40 S-40 4 3 4 4 3 18 72
41 S-41 3 2 3 2 3 13 52
42 S-42 3 1 2 4 3 13 52
43 S-43 2 2 3 4 4 15 60
44 S-44 4 2 4 3 4 17 68
45 S-45 3 3 3 3 3 15 60
46 S-46 4 3 3 3 4 17 68
47 S-47 4 2 4 4 4 18 72
48 S-48 3 3 4 4 4 18 72
∑ 48
140 114 156 161 166 737 2952
Table 4. The Result of the Students’ Achievement in the First Cycle.
The average of the students result =
t ftheStuden
TheNumbero rcentage
TheTotalPe ×100
= 100
48 2952 ×
= 61.50 According to the analysis, the result of the students’ achievement was
higher than pre-test. The average of the students’ achievement in the first cycle was 61.50. It showed that by using STAD technique, the teaching and learning
process was successful. Based on the evaluation, there were some students who still made some
mistakes in terms of grammar. They still had difficulties constructing sentences in simple present tense. Consequently, the writer conducted the next cycle by
emphasizing on grammar.
4.1.2.4 Fourth Activity Cycle II
The fourth activity was the implementation of the whole range of activity cycle II. It was conducted on December 13
th
and 15
th
, 2008. In general, the procedure of the teaching and learning in this cycle was the same as the previous
cycle. The main focus of the treatment was to eliminate students’ difficulties in constructing sentences in simple present tense. In this activity, the students were
doing the same activities like in cycle 1. The difference was the title given. The writer still used STAD technique in this activity.
The result of the students’ achievement in the third activity cycle II can be seen as follows:
The Result of the Students’ Achievement in the Second Cycle Component of Writing Scoring
No. Students
Code F G V C S
Score
1 S-01 3 3 4 4 3 17 68
2 S-02 4 4 5 4 5 22 88
3 S-03 3 2 3 3 4 15 60
4 S-04 3 2 4 4 4 17 68
5 S-05 3 2 3 3 4 15 60
6 S-06 4 2 3 4 3 16 64
7 S-07 3 3 3 3 4 16 64
8 S-08 4 4 3 4 5 20 80
9 S-09 3 2 4 4 3 16 64
10 S-10 4 3 3 3 3 16 64
11 S-11 3 4 3 3 4 17 68
12 S-12 3 3 4 4 3 17 68
13 S-13 4 2 4 3 4 17 68
14 S-14 3 3 3 4 3 16 64
15 S-15 4 3 4 4 3 18 72
16 S-16 3 3 4 3 3 16 64
17 S-17 3 2 3 3 3 14 56
18 S-18 5 2 3 4 3 17 68
19 S-19 4 4 3 3 4 18 72
20 S-20 3 2 3 3 3 14 56
21 S-21 3 4 3 4 4 18 72
22 S-22 4 3 3 3 3 16 64
23 S-23 3 2 3 3 2 13 54
24 S-24 5 3 3 3 4 18 72
25 S-25 3 2 3 3 3 14 56
26 S-26 5 3 4 3 3 18 72
27 S-27 3 2 4 4 3 16 64
28 S-28 4 3 4 3 4 18 72
29 S-29 3 3 4 2 3 15 60
30 S-30 4 2 4 4 5 19 76
31 S-31 4 3 3 3 4 17 68
32 S-32 3 2 2 3 4 14 56
33 S-33 3 2 4 4 4 17 68
34 S-34 4 3 4 4 4 19 76
35 S-35 5 3 5 4 4 21 84
36 S-36 3 2 3 3 4 15 60
37 S-37 4 3 3 4 4 18 72
38 S-38 3 3 4 3 4 17 68
39 S-39 2 2 4 4 5 17 68
40 S-40 3 3 4 4 4 18 72
41 S-41 2 3 3 2 3 13 52
42 S-42 3 1 3 4 4 15 60
43 S-43 3 2 4 4 4 17 68
44 S-44 4 3 4 4 5 20 80
45 S-45 3 3 3 3 5 17 68
46 S-46 4 4 3 3 4 18 72
47 S-47 5 3 4 4 4 20 80
48 S-48 4 3 4 4 5 20 80
∑ 48
168 130 168 166 180 812 3250
Table 5. The Result of the Students’ Achievement in the Second Cycle.
The average of the students result =
t ftheStuden
TheNumbero rcentage
TheTotalPe ×100
= 100
48 3250 ×
= 67.70 From this activity, the students’ achievement was higher than treatment I.
It showed that there was an increase from treatment I to treatment II that was from 61.50 to 67.70. Therefore, it can be concluded that the third activity was
successful.
4.1.2.5 Fifth Activity Post-Test
