Data Description Research Finding
Table 4.3 The Calculation Result of both Experimental and Controlled Classes
No. X
Y xX-M
1
yY-M
2
x
2
y
2
1. +19
+40 -21.36
8.96 456.2496
80.2816 2.
+43 +33
2.64 1.96
6.9696 3.8416
3. +46
+17 5.64
14.04 31.8096
197.1216 4.
+49 +33
8.64 1.96
74.6496 3.8416
5. +60
+17 19.64
14.96 385.7296
197.1216 6.
+16 +37
-24.36 5.96
593.4036 35.5416
7. +37
+33 -3.36
1.96 11.2896
3.8416 8.
+43 +30
2.64 -1.04
6.9696 1.0816
9. +40
+23 -0.36
-8.04 0.1296
64.6416 10.
+33 +40
-7.36 8.96
54.1696 80.2816
11. +34
+30 -6.36
-1.04 40.4496
1.0816 12.
+40 +30
-0.36 -1.04
0.1296 1.0816
13. +50
+30 9.64
-1.04 92.9296
1.0816 14.
+20 +33
-20.36 1.96
414.5296 3.8416
15. +50
+26 9.64
-5.04 92.9296
25.4016 16.
+67 +23
26.64 -8.04
709.6896 64.6416
17. +60
+36 19.64
4.96 385.7296
24.6016 18.
+49 +30
8.64 -1.04
74.6496 1.0816
19. +33
+40 -7.36
8.96 54.1696
80.2816 20.
+47 +23
6.64 -8.04
44.0896 64.6416
21. +43
+43 2.64
11.96 6.9696
143.0416 22.
+30 +39
-10.36 7.96
107.3296 63.3616
23. +30
+17 -10.36
14.04 107.3296
197.1216 24.
+23 +40
-17.36 8.96
301.3696 80.2816
25. +47
+33 6.64
1.96 44.0896
3.8416 =
� −6
86.6
4097.754 1401.42
After determining the mean of both experimental and controlled class X and Y,
to determine the Standard Deviation of experimental class X used a
formula as follow: SD
1
=
2
1
=
4097 .75 25
= 163.91 = 12.80
Meanwhile, to determine the Standard Deviation controlled class used a formula as following:
SD
2
=
2
2
=
1401 .42 25
= 56.06 = 7.49
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:
�
1
=
�
1 1
−1
=
12.80 25−1
=
12.80 4.90
= 2.61 Meanwhile, to determine the standard Error Mean of controlled class Y
used a formula as follow: �
2
=
�
2 2
−1
=
7.49 25−1
=
7.49 4.90
= 1.53 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 controlled class Y, used a formula as follow:
�
1
−
2
= �
1
2
+ �
2
2
= 2.61
2
+ 1.53
2
= 6.81 + 2.34
= 9.15 = 3.02
Then, the writer calculated t
observe
: �
�
=
1
−
2
�
1 − 2
=
40.36 −31.04
3.02
=
9.32 3.02
= 3.09
Finally, the writer calculated t
table
in significance level of 5 and with degree of freedom df:
df = N
1
+ N
2
– 2 = 25 + 25
– 2 = 50
– 2 = 48
Based on the degree of freedom, the writer gained the t-table: Degree of significance 5 = 1.67
From the calculation toward 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
= 3.09 1.67 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 hypothesis as tentative assumption below:
1. If t
o
t
t
= the alternative hypothesis Ha is accepted and the null hypothesis H
is rejected 2.
If t
o
t
t
= the alternative hypothesis Ha is rejected and the null hypothesis H
is accepted. From the data calculation, it was obtained the value of t
o
was 3.09 and the degree of freedom df was 48. In this research, the writer used the degree of
significance 5 where the value of degree of significance 5 is 1.67. By comparing the value t
o
= 3.09 and t
table
on the degree of significance 5 is 1.67, the writer made a conclusion of hypothesis that t
o
was higher than t
table
, namely 1.67 3.09. It means that the alternative hypothesis H
a
was accepted and the null hypothesis H
o
was rejected, so that using mind mapping technique is effective on students’ writing descriptive text.