∑x X = ──
n
_ 1345
X = ───
36 _
X = 37,36
Based on that calculation, the mean score of the class in pre-test was 37,36. In the next step is calculating percentage score which passes the Minimum
Mastery Criterion KKM using formula as: F
P = ── X 100
N 3
P = ── X 100 36
300 P = ──
36 P = 8.33
From the calculation of pre-test above, there were three who passed the Minimum Mastery Criterion KKM. The mean of the pre-test is 37,36 and the
percentage of the students who pass the KKM is 8.33. Next, after finding the pre-test score, the writer calculated the result of post-
test 1. At the first cycle of CAR, the writer calculates the mean of students’ score,
the percentage of the students who pass the Minimum Mastery Criterion KKM, and improve
of students’ score in teaching vocabulary from the pre-test to the post-test 1 by using the formula as:
_ ∑x
X = ── n
_ 2105
X = ───
36
_ X = 58.47
Based on the calculation above, the mean of the student s’ score in post-test
1 was 58.4 7. the improvements of the students’ score from pre-test to post-test 1
derived 21.11. and the percentage from pre-test to postt-test 1 is 56.45. to know that improvement into percentage, the writer calculated is as follows:
y1 - y P =
─── X 100 y
58.47 – 37.36
P = ───────── X 100
37.36 21.11
P = ──── X 100
37.36 P = 56.50
The percentage of students who pass the Minimum Mastery Criterion KKM through the formula:
F P = ── X 100
N 12
P = ── X 100 36
P = 33.33
So based on the calculation, the writer knows that the mean of the students at firrst cycle is 58.4
7. Moreover, the improvement of students’ score of vocabulary fromre-test to first post-testis 56.50. Since there are twenty students
who get 60 score and above, it shows that 33.33 students can pass the Minimun Mastery Criterion KKM.
