b. Post Test
In this test, there was only 1 student from experiment class Using The Jakarta Post Newspaper Articles who got the score under 70 and
there were 8 students from controlled class Using the textbooks who got the score under 70 and these scores are characterized as good score.
Besides that, there were 10 students from experiment class Using The Jakarta Post Newspaper Articles who got the score above 80, but
there were only 4 students from controlled class Using the textbooks who got the score above 80 and these scores are characterized as a
excellent score see table 2 and table 3, p. 35 and 36.
Table 3.1 The Criteria of Students’ Pre Test and Post Test Scores are as Follow:
No. Scores
Characteristic 1.
≤ 50 Weak
2. 51-60
Enough
3. 61-70
Good
4. 71-80
Very good
5. 81-90
Excellent
The writer also used statistic calculation of t-test to make the data clearly. The t-test formula with degree of significance 5 looks like as
follows.
Table 3.2 The Result of the Calculation of the Test Both Experiment Class and
Controlled Class
Students’ Identification
Number X
Y X
y x
2
y
2
1 28
20 1.58
-2 2.49
4 2
24 24
-2.42 2
5.85 4
3 28
24 1.58
2 2.49
4 4
24 24
-2.42 2
5.85 4
5 28
20 1.58
-2 2.49
4 6
24 20
-2.42 -2
5.85 4
7 32
24 5.58
-2 31.13
4 8
24 20
-2.42 -2
5.85 4
9 28
20 1.58
-2 2.49
4 10
36 20
9.58 -2
91.77 4
11 32
24 5.58
2 31.13
4 12
32 24
5.58 2
31.13 4
13 24
16 -2.42
-6 5.85
36 14
32 24
5.58 2
31.13 3
15 24
28 -2.42
6 5.85
36 16
24 20
-2.42 -2
5.85 4
17 28
20 1.58
-2 2.49
4 18
24 16
-2.42 -6
5.85 36
19 24
28 -2.42
6 5.85
36 20
28 24
1.58 2
2.49 4
21 24
12 -2.42
-10 5.85
100 22
20 28
-6.42 6
41.21 36
23 24
24 -2.42
2 5.85
4 24
20 24
-6.42 2
41.21 4
25 32
24 5.58
2 31.13
4 26
20 24
-6.42 2
41.21 4
27 32
20 5.58
-2 31.13
4 28
20 20
-6.42 -2
41.21 4
�
=
740 MX=
740 28
= 26.42
�
=
616
MY =
616 28
= 22 -
-
522.6 368
Note: X
= the students’ gained scores of the experiment class Y
= the students’ gained scores of the experiment class x
= X ‒ MX
y = Y
‒ MY a.
Determining Mean 1 M1 with formula: M
1 =
X N
1
=
740 28
= 26.42
b. Determining Mean 2 M2 with formula:
M
2
=
X N
2
=
616 28
= 22
c. Determining Standard of Deviation Score of Variable X SD1,
with formula: SD
1 =
X
2
N
1
=
522.6 28
=
18.66
= 4.31
d. Determining Standard of Deviation Score of Variable Y, with
formula: SD
2 =
Y
2
N
2
=
368 28
= 13.14
= 3.62
e. Determining Standard Error of Mean of Variable X, with formula:
SE
M1
=
SD
1
N
1
−1
=
4.31 28−1
=
4.31 5.19
= 0.83
f. Determining Standard Error of Mean of Variable Y, with formula:
SE
M2
=
SD
2
N
2
−1
=
3.62 28−1
=
3.62 5.19
= 0.69
g. Determining Standard Error of difference of Mean of Variable X
and Variable Y, with formula: SE
MI-M2
= ��
�1 2
+ ��
�2 2
= 0.83
2
+ 0.69
2
= 0.69 + 0.47
=
1.15
= 1.07
h. Determining t
o
with formula:
t
o
=
M
1
−M
2
SE
M 1 −M 2
=
26.42 −22
1.07
=
4.42 1.07
= 4.13
i. Determining t-table in significance level 5 with Degree of
Freedom df:
df = N1+N2 – 2
= 28 + 28 – 2
=
54
Because the value of 54 is not mentioned in the table, the writer used the closer value to 54 is 60 as degree of freedom. The
writer gained t-
table
; Degree of significance 5 = 2.01 see appendix
t
table
= 2.01
j. The comparison between t
o
and t-table From the calculation above, it is clear that the score of the
experiment class is higher than the score of controlled class. The writer also got the result of the comparison between t
o
and t-
table
; 4.13 2.01
= t
o
t
table
From the result of statistic calculation, it was found that the value of the t
o
was 4.13 and the degree of freedom df was 54. In this research, the writer used the degree of significance 5. It can
be seen that the df and degree of significance 5=2.01. By comparing the value of t
o
= 4.13 and t-table on the degree of significance 5 = 2.01, the writer summarized that t
o
was higher than t-table.
10. The Testing of The Hypotheses
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