CHAPTER IV RESEARCH FINDINGS
This chapter presents and discusses data description, test of hypothesis, and data discussion.
A. Data Description
To know the result of the test, the writer makes the table of student’s score for each class as follows :
Table 4.1 The Test Scores of the Experimental Class Students
Student Score
Student Score
Student Score
Student Score
1 75
11 80
21 77
31 77
2 59
12 76
22 61
32 64
3 86
13 72
23 76
33 92
4 48
14 58
24 50
34 72
5 75
15 70
25 61
35 85
6 58
16 85
26 73
36 78
7 59
17 65
27 70
37 65
8 63
18 48
28 63
38 70
9 62
19 64
29 73
39 50
10 86
20 54
30 80
40 90
Number of Class C C
= 1 + 3.322 log N = 1 + 3.322 log 40
= 1 + 3.322 1.6 = 1 + 5.31
= 6.31
Ratio R is the highest mark H minus the lowest mark L plus 1 R
= H – L + 1 = 92 - 48 + 1
= 44 + 1 = 45
Interval I is obtained from ratio R divided number of class C I
= C
R = 45
6.31 = 7.13
= 7 Next, after finding the result of formula explained previously, the writer divided
the students’ score into several classes equal with their interval.
Table 4.2 Interval of the Students’ Score of Experimental Class
Score
90-96 2
83-89 4
76-82 7
69-75 9
62-68 7
55-61 6
48-54 5
N=40 From the table above, it showed 7 for the interval of class, it was from 90-
96 to 48-54. most of students got 69-75, it can be seen from the frequency of the table was 9.
Table 4.3 The Test Scores of the Controlled Class Students
Student Score
Student Score
Student Score
Student Score
1 58
11 45
21 73
31 65
2 50
12 48
22 58
32 63
3 86
13 82
23 76
33 54
4 46
14 50
24 50
34 45
5 76
15 90
25 87
35 72
6 75
16 54
26 78
36 50
7 48
17 88
27 76
37 76
8 50
18 79
28 81
38 82
9 40
19 58
29 65
39 60
10 63
20 63
30 84
40 72
Number of Class C C
= 1 + 3.322 log N = 1 + 3.322 log 40
= 1 + 3.322 1.6 = 1 + 5.31
= 6.31 Ratio R is the highest mark H minus the lowest mark L plus 1
R = H – L + 1
= 90 – 40 + 1 = 50 + 1
= 51 Interval I is obtained from ratio R divided number of class C
I =
C R
= 51 6.31
= 8.08 = 8
Next, after finding the result of formula explained previously, the writer divided the students’ score into several classes equal with their interval.
Table 4.4 Interval of the Students’ Score of control Class
Score
88-95 2
80-87 6
72-79 10
64-71 2
56-63 7
48-55 9
40-47 4
N
2
=40
From the table above, it showed took 8 for the interval of class for control class. It was from 88-95 to 40-47. most of students got 72-79, it can be seen from
the frequency of the table was 10.
Table 4.5 The Result Calculation of Experimental Class
Score X
x’ ’
2
’ ’
2
90-96 2
+3 9
6 18
83-89 4
+2 4
8 16
76-82 7
+1 1
7 7
69-75 9
M’ 72 62-68
7 -1
1 -7
7 55-61
6 -2
4 -12
24 48-54
5 -3
9 -15
45
N = 40 ’ =
- 13 ’
2
= 117
From the table of result calculation above, the writer gained the total frequency of the students’ score in experiment class were -13, and the double
value of frequency students’ score in experiment class were 117 from the total number of experiment class was 40 students.
Table 4.6 The Result Calculation of Controlled Class
Score f
Y ’
’
2
’ ’
2
88-95 2
+2 4
4 8
80-87 6
+1 1
6 6
72-79 10
M’75.5 64-71
2 -1
1 -2
2 56-63
7 -2
4 -14
28 48-55
9 -3
9 -27
81 40-47
4 -4
16 -16
64
N = 40 fy’=
-49 fy’
2
= 189
From the table of result calculation above, the writer gained the total frequency of the students’ score in control class were -49, and the double value of
students’ score in experiment class were 189 from 40 students, the total number of control class.
After calculate the data given above, finally the writer continued the next step, which the steps comprise of Determining mean of experiment and control
class, Determining of Standard Deviation of Variable experiment and control class, Determining of Standard Error Mean of variable experiment and control
class, Determining of Standard Error Mean Difference of experiment and control class, and Determining t
o
, the value of t observation. To get the description about the comparative scores between the
calculation scores of experimental class see table 4.5 and the calculation scores of control class see table 4.6 were further analyzed by using the formula below :
1. Determining Mean I with formula :
M
1
= M’ + i N
fx Σ
= 72 + 7 40
13 −
= 72-2,275 = 69,725
2. Determining Mean II with formula :
M
2
= M’ + i N
fx Σ
= 75.5 + 8 40
5 ,
75 −
= 75,5-15,1 = 60,4 3.
Determining of Standard Deviation of Variable I : SD
1
= i
2 2
2
N fx
N fx
Σ −
Σ
= 7
2 2
40 13
40 117
− −
= 7 1600
169 925
, 2
− = 7
8 ,
2
= 7 x 1,67 = 11,69
4. Determining of Standard Deviation of Variable II :
SD
2
= i
2 2
2
N fy
N fx
Σ −
Σ
= 8
2 2
40 49
40 189
− −
= 8 1600
2401 725
, 4
− = 8
5 ,
1 725
, 4
− = 8
225 ,
3 = 8 x 1,795
= 14,4 5.
Determining of Standard Error Mean of Variable I : SE
M1
= 1
1 1
− N
SD
= 1
40 69
, 11
− =
24 ,
6 69
, 11
= 1,87
6. Determining of Standard Error Mean of Variable II :
SE
M2
= 1
2 2
− N
SD
= 1
40 4
, 14
− =
24 ,
6 4
, 14
= 2,3
7. Determining of Standard Error Mean Difference of M
1
and M
2
: SE
M1-M2
=
2 2
2 1
M M
SE SE
+ =
2 2
3 ,
2 87
, 1
+ =
29 ,
5 496
, 3
+ 786
, 8
= 2,964 8.
Determining to with formula : t
o
= 964
, 2
325 ,
9 964
, 2
4 ,
60 725
, 69
2 1
2 1
= −
= −
− M
M
SE M
M = 3,146
9. Determining t-table in significance level 5 and 1 with df :
= N
1
+ N
2
– 2 = 40 + 40 – 2 = 78 The writer gained t-table :
S.L. 5 = 1,99 S.L. 1 = 2,64
10. The Comparison between t-score with t-table
t-score = 1,99 3,164 2,64
B. Test of Hypothesis
