see whether the datasample in both classes was homogenous or heterogeneous.
Hypothesis: H
: the condition of experimental class is not different from controlled class.
H
1
: the condition of experimental class is different from controlled class.
The criteria of the test: α = 0.05
H :
�
1−� −1
F
�
1 2
� 1−1 2−1
H
1
: F ≥
�
1 2
� 1, 2
The formula used to can be seen as follows: F =
ℎ ℎ� ℎ � �� �
ℎ � � �� �
Or F =
1 2
2 2
The calculation can be seen as follows: F =
=
198.94 114.64
= 1.73 N1-1 = 35-1=34
N2-1 = 35-1=34
�
0.05 34,34
= 1.84 It can be seen that F F
12 α n1-1n2-1
1.73 1.84. Based on the criteria, it can be concluded that H
is accepted and the sample in experimental class and controlled class were homogenous.
3. Hypothesis Testing
To test the hypothesis whether there is significant different to classes, the writer calculated the data. The experimental class was X and the controlled
class was Y. The procedures of calculated were as follows:
Table 4.10 Standard Deviation Table
S tud
en t
Experimental Class X Controlled Class X
Pre- Test
Post- Test
Gained Score
X X
2
Pre- Test
Post- Test
Gained Score
Y Y
2
1 50
60 10
100 65
70 5
25 2
70 75
5 25
65 60
-5 25
3 50
75 25
625 60
60 4
55 85
30 900
65 55
-10 100
5 45
80 35
1225 75
80 5
25 6
45 60
15 225
40 30
-10 100
7 35
30 -5
25 30
35 5
25 8
75 85
10 100
80 90
10 100
9 70
85 15
225 40
50 10
100 10
55 80
25 625
80 85
5 25
11 60
60 65
80 15
225 12
60 70
10 100
50 50
13 70
80 10
100 50
45 -5
25 14
55 70
15 225
65 80
15 225
15 65
75 10
100 70
75 5
25 16
40 80
40 1600
50 65
15 225
17 50
60 10
100 55
60 5
25 18
55 85
30 900
60 85
25 625
19 30
60 30
900 75
70 -5
25 20
55 80
25 625
75 90
15 225
21 75
95 20
400 65
60 -5
25 22
55 60
5 25
65 70
5 25
23 50
70 20
400 70
80 10
100 24
55 50
-5 25
50 60
10 100
25 45
85 40
1600 65
55 -10
100 26
65 75
10 100
45 60
15 225
27 60
75 15
225 35
35 28
65 65
65 55
-10 100
29 65
75 10
100 60
85 25
625 30
65 75
10 100
45 65
20 400
31 45
80 35
1225 70
65 -5
25 32
55 80
25 625
60 80
20 400
33 70
90 20
400 65
60 -5
25 34
55 65
10 100
75 85
10 100
35 60
70 10
100 45
55 10
100
∑X =
1975 ∑X
1
= 2545
∑X = 570
∑X
2
= 14150
∑Y =
2095 ∑Y
1
= 2285
∑Y = 190
∑Y
2
= 4500
Based on the data on the table 4.10 above, we can apply those data into the formula of t-test to get t
table
value was expressed as follows: t =
�
− √
�� �
2
+
�� �
2
The calculation can be seen as follows: a. Determining Mean of Variable X with formula:
=
∑
=
570 35
= 16.28 b. Determining Mean of Variable Y with formula:
=
∑
=
190 35
= 5.42 c. Determining of Standard Deviation score of Variable X, with formula:
�� = √ ∑
2
−
∑
2
− 1
�� =
√
14150−
570 2 35
35−1
�� =
√
14150−
324900 35
34
�� =
√
14150−9282.85 34
�� =
√
4867.15 34
�� = √143.15 �� = 11.96
d. Determining of Standard Deviation score of Variable Y, with formula: �� = √
∑
2
−
∑
2
− 1
