36 72
72 37
68 72
4 38
64 64
39 76
68 -8
40 76
80 4
∑
2736 2964
228
̅ Mean
68.4 74.1
5.7
The data showed the score in controlled class from 40 students in the class, the mean of pre-test gained was 68.4 and the mean of post-test was 74.1. The
mean of gained score was 5.7. The smallest score in the pre-test was 40 and the highest score was 88. After the application of Grammar Translation Method as a
as a treatment given in teaching reading narrative text, the students took the post- test. The data showed in post-test that the smallest score was 60 and the highest
score was 88.
2. Analysis of Pre-test and Post-test
1. Normality of the Data Before analyzing the hypothesis, the normality of the data should be
analyzed as well. This analysis was used to see whether the data got in the research has been normally distributed or not. Lyllifors formula was used 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 could 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
a. Normality of Experimental Class 1. Normality of Pre-test
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 showed that T
max
T
table
0.123 0.161. Conclusion: In the significant degree of 0.05, H
o
was accepted. It means that the data was normally distributed.
2. Normality of Post-test 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 showed that T
max
T
table
0.098 0.161. Conclusion: In the significant degree of 0.05, H
o
is accepted. It means that the data is normally distributed.
b. Normality of Controlled Class
1. Normality of Pre-test 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 closer value to n = 40 that is n = 30
H
1
: T 0.161 H
o
: T 0.161 The result showed that T
max
T
table
0.107 0.161. Conclusion: In the significant degree of 0.05, H
o
is accepted. It means that the data is normally distributed.
2. Normality of Post-test 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 closer value to n = 40 that is n = 30
H
1
: T 0.161 H
o
: T 0.161 The result showed that T
max
T
table
0.135 0.161.
Conclusion: In the significant degree of 0.05, H
o
was accepted. It means that the data was normally distributed.
2. Homogenity of the Data Based on the calculation of normality, the result that all data in pre-test
and post-test of both experimental class and controlled class have been distributed normally. The next step of the calculation was finding the homogenity of the data.
The purpose of this calculation was to see whether the datasample in both classes was homogenous or heterogenous.
Hypothesis: H
o
: The condition of experimental class is not different from controlled class. H
1
: The sample of experimental 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 could be seen as follows:
or
The calculation could be seen as follows:
n1-1 = 40-1 = 39 n2-1 = 40-1 = 39
F
0.05n1-1, n2-1
= F
table
F
0.05n1-1, n2-1
= 1.84 F
table
From the calculation, it could be seen that F F
αn1-1, n2-2
0.29 1.84 . Based on the criteria, it could be concluded that H
o
was accepted. It means that the sample in experimental class and controlled class were homogenous.
B. Hypothesis Testing
