h. Determining with formula:
= i. Determining t-table in significance level 5 with degree of
freedom df: Df = N1 + N2 – 2
Variable X : Experiment Class Teaching the Present
Continuous Tense through ALM Variable Y
: Controlled Class Teaching the Present Continuous Tense through GTM
j. The Hypotheses The statistic hypothesis stated:
Null Hypothesis Ho: There is no effectiveness between teaching the present continuous tense through Audio-Lingual Method ALM and
Grammar Translation Method GTM for the second grade at Al-Binaa of Islamic Boarding School Bekasi.
Alternative Hypothesis Ha: there is effectiveness between teaching the present continuous tense through Audio-Lingual Method ALM and
Grammar Translation Method GTM for the second grade at Al-Binaa of Islamic Boarding School Bekasi.
B. Research Finding
1. Description of Data
As mention before, the writer took the experiment. She got the data from pre-test and post-test of teaching the present continuous tense
observation.
In this part, the writer gives the report concerning the data description according to pre-test and post-test from experiment and
control class. The writer explains their scores, as follows:
Table 2.1 The Students’ Scores of Experiment Class VIII. G
Students N1
Pre-Test Post-Test
Gained Scores
1. 68
96 28
2. 81
84 3
3. 93
93 4.
85 96
11 5.
100 97
- 3 6.
68 74
6 7.
85 85
8. 80
96 16
9. 54
79 25
10. 62
89 27
11. 80
82 2
12. 66
93 27
13. 89
97 8
14. 81
93 12
15. 85
93 8
16. 89
85 - 4
17. 79
81 2
N 1345
1513 168
Statistical Data of Pre-test of experiment class Frequency Distribution
High Score H and Lowest Score L H = 100
L = 54 Range
R = H – L
= 100 – 54
= 46 Number of Class K
K = 1 + 3.3 log N = 1 + 3.3 log 17
= 1+ 8.814 = 9.814
= 10 Interval
I =
R K
=
46 10
= 4.6 = 4.6 + 1
= 5.6 = 6
Table 2.2 Interval of pre-test of the students of the experiment class
No. Interval Score
Number of Students
1. 94
– 100 1
2. 87
– 93 3
3. 80
– 86 7
4. 73
– 79 1
5. 66
– 72 3
6. 59
– 65 1
7. 52
– 58 1
Total 17
Mean M
x
=
∑X N
=
1345 17
= 79.11 Median
N = 2n + 1 54 62 66 68 68 79
80 80 81 81 85 85 85 89 89 93 100
N = 2n + 1 17 = 2n + 1
17 – 1 = 2n
2n = 16 n = 8
The result of median is the number 8 + 1 = 9
= 81 Statistical Data of Post-test of experiment class
Frequency Distribution High Score H and Lowest Score L
H = 97 L = 74
Range R = H
– L = 97
– 74 = 23
Number of Class K K = 1 + 3.3 log N
= 1 + 3.3 log 17 = 1 + 1.060
= 5. 060 = 5
Interval
I =
R K
=
23 5
= 4.6 = 5
Table 2.3 Interval of post-test of the students of the experiment class
No. Interval Score
Number of Students
1. 92
– 97 9
2. 86
– 91 1
3. 80
– 85 5
4. 74
– 79 2
Total 17
Mean M
x
= ∑x
N
=
17
= 89 Median
N = 2n + 1 74 79 81 82 84 85
85 89 93 93 93 93 96 96 96 97 97
N = 2n + 1 17 = 2n + 1
17 – 1 = 2n
2n = 16 n
= 8 The result of median is the number
8 + 1 = 9 = 93
From the data of experiment class above, the writer concludes that the lowest score of the pre-test is 54 and the high score of pre-test
is 100. So, average score of the pre-test is 79.11. The lowest score of the pre-test is 74 and the high score of post-test is 97. And then, the
average score for post-test 89 and for gained score 9.9. It means that the post-test score is higher than pre-test.
The following table is the result of the students in control class through Grammar Translation Method in teaching present continuous
tense. Table 2.4
The Students’ Scores of Control Class VIII. H
Students N1
Pre-Test Post-Test
Gained Scores
1. 76
84 8
2. 79
89 10
3. 90
97 7
4. 97
96 - 1
5. 92
97 5
6. 71
88 17
7. 75
76 1
8. 80
84 4
9. 53
64 11
10. 79
89 10
11. 55
60 5
12. 97
97 13.
63 72
9 14.
93 93
15. 79
97 18
16. 76
85 9
N 1255
1368 113
Statistical Data of Pre-test of controlled class Frequency Distribution
High Score H and Lowest Score L H = 97
L = 53 Range
R = H – L
= 97 – 53
= 44
Number of Class K K = 1 + 3.3 log N
= 1 + 3.3 log 16 = 1 + 8.29
= 9.29 = 9
Interval
I =
R K
=
44 9
= 4.89 = 5
Table 2.5 Interval of pre-test of the students of the control class
No. Interval Score
Number of Students 1.
92 – 97
4 2.
86 – 91
1 3.
80 – 85
1 4.
74 – 79
6 5.
68 – 73
1 6.
62 – 67
1 7.
56 – 61
- 8.
50 - 55 2
Total 16
Mean M
x
= ∑x
N =
16
= 78.43 Median
N = 2n + 1 53 55 63 71 75 76
76 79 79 79 80 90 92 93 97 97
N = 2n + 1 16 = 2n + 1
16 – 1 = 2n
2n = 15 n
= 7.5 = 7
The result of median is the number 7 + 1 = 8
= 79 Statistical Data of Post-test of controlled class
Frequency Distribution
High Score H and Lowest Score L H = 97
L = 64 Range
R = H – L
= 97 – 60
= 37 Number of Class K
K = 1 + 3.3 log N = 1 + 3.3 log 16
= 1 + 3.973 = 4.973
= 5 Interval
I =
R K
=
37 5
= 7.4 = 7
Table 2.6 Interval of post-test of the students of the control class
No. Interval Score
Number of Students
1. 91
– 100 6
2. 81
– 90 6
3. 71
– 80 2
4. 61
– 70 2
Total 16
Mean M
x
= ∑x
N =
16
= 86.56 Median
N = 2n + 1 60 64 72 76 84 84
85 88 89 89 93 96 96 97 97 97 97
N = 2n + 1 16 = 2n + 1
16 – 1 = 2n
2n = 15
n = 7.5 = 7
The result of median is the number 7 + 1 = 8
= 88 From the data of controlled class above, the writer concludes that
the lowest score of the pre-test is 53 and the high score of pre-test is 97. So, average score of the pre-test is 78.43. The lowest score of the post-test
is 64 and the high of post-tests core is 97. And then, the average score for post-test 85.5 and for gained score 7.06. It means that the post-test score is
higher than pre-test
2. The Analysis of Data
