3.12 Treatment
The two groups are given different treatment. The experimental group got a treatment by using British Council Animation in teaching writing narrative. On
the other hand, the control group was taught by using conventional technique. It means that the students of control group got the explanation about the material of
writing narrative orally.
3.13 Post Test
As stated by Brown 2004: 3, a test is a method of measuring a person’s ability, knowledge, or performance in a given domain. After the try out and pre test is
collected, then it will be measured by the post test. The aim of conducting post test was to know the result of the students’ achievement after the treatments. Then
the data of post test were analyzed to see whether there was significant difference of students’ result in writing narrative achievement between those who were
taught using British Council Animation and those who were taught without this media.
3.14 Technique of Data Analysis
3.14.1 t-test statistical analysis
In doing this research the t-test table is used to interpret the t obtained. If the t
value
is higher than t
table
, it means that there is no significant difference between the two means. We can use the following formula:
t
2 1
n 1
n 1
s x
x
+ −
=
2 1
In which,
If the value of t
value
t
table
it is concluded that t
value
is significant, there is a difference between experimental and control group.
3.14.2 Normality
In order to prove the pre test post test of each group to be normally distributed, it is used the normality methods. We can use the following formula:
If the value of x
2 value
x
2 table
α , the pre test post test for each group is said to be normally distributed.
3.14.3 Homogeneity
To find out homogeneity of data. The writer uses the “homogeneity formula” as follows:
In which : Vb
: variant of the pre test post test of experimental group Vk
: variant of the pre test post test of controlled group If the F
value
F
table
, we can conclude that data of the pre test post test is homogeny
2 n
n s
1 n
s 1
n s
2 1
2 2
2 2
1 1
− +
− +
− =
∑
=
− =
k 1
i i
O
i i
E E
2 2
χ
VK Vb
F =
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