Table 4.4
The Students’ Score of Control Class
Students Pre-test
Post-test Gained
1 75
78 3
2 45
65 20
3 42
55 13
4 75
80 5
5 80
76 -4
6 68
72 4
7 65
70 5
8 76
83 7
9 73
80 7
10 64
65 1
11 56
64 8
12 78
80 2
13 75
81 6
14 45
58 13
15 52
74 22
16 45
50 5
17 60
70 10
18 65
70 5
19 48
50 2
20 71
75 4
21 50
64 14
22 70
78 8
23 50
50 24
52 73
21
∑ = 1480 ∑ = 1661
∑ = 181 � = 61.67
� = 69.21 � = 7.54
Based on the result of pre-test and post-test from the controlled class, it could be seen from 24 students in the controlled class, the mean of pre-test
was 61.67 and the mean of post test was 66.4. So, the average of gain score was only 7.54. In the pre-test of controlled class, the lowest score was 42 and
the highest score was 80. Then in the post-test, the lowest score in the controlled class was 50 and the highest score was 83. In this class there was
six students who got the score above 75 in the pre-test, then it increases become nine students who got score above 75. It has good enough inflation
score. From the data above, number of class interval of post-test in
experimental class as follows: K = 1 + 3.322 logn
K = 1 + 3.322 log24 K = 1 + 4.59
K = 5.59 = 6 Interval as follows:
I = The highest score – The lowest score
Number of Class I = 80-42 = 6.33 = 6
6
Table 4.5
Class interval of pre-test in controlled class
Class Interval Tally
Frequence
42-47 IIII
4 48-53
IIIII 5
54-59 I
1 60-65
IIII 4
66-71 III
3 72-80
IIIIIII 7
From the data above, number of class interval of post-test in experimental class as follows:
K = 1 + 3.322 logn K = 1 + 3.322 log24
K = 1 + 4.59 K = 5.59 = 6
Interval as follows: I = The highest score
– The lowest score Number of Class
I = 83-50 = 5.5 = 6 6
Table 4.6
Class interval of post-test in controlled class
Class Interval Tally
Frequence
50-55 IIII
4 56-61
I 1
62-67 IIII
4 68-73
IIIII 5
74-79 IIIII
5 80-85
IIIII 5
Therefore, based on the both scores from experimental and controlled calss, the average of scores from the students who got treatment using picture
sequences on writing narrative text was higher than the students from controlled class.
Tabel 4.7
The comparision of Experiment and Control Class
Student X
Y X
Y X
Y
1 15
3 1.67
-4.54 2.78
20.63 2
12 20
-1.33 12.46
1.78 155.21
3 26
13 12.67
5.46 160.44
29.79 4
4 5
-9.33 -2.54
87.11 6.46
5 14
-4 0.67
-11.54 0.44
133.21 6
6 4
-7.33 -3.54
53.78 12.54
7 10
5 -3.33
-2.54 11.11
6.46 8
16 7
2.67 -0.54
7.11 0.29
9 10
7 -3.33
-0.54 11.11
0.29 10
22 1
8.67 -6.54
75.11 42.79
11 6
8 -7.33
0.46 53.78
0.21 12
20 2
6.67 -5.54
44.44 30.71
13 -3
6 -16.33
-1.54 266.78
2.38 14
5 13
-8.33 5.46
69.44 29.79
15 28
22 14.67
14.46 215.11
209.04 16
13 5
-0.33 -2.54
0.11 6.46
17 4
10 -9.33
2.46 87.11
6.04 18
23 5
9.67 -2.54
93.44 6.46
19 10
2 -3.33
-5.54 11.11
30.71 20
27 4
13.67 -3.54
186.78 12.54
21 7
14 -6.33
6.46 40.11
41.71 22
16 8
2.67 0.46
7.11 0.21
23 14
0.67 -7.54
0.44 56.88
24 15
21 1.67
13.46 2.78
181.13
∑ =
320
∑ =
181
∑ =
0.00
∑ =
0.00
∑ =
1489.33
∑ =
1021.96
�
= 13.33
�
= 7.54
From the table above, the writer got ∑ = 320 by adding all score in
variable , while ∑ = 181 by adding all score in variable .
