32
Criteria of the test: In the significant degree of 0.05, the value in the table of Lilyfors shows:
T
0,0524
= 0.173 H
= T 0.173 H
1
= T 0.173 The result showed that T
max
T
table
0.165 0.173, it means that the data is normally distributed.
c. Normality of Pre-test in the Control Class
Hypotheses H
o
= Data of X is normally distributed H
1
= Data of X is not normally distributed
Table 4.10
Calculation of Pre-test Normality in Controlled Class
x f
fx x2
fx2 p=fn
z=xi- xs
∅ ∑p
T= ∅-∑p
42 1
42 1764
1764 0.041667 -1.72812 0.04198 0.041667
0.0003165 45
3 135
2025 6075
0.125 -1.48262 0.06909 0.166667
0.0975793 48
1 48
2304 2304 0.041667
-1.23712 0.10802 0.208333 0.1003127
50 2
100 2500
5000 0.083333 -1.07346 0.14153 0.291667
0.1501337 52
2 104
2704 5408 0.083333
-0.90979 0.18147 0.01123
0.1702363
56 1
56 3136
3136 0.041667 -0.58246 0.28013 0.416667
0.1365378 60
1 60
3600 3600 0.041667
-0.25513 0.39931 0.458333 0.0590207
64 1
64 4096
4096 0.041667 0.072206 0.52878 0.5
0.0287809 65
2 130
4225 8450 0.083333 0.154039 0.56121 0.583333
0.0221229 68
1 68
4624 4624 0.041667 0.399538 0.65525
0.625 0.0302515
70 1
70 4900
4900 0.041667 0.563204 0.71335 0.666667 0.0466853
71 1
71 5041
5041 0.041667 0.645037 0.74055 0.708333 0.0322151
73 1
73 5329
5329 0.041667 0.808703 0.79066 0.75
0.0406571 75
3 225
5625 16875
0.125 0.972369 0.83457 0.875
0.0404334 76
1 76
5776 5776 0.041667 1.054202
0.8541 0.916667 0.0625618
78 1
78 6084
6084 0.041667 1.217868 0.88836 0.958333 0.0699703
80 1
80 6400
6400 0.041667 1.381535 0.91644 1
0.0835573
1073 24
1480 70133
94862 63.12 1.41 87,06 4125.47 5580.12
� = ∑ �
� − [ ∑ �
� ] =
− ⌈ ⌉
= . − [ . ]
= . −
. =
. S
= √ .
S =
. S
= 12.22 S
2
= 149.39 M
= 63.12 T
max
= 0.170 T
table
= 0.173
Criteria of the test: In the significant degree of 0.05, the value in the table of Lilyfors shows:
T
0.0524
= 0.173 H
= T 0.173 H
1
= T 0.173 The result showed that T
max
T
table
0.170 0.173, it means that the data is normally distributed.
d. Normality of Post-test in the Control Class
Hypotheses H
o
= Data of X is normally distributed H
1
= Data of X is not normally distributed
Table 4.11
Calculation of Post-test Normality in Controlled Class
x f
fx x2
fx2 p=fn
z=xi- xs
∅ ∑p
T= ∅-∑p
50 3
150 2500
7500 0.125
-1.99475 0.023035 0.125
0.10196 55
1 55
3025 3025 0.041667
-1.50262 0.066468 0.166667 0.1002
58 1
58 3364
3364 0.041667 -1.20735 0.113649 0.208333
0.09468 64
2 128
4096 8192 0.083333
-0.6168 0.268684 0.291667 0.02298
65 2
130 4225
8450 0.083333 -0.51837 0.302099
0.375 0.0729
70 3
210 4900
14700 0.125
-0.02625 0.48953
0.5 0.01047
72 1
72 5184
5184 0.041667 0.170604 0.567732 0.541667 0.026066 73
1 73
5329 5329 0.041667 0.269029 0.606046 0.583333 0.022713
74 1
74 5476
5476 0.041667 0.367454 0.64336
0.625 0.01836
75 1
75 5625
5625 0.041667 0.465879 0.679349 0.666667 0.012682 76
1 76
5776 5776 0.041667 0.564304 0.713727 0.708333 0.005393
78 2
156 6084
12168 0.083333 0.761155 0.776718 0.791667 0.01495
80 3
240 6400
19200 0.125 0.958005
0.83097 0.916667 0.0857
81 1
81 6561
6561 0.041667 1.05643 0.854614 0.958333
0.10372 83
1 83
6889 6889 0.041667 1.253281 0.894948
1
0.10505 1054
24 1661
75434 117439
70.27 1.60 110.73 5028.93 7829.27
� = ∑ �
� − [ ∑ �
� ] =
− ⌈ ⌉
= , −
. =
. S
= √ .
