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80 Table 13. Frequency Distribution of A
2
B
2
Class Limit f
i
X
i
f
i
X
i
50 - 54 2
52 104
55 - 59 4
57 228
60 - 64 6
62 372
65 - 69 4
67 268
70 - 74 2
72 144
Sum 18 310
1116
Figure 10. Histogram and Polygon Data A
2
B
2
B. Normality and Homogeneity Test
Before analyzing the data using inferential analysis, normality and homogeneity test must be done. The normality test is to know that the sample
is in normal distribution and the homogeneity test is to know that the data are homogenous. Each test is presented in the following section.
1. Normality Test The sample is in normal distribution if L
L obtained is lower than L
t
L table, α = 0.05. L stands for Lilliefors.
2 4
6
2 4
1 2
3 4
5 6
7 8
9
Interval
F requ
enc y
49.5 54.5 59.5 64.5 69.5 74.5
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81 Table 14. The Summary of Normality Test using Lilliefors
No. Data
The Number
of Sample
L Obtained Lo
L Table Lt
Alfa α
Distribution of
Population 1
2 3
4 5
6 7
8 A
1
A
2
B
1
B
2
A
1
B
1
A
1
B
2
A
2
B
1
A
2
B
2
36 36
36 36
18 18
18 18
0.146 0.095
0.091 0.071
0.147 0.129
0.097 0.114
0.148 0.148
0.148 0.148
0.200 0.200
0.200 0.200
0.05 0.05
0.05 0.05
0.05 0.05
0.05 0.05
Normal Normal
Normal Normal
Normal Normal
Normal Normal
2. Homogeneity Test Homogeneity test is done to know that the data are homogeneous. If
χ
o 2
is lower than χ
t 2
0.05
, it can be concluded that the data are homogeneous.
Table 15. The Homogeneity Test Sample df
1df s
1 2
log s
1 2
df log s
1 2
1 2
3 4
17 17
17 17
0.059 0.059
0.059 0.059
18.30 55.91
40.41 39.32
1.262 1.748
1.606 1.595
21.462 29.708
27.080 27.109
68 0.235 105.589
χ
2
= 2,303{B – ΣlogS
i
x n-1} = 2,303107.801 – 105.589
= 5.09
Based on the result of calculation above, it can be seen that the χ
2
5.09 is lower than
χ
t
at the level of significance α 5 = 7.81. So χ
2
χ
t
5.09 7.81 and the data are homogeneous.
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82
C. Hypothesis Test
Hypothesis test can be done after the result of normality and homogeneity test are fulfilled. The test is done by using multifactor analysis of
variance 2 x 2. H
o
is rejected if F
o
F
t
meaning that there are a difference and an interaction. If H
o
is rejected the analysis is continued to know which group is better using Tukey test. The multifactor analysis of variance 2 x 2 and
Tukey test are described as below. 1. Summary of a 2 x 2 Multifactor Analysis of Variance
Table 16. Multifactor Analysis of Variance Source of Variance
SS df
MS F
o
F
t 0,05
Between columns 284.014
1 284.014
7.38 4.00
Between rows 1596.125
1 1596.125
41.47 4.00
Columns by rows interaction
3916.125 1 3916.125
101.75 4.00 Between groups
5796.264 3
1932.088 -
- Within group
2617.056 68
38.486 -
- Total 8413.319
71 -
- -
The table above shows that: a. Because F
o
between columns 7.38 is higher than F
t
at the level of significance
α = 0.05 4.00 and α = 0.01 7.08, the difference between columns is significant. It can be concluded that the methods
of teaching differ significantly from one another in their effect on the reading comprehension of the subject in the experiment.
b. Because F
o
between rows 41.47 is higher than F
t
at the level of significance
α = 0.05 4.00 and α = 0.01 7.08, the difference between rows is significant. It can be concluded that the students
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83 having high intelligence and those having low intelligence are
significantly different in their reading comprehension. c. Because F
o
interaction between group 101.75 is higher than F
t
at the level of significance
α = 0.05 4.00 and α = 0.01 7.08, there is interaction between the two variables, teaching methods and
intelligence. It means the effect of teaching methods depends on the degree of intelligence to teach reading.
