Table 4.8 The comparison score of experimental class and controlled class
Students X
Y X-MX
Y-MX X-MX
2
Y-MX
2
1 42
10 8
-12 64
144 2
19 1
-15 -21
225 441
3 44
6 10
-16 100
256 4
53 20
19 -2
361 4
5 47
18 13
-4 169
16 6
36 29
2 7
4 49
7 58
25 24
3 576
9 8
43 25
9 3
81 9
9 37
25 3
3 9
9 10
52 31
18 9
324 81
11 33
-1 -22
1 484
12 35
10 1
-12 1
144 13
40 28
6 6
36 36
14 52
25 18
3 324
9 15
44 15
10 -7
100 49
16 20
30 -14
8 196
64 17
4 22
-30 900
18 34
10 -12
144 19
37 26
3 4
9 16
20 22
1 -12
-21 144
441 21
27 42
-7 20
49 400
22 20
35 -14
13 196
169 23
20 2
-14 -20
196 400
24 6
33 -28
11 784
121 25
6 4
-28 -18
784 324
26 22
27 -12
5 144
25 27
29 26
-5 4
25 16
28 27
38 -7
16 49
256 29
29 27
-5 5
25 25
30 40
39 6
17 36
289 31
43 25
9 3
81 9
32 41
37 7
15 49
225 33
53 33
19 11
361 121
34 41
33 7
11 49
121
N=34 Ʃ χ=1156
Ʃ γ=758 Ʃ χ=0
Ʃ γ=0 Ʃ χ
2
=6452 Ʃ γ
2
=4906 MEAN
Mχ=34 Mγ=22.294
Mχ
2
=13.983 Mγ
2
=12.186
Then the writer calculated the data based on the step of t-test. The formulatin as follow:
1. Determining Mean of variable X, with formula: M
x
=
M
x
= M
x
= 34 2. Determining Mean of variable Y, with the formula:
M
y
=
M
y
= M
y
= 22.29 3. Determining Standard of Deviation Score of variable X, with the formula :
SD
x
=
√
SD
x
=
√
SD
x
= √
SD
x
= 13.98 4. Determining Standard of Deviation Score of variable Y, with the formula:
SD
y
=
√
SD
y
=
√
SD
y
= √
SD
y
= 12.186 5. Determining Standard Error of Mean of variable X, with the formula:
SE
x
=
√
SE
x
=
√
SE
x
=
√
SE
x
=
SE
x
= 2.398 6. Determining Standard Error of Mean of variable Y, with the formula:
SE
y
=
√
SE
y
=
√
SE
y
=
√
SE
y
=
SE
y
= 2.090 7. Determining Standard Error of Different Mean of variable X and variable Y,
with the formula: SE
MX-My
=
√
SE
MX-My
=
√ SE
MX-My
=
√ SE
MX-My
=
√
SE
MX-My
=
3.181 8. Determining t
o
with the formula:
t
o
=
t
o
=
t
o
=
t
o
=
3.68 9. Determining t-table in significance level 5 with degree of freedom df:
df = N
1
+ N
2
– 2 df = 34 + 34
– 2 df = 68-2
df = 66 a. Finally the writer got the degree of freedom df is 34+34= 68-2= 66.
From the t table t
table
, df = 66 the writer took degree of significance 5 0.05 to interpret the t
o
that have been gained, that is : In degree of significance 5, t
t
= 2.00 b. The t
o
= 3.68, so the result of t
t
: t
o
in degree of significance 5 is, t
o
: t
t
= 3.68 =2.00 t
o
t
t
It means that t
o
is higher than t
t
in degree of significance 5.
C. Data Interpretation
This part covers the discussion of the teaching writing by using Think Pair Share Technique and then interprets the result of data. The
discussion is based on the research question, which was whether there was any significant difference on
students’ writing ability between the students who was taught of Descriptive writing using Think Pair Share Technique
and Conventional method. In this research, there were 34 students in experimental and 34 in
the controlled class. Therefore, the degree of freedom df is 34+34= 68- 2= 66. The writer took the degree of significance in level 5 to give the
interpretation of the result of this research. With df = 66 the degree of significance in level 5 = 2.00.
From the calculation by using SPSS and manual formula the result of t-test is same, it can be seen in the table 4.6 and 4.8. The data shows
that the experimental class has the higher score than controlled class in pre-test, post-test and gained score. So, it was a significant difference from
measurement score in the experimental and the controlled class Mx= 34, My= 22.29, SDx= 13.98, SDy= 12.186, t 66= 2.00, p=0.00.
Then, the writer compares the t
o
with the value of t
t ,
the result is t :
t
t
= 3.682.00. The comparison between t
o
and t
table
indicates that t
o
is higher than t
table
in the degree of significance of 5. As it is known that t
o
is higher than t
table
in the degree of significance 5, it means that H
a
is accepted in degree of significance 5. In another words, it means that H
o
is rejected. It can be concluded that there is a significant difference in result between students
’ ability of SMAN 2 Kabupaten Tangerang grade X in learning descriptive writing using Think Pair Share Technique and
using conventional method. Which is concluded that the use of Thnink Pair Share
Technique influences the students’ writing ability.
53
CHAPTER V CONCLUSION and SUGGESTION
A. CONCLUSSION
Based on the research conducted in grade X1 and grade X5 at SMAN 2 Kabupaten Tangerang, in academic year 20132014, it can be concluded that
teaching Descriptive Writing using Cooperative Learning Think Pair Share Technique to the students of SMAN 2 Kabupaten Tangerang grade X gives
positive influence to the students’ achievement.
The result of the analysis in the research can be seen in the previous chapter, that t
o
value is 3.68 and t
table
value is 2.00 in degree of significant 5. It can be concluded that t
o
is higher than t
t
t
o
: t
t
= 3.6 2.00. So, the hypothesis H
a
is accepted because t
o
t
t
, it can be infered that teaching descriptive writing by using think pair share technique gives the positive
influence to the tenth grade students’ achievement on SMAN 2 Kabupaten Tangerang.
B. SUGGESTION
In this part, the writer would like to give some suggestion for the teacher, as follow:
1. First the teacher should deliver material clearly and also should pay attention to the student’s activity while teaching and learning processes,
in this case the teacher should focus on teaching and learning strategy not only focus on transfering information.
2. Second, the teacher should be more creative providing the topic which can motivate students to write.
3. The last is Think Pair Share technique can be applied in English learning process as one of the innovation in teaching and learning
process, particularly to improve students ’ ability in writing descriptive
text.
54
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