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1. Analyzing the Significance of Students’ Speaking Improvement
The result of the analysis was taken from the pretest result on
March, 09
th
2015 and post test result on March, 20
th
2015. Data were
analyzed by ttest formula and the result can be seen on table V.
After the result was found, the t
obs
had been compared with the t
cri,
with α : 5 alpha: 0,05 t
obs
on degree of freedom is 30 see table VII, it
showed that t
cri
= 2,042. Now, looking back to the hypotheses:
H : Student’s speaking skill cannot significantly improve after
ThinkPairShare Technique using flashcards is implemented.
H
a
: Students’ speaking skill can significantly improve after
ThinkPairShare Technique using flashcards is implemented.
Will be taken if the t
cri
is smaller than t
obs
, it means that H
is rejected and H
a
is accepted.
By looking at the result that t
cri
is smaller than t
obs
2,042 15,7 it
concluded that H is rejected and H
a
is accepted.
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2. Analyzing the Significance Difference between Experimental and
Control Class
The result of the analysis was taken from the post test result of
experimental class and post test result of control class which did not apply
any treatment in speaking lesson. To make sure that both the control and
experimental class had not too much difference in speaking skill, the pretest
result from control class was needed to take. The control class 8B consists
of 30 students, the pretest result see table III was taken before the teacher
implements the speaking teaching technique based on the lesson plan that
constructed by teacher himself. Based on the table III, it shows that the mean
of the students’ speaking score was 12,4 points.
Then, the researcher implemented the speaking teaching technique
based on the lesson plan that constructed by the teacher see appendix VI.
The material was based on the theme ‘What is Your Favorite Story’ and the
teaching goal is to make students be able to response and create the
expression of asking for and giving agreement.
After that, researcher took post test of the control group and the
mean of the score was 12,6 points see table IV. It showed that their
speaking score was increased 0,2 points by the mean of their pretest score
which was 12,4 points.
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To analyze the significance different between experimental and
control class, researcher took from post test result from both sample. Both
had been compared by using formula. The data were analyzed by using ttest
formula and the result can be seen on table VI.
Before testing the significance, the hypothesis needs to be
constructed first:
H : There is no different in the speaking skill between experimental group
that apply ThinkPairShare technique using flashcard and control group
that does not apply ThinkPairShare technique
H
a
: There is significant different in speaking skill between experimental
group that apply ThinkPairShare Technique and the control group that
does not apply ThinkPairShare technique using flashcards.
Or can be written as:
H : µ
1
= µ
2
H
a
: µ
1 ≠
µ
2
To decide the ttest formula, the variance of the sample needed to
be tested that they are homogeny or not. Testing the sample homogeny used
F test by using the following formula:
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F = Highest Variance
Smallest Variance
Based on the table VI, it can be seen that the highest variance is 46
and the smallest one is 24. It can be said, 46 : 24 is 1,92. F
obs
needs to be
compared with F
cri
see table VIII with upper degree of freedom = 32 – 1 is
31 and lower degree of freedom = 30 – 1 was 29. By deciding the alpha α :
5, it can be seen that F
cri
= 1,85. In this case, the rule was taken ‘if F
obs
is
smaller or equal with F
cri,
it means that H
o
is accepted and H
a
is rejected’. If
H is accepted, it means that the variance is homogeny.
After test by using F formula, it was found that F
obs
was bigger
than F
cri
1,92 1,85, it means that the variance was not homogeny.
Because the nominal sample between experiment and control group was
different, the formula used for ttest was using separated variance see table
VI.
After the result was found, the t
obs
had been compared with the t
cri.
Because the nominal of two sample groups were different, the researcher
used substitute t
cri.
Where degree of freedom of experimental class = n
1
– 1
and degree of freedom of control class = n
2
– 1 with α: 5:
n
1
= 32; degree of freedom = 31, so t
cri
= 2,042
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n
2
= 30; degree of freedom = 29, so t
cri
= 2,045
Then, this t
cri
had been calculated from the difference value of t
cri
by dividing them to 2:
2,0452,042:2 = 0,0015
Next step was adding them with the smallest t
cri
:
2,042 + 0,0015 = 2,043
It concluded that substitute t
cri
is 2,043 which was smaller than t
obs
2,043 2,61. It means that H was rejected and H
a
was accepted.
C. Discussion