2. Tukey Test The finding of q is found by dividing the difference between the
means by the square root of the ratio of the within group variation and the sample size.
q = n
s
w
x -
x
2 j
i
Data
Teaching Method
Intelligence
Contextual Teaching and Learning A
1
Grammar Translation Method A
2
Sum Group 1
Group 2 High B
1
Data = 18 Data = 18
Data = 36 ΣX = 1354 X = 75.2 ΣX = 1017 X = 56.5 ΣX = 2371 X = 65.9
Group 3 Group 4
Low B
2
Data = 18 Data = 18
Data = 36 ΣX = 919 X = 51.1 ΣX = 1113 X = 61.8 ΣX = 2031 X = 56.4
Total Data = 36
Data = 36 Data = 72
ΣX = 2273 X = 63.1 ΣX = 2130 X = 59.2 ΣX = 4403 X = 61.2 a.
Between columns q
A
= n
ance error vari
X -
X
2 1
A A
= 36
38.486 2
. 59
1 .
63 −
= 03
. 1
9 .
3 = 3.84
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84 Because between columns q
A
=3.84 is higher than q
t
at the level of significance
α = 0.05 2.86, contextual teaching and learning method differs significantly from grammar translation method for
teaching reading. The mean score of the students who are taught using CTL A
1
63.1 is higher than A
2
59.2, so it can be concluded that contextual teaching and learning method is more effective to teach
reading than grammar translation method. b. Between rows
q
B
= n
ance error vari
X -
X
2 1
B B
= 36
38.486 4
. 56
9 .
65 −
= 03
. 1
4 .
9 = 9.11
Because between rows q
B
= 9.11 is higher than q
t
at the level of significance
α = 0.05 2.86, the students having high intelligence differs significantly from the students having low intelligence. The
mean score of the students having high intelligence B
1
65.9 is higher than B
2
56.4, so it can be concluded that the students having high intelligence have better reading comprehension than those having
low intelligence. c.
Between cells A
1
B
1
and A
2
B
1
q
HI
= n
ance error vari
X -
X
1 2
1 1
B A
B A
= 18
38.486 5
. 56
2 .
75 −
= 46
. 1
7 .
18 = 12.8
Because between cells A
1
B
1
and A
2
B
1
= 12.8 is higher than q
t
at the level of significance
α = 0.05 2.97, contextual teaching and
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85 learning method differs significantly from grammar translation method
for students having high intelligence. The mean score of the students having high intelligence who are taught using CTL A
1
B
1
75.2 is higher than A
2
B
1
56.5, so it can be concluded that contextual teaching and learning method is more effective than grammar
translation method to teach reading for students having high intelligence
d. Between cells A
1
B
2
and A
2
B
2
q
LI
= n
ance error vari
X -
X
2 2
2 1
B A
B A
= 18
38.486 8
. 61
1 .
51 −
= 46
. 1
3 .
5 =
3.65 Because between cells q
LI
= 3.65 is higher than q
t
at the level of significance
α = 0.05 2.97, grammar translation method differs significantly from contextual teaching and learning method for
students having low intelligence. The mean score of the students having low intelligence who are taught using GTM A
2
B
2
61.8 is higher than A
1
B
2
51.1, so it can be concluded that grammar translation method is more effective than contextual teaching and
learning for students having low intelligence. e. Based on the result of analysis on the above points c and d, that is
contextual teaching and learning method is more effective than grammar translation method for the students having high intelligence
and grammar translation method is more effective than contextual teaching and learning method for the students having low intelligence,
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86 so it can be summarized that there is an interaction between teaching
methods and students’ intelligence in teaching reading comprehension.
D. Discussion of the Result of the Study
