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67
63 - 66 62.5 - 66.5
64.5 lllll
5 33.33
322.50 20801.25
67 - 70 66.5 - 70.5
68.5 lllll
5 33.33
342.50 23461.25
71 - 74 70.5 - 74.5
72.5 ll
2 13.33
145.00 10512.50
15 100
987.50 65287.75
Figure 4.8. Histogram and Polygon Data A
2
B
2
B. Normality and Homogeneity Test
Before analyzing the data using inferential analysis, normality and homogeneity test must be done. The normality test is to know that the sample
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68 is in normal distribution and the homogeneity test is to know that data are
homogeneous. Each test is presented in the following section: 1. 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
05 .
=
a L stands for Lilliefors.
Table 4.9. Normality Test.
No Data
The Number
of Sample
L Obtained
L
o
L Tabel
L
t
Alafa a
Distribution of
Population
1 A
1
30 0.125
0.161 0.05
Normal 2
A
2
30 0.124
0.161 0.05
Normal 3
B
1
30 0.137
0.161 0.05
Normal 4
B
2
30 0.097
0.161 0.05
Normal 5
A
1
B
1
15 0.135
0.220 0.05
Normal 6
A
1
B
2
15 0.125
0.220 0.05
Normal 7
A
2
B
1
15 0.151
0.220 0.05
Normal 8
A
2
B
2
15 0.105
0.220 0.05
Normal
Based on the table above, it can be concluded that: 1. Normality test of scores of the students who are taught by using Test Taking
Teams A
1
.
commit to user
69 Based on the computation result of score of the students who are taught
using group test taking teams, the highest score of Fzi - Szi or L
o
is 0.125. From the table of critical value of Lillifors test with the students
number N of 30 at the significance level α = 0.05 the score of L
t
is 0.161. Because L
o
0.125 is lower than L
t
0,161, it can be concluded that the sample is in normal distribution.
2. Normality test of scores of the students who are taught by using Direct Instruction Model A
2
. Based on the computation result of score of the students who are taught
using group direct instruction model, the highest score of Fzi - Szi or L
o
is 0.124. From the table of critical value of Lillifors test with the students number N of 30 at the significance level α = 0.05 the score of L
t
is 0.161. Because L
o
0.124 is lower than L
t
0.161, it can be concluded that the sample is in normal distribution.
3. Normality test of scores of the students having high interest who are taught by using TTT and DIM B
1
. Based on the computation result of score of the students having high
interest, the highest score of Fzi - Szi or L
o
is 0.137. From the table of critical value of Lillifors test with the students number N of 30 at the
significance level α = 0.05 the score of L
t
is 0.161 because Lo 0.137 is lower than L
t
0.161, it can be concluded that the sample is in normal distribution.
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70 4. Normality test of scores of the students having low interest who are taught by
using TTT and DIM B
2
. Based on the computation result of score of the students having high interest,
the highest score of Fzi - Szi or L
o
is 0.097. From the table of critical value of Lillifors test with the students number N of 30 at the significance
level α = 0.05 the score of L
t
is 0.161 because L
o
0.116 is lower than Lt 0.161, it can be concluded that the sample is in normal distribution.
5. Normality test of scores of the students or the group having high interest who are taught by using Test Taking Teams A
1
B
1
. Based on the computation result of score of the students who are taught using
test taking teams, the highest score of Fzi - Szi or L
o
is 0.135. From the table of critical value of Lillifors test with the students number N of 15 at
the significance level α = 0.05 the score of L
t
is 0.220 because L
o
0.135 is lower than L
t
0.220, it can be concluded that the sample is in normal distribution.
