B. Testing Hypotheses
Before analyzing the data by using inferential analysis, normality and homogeneity test were conducted. The normality test is used to know whether the
samples in this research are in normal distribution and the homogeneity test is used to know whether the data are homogeneous. The result of normality and
homogeneity tests can be seen as follows: 1.
Normality Test a.
Normality test of the data of students who are taught using Guided Imagery A
1
Based on the statistical calculation of the students’ scores who are taught using Guided Imagery, it is known that the highest value of L obtained L
o
is 0.082. The critical value of
Liliefors
test with number of students n = 34 at the significance level α = 0.05 L
t
is 0.152. Because L
o
is lower than L
t
, it can be concluded that the sample is in normal distribution.
b. Normality test of the data of students who are taught using Direct
Instruction A
2
Based on the statistical calculation of the students’ scores who are taught using Direct Instruction, it is known that the highest value of L obtained L
o
is 0.097. The critical value of
Liliefors
test with number of students n = 34 at the significance level α = 0.05 L
t
is 0.152. Because L
o
is lower than L
t
, it can be concluded that the sample is in normal distribution.
c. Normality test of the data of students having high creativity B
1
Based on the statistical calculation of the students’ scores who have high creativity, it is known that the highest value of L obtained L
o
is 0.070. The critical value of
Liliefors
test with number of students n = 34 at the significance level α = 0.05 L
t
is 0.152. Because L
o
is lower than L
t
, it can be concluded that the sample is in normal distribution.
d. Normality test of the data of students having low creativity B
2
Based on the statistical calculation of the students’ scores who have low creativity, it is known that the highest value of L obtained L
o
is 0.076. The critical value of
Liliefors
test with number of students n = 34 at the significance level α = 0.05 L
t
is 0.152. Because L
o
is lower than L
t
, it can be concluded that the sample is in normal distribution.
e. Normality test of the data of students having high creativity who are
taught using Guided Imagery A
1
B
1
Based on the statistical calculation of the students having high creativity who are taught using Guided Imagery, it is known that the highest value of L obtained L
o
is 0.089. The critical value of
Liliefors
test with number of students n = 17 at the significance level α = 0.05 L
t
is 0.206. Because L
o
is lower than L
t
, it can be concluded that the sample is in normal distribution.
f. Normality test of the data of students having high creativity who are
taught using Direct Instruction A
2
B
1
Based on the statistical calculation of the students having high creativity who are taught using Direct Instruction, it is known that the highest value of L obtained
L
o
is 0.135. The critical value of
Liliefors
test with number of students n = 17 at the significance level α = 0.05 L
t
is 0.206. Because L
o
is lower than L
t
, it can be concluded that the sample is in normal distribution.
g. Normality test of the data of students having low creativity who are
taught using Guided Imagery A
1
B
2
Based on the statistical calculation of the students having low creativity who are taught using Guided Imagery Strategy, it is known that the highest value of L
obtained L
o
is 0.151. The critical value of
Liliefors
test with number of students n = 17
at the significance level α = 0.05 L
t
is 0.206. Because L
o
is lower than L
t
, it can be concluded that the sample is in normal distribution. h.
Normality test of the data of students having low creativity who are taught using Direct Instruction A
2
B
2
Based on the statistical calculation of the students having low creativity who are taught using Direct Instruction, it is known that the highest value of L obtained
L
o
is 0.183. The critical value of
Liliefors
test with number of students n = 17 at the significance level α = 0.05 L
t
is 0.206. Because L
o
is lower than L
t
, it can be concluded that the sample is in normal distribution.
The table 4.9 below summarizes the result of normality calculation of the eight groups of data explained above:
Table 4.9 The Result of Normality Test
No. Data
No. of sample L
o
L
t
α Status
1. A
1
34 0.082
0.152 0.05
Normal 2.
A
2
34 0.097
0.152 0.05
Normal 3.
B
1
34 0.070
0.152 0.05
Normal 4.
B
2
34 0.076
0.152 0.05
Normal 5.
A
1
B
1
17 0.089
0.206 0.05
Normal 6.
A
2
B
1
17 0.135
0.206 0.05
Normal 7.
A
1
B
2
17 0.151
0.206 0.05
Normal 8.
A
2
B
2
17 0.183
0.206 0.05
Normal
2. Homogeneity Test
To check the homogeneity, a set of statistical calculation was conducted. The data were divided into four groups in which each group consists of 17
students’ scores. The table 4.10 summarizes the result of homogeneity test:
Table 4.10 Summary of Homogeneity Test
Sample df =n-1
1df s
i 2
Log s
i 2
Log s
i 2
df
1 16
0.063 89.632
1.952 31.240
2 16
0.063 58.368
1.767 28.259
3 16
0.063 128.816
2.110 33.760
4 16
0.063 93.654
1.972 31.544
Sum 124.802
o 2
=
ln
10 B –
log
s
i 2
df = 2.303125.868-124.802
= 2.455 Based on the result of calculation above, it can be seen that the
o 2
2.455 is lower than
t 2
7.815 at the level of significance = 0.05. Therefore, it can be
concluded that the data are homogeneous.
C. Hypothesis Test
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