B. Normality of the Data
Before analyzing the hipotheses, the writer had to analyze the normality of the data. This analysis is used to see whether the data got in the research
has been normally distributed or not. When it is normally distributed, its dissemination is also noral and can represent the population. Because the
object of this study belongs to little sample, it is recomended using Lillyfors. In this formula, the data was transformed into the basic value. The maximum
dispute T got from the calculation must be in absolute value +. The result of normality can be seen by comparing the value of T
max
to T
table.
a. Normality of Pre-test in the Experimental Class
Hypotheses H
o
= Data of X is normally distributed H
1
= Data of X is not normally distributed
Table 4.8
Calculation of Pre-test Normality in Experimental Class
x f
fx x2
fx2 p=fn
z=xi- xs
∅ ∑p
T= ∅-∑p
35 1
35 1225
1225 0.041667 -2.18431
0.01447 0.041667 0.0272
36 1
36 1296
1296 0.041667 -2.10308 0.017729 0.083333
0.0656 49
1 49
2401 2401 0.041667
-1.04703 0.147544 0.125 0.022544
52 2
104 2704
5408 0.083333 -0.80332 0.210894 0.208333 0.002561
54 1
54 2916
2916 0.041667 -0.64085 0.260809
0.25 0.010809 56
1 56
3136 3136 0.041667
-0.47838 0.316189 0.291667 0.024522 58
1 58
3364 3364 0.041667
-0.31591 0.376034 0.333333 0.042701 62
1 62
3844 3844 0.041667 0.009026 0.503601
0.375
0.128601
63 1
63 3969
3969 0.041667 0.090261 0.53596 0.416667 0.119293
64 1
64 4096
4096 0.041667 0.171496 0.568083 0.458333 0.10975
65 2
130 4225
8450 0.083333 0.25273 0.599762 0.541667 0.058095
68 1
68 4624
4624 0.041667 0.496435 0.690206 0.583333 0.106873 70
2 140
4900 9800 0.083333 0.658904 0.745021 0.666667 0.078355
73 2
146 5329
10658 0.083333 0.902609 0.816633 0.75 0.066633
75 1
75 5625
5625 0.041667 1.065078 0.85658 0.791667 0.064913
76 2
152 5776
11552 0.083333 1.146313 0.874167 0.875
0.00083 78
2 156
6084 12168 0.083333 1.308782 0.904696 0.958333
0.05364 80
1 80
6400 6400 0.041667 1.471252 0.929388
1 0.07061
1114 24
1528 71914
100932 61.89 1.33 84.89 3995.22
5607.33
� = ∑ �
� − [ ∑ �
� ] =
− ⌈ ⌉
= . − [ . ]
= . −
. =
. S
= √ .
S =
. S
= 12.31 S
2
= 151.63 M
= 61.89 T
max
= 0.128 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
0.128 0.173, it means that the data is normally distributed.
b. Normality of Post=test in the Experimental Class
Hypotheses H
o
= Data of X is normally distributed H
1
= Data of X is not normally distributed
Table 4.9
Calculation of Post-test Normality in Experimental Class
x f
fx x2
fx2 p=fn
z=xi- xs
∅ ∑p
T= ∅-∑p
63 2
126 3969
7938 0,083333 -1,94409 0,025942 0,083333
0,05739 65
1 65
4225 4225 0,041667
-1,69189 0,045334 0,125
0,07967 70
3 210
4900 14700
0,125 -1,06137 0,144261
0,25 0,10574
72 1
72 5184
5184 0,041667 -0,80916 0,209211 0,291667
0,08246 75
5 375
5625 28125 0,208333
-0,43085 0,333288 0,16783
0,165458
78 2
156 6084
12168 0,083333 -0,05254 0,479048 0,541667
0,06262 80
4 320
6400 25600 0,166667 0,199664 0,579128 0,708333
0,12921 82
1 82
6724 6724 0,041667 0,451871 0,674319 0,791667
0,11735 85
1 85
7225 7225 0,041667 0,830181 0,796782 0,833333
0,03655 86
2 172
7396 14792 0,083333 0,956284 0,830536
0,875 0,04446
90 1
90 8100
8100 0,041667 1,460698 0,927951 0,916667 0,011284 95
1 95
9025 9025 0,041667 2,091215 0,981746
1 0,01825
941 24 1848
74857 143806
78,4 2
154 6238,083 11983,83
� = ∑ �
� − [ ∑ �
� ] =
− ⌈ ⌉
= . − [ ]
= . −
= .
S = √ .
S = .
S = 7.93
S
2
= 62.92 M
= 78.42 T
max
= 0.165 T
table
= 0.173
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
