2. Normality of Post-test Hypothesis:
H
o
: Data of Y is normally distributed. H
1 :
Data of Y is not normally distributed.
Table 4.6 Normality of Post-test
∑ [
∑ ]
[ ]
= 5877.20 –[76.58]
2
= 5877.20 – 5864.49 = 12.71
S
2
= 12.71 SD =
√
M
y
= 76.58 L
ob
= 0.155467 L
tab
= 0.161
Criteria of the test: In the significant degree of 0.05, the value in the table of Lillyfors shows:
T
0,0530
= 0.161 Because n=39 is not mentioned in the table of Lillyfors, the writer used the closer value to n=39 that is n=30
H
1
: T 0.161 H
o
: T 0.161 The result showed that T
max
T
table
0.155467 0.161. Conclusion: In the significant degree of 0.05, H
o
is accepted. It means that the data is normally distributed.
2. Homogeneity of the Data
Based on the calculation of normality, the writer got the result that all data in pre-test and post-test of both experiment class and controlled class have been
normally distributed. The next step of the calculation was finding the homogeneity of the data. The purpose of this calculation was to see whether the
data sample in both classes were homogenous or heterogeneous.
Hypothesis: H
o
: The condition of experiment class is not different from controlled class. H
1
: The sample of experiment class is different from controlled class.
The criteria of the test: α = 0.05
H : F
αn1-1, n2-2
F F
αn1-1, n2-2
H
1
: F F
αn1-1, n2-2
The formula used can be seen as follows:
or
The calculation can be seen as follows:
n1-1 = 39-1 = 38 n2-1 = 39-1 = 38
F
0.05n1-1, n2-1
= 1.69 F
table
From the calculation, it can be seen that F F
αn1-1, n2-2
0.26 1.69. Based on the criteria, it can be conclude that H
is accepted. It means that the sample in experiment class and controlled class were homogenous.
3. Hypothesis Testing
In this part, the writer calculated the data to test the hypothesis that whether there is significant different between students’ writing skill in narrative text in
experiment class which pictures series and students’ writing skill in narrative text
in control class without pictures series. The writer calculated the data using T-test
formula. Two classes were compared, the experiment class was X variable and the controlled class was Y variable.
The formula of T-test was expressed as follows:
The calculation can be seen as follows: 1.
Determining Mean of variable X:
∑
2. Determining Mean of variable Y:
∑
3. Determining Standard of Deviation Score of Variable X:
√
∑
√ √
4. Determining Standard of Deviation Score of Variable Y:
√∑ √
√
5. Determining Standard Error Mean of Variable X:
√ √
√
6. Determining Standard Error Mean of Variable Y:
√ √
√
