33
CHAPTER IV RESEARCH FINDINGS AND INTERPRETATION
A. Data Description
In the data description, the writer described the data which had been collected. Those data are the students’ pretest and posttest scores of writing a
descriptive text. The writer provided a table to present the pretest and posttest scores of both experimental and controlled classes.
Table 4.1 Students’ score of experimental class
Students N1
Pre-test Post-test
Gained Score
S1 50
63 +13
S2 47
60 +13
S3 47
57 +10
S4 40
43 +3
S5 43
53 +10
S6 47
43 −3
S7 40
47 +7
S8 53
73 +20
S9 37
50 +13
S10 37
50 +13
S11 47
43 −4
S12 60
80 +20
S13 47
47 S14
50 50
S15 50
77 +27
S16 60
57 −3
∑X 139
Table 4.2 Students’ score of controlled class
Students N1
Pre-test Post-test
Gained Score
S1 30
37 +7
S2 40
37 −3
S3 40
37 −3
S4 40
47 +7
S5 30
30 S6
57 50
−7 S7
30 37
+7 S8
30 30
S9 43
43 S10
40 47
+7 S11
40 43
+3 S12
43 47
+4 S13
37 43
+6 S14
27 27
S15 30
27 −3
S16 27
30 +3
∑Y 28
B. Data Analysis
Based on the pre-test and post-test result scores of both experimental and controlled classes, the writer analyzed the data by using some formulations to
know the “t” value by using the degree of significant level 5 and to know the significant effectiveness of using
clustering technique to students’ descriptive writing.
First of all, the writer determined the mean of X or experimental class and mean of Y or controlled class. To determine the mean of X by using this formula:
M
1
=
∑
= = 8.7
Meanwhile, the writer also determined the mean of Y or controlled class by using the following formula:
M
2
=
∑
= = 1.75
To make clearer, the writer provided the table to show the result of mean of experimental and controlled classes as following:
Table 4.3 The calculation result of both experimental and controlled classes
No. X
Y x X-M
1
y Y-M
2
x
2
y
2
1. +13
+7 4.3
5.25 18.49
27.5625 2.
+13 −3
4.3 -4.75
18.49 22.5625
3. +10
−3 1.3
-4.75 1.69
22.5625 4.
+3 +7
-5.7 5.25
32.49 27.5625
5. +10
1.3 -1.75
1.69 3.0625
6. −3
−7 -11.7
-8.75 136.89
76.5625 7.
+7 +7
-1.7 5.25
2.89 27.5625
8. +20
11.3 -1.75
127.69 3.0625
9. +13
4.3 -1.75
18.49 3.0625
10. +13
+7 4.3
5.25 18.49
27.5625 11.
−4 +3
-12.7 1.25
161.29 1.5625
12. +20
+4 11.3
2.25 127.69
5.0625 13.
+6 -8.7
4.25 75.69
18.0625 14.
-8.7 -1.75
75.69 3.0625
15. +27
−3 18.3
-4.75 334.89
22.5625 16.
−3 +3
-11.7 1.25
136.89 1.5625
∑X=139 ∑Y=28 ∑x = 0
∑y = 0 ∑x
2
= 1289.44
∑y
2
= 293
After determining the mean of both experimental and controlled class X and Y, the writer determined the Standard Deviation of experimental class X
used a formula as follow:
SD
1
=
√
∑
=
√ √
= 8.98 Meanwhile, to determine the Standard Deviation of controlled class Y used
a formula as follow:
SD
2
=
√
∑
=
√ √
= 4.3 Then, the result of both standard deviation X and Y the writer calculated
the Standard Error Mean of experimental class X uses a formula as follow:
=
√
√
=
√
=
2.3 Meanwhile, to determine the Standard Error Mean of controlled class Y
used a formula as follow:
=
√
=
√
=
√
=
1.1
After determining the standard deviation of both experimental and controlled class, the writer calculated the difference of standard error between
mean of experimental class X and mean of controlled class Y, used a formula as follow:
– = √
2
+
2
√ 2.3
2
+ 1.1
2
= √5.29 + 1.21
= √6.5 = 2.5
Then, the writer calculated t
observed
:
t
o
=
t
o
=
t
o
=
t
o
=
2.78
Finally, the writer calculated the t
table
in significance level of 5 and with degree of freedom df:
df = N
1
N
2
2 df = 16+16
– 2 = 30
Based on the degree of freedom df, the writer gained the t-table: Degree of significance 5 = 2.04
From the calculation towards the result of pre-test and post-test of both experimental and controlled classes, the result shows that the obtained score of
experimental class is higher than that of controlled class. Besides, the writer obtained the comparison between t
o
and t
t
for the degree of significance 5: Degree of significance 5: t
o
t
t
= 2.78 2.04 The data which have been calculated by using t-test, based on the result pre-
test and post-test of both experimental and controlled classes is to prove the research hypotheses as the tentative assumption below:
1. If t
o
t
t :
the alternative hypotheses Ha is accepted and the null hypotheses Ho is rejected. It means that there is a significant difference
between the result of students’ descriptive writing by using clustering technique and students’ descriptive writing without using clustering
technique for the second grade students of MTs. Nurul Hidayah Jeruk Purut. 2.
If t
o
t
t :
the alternative hypotheses Ha is rejected and the null hypotheses Ho is accepted. It means that there is no significant difference between
student s’ descriptive writing by using clustering technique and students’
descriptive writing without using clustering technique for the second grade students of MTs. Nurul Hidayah Jeruk Purut.
From the data calculation, the value of t
o
is 2.78 and the degree of freedom df is 30. In this research, the writer used the degree of significance 5 where the
value of degree of significance 5 is 2.04. By comparing the value of t
o
= 2.78 and t-table on the degree of significance 5 = 2.04 the writer made a conclusion of the hypotheses that t
o
is higher than t- table; 2.04 2.78. It meant that the alternative hypotheses Ha is accepted and
the null hypotheses is rejected Ho. Therefore, the using of clustering technique is effective to students’ descriptive writing.
C. Data Interpretation
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