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B. Normality and Homogeneity Test
1. Normality Test a. Normality test of scores of the students who were taught using process
approach A
Based on the result of the calculation of the students’ scores who were taught using the process approach, the highest value of L
o
L obtained was 0.1234. From the table of critical value of Liliefors test
with the student’s number n = 28 at the significance level α = 0.05,
the score of L
t
was 0.161. Because L
o
was lower than L
t
or L
o
0.1234 L
t
0.161, it can be concluded that the data were in normal distribution.
b. Normality test of scores of the students who were taught using product approach A
Based on the result of the calculation of the students’ scores who
were taught using product approach, the highest value of L
o
L obtained was 0.1311. From the table of critical value of Liliefors test
with the student’s number n = 28 at the significance level α = 0.05,
the score of L
t
was 0.161. Because L
o
was lower than L
t
or L
o
0.1311 Lt 0.161, it can be concluded that the data were in normal distribution.
c. Normality test of scores of the students having high creativity B
Based on the result of the calculation of the students’ scores having high creativity, the highest value of L
o
L obtained is 0.1503.
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From the table of critical value of Liliefors test with the student’s number N = 28 at the significance
level α = 0.05, the score of L
t
was 0.161. Because L
o
was lower than L
t
or L
o
0.1503 Lt 0.161, it can be concluded that the data were in normal distribution.
d. Normality test of scores of the students having low creativity B
Based on the result of the calculation of the students’ scores of the students having low creativity, the highest value of L
o
L obtained was 0.1245. From the table of critical value of Liliefors test with the
student’s number n = 28 at the significance level α = 0.05, the score
of L
t
was 0.161 because L
o
is lower than L
t
or L
o
0.1245 L
t
0.161, it can be concluded that the data were in normal distribution. e. Normality test of scores of the students having high creativity who
were taught using process approach A B
Based on the result of the calculation of the students’ scores having high creativity who were taught using process approach, the
highest value of L
o
L obtained was 0.1251. From the table of critical value of Liliefors test with the student’s number n = 14 at the
significance level α = 0.05, the score of L
t
was 0.227. Because L
o
was lower than L
t
or L
o
0.1251 L
t
0.227, it can be concluded that the data were in normal distribution.
f. Normality test of scores of the students having low creativity who
were taught using process approach A B
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Based on the result of the calculation of the students’ scores having low creativity who were taught using process approach, the
highest value of L
o
L obtained was 0.0636. From the table of critical value of Liliefors test with the student’s number n = 14 at the
significance level α = 0.05, the score of L
t
was 0.227. Because L
o
was lower than L
t
or L
o
0.0636 L
t
0.227, it can be concluded that the data were in normal distribution.
g. Normality test of scores of the students having high creativity who were taught using product approach A
B Based on the result of the calculation of the students’ scores
having high creativity who were taught using product approach, the highest value of L
o
L obtained was 0.1157. From the table of critical value of Liliefors test with the student’s number N = 14 at the
significance level α = 0.05, the score of L
t
was 0.227. Because L
o
is lower than L
t
or L
o
0.1157 L
t
0.227, it can be concluded that the data were in normal distribution.
h. Normality test of scores of the students having low creativity who were taught using product approach A
B Based on the result of the calculation of the students’ scores
having low creativity who were taught using product approach, the highest value of L
o
L obtained was 0.1429. From the table of critical value of Liliefors test with the student’s number N = 14 at the
significance level α = 0.05, the score of L
t
was 0.227. Because L
o
was
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lower than L
t
or L
o
0.1429 L
t
0.227, it can be concluded that the data were in normal distribution.
Table 4.9 Summary of Normality Test No.
Data No
of Sample
L
o
L
t
α Status
1. A
1
28 0.1234
0.161 0.05
Normal 2.
A
2
28 0.1311
0.161 0.05
Normal 3.
B
1
28 0.1503
0.161 0.05
Normal 4.
B2 28
0.1245 0.161
0.05 Normal
5. A
1
B
1
14 0.1251
0.227 0.05
Normal 6.
A
1
B
2
14 0.0636
0.227 0.05
Normal 7.
A
2
B
1
14 0.1157
0.227 0.05
Normal 8.
A
2
B
2
14 0.1429
0.227 0.05
Normal
2. Homogeneity test Homogeneity test is conducted to know whether data are
homogeneous or not. The data can be said as homogeneous if χ
² is lower than
χ
t
²
0.05.
The result of the analysis is as follows:
Table 4.10 Summary of Homogeneity Test
Sample df
1df s
² log s
² df log s
²
1 13
0.0769231 61.1428571
1.786345729 23.22249448
2 13
0.0769231 37.7582418
1.577011763 20.50115292
3 13
0.0769231 47.1703297
1.673668912 21.75769585
4 13
0.0769231 140.884615
2.148863571 27.93522642
∑ 52
93.41656966
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B log s² =
log 71.739² = 1.855755
B = log s² ∑ ni-1 = 1.855752 = 96.49928
χ² = ln10 {B-
∑n-1 log s ² } =
= 2.306 { 96.49928 – 93.416} = 7.10873
Based on the result of homogeneity test above, it can be seen that the score of
χ² was 7.098. From the table of Chi-Square distribution with the significance level α = 0.05, the score of χ
t
²
0.953
7.81. Because χ² 7.098
was lower than χ
t
²
0.953
7.81 or χ² χ
t
² 7.098 7.81, it can be concluded that the data are homogeneous.
C. Hypothesis Testing
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