� = .
S = 10.16
S
2
= 103.27 M
= 70.27 T
max
= 0.105 T
table
= 0.173
1 ∑
= =
.
∑ =
= . Criteria of the test:
In the significant degree of 0.05, the value in the table of Lilyfors shows: T
0.0524
= 0.173 H
= T 0.173 H
1
= T 0.173 The result showed that T
max
T
table
0.105 0.173, it means that the data is normally distributed.
From the result of statistical calculation, it can be seen that normality of pre-test is 0.128, post test 0.165 in the experimental class, while T
table
= 0.173 and normality of pre test 0.170, post test 0.105 in the control class.
Since the population is normally chosen, distributed normally.
C. Data Analysis
After getting the normality of experiment and control class, the writer made an analysis of data from the result of comparing between both of class
VIII A and VIII B see Table 4.3. Afterwards, the writer calculated them based on the step of the t-test formula, as follow:
1. Determining Mean X
Mean variable X = 2.
Determining Mean Y
Mean variable Y =
3. Determining of Standard Deviation of variable X
� = √
∑ = √
. = √ . = .
� =
� √ −
= .
√ − =
. √
= ,
. = .
= √ . + .
= 2.13
� − �
= √� + �
= √ . + .
4. Determining of Standard Deviation of variable Y
5. Determining Standard of Error Mean of variable X
6. Determining Standard of Error Mean of variable Y
7. Determining Standard of Error Mean difference of M
x
dan M
y
8. Determining t
o
:
9. Determining t-table in significant level 5 and 1 with df
�� = +
− =
+ − =
At the degree of significant of 5 = 1.7 At the degree of significant of 1 = 2.4
� = √
∑ = √
. = √ .
= .
�
�
= −
� − �
= . − .
. =
. .
= . �
= �
√ − =
. √ −
= .
√ =
. .
= . 37
10. The comparison between t-score with t-table
t-score = 1.7 2.72 2.4 D.
The Testing of The Hypothesis
The research was held to answer the question wheter using picture sequence is effective on students’ writing of narrative text at the eight grade
students of SMP Islam Al Syukro Universal Ciputat. In order to provide the answer for the question above, the Alternative Hypothesis H
a
and Null Hypothesis H
were proposed as follows: 1.
Null Hypothesis H : picture sequence is not effective in the teaching of
narrative text writing 2.
Alternative Hypothesis H
a
: picture sequence is effective in the teaching of narrative text writing
To prove the hypothesis, the obtained data from experiment class and control class were calculated by using t
test
formula with assumption as follows:
1. If t
o
t
table
in significant degree of 1, the Null Hypothesis H is
accepted and the Hypothesis Alternative H
a
is rejected. It means that there is no significant effect of picture sequence on students writing of
narrative text. 2.
If t
o
t
table
in significant degree of 1, the Null Hypothesis H is
rejected and the Hypothesis Alternative H
a
is accepted. It means that there is significant effect of picture sequence on students’ writing of
narrative text.
The hypothesis criteria above states that: if t
o
t
t
= Ha is accepted and H
is rejected, and if t t
t
= H
a
is rejected and H is accepted. H
a
is Hypothesis Alternative, H
is Null Hypothesis, t
o
is t observation and t
t
is t test.
The result of the statistic calculation indicates that the value of t is 2.72
which is higher that ttable t
t
at significance level 5 = 1.7 and t table t
t
at
significance level 1 2.4 it means that the Null Hypothesis H is rejected
and the Hypothesis Alternative H
a
is accepted.
E. Data Interpretation
The writer took pre-test and post-test with the same test and got the data result as the table above. The average score of experimental class that is
taught by using picture sequence is known was 13.33. Meanwhile the average
score of control class that is taught with using conventional technique is
known 7.54.
Based on the result of controlled class scores, we can conclude that there are some reasons why the condition happened. After the pre-test held, the
controlled class did not get the treatment in the teaching of narrative text writing through picture sequence. Therefore, the students in the controlled
class did not have a chance to increase their writing skill especially in writing narrative text through picture sequence. They still get difficulty in writing
narrative text because they did not know how to organize their idea in order to create a good story. As the result, their writing skill, especially in writing
narrative text still low and their writing scores were not significantly improved.
In addition, based on the result of experimental class scores, after having treatment of picture sequence in the teaching of narrative text writing, their
narrative text writing scores were increased. The improvement happened because they got the benefit of using picture sequence. Hence, they can write
narrative text easier and more focus because what they want to write was based on the picture sequence. Furthermore, based on the calculation above, it
showed that there is obvious difference on students’ writing of narrative text,
it can be concluded that the difficulty of experimental class students was solved by having treatment of using picture sequence in the teaching of
