89
The summary of the normality using critical value of Liliefors shows that all of the values L
obtain
gained are lower than L
table
. Therefore, it can be concluded that all of the groups of data are in normal distribution.
2. Homogeneity
The homogeneity test is used to know whether the data are homogenous or not. The result of homogeneity test is as follows:
Table. 4.10. The summary of homogeneity test
X
1
X
2
X
3
X
4
Total N
15 15
15 15
60
X
1235 1068
1083 1105
X
2
102207 76490
78845 81769
Si
2
37.5238095
32.028574 46.6
26.238092
S
2
35.597619
Log s
2
1.55142095
B
86.8795733
LN 10 2.302585
χ
2
1.24177331
χ
t 2
7.81
Based on the computation above, the value of chi-square observation
χ
2
is 1.24177 while the table value of the chi- square χ
t 2
at the level
of significance α=0.05 is 7.81. Because χ
2
is lower than χ
t 2
, it can be concluded that the data are homogeneous.
C. Hypothesis Testing
After the normality test and homogeneity test are conducted, the next step is analyzing data by using ANOVA. The data are analyzed to check if
90
there is significant difference of means among groups. The result of ANOVA test must be stated as F
o
F
t
so that the null hypothesis Ho is rejected. The result of ANOVA 2 x 2 is described as follows:
1. The summary of ANOVA 2x2
Table 4.11. The summary of ANOVA 2x2
Source of Variance SS
df MS
F F
t0.05
Between Columns 220.417
1 220.417
6.19169 4.00
Between Rows 350.417
1 350.417
9.84382 Columns by Rows
Interaction 595.35
1 595.35
16.7244 Between Groups
1166.18 3
388.728 Within Groups
1993.47 56 35.5976
Total 3159.65
59
Table 4.12. The summary of the Mean Scores
Motivation B Teaching Material A
CBC A
1
Textbook A
2
High Motivation B
1
= 82.33 = 72.2
= 77.4
Low Motivation B
2
= 71.2 = 73.67
= 72.3 = 76.97
= 74.07
From the tables above, it can be concluded that: a.
The impact of employing teaching materials upon the students‟ writing skill
F
o
between columns 6.19 is higher than F
table
at the level of significance α=0.05 4.00. Consequently, H
stating that there is no
