B. Research Findings
1. Description of Data
The writer enters experiment class amount 2 days, at the second days. She uses lesson plan for teaching vocabulary. After that the writer givens the experiment
and control class the test, and the test has the similar questions, and gave in the final meeting.
After she has finished the field research, the writer got the score as follows:
Table I The scores of experiment class
STUDENTS SCORES
1 50 2 50
3 55 4 65
5 65 6 65
7 70 8 70
9 70 10 70
11 75 12 75
13 80 14 85
15 85
16 85 17 90
18 90 19 90
20 90 21 90
22 90 23 90
24 95 25 100
N1 = 25 ∑ X1 = 1940
M1 = 77,6
Table II The scores of control class
STUDENTS SCORES
1 45 2 45
3 45 4 50
5 50 6 50
7 55 8 55
9 55 10 60
11 65 12 65
13 60 14 60
15 65 16 70
17 70 18 70
19 75 20 75
21 80 22 80
23 85 24 90
25 95 N
1
= 25 ∑ X
2
= 1615 M
1
= 64,5
2. Analysis of Data
After writing the comparison between the score of the experiment class and the control class, the writer calculates deviation and squared deviation for two classes
as follows:
Table III The comparison of scores of each student in the experiment class and control
class No
X
1
X
2
x
1
x
2
x
1 2
x
2 2
1 50
45 - 27.6
- 19.6 761. 76
384. 16 2
50 45
- 27.6 - 19.6
761. 76 384. 16
3 55
45 - 22.6
- 19.6 510. 76
384. 16 4
65 50
-12. 6 - 14. 6
158. 76 213. 16
5 65
50 -12. 6
- 14. 6 158. 76
213. 16 6
65 50
-12. 6 - 14. 6
158. 76 213. 16
7 70
55 - 7. 6
- 9. 6 57. 76
92. 16 8
70 55
- 7. 6 - 9. 6
57. 76 92. 16
9 70
55 - 7. 6
- 9. 6 57. 76
92. 16 10
70 60
- 7. 6 - 4. 6
57. 76 21. 16
11 75
65 - 2. 6
- 4. 6 5. 76
21. 16 12
75 65
- 2.6 - 4. 6
54. 76 21. 16
13 80
60 2. 4
0. 4 54. 76
0. 16 14
85 60
7. 4 0. 4
54. 76 0. 16
15 85
65 7. 4
0. 4 54. 76
0. 16 16
85 70
7. 4 5. 4
54. 76 29. 16
17 90
70 12. 4
5. 4 153. 76
29. 16 18
90 70
12. 4 5. 4
153. 76 29. 16
19 90
75 12. 4
10. 4 153. 76
108. 16 20
90 75
12. 4 10. 4
153. 76 108. 16
21 90
80 12. 4
15. 4 153. 76
237. 16 22
90 80
12. 4 15. 4
153. 76 237. 16
23 90
85 12. 4
20. 4 153. 76
416. 16
24 95
90 17. 4
25. 4 302. 76
645. 16 25
100
∑ X
1
= 1940
95
∑ X
2
= 1615 22. 4
∑ X
1
=0 30. 4
∑ X
2
= 0 501. 76
∑ X
1 2
= 4806 924. 16
∑ X
2 2
= 4896
From the table on the page above, the writer gets the score as: ∑ X
1
= 1940; ∑
X
2
= 1615 ; 0 ; ∑ X
2
= - 13 ; ∑ X
1 2
= 4806; ∑ X
2 2
= 4896, and N
1
and N
2
are= 25, after wards she calculates them based on the steps of the t –test formula as below:
a. M
1
= : ∑ X
1
= 1940 = 77, 6
b. M
2
= : ∑ X
2
= 1615 = 64,6
c. t =
] .
][ 2
[ ]
][ [
2 2
2 2
1 2
2 2
1 2
1
N N
N N
N N
X X
M M
− +
+ Σ
+ Σ
−
= 77. 6 – 64. 6
] 25
. 25
][ 2
25 25
] 25
25 ][
4896 4806
[ −
+ +
+
= 13
3000 485100
= 13 17
. 16
= .
4 13
= 3. 250
d. df = N
1
+ N
2
- 2 = 25 + 25 – 2 = 48
There is no degree of freedom for 48, so the writer uses the closer df and it is 50.
e. In degree of significance 5 from 50 in t – table = 2. 009
In degree of significance 1 from 50 in t – table = 2. 678 f.
The writer compared t
o
to t
t
that if t
o
t
t ;
Ho is rejected and Ha is accepted, but when t
o
t
t
it means that Ho is accepted and Ha is rejected.
t
o :
t
t =
3. 25 2.00 in degree of significance 5
t
o :
t
t =
3. 25 2.67 in degree of significance 1
3. Test of Hypothesis
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