6. Normality test of scores of the students or the group having high interest who are taught by using Direct Instruction Model A
2
B
1
. Based on the computation result of score of the students or group having low
interest who are taught using group test taking teams, the low score of Fzi - Szi or L
o
is 0.125. From the table of critical value of Lillifors test with the students number N
of 15 at t he significance level α = 0.05 the score of L
t
is 0.220 because L
o
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71 0.125 is lower than Lt 0.220, it can be conclude that the sample is in
normal distribution. 7. Normality test of scores of the students or the group having low interest who
are taught by using Test Taking Teams A
1
B
2
. Based on the computation result of score of the students or the group having
high interest who are taught by using group direct instruction model, the high score of Fzi - Szi or L
o
is 0.151. From the table of critical value of Lillifors test with the students number N of 15 at the significance level α =
0.05 the score of L
t
is 0.220. Because L
o
0.151 is lower than L
t
0,220, it can be concluded that the sample is in normal distribution.
8. Normality test of scores of the students or group having low interest who are taught by using Direct Instruction Model A
2
B
2
. Based on the computation result of score of the students or group having low
interest who are taught by using group direct instruction, the highest score of Fzi - Szi or L
o
is 0.105. From the table of critical value of Lillifors test with the students number N
of 15 at the significance level α = 0.05 the score of L
t
is 0220. Because L
o
0.105 is lower than L
t
0.220 it can be concluded that the sample is normal distribution.
2. Homogeneity Test Homogeneity test is done to know that the data are homogenous. If
2 o
c is lower than
2 t
c 0.05, so the data is homogeneous. Based on the computation result of the homogeneity test,
2 o
c 2.548 and
2 t
c 7.81 at
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72 the level of significance 0.05. It can be concluded that the data are
homogenous. To make clearer, the computation of the homogeneity test can be seen as
follows: Table 4.10. The Result of Homogeneity Test.
Table 4.11. The Homogeneity Test.
sample df
1df s²
logs² dflogs²
1 14
0.0714 23.1143
1.3639 19.0943
2 14
0.0714 27.4286
1.4382 20.1348
3 14
0.0714 12.4952
1.0967 15.3544
4 14
0.0714 26.9238
1.4301 20.0219
74.6055
NO GROUP 1
GROUP 2 GROUP 3
GROUP 4 X²
X² X²
X² X
1
X
2
X
3
X
4
1 70
50 60
55 4900
2500 3600
3025 2
75 53
63 60
5625 2809
3969 3600
3 75
55 63
60 5625
3025 3969
3600 4
75 57
63 63
5625 3249
3969 3969
5 77
60 65
65 5929
3600 4225
4225 6
77 60
65 65
5929 3600
4225 4225
7 80
60 63
67 6400
3600 3969
4489 8
80 63
67 63
6400 3969
4489 3969
9 80
63 65
65 6400
3969 4225
4225 10
80 65
67 70
6400 4225
4489 4900
11 83
65 67
70 6889
4225 4489
4900 12
85 65
70 70
7225 4225
4900 4900
13 85
65 70
70 7225
4225 4900
4900 14
85 67
70 73
7225 4489
4900 5329
15 87
67 73
73 7569
4489 5329
5329 ∑X
1194 915
991 989
∑X²
95366 56199
65647 65585
∑X² 1425636
837225 982081
978121 ∑X²n
95042.4 55815
65472.07 65208.07
S
i 2
23.114 27.429
12.495 26.924
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73
2 2
log log
1
i i
i
s df
s x
n S
= -
å
=74.6055 548
. 2
6055 .
74 712
. 75
3026 .
2 log
1 10
ln
2 2
= -
= -
- =
å
x s
x n
B x
i i
o
c
Because is lower than it can be concluded that the data are
homogeneous C. Hypothesis Test
After the result of normality and homogeneity test are calculated, hypothesis test can be done. The data analysis is done by using multifactor
analysis of variance ANOVA 2 x 2. H
o
is rejected if F
o
F
t.
