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Ta ble 4.9. Norma lity Test
No Data
The Number of Sample
Distribution of Sample
1 24
0.0960 0.173
0.05 Normal
2 24
0.1423 0.173
0.05 Normal
3 24
0.117 0.173
0.05 Normal
4 24
0.104 0.173
0.05 Normal
5 12
0.1976 0.243
0.05 Normal
6 12
0.1466 0.243
0.05 Normal
7 12
0.1700 0.243
0.05 Normal
8 12
0.1127 0.243
0.05 Normal
The summary of normality test shows that all the values of
L
o
is lower than
L
t
, so it can be concluded that all data are in normal distribution
2. Homogeneity Test
Homogeneity test is done to know that the data are homogeneous. If , it can be concluded that the data are homogeneous.
Ta ble 4.10. The Summar y of Homogeneity Test
Sampel n-
1 df 1df
s
i 2
Log s
i 2
df log s
i 2
1 11
0.090901 4.20
0.6232 6.8552
2 11
0.090901 10.81
1.0338 11.3718
3 11
0.090901 17.60
1.2455 13.7005
4 11
0.090901 11.36
1.0553 11.6083
44 0.363604
43.5358
= In10
{B-
n
i
1log s
i 2
} =
2. 302645.799 43.5358 = 5.211
Because
x
o 2
5.211 is lower than
x
t 2
0.05
7.81, it can be concluded that the
data are homogeneous.
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C. Hypothesis Testing
ANOVA test Multifactor Analysis of Variance
Before the data are analyzed using ANOVA test, the data are divided into four groups, they are: 1 The data of reading test of the students or the group
having high interest who are taught by using Clustering . 2 The data of
reading test of the students or the group having low interest who are taught by using Clustering
. 3 The data of reading test of the students or the group having high interest who are taught by using Direct Instruction
. 4 The data of reading test of the students or the group having low interest who are
taught by using Direct Instruction .
Ta ble 4.11. Summary of mea n scores
Teaching technique
Total Rows
Clustering A1 A
1
B
1
A
2
B
1
Hig h Inter
est B1
Low Inter
est B 2
A
1
B
2
A
2
B
2
Total Colums
1 ___
X
3 ___
X
1 ___
X
r
2 ___
X
4 ___
X
2 ___
X
r
1 ___
X
c
2 ___
X
c t
___
X
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Ta ble 4.12. The summary of a 2x2 multifa ctor a na lysis of varia nce
Summary SS
df MS
F
o
F
t 0.05
Between columns Clustering 346.82
1 346.82
31.529 4.00
Between rows Learning Interest 63.12
1 63.12
5.738 Columns by rows interaction
50 1
50 4.545
Between groups 459.84
3 153.28
Within groups 484.16
44 11
Total 1403.94
47
Based on the computation result of ANOVA test, it can be concluded that:
1 Because
F
o
between columns
31.529
is bigger that
F
t
at the level of significance Therefore, the null
hypothesis H stating that there is no significant difference in reading
comprehension between the students who are taught by using clustering technique and students who are taught by using direct instruction is rejected. It can be
concluded that teaching reading using Clustering to the students is significantly different from the one using Direct Instruction. The mean score of students taught
using Clustering
25.923
is higher than the one of those taught using Direct Instruction
20.962
. So, teaching reading using clustering is more effective than the one using direct instruction.
2 Because
F
o
between rows
5.738
is bigger that
F
t
0.05 4.00, the difference between rows is significant. Therefore, the null hypothesis H
stating that there is no significant difference in reading comprehension between the students who have low level of interest and students
who have high level of interest is rejected. It can be concluded that students having high learning interest taught using Clustering is significantly different
from those having low learning interest. The mean score of students having high learning interest
26.542 is higher than the one of those of having low learning
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interest
24.250
. So, reading achievement of the students having high interest is better than the one of those having low interest.
3 Because
F
o
interaction
4.545
is bigger than
F
t
0.05 4.00. Therefore, the null hypothesis H stating that there is no interaction
between teaching techniques interest in reading comprehension is
rejected. ing
comprehension depends on the student .
From the hypothesis testing above can be seen that there is interaction effect between the two variables, teaching technique and interest, so calculation
must be continued to Tukey test. It shows as follows:
1. Clustering compared with Direct Instruction A
