2. Post-test Result
a. Test of Data Normality
The data normality test was employed by using Kolmogorov Smirnov test to know whether the sample of experiment and control
class derived from population that has normal distribution or not. The result of the analysis is presented in the following table.
Table 4.6 Tests of Normality
Kolmogorov-Smirnov
a
Shapiro-Wilk Statistic
df Sig.
Statistic df
Sig. EXPERIMENT
CLASS .158
25 0.109
.917 25
0.043 CONTROL CLASS
.153 25
0.133 .949
25 0.239
In the test, the level of significance was set up at 0.05. Based on Table 4.1, it shows that significance score of Experiment Class is
0.109 and significance score of Control Class is 0.133. It means that the significant score of Experiment class and Control class are higher
than 0.05, or 0.109 0.05 and 0.133 0.05. The conclusion is the sample taken for experiment class and control class derived from the
population that has normal distribution.
b. Test of Data Homogeneity
The data homogeneity test was accomplished after conducted normal distribution test. The data homogeneity test was employed by
using Levene test to know whether the sample of experiment and control class derived from population that has homogenous or not.
The following table is the description of the test result.
Table 4.7 Test of Homogeneity of Variances
Levene Statistic df1
df2 Sig.
2.124 5
17 0.112
In the Table 4.4, the level of significance score was set up at 0.05. Based on Table 4.4, it shows that significance score of
homogeneity is 0.112. It means that the significant score is higher than 0.05, or 0.112 0.05. The conclusion is the sample of
experiment class and control class derived from homogenous population because both of them have the homogenous variance.
C. The Analysis of Data
In analyzing the data from the result of pre-test and post-test, the writer used statistic calculation of the t-test formula with degree of significance 5
as follows.
1. Determining mean X with the formula:
∑
= = 14.8
2. Determining mean Y with the formula:
∑
= = 10.4
3. Determining the standard deviation of variable X:
= √
∑
= √
= √
= 4.09
4. Determining the standard deviation of variable Y:
= √
∑
= √
= √
= 4.83
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