F
o
is higher than F
t
. It means that there is a significant difference. After knowing that H
o
is rejected, the further analysis is done to know the difference between the two groups Group A and group B and cells using Tukey test. Then, to know
which group is better, the mean scores of the groups and cells are compared. Both ANOVA 2 x 2 and Tukey tests are presented as below:
a. Summary of a 2 x 2 Multifactor Analysis of varianc
Table 4.12. Multifactor Analysis of Variance
Source of Variance SS
df MS
Fo Ft0.05
Between Columns Stress 277.35
1 277.35
12.332 4.016
Between Rows Task
1316.02 1
1316.02 58.514
4.016 Columns By Rows
Interaction 1278.82
1 1278.82
56.860 4.016
Between Groups
2872.18 3
957.39
Within Groups 1259.47
56 22.49
81 .
7
2 05
. ,
3
=
t
c
2
c
2 t
c
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74
TOTAL 4131.65
59
Based on the table above, it can be concluded that: 1 Because F
o
between columns
12.332
is higher than F
t
at the level of significance α = 0.05 4.016, the difference between columns is significant.
Because the mean of A
1
70.30 is higher than that of A
2
66.00, it can be concluded that Test Taking Teams is more effective than Direct Instruction
Model to teach reading. 2 Because F
o
between rows
58.514
is higher than F
t
at the level of significance α = 0.05 4.016, the difference between rows is significant. It
can be concluded that the achievement of students who have high and those who have low interest are significantly different. Then, because the mean of
B
1
72.83 is higher than B
2
63.47, it can be concluded that the students having high interest have better reading competence than those having low
interest.
3 Because F interaction columns by row 56.860 is higher than F
t
at the level of significance
a = 0.05 4.016, it can be concluded that there is interaction between teaching models and students’ interest. It means that the
effectiveness of teaching models depend on the levels of students’ interest. b. Summary of Tukey Test
The Summary of Tukey test result is presented below: Table 4.13. Summary of Tukey Test
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75 Between group
q
o
q
t
Meaning Category
A
1
- A
2
4.966 2.89
qo qt Significant
B
1
- B
2
10.818 2.89
qo qt Significant
A
1
B
1
- A
2
B
1
11.052 3.01
qo qt Significant
A
1
B
2
- A
2
B
2
4.029 3.01
qo qt Significant
1 Because q
o
between columns A
1
- A
2
4.966 is higher than q
t
at the level of significance α = 0.05
2.89 , applying test taking teams differs
significantly from direct instruction model to teach reading. Because the mean of A
1
70.30 is higher than that of A
2
66.00, it can be concluded that test taking teams is more effective than direct instruction model to
teach reading. 2 Because q
o
between rows B
1
- B
2
10.818 is higher than q
t
at the level of significance α = 0.05 2.89, it can be concluded that the students who
have high interest and those who have low interest are significantly different in their reading competence. Because the mean of B
1
72.83 is higher than B
2
63.47, it can be concluded that the students who have high interest have better reading competence than those who have low interest.
3 Because q
o
between cells A
1
B
1 –
A
2
B
1
11.052 is higher than q
t
at the level of significance α = 0.05 3.01, applying test taking teams differs
significant from direct instruction model for students who have high interest. Because the mean of A
1
B
1
79.60 is higher than that of A
2
B
1
66.07, it can be concluded that test taking teams is more effective than direct instruction model to teach reading for students having high interest.
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76 4 Because q
o
between cells A
1
B
2 –
A
2
B
2
4.029 is higher than q
t
at the level of significance α = 0.05 3.01, applying direct instruction model differs
significantly from test taking teams for students who have low interest. Because the mean of A
2
B
2
65.93 is higher than that of A
1
B
2
61.00, it can be concluded that direct instruction model is more effective than test
taking teams to teach reading for students who have low interest. Because Test Taking Team is more effective than Direct Instruction
Model for students having high interest and Direct Instruction Model is more effective than Test Taking Teams for students who have low interest, it can be
concluded that there is an interaction between the teaching model and students’ interest in teaching reading competence.
D. Discussion of the Result of the Study
