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87
B. Normality and Homogeneity test
Before analyzing the data for testing hypothesis, analyzing the normality and homogeneity test must be done. The normality test is to know the sample is
in normal distribution and the homogeneity test is to know the data are homogenous. Each test is presented as follows.
1. Normality Test
The sample is on normal distribution if L obtained is lower than
a. Normality test of scores of the students who are taught using clustering A1.
Based on the calculation, the highest values of FZ
i
- SF
i
is 0.0960, or
L
o
is 0.0960 and
L
t
= 0.173. From the table of critical value of Lilliefors test with
N=24 at the significant level = 0.05, the score of
L
t
is 0.173. Because
L
o
is lower than
L
t
or
L
o
0.0960
L
t
0.173, it can be concluded that the data is in normal distribution.
b. Normality test of scores of the students who are taught using direct
instruction A2.
Based on the calculation, the highest values of FZ
i
- SF
i
is 0.1423, or
L
o
is 0.1423 and
L
t
= 0.173. From the table of critical value of Lilliefors test with
N=24 at the significant level = 0.05, the score of
L
t
is 0.173. Because Lo is lower than
L
t
or
L
o
0.1423
L
t
0.173, it can be concluded that the data is in normal distribution.
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c. Normality test of scores of the students who have high interest who are
taught using clustering and direct instruction B1.
Based on the calculation, the highest values of FZ
i
- SF
i
is 0.117, or
L
o
is 0.117 and
L
t
= 0.173. From the table of critical value of Lilliefors test with
N=24 at the significant level = 0.05, the score of
L
t
is 0.173. Because
L
o
is lower than
L
t
or
L
o
0.117
L
t
0.173, it can be concluded that the data is in normal distribution.
d. Normality test of scores of the students who have low interest who are taught
using clustering and direct instruction B2.
Based on the calculation, the highest values of FZ
i
- SF
i
is 0.104, or
L
o
is 0.104 and
L
t
= 0.173. From the table of critical value of Lilliefors test with
N=24 at the significant level = 0.05, the score of
L
t
is 0.173. Because
L
o
is lower than
L
t
or
L
o
0.104
L
t
0.173, it can be concluded that the data is in normal distribution.
e. Normality test of scores of the students who have high interest who are
taught using clustering A
1
B
1
. Based on the calculation, the highest values of FZ
i
- SF
i
is 0.1976, or
L
o
is 0.1976 and
L
t
= 0.243. From the table of critical value of Lilliefors test with
N=12 at the significant level = 0.05, the score of
L
t
is 0.243. Because
L
o
is lower than
L
t
or
L
o
0.1976
L
t
0.243, it can be concluded that the data is in normal distribution.
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89
f. Normality test of scores of the students who have low interest who are taught
using clustering A
1
B
2
. Based on the calculation, the highest values of FZ
i
- SF
i
is 0.1466, or
L
o
is 0.1466 and
L
t
= 0.243. From the table of critical value of Lilliefors test with
N=12 at the significant level = 0.05, the score of
L
t
is 0.243. Because
L
o
is lower than
L
t
or
L
o
0.1466
L
t
0.243, it can be concluded that the data is in normal distribution.
g. Normality test of scores of the students who high interest who are taught
using direct instruction A
2
B
1
. Based on the calculation, the highest values of FZ
i
- SF
i
is 0.1700, or
L
o
is 0.1700 and
L
t
= 0.243. From the table of critical value of Lilliefors test with
N=12 at the significant level = 0.05, the score of
L
t
is 0.243. Because
L
o
is lower than
L
t
or
L
o
0.1700
L
t
0.243, it can be concluded that the data is in normal distribution.
h. Normality test of scores of the students who have low interest who are taught
direct instruction A
2
B
2
. Based on the calculation, the highest values of FZ
i
- SF
i
is 0.1127, or
L
o
is 0.1127 and
L
t
= 0.243. From the table of critical value of Lilliefors test with
N=12 at the significant level = 0.05, the score of
L
t
is 0.243. Because
L
o
is lower than
L
t
or
L
o
0.1147
L
t
0.243, it can be concluded that the data is in normal distribution.
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90
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
Kategori