Y Pre-test
Post-test Gain Score
20 56
64 8
21 76
68 -8
22 72
76 4
23 64
64 24
76 72
-4 25
60 64
4 26
72 76
4 27
76 80
4 28
52 60
8 29
76 68
-8 30
68 76
8 31
76 76
32 64
72 8
33 72
80 8
34 68
72 4
35 56
64 8
36 64
76 12
37 60
64 4
38 72
72 39
60 68
8 40
76 76
∑
2668 2788
120
̅ Mean 66.70
69.70 3
From the description of score in controlled class above, it could be seen that from 40 students in the class, the mean of pre-test was 66.80 and the mean of
post-test was 69.85. The writer got the mean of gain score was 3.05. The smallest score in the pre-test was 52 and the highest score was 76. After the writer giving
the treatment without using English teens magazine, the writer gave the students post-test. The data showed in post-test that the smallest score was 56
and the highest score was 80. It can be summerized that the lowest and the highest score
in post-test were also higher than pre-test.
B. Analysis of Pre-test and Post-test
1. Normality of the Data Before analyzing the hypothesis, the writer had to analyze the normality of
the data. This analysis is used to see whether the data got in the research has been normally distributed or not. The writer used the Lyllifors formula to test the
normality. In this formula, the data was transformed into the basic value. The maximum dispute T got from the calculation must be in absolute value +. The
result of normality can be seen by comparing the value of T
max
to T
table
. The criteria of hypothesis is :
H
1
: T T
table
H
o
: T T
table
Hypothesis : H
o
: Data of X is normally distributed. H
1 :
Data of X is not normally distributed. Criteria of the test :
In the sigificant degree of 0.05, the value in the table of Lillyfors shows: T
0.0530
= 0.161 Because n = 30 is not mentioned in the table of Lillyfors, the writer used the closer value to n = 40 that is n = 30
H
1
: T 0.161 H
o
: T 0.161 The result of pre-test normality in experiment class showed that T
max
T
table
0.0985 0.161. Conclusion : In the significant degree of 0.05, H
o
is accepted. It means that the data is normally distributed. The result of post-test
normality in experiment class showed that T
max
T
table
0.1108 0.161. Conclusion : In the significant degree of 0.05, H
o
is accepted. It means that the data is normally distributed.
The result of pre-test normality in controlled class showed that T
max
T
table
0.1177 0.161. Conclusion : In the significant degree of 0.05, H
o
is accepted. It means that the data is normally distributed. The result of post-test
normality in controlled class showed that T
max
T
table
0.1352 0.161.
Conclusion : In the significant degree of 0.05, H
o
is accepted. It means that the data is normally distributed.
2. Homogenity of the Data Based on the calculation of normality, the writer got the result that all data
in pre-test and post-test of both experiment class and controlled class have been distributed normaly. The next step of the calculation was finding the homogenity
of the data. The purpose of this calculation was to see whether the data sample in both classes was homogenous or heterogenous.
Hypothesis: H
o
: The condition of experiment class is not different from controlled class. H
1
: The sample of experiment class is different from controlled class. The criteria of the test:
α = 0.05 H
o
: F
αn1-1, n2-2
F F
αn1-1, n2-2
H1: F F
αn1-1, n2-2
The formula used can be seen as follows:
or
The calculation can be seen as follows:
n1-1 = 40-1 = 39 n2-1 = 40-1 = 39
F
0.05n1-1, n2-1
= 1.84 F
table
From the calculation, it can be seen that F F
αn1-1, n2-2
1.105 1.84 . Based on the criteria, it can be conclude that H
o
is accepted. It means that the sample in experiment class and controlled class were homogenous.
