Equal variance
s not assumed
1.666 57.567 .101 4.600
2.762 -.929 10.129
Based on table 4.10 M
1
was 72.87 and M
2
was 68.27. If it was compared with mean score in pre-test the result showed the improvement 12.94 points in
experimental class and 13.34 poin in controlled class. It meant that controlled class had greater improvement of mean score 0.4 poin. It also showed that SD
1
was 11.150 and SD
2
was 10.221. Then Standard error mean in experimental class SE
M1
result was 2.036 and in controlled class SE
M2
was 1.866. In table 4.11 it showed the t-test result at post-test. Based on Levenes Test
above showed p=0.598, it was greater than 0.05 that meant the post-test data was homogenous. Since the data was homogenous, then the t-test could be showed at
equal variance assumed. The t-test of post-test was 0.101 p=0.101, It meant that p was greater than 0.05 p0.05. Based on the result it meant that null hypothesis
was accepted, there were no significance differences between experimental class and controll class in post test. The table also showed degree of freedom df
presented 58 with significant level 5, it meant the t
table
was 1.672. While based on table t
was 1.666, it meant t t
table
1.6661.672.
3. Gained Score
Tabel 4.12
Group Statistics
Class N
Mean Std. Deviation
Std. Error Mean Gained
Eksperiment 30
14.27 9.078
1.657 Control
30 12.67
6.915 1.262
Tabel 4.13
Independent Samples Test
Levenes Test for Equality of
Variances t-test for Equality of Means
F Sig.
t Df
Sig. 2-
taile d
Mean Differe
nce Std.
Error Differe
nce 95
Confidence Interval of the
Difference Lower Upper
Gained Equal variances
assumed 2.151 .148
.768 58 .446
1.600 2.083 -2.571
5.771 Equal
variance s not
assumed .768 54.177 .446
1.600 2.083 -2.577
5.777
Based om table 4.12 it was presented the mean of experimental class M
1
was 14.27 then in controlled class M
2
was 12.67. Furthermore, the standard deviasion of experimental class SD
1
was 9.078 and controlled class was SD
2
was 6.915. Then Standard error mean SE
M1
in experimental class was 1.657 and in controlled class was 1.262.
Table 4.12 showed that significant level on Levene’s test was 0.148,
therefore the writer use 5 0.05 for the significance level. It was showed that p was greater then 0.05 p0.05, it meant the data was homogenous. Since it was
homogenous the result showed on Equal variance assumed. The result showed that p=0.446, it was greater than significance level 0.05 p0.05. It meant the
null hypothesis was still accepted, there were no significant differences between students who were taught reading comprehension with blended learning model
and without blended learning. The table also showed the degree of freedom df was 58 and the significant level 5 was 1.672, the t
was 0.768. Therefore the t t
table
0.7681.672. Moreover, the writer also used manual calculation in analyzing the data as
a comparison with calculating through SPSS program. It started in calculating the gained total gained score in experimental and controlled classes. It was showed
that the experimental class had higher total gained score 428 while the controlled class had 380. Then the writer calculated the total of X
2
= 2451.73067 and Y
2
= 1386.66667, for the details it was attached at appendix. Furthermore, the manual
calculating process was presented as follows:
a. Determining Mean Variable X: b. Determining Mean Variable Y: M
1
=
∑
M
2
=
∑
