53
: The student who passed the KKM 70 41
72 75
85 42
48 55
65 43
60 65
75 44
60 75
80 45
52 70
85 46
48 60
75 47
44 50
60 48
48 60
65 49
60 65
70
52.81 64.70
75.71 X =
∑ x n
M ean :
54
Figure 2.2
In analyzing numerical data, the writer compared the test result between pretest and post test of each cycle. Next, the writer gave students the pretest which it
had been done before CAR. The mean score of the pretest was calculated such following:
X
=
∑ x n
X =
2588 49
X
=
52.81
55
Based on that calculation, the mean score of the class in pretest was 52.81. On the other case, to know the class percentage whose passed the KKM using the
formula as:
P = 6.12
From that calculation, there were three students who passed the KKM, so after dividing with the number of students in the class and altering that into percentage, it
could be derived about 6.12 students whom passed the KKM. Next, after scoring the pretest 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 100
P =
F N
X 100
P =
3 49
X
=
∑ x n
X
=
3170 49
X =
64.70
56
It was known that the mean score of the class in the post test 1 derived 64.70. It gained any improvement 11.89 from the pretest or having 22.51 from the pretest
to the posttest 1 result. To know that improvement into percentage, the writer calculated as following:
P = 22.5
P =
22.51
In the 1
st
cycle of post test 1, there were 19 students who passed the KKM. If it was calculated into class percentage, it was derived 38.77 through the formula:
P = 38.77
P =
y1 - y y
X 100
64.70 – 52.81 52.81
P =
X 100
11.89 52.81
P =
X 100
X 100
P =
F N
X 100
P =
19 49
57
In the end of cycle two, the mean of students’ score in vocabulary post test 2 gained 75.71. It was derived from:
To know the improvement from the pretest to post test 2 into percentage, after getting the mean score 75.71, the writer made a percentage calculation as following:
P = 43.36
Based on that computation, it could be seen that the post test 2 had 43.36 improvement from the pretest or 20.85 43.36 – 22.51 improvement from the
X
=
∑ x n
X
=
3710 49
X =
75.71
P =
y2 - y y
X 100
P =
75, 71 – 52,81 52,81
X 100
P =
22,90 52,81
X 100
58
pretest 1. Meanwhile, to know the percentage of the class that passed the KKM, it could be calculated as following:
P = 69.38
In the end of cycle two, the result of the post test showed that there were 34 students or 69.38 who passed the KKM. It improved from the pretest which gained
only 6.12 and in the pretest 1 which had any improvement become 38.77. Therefore, based on the class percentage result from the pretest to the post test 2 in
the second cycle improved about 63.26 69.38 – 6.12. It proved that the target of CAR success in which minimum 75 students passed the KKM could be achieved.
D. Interpretation of Test Result
