21
BAB IV RESEARCH FINDING AND INTERPRETATION
A. Data Description
The data of students’ achievement divided into two kinds, namely the data in experiment class and the data in control class, which were gained from pre-test
and post-test that were applied in both of class.
The test was arranged in a blank paper for pre-test and post-test. To know the result of the test, it will be presented on the table below:
Table 4.1 The Test Result of Experiment Class
Students Pre-Test Score
Post-Test Score Gained Score
1. 33
80 47
2. 60
87 27
3. 40
60 20
4. 27
93 66
5. 47
80 33
6. 67
93 26
7. 47
80 33
8. 40
47 7
9. 33
80 43
10. 67
87 20
11. 53
87 34
22
12. 47
87 40
13. 47
93 46
14. 47
60 13
15. 47
80 33
16. 40
87 47
17. 67
80 13
18. 33
73 40
19. 40
67 27
20. 27
80 53
21. 40
73 33
22. 13
80 67
23. 47
80 33
24. 67
93 26
25. 60
87 27
26. 33
87 54
27. 47
60 13
28. 67
73 6
29. 67
80 13
30. 57
87 30
31. 60
80 20
32. 60
87 27
33. 47
60 13
1574 2608
1030 M
47.7 79
As mentioned in the table, it can be clarified that, the mean score of pre- test in experiment class was 47.7, while the mean score of post-test was 79. The
total gained score in this class was 1030. It can be concluded there was significant difference in the pre-test and post-test.
23
Table 4.2 The Result of Control Class
Students Pre-Test
Post-Test Gained Score
1. 33
33 2.
53 47
-6 3.
47 53
6 4.
73 60
-13 5.
53 47
-6 6.
47 33
-14 7.
27 60
33 8.
67 73
6 9.
27 47
20 10.
67 47
-20 11.
13 47
34 12.
40 40
13. 47
47 14.
20 60
40 15.
60 40
-20 16.
40 40
17. 27
47 20
18. 27
53 26
19. 40
47 7
20. 40
53 13
21. 40
67 27
22. 33
40 7
23. 47
53 6
24. 27
53 26
25. 27
47 20
26. 27
47 20
27. 33
53 20
28. 33
60 27
24
29. 40
33 -7
30. 33
60 27
31. 40
33 -7
32. 47
53 6
33. 33
53 20
∑ 1308
1626 318
M 39.6
49.3 The table II above informs that the mean score of pre-test in control class
was 39.6, while the mean score of post-test here is 49.3. There is still improvement score between pre-test and post-test but not as significance as in the
experiment class. The total gained score 318. From the tables presented above, the writer can conclude the experiment
class learning vocabulary using guessing game got the higher score than in the control class learning vocabulary without using guessing game.
As mentioned before, 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.
Table 4.3 Standard Deviation Table
Students X1
gained Score
X2 gained
score X
1
X2 X
1 2
X
2 2
1. 47
15.8 -9.6
249.64 92.16
2. 27
-6 -4.2
-15.6 17.64
243.36 3.
20 6
-11.2 -3.6
125.44 12.96
4. 66
-13 34.8
-22.6 1211.04
510.76 5.
33 -6
1.8 -15.6
3.24 243.36
6. 26
-14 -5.2
-23.6 27.04
556.96 7.
33 33
1.8 23.4
3.24 547.56
8. 7
6 -24.2
-3.6 585.64
12.96
25
9. 43
20 11.8
10.4 139.24
108.16 10.
20 -20
-11.2 -29.6
125.44 876.16
11. 34
34 2.8
24.4 7.84
595.36 12.
40 8.8
-9.6 77.44
92.16 13.
46 14.8
-9.6 219.04
92.16 14.
13 40
-18.2 30.4
331.24 924.16
15. 33
-20 1.8
-29.6 3.24
876.16 16.
47 15.8
-9.6 249.64
92.16 17.
13 20
-18.2 10.4
331.24 108.16
18. 40
26 8.8
16.4 77.44
268.96 19.
27 7
-4.2 -2.6
17.64 6.76
20. 53
13 21.8
3.4 475.24
11.56 21.
33 27
1.8 17.4
3.24 302.76
22. 67
7 35.8
-2.6 1281.64
6.76 23.
33 6
1.8 -3.6
3.24 12.96
24. 26
26 -5.2
16.4 27.04
268.96 25.
27 20
-4.2 10.4
17.64 108.16
26. 54
20 22.8
10.4 519.84
108.16 27.
13 20
-18.2 10.4
331.24 108.16
28. 6
27 -25.2
17.4 635.04
302.76 29.
13 -7
-18.2 -16.6
331.24 275.56
30. 30
27 -1.2
17.4 1.44
302.76 31.
20 -7
-11.2 -16.6
125.44 275.56
32. 27
6 -4.2
-3.6 17.64
12.96 33.
13 20
-18.2 10.4
331.24 108.16
N= 33 ∑X1=1030 ∑X2=318 ∑X
1
= 0.4 ∑X
1
= 1.2 ∑X
1 2
= 7903.52 ∑X
2 2
= 8465.68 Based on the data presented in the table, it shows that the lowest gained
score was 0 and the highest was 67. Furthermore, after decided the calculation of the text both pre-test and post-test, the writer used t-test formula according Anas
26
Sudijono, to find out the effectiveness of teaching vocabulary by using guessing game.
1
M1
= M2 =
= =
= 31.2 = 9.6
t
o
=
√[ ][
]
t
o
=
√
{ } { } { } { }
t
o
=
√
{ } { } { } { }
t
o
=
√
t
o
=
√
t
o
=
1
Anas Sudijono, Pengantar Statistik Pendidikan, Jakarta: PT. Raja Grafindo Persada, p. 308
27
t
o
= 5.48
Determining t-table in significance level 5 with df: d
= N1 + N2 – 2
= 33 + 33 – 2
= 64 The value of t
table
is 1.67. From the result of the statistic calculation, it can be seen that the value of t
o
is 5.48 and the degree of freedom in the table of significance that on the d
= 64 The comparison between t
observation
t
o
is higher than t
t
t
table
B. Test of Hypothesis
Kategori