63
The writer did not take the total number of the population as the result she took 80 students only. That was taken from the class of VII-1 and VII-2, the VII-1 as the experiment
class and VIII-2 as the control class.
4. Technique of Collecting Data
The writer gave the students test namely pre-test and post-test. Pre-test was given before she began the teaching learning process in two classes. The pre-test is similar in form
to post-test that was given after teaching learning process finished, in order to get objective data of students achievement in reading.
The writer gave the test to eighty students of the first year, which was cosists of 30 items; 10 item are essay form; 10 item are matching and 10 item are translated.
5. Technique of Data Analysis
The technique of data analysis which is used in this research, is: t
Ο
= M
I
– M
2
SE
MI – M2
t
Ο
= t - observation M
Ι
= Mean of Variable X variable I M
2
= Mean of Variable Y variable II SE
MΙ
= Standard Error of Mean Variable X SE
M2
= Standard Mean of Variable Y
B. RESEARCH FINDING
1. Data Description
To know the result of the test, the writer makes the table of students’ score for each class as follow:
Table 3.1 The Test Score of the Experiment Class VII-1
64
Students Pre-test
Post-test 1
60 76
2 76
86 3
63 80
4 50
56 5
60 73
6 70
86 7
56 70
8 76
90 9
53 66
10 70
80 11
53 60
12 66
76 13
50 66
14 56
70 15
60 76
16 50
63 17
73 83
18 53
63 19
56 73
20 80
90 21
53 53
22 90
93 23
86 86
24 53
70 25
56 76
26 56
73 27
73 83
28 56
63 29
70 80
30 63
76 31
56 70
32 66
80 33
63 76
34 66
83 35
50 60
36 83
93
65
37 66
80 38
53 93
39 50
83 40
73 86
N = 40
3040
1
= X
M
1
=
N X
1
Σ
=
40 3040
= 76.00 Table 3.2
The Test Score of the Control Class VII-2 Students
Pre-test Post-test
1 60
66 2
53 63
3 50
50 4
63 70
5 56
66 6
76 83
7 46
50 8
53 56
9 83
86 10
56 60
11 63
70 12
63 70
13 60
63 14
50 53
15 70
76 16
73 80
17 50
53 18
56 60
19 63
76 20
56 60
21 60
63 22
46 46
23 80
90 24
60 70
66
25 50
66 26
53 56
27 56
60 28
66 73
29 50
50 30
53 56
31 56
66 32
76 83
33 50
50 34
63 76
35 50
56 36
70 80
37 46
53 38
66 70
39 46
46 40
53 56
N = 40
2577
2
= X
M
2
=
N X
2
=
40 2577
= 64.42 Table 3.3
Table Score of Result Evaluation from Experiment Class Score
f 90 – 94
5 85 – 89
4 80 – 84
9 75 – 79
6 70 – 74
7 65 – 69
2 60 – 64
5 55 – 59
1 50 – 54
1 45 – 49
N
1
= 40
67
Table 3.4 Table Score of Result Evaluation from Control Class
Score f
90 – 94 1
85 – 89 1
80 – 84 4
75 – 79 3
70 – 74 6
65 – 69 4
60 – 64 7
55 – 59 5
50 – 54 7
45 – 49 2
N
2
= 40
2. Data Analysis
From the data description above the writer analyze the score by making the table of distribution of frequency from the score of student who was taught by using comic Variabel
X and taught without using comic Variable Y as follow: Table 3.5
Table of the Distribution of Frequency from Experiment Class Score
f x
x fx
fx
2
90 – 94 5
4
20 80
85 – 89 4
3 12
36 80 – 84
9 2
18 36
75 – 79 6
1 6
6 70 – 74
7 M’
65 – 69 2
-1 -2
2 60 – 64
5 -2
-5 10
68
55 – 59 1
-3 -3
9 50 – 54
1 -4
-4 16
45 – 49 -5
N
1
= 40 Σ
fx’ = 42 Σ
fx’
2
= 195
Table 3.6 Table of the Distribution of Frequency from Control Class
Score f
y y
fy fy
2
90 – 94 1
4 4
48 85 – 89
1 3
3 36
80 – 84 4
2 8
28 75 – 79
3 1
3 5
70 – 74 6
M’ 65 – 69
4 -1
-4 4
60 – 64 7
-2 -14
24 55 – 59
5 -3
-15 18
50 – 54 7
-4 -28
32 45 – 49
2 -5
-10 N
2
= 40 Σ
fy’ = -53 Σ
fy’
2
= 283
After making the table of distribution of frequency the writer calculates the score by using the following steps:
a. Determining mean of variable X, with the formula: M
1
= M’ + i
N fx
= 72 + 5
40 42
= 72 + 5.250 = 77.250
b. Determining mean of variable Y, with the formula:
69
M
2
= M’ + i
N fy
= 72 + 5
40 53
−
= 72 + 5
325 .
1 −
= 72 – 6.625 = 65.375
c. Determining standard deviation of variable X. with the formula: SD
1
= i +
N fx
N fx
2 2
−
= 5 +
40 42
40 195
2
−
= 5
103 .
1 875
. 4
−
= 5
772 .
3
= 5. 1.942 = 9.710
d. Determining standard deviation of variable Y, with the formula: SD
2
= i
N fy
N fy
2 2
−
= 5
40 53
40 283
2
− −
= 5
2
325 .
1 075
. 7
−
= 5
756 .
1 075
. 7
−
= 5
319 .
5
= 5
×
2.306
70
= 11.530 e. Determining mean of variable X variable 1, with the formula:
SE
M1
=
1
1
− N
SD
=
1 40
710 .
9 −
=
39 710
. 9
=
245 .
6 710
. 9
= 1.555 f. Determining mean of standard error of variable Y, with the formula:
SE
M2
=
1
2
− N
SD
=
1 40
530 .
11 −
=
39 530
. 11
=
245 .
6 530
. 11
= 1.846 g. Determining standard error of different mean between variable X and variable Y, with
the formula: SE
M1- M2
=
2 2
2 1
M M
SE SE
+
=
2 2
846 .
1 555
. 1
+
=
408 .
3 428
. 2
+
=
826 .
5
71
= 2.414 h. Determining t
o
with the formula: t
o
=
2 1
2 1
M M
SE M
M
−
−
=
414 .
2 375
. 65
250 .
77 −
=
414 .
2 875
. 11
= 4.921 i. Determining t-table in significant level 5 and 1 with df:
df = N1 + N2 - 2 = 40 + 40 - 2
= 78 df = 78 see table of ”t” values at the degree of significance of 5 and 1.
Because the value is not mentioned in the table, the writer uese the closer value to 78 that is 80 as degree of freedom df.
t-table t
t
at significance 5 = 1.99 t-table t
t
at significance 1 = 2.64 1.99 4.921 2.64
Based on the analysis of the result in the table above, it can be see that t
o
is higher than t
t
, it can be interpreted that there is a significant different between teaching reading using comic and teaching reading without using comic.
3. Hypothesis testing
