Data of the Students’ Achievement in Learning Passive Voice of Simple
9 60
75 15
10 65
70 5
11 60
70 10
12 40
65 25
13 75
80 5
14 65
70 5
15 70
80 10
16 50
60 10
17 45
75 30
18 60
60 19
35 50
15 20
50 60
10 21
65 70
10 22
55 65
10 23
50 65
15 24
30 55
25 25
40 55
15 Note: students who got lowest score less from KKM
Table 4.3 above informs that in control class, in pre-test there were two students who got score 30, one students who got score 35, two students who got score
40, one students who got score 45, fifth students who got score 50, four students who got score 55, three students who got score 60, three students who got score 65, one
student who got score 70 and one student who got score 75. In post-test, their scores were higher than pre-test. It can be seen that the
highest score in post-test was 80 and there were three students who got it.Then there were nineteen student who got the lowest score less from KKM. One students who
got score 40, one student who got score 50, and two students who got score 55, four students who got score 60, six students who got score 65 and fifth students who got
score 70. From the table presented above the writer can conclude the experiment class
learning passive voice of simple present tense using gamearrange word game got
the higher score than the control class without using arrange words game technique when they learnt passive voice of simple present tense.
As mentioned before, that in analyzing the data from the result of pre-test and post-test, the writer used statistic calculation of the t-test formula with the degree of
significance 5 and 1. Ratio Scale of Gained Score
0, 5, 5, 5, 5, 5, 10, 10, 10, 10,10,10, 10, 15, 15, 15, 15, 15, 15, 20, 20, 20, 25, 25, 35 The higher score H : 30
The lowest score L : 0 Range = H
– L = 3 – 0 = 30 Class = 1+3.3 log n
=1+3.3 log 25 =5.61
Interval = =
Table 4.4 Table of Frequency of Gained Score of Control Class
Interval Gained score of Control Class
Frequency Persentage
—4 1
4 5
—9 4
16 10
—14 8
32 15
—19 6
24 20
—24 2
8 25
—30 4
14 From the table above, it can be shown that most of students got score in the
interval 10 —14. In other word there were 32 students who got score between 10
into 14. The students who got gained score in the interval 15 —19 were six students,
students who got gained score in the interval 20 —24 were two students, students who
got gained score in the interval 25 —30were four student. In the control calss the
lowest gained score was 0 and the higher gained score was 30. Table 4.5 Standard Deviation Table
Students X
Y x
y x²
y² 1
35 25
14.6 11.6
213.16 134.56
2 20
15 -0.4
1.6 0.16
2.56 3
15 20
-5.4 6.6
29.16 43.56
4 25
10 4.6
-3.4 21.16
11.56 5
10 15
-10.4 1.6
108.16 2.56
6 20
10 -0.4
-3.4 0.16
11.56 7
20 20
-0.4 6.6
0.16 43.56
8 30
5 9.6
-8.4 92.16
70.56 9
35 15
14.6 1.6
213.16 2.56
10 10
5 -10.4
-8.4 108.16
70.56 11
20 10
-0.4 -3.4
0.16 11.56
12 5
25 -15.4
11.6 237.16
134.56 13
15 5
-5.4 -8.4
29.16 70.56
14 20
5 -0.4
-8.4 0.16
70.56 15
10 10
-10.4 -3.4
108.16 11.56
16 20
10 -0.4
-3.4 0.16
11.56 17
30 30
9.6 16.6
92.16 275.56
18 25
4.6 -13.4
21.16 179.56
19 25
15 4.6
1.6 21.16
2.56 20
10 10
-10.4 -3.4
108.16 11.56
21 20
10 -0.4
-3.4 0.16
11.56 22
15 10
-5.4 -3.4
29.16 11.56
23 20
15 -0.4
1.6 0.16
2.56 24
25 25
4.6 11.6
21.16 134.56
25 30
15 9.6
1.6 92.16
2.56 N=25
510 335
1546.0 1336.0
This table shows that the lowest gained from experiment class X was 5 and from the control class Y was 0, and the highest gained from X was 35 and from
Y was 30. While the sum of gained scoreX was 510 and Y was 335. The sum of squared of gained score X was 1546.0 and Y was1336.0.
The writer calculated them based on the step of the t-test. The formulation as follows:
a. Determining Mean of variable X, with formula:
b. Determining Mean of variable Y, with formula:
c. Determining of Standard of Deviation score of Variable X, with formula:
√
√
=
√
= 7.86
d. Determining Standard Deviation Score of Variable Y, with formula:
√
√
√
e. Determining Standard Error Mean of Variable X, with formula:
√ √
√
f. Determining Standard Error Mean of Variable Y, with formula:
√ √
√
g. Determining Standard Error Mean of Difference Mean of variable X and Mean
of Variable Y, with formula:
√
+ √
h. Determining t0 with formula:
i. Determining t- table in significant level 5 and 1 with df.
Because the value of 48 is not mentioned in the table, the writer as uses the closer value to 48 which is 50 as degree of freedom df.
T
table
t
t
at significance 5= 2.01 T
table
t
t
at significance 1= 2.70