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B. Prerequisite Testing
The statistic analysis needs several requirements that have to be met, namely, normality testing and homogeneity testing.
1. Normality Testing
Liliefors testing is used to compute the normality of the data. The normality test is to check whether the data are in normal distribution or not. If L
o
L obtained is lower than L
t
L table at the level of significance α=0.05 on Liliefors, the sample is in normal distribution. The formula used in this testing is:
a Cell A
1
B
1
. In this cell, that contains 15 students having high learning motivation who are
taught using internet-based material n=15, the highest value of FZ
i
-SF
i
or L
o
is 0.132. L
t
at the level of significance α=0.05 is 0.220. Because L
o
is lower than L
t
0.132 0.220, it can be concluded that the sample is in normal distribution.
b Cell A
2
B
1
. In the cell A
2
B
1
, that contains 15 students having high learning motivation who are taught using textbook material n=15, the highest value of FZ
i
-SF
i
or L
o
is 0.085. L
t
at the level of signif icance α=0.05 is 0.220. Because L
o
is
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lower than L
t
0.085 0.220, it can be concluded that the sample is in normal distribution.
c Cell A
1
B
2
. In the cell A
1
B
2
, that contains 15 students having low learning motivation who are taught using internet-based material n=15, the highest value of FZ
i
- SF
i
or L
o
is 0.170. L
t
at the level of significance α=0.05 is 0.220. Because L
o
is lower than L
t
0.170 0.220, it can be concluded that the sample is in normal distribution.
d Cell A
2
B
2
. In the cell A
2
B
2
, that contains 15 students having low learning motivation who are taught using textbook material n=15, the highest value of FZ
i
-SF
i
or L
o
is 0.0796. L
t
at the level of significance α=0.05 is 0.220. Because L
o
is lower than L
t
0.07960.220, it can be concluded that the sample is in normal distribution.
e Cell A
1
. In the cell A
1
, that contains 30 who are taught using internet-based material n=30, the highest value of FZ
i
-SF
i
or L
o
is 0.080. L
t
at the level of significance α=0.05 is 0.161. Because L
o
is lower than L
t
0.080 0.161, it can be concluded that the sample is in normal distribution.
f Cell A
2
. In the cell A
2
, that contains 30 students who are taught using textbook material n=30, the highest value of FZ
i
-SF
i
or L
o
is 0.0907. L
t
at the level of
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significance α=0.05 is 0.161. Because L
o
is lower than L
t
0.0907 0.161, it can be concluded that the sample is in normal distribution.
g Cell B
1
. In the cell B
1
, that contains 30 students who are having high learning motivation n=30, the highest value of FZ
i
-SF
i
or L
o
is 0.156. L
t
at the level of significance α=0.05 is 0.161. Because L
o
is lower than L
t
0.156 0.161, it can be concluded that the sample is in normal distribution.
h Cell B
2
. In the cell B
2
, that contains 30 students who are having low learning motivation n=30, the highest value of FZ
i
-SF
i
or L
o
is 0.0589. L
t
at the level of significance α=0.05 is 0.161. Because L
o
is lower than L
t
0.0589 0.161, it can be concluded that the sample is in normal distribution.
Table 14. The Normality Test No.
Data The
Number of sample
L Obtained
L
o
L Table
L
t
Alfa Distribution
of Population
1 2
3 4
5 6
7 8
A
1
B
1
A
2
B
1
A
1
B
2
A
2
B
2
A
1
A
2
B
1
B
2
15 15
15 15
30 30
30 30
0.1321 0.0850
0.1704 0.0796
0.0800 0.0907
0.1560 0.0589
0.220 0.220
0.220 0.220
0.161 0.161
0.161 0.161
0.05 0.05
0.05 0.05
0.05 0.05
0.05 0.05
Normal Normal
Normal Normal
Normal Normal
Normal Normal
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2. Homogeneity Testing
Kategori