commit to user 112
1. The Normality Test
The sample is in normal distribution if L
o
L obtained is lower than L
t
L table at the level of significance
α = 0.05 on Lilliefors. The formula used in this testing is:
晐 ∑ 莐
∑ 莐 n
n 1 莐 莐
晐 a.
Column A
1
. In this column, that contains 30 students who were taught by using Note-
Taking Pairs technique n = 30, the highest value of Fz
i
– sz
i
or L
o
is 0.1340. L
t
at the level of significance α = 0.05 is 0.161. Because L
o
is lower than L
t
0.1340 0.161, it can be concluded that the sample is in normal distribution.
b. Column A
2
. In this column, that contains 30 students who were taught by using DR-TA
technique n = 30, the highest value of Fz
i
– sz
i
or L
o
is 0.0879. L
t
at the level of significance α = 0.05 is 0.161. Because L
o
is lower than L
t
0.0879 0.161, it can be concluded that the sample is in normal distribution.
c. Row B
1
. In this row, that contains 30 students having high linguistic intelligence n =
30, the highest value of Fz
i
– sz
i
or L
o
is 0.1562. L
t
at the level of significance α = 0.05 is 0.161. Because L
o
is lower than L
t
0.1562 0.161, it can be concluded that the sample is in normal distribution.
commit to user 113
d. Row B
2
. In this row, that contains 30 students having low linguistic intelligence n =
30, the highest value of Fz
i
– sz
i
or L
o
is 0.1550. L
t
at the level of significance α = 0.05 is 0.161. Because L
o
is lower than L
t
0.1550 0.161, it can be concluded that the sample is in normal distribution.
e. Cell A
1
B
1
In this cell, that contains 15 students having high linguistic intelligence who were taught by using Note-Taking Pairs technique n = 15, the highest value
of Fz
i
– sz
i
or L
o
is 0.1790. L
t
at the level of significance α = 0.05 is 0.220. Because L
o
is lower than L
t
0.1790 0.220, it can be concluded that the sample is in normal distribution.
f. Cell A
1
B
2
. In this cell, that contains 15 students having low linguistic intelligence who
were taught by using Note-Taking Pairs technique n = 15, the highest value of Fz
i
– sz
i
or L
o
is 0.2179. L
t
at the level of significance α = 0.05 is 0.220. Because L
o
is lower than L
t
0.2179 0.220, it can be concluded that the sample is in normal distribution.
g. Cell A
2
B
1
In this cell, that contains 15 students having high linguistic intelligence who were taught by using DR-TA technique n = 15, the highest value of Fz
i
– sz
i
or L
o
is 0.1015. L
t
at the level of significance α = 0.05 is 0.220. Because L
o
is lower than L
t
0.1015 0.220, it can be concluded that the sample is in normal distribution.
commit to user 114
h. Cell A
2
B
2
. In this cell, that contains 15 students having low linguistic intelligence who
were taught by using DR-TA technique n = 15, the highest value of Fz
i
– sz
i
or L
o
is 0.1622. L
t
at the level of significance α = 0.05 is 0.220. Because L
o
is lower than L
t
0.1622 0.220, it can be concluded that the sample is in normal distribution.
The normality test of those data is concluded in Table 4.9. below.
Table 4.9. The Normality Test
No. Data Number of
Sample L obtained
L
o
L table L
t
Alpha α
Distribution of Sample
1 A
1
30 0.1340
0.161 0.05
normal 2
A
2
30 0.0879
0.161 0.05
normal 3
B
1
30 0.1562
0.161 0.05
normal 4
B
2
30 0.1550
0.161 0.05
normal 5
A
1
B
1
15 0.1790
0.220 0.05
normal 6
A
1
B
2
15 0.2179
0.220 0.05
normal 7
A
2
B
1
15 0.1015
0.220 0.05
normal 8
A
2
B
2
15 0.1622
0.220 0.05
normal
2. The Homogeneity Test
Kategori