The calculation of the GPA explained on the following tables:
Table 4.3 The GPA of Students from SMA
No Student
s English Scores
T ot
al G
PA
R eadi
ng 1
L ist
ening 1
Sp eaki
ng 1
G ra
m m
ar 1
V ocabular
y 1
1 A
71.1 54.7
75.5 67.8
65 20
2 B
D B
C C
6 6
4 4
2 B
73.1 75.5
80.5 62.8
76 30
3 B
B A
C B
6 6
8 4
6
3 C
67.2 83.7
72.6 71.5
70 30
3 C
A B
B B
4 8
6 6
6
4 D
74.7 79.3
78.5 62.9
74 28
2.8 B
B B
C B
6 6
6 4
6
5 E
53.5 63.3
65 55
65 14
1.4 D
C B
D C
4 6
4
6 F
83.5 78.6
78.8 71.4
76 32
3.2 A
B B
B B
8 6
6 6
6
7 G
70.1 83.7
68.5 78.1
74 30
3 B
A C
B B
6 8
4 6
6
8 H
76.3 82.2
71.3 68.8
72 30
3 B
A B
C B
6 8
6 4
6
9 I
77.8 86.9
82 73
83.8 36
3.6 B
A A
B A
6 8
8 6
8
11 J
76.7 70.7
76.8 66.7
64.8 26
2.6 B
B B
C C
6 6
6 4
4
11 K
78 85
80 72.6
82 36
3.6 B
A A
B A
6 8
8 6
8
12 L
73.3 83.3
74.7 74.3
74.8 32
3.2 B
A B
B B
6 8
6 6
6
13 M
75.8 80.4
67.1 78.9
77.8 30
3 B
A C
B B
6 8
4 6
6 The table above informed that there was only one student who got A in
Reading, seven students got A in Listening, three students got A in speaking, two students got A in Vocabulary, and no one got A in Grammar. There were nine
students who reached the rate 3 of GPA. The highest GPA was 3.6 which reached by two students I and K, while the lowest GPA was 1.4.
Table 4.4 The GPA of Students from MA
No Student
s English Scores
T ot
al G
PA
R eadi
ng 1 L
ist ening 1
Sp eaki
ng 1 G
ra m
m ar 1
V ocabular
y 1
1 A
73.3 82.5
78 70.1
70 30
3.2 B
A B
B B
6 8
6 6
6
2 B
72.7 74.3
73.8 70.8
70 30
3 B
B B
B B
6 6
6 6
6
3 C
62.9 59.6
71.4 64
65 18
1.8 C
D B
C C
4 6
4 4
4 D
67.1 74.5
73.9 74.6
70 28
2.8 C
B B
B B
4 6
6 6
6
5 E
74.3 78.8
72.4 58
74.3 24
2.4 B
B B
D B
6 6
6 6
6 F
74.4 68.3
70.5 68.8
70.2 26
2.6 B
C B
C B
6 4
6 4
6
7 G
78.4 85.3
84.6 7
76.4 34
3.4 B
A A
B B
6 8
8 6
6
8 H
64.5 65.3
70.4 67.8
72 24
2.4 C
C B
C B
4 4
6 4
6
9 I
74.8 84.3
71.8 67
79.4 30
3 B
A B
C B
6 8
6 4
6
10 J
76.4 83.2
74.6 65.4
75.6 30
3 B
A B
C B
6 8
6 4
6
11 K
74.3 74.2
74.4 65.6
70.2 28
2.8 B
B B
C B
6 6
6 4
6
12 L
74.3 70.1
70.6 66.8
70.6 28
2.8 B
B B
C B
6 6
6 4
6
13 M
73.6 76
70.5 51.6
65.6 22
2.2 B
B B
D C
6 6
6 4
14 N
64.2 78.3
70.7
58.2
60.2 20
2 C
B B
D C
4 6
6 4
15 O
74.4 85
75 69.4
75.8 30
3 B
A B
C B
6 8
6 4
6 As showed by the table above that there were five students who got A in
Listening, one student got A in Speaking, and no student got A in Reading, Grammar, and Vocabulary. There were six students who reached the rate 3 of
GPA. The highest GPA was 3.4 which reached by a student G, and the lowest GPA was 1.8.
B. Data Analysis
1. Mean and Deviation Standard
After completing the data, the writer analyzed the means and the deviation standard. The calculation of mean got from the total GPA divided by the number
of students. The calculation of deviation standard got from the GPA divided by the mean as showedby the formula
X
1
-
then be multiplied as showed by the formula
x
1
-
2
, and the total of
x
1
-
2
divided by the 1 subtracted from the number of students as showed by the formula
. The calculation of deviation
standard can simply use the formula ”=STDEV”. The result of the mean and deviation standard were explained on the
following tables.
Table 4.5 Mean and Deviation Standard of Students from SMA
No Students
GPA X
1
- x
1
-
2
1 A
2 -0.08
0.77 2
B 3
0.12 0.01
3 C
3 0.12
0.01 4
D 2.8
-0.08 0.01
5 E
1.4
-0.48 2.18
6 F
3.2 0.32
0.1 7
G 3
0.12 0.01
8 H
3 0.12
0.01 9
I 3.6
0.72 1.52
10 J
2.6 -0.28
0.08 11
K 3.6
0.72 0.52
12 L
3.2 0.32
0.1 13
M 3
0.12 0.01
N= 13 37.4
4.36 =∑
= 2.88
S = √
∑
= 0.60
The result of the calculation above showed that most students from SMA reach the rate 3 of GPA even more, but the lowest GPA reachthe rate 1.4 that why
the mean of GPA only reach 2.88. The result of deviation standard was 0.60
Table 4.6 Mean and Deviation Standard of Students from MA
No Students
GPA X
1
- x
1
-
2
1 A
3.2 0.51
0.26 2
B 3
0.31 0.09
3 C
1.8 -0.89
0.8 4
D 2.8
0.11 0.01
5 E
2.4 -0.29
0.9 6
F 2.6
-0.09 0.01
7 G
3.4 0.71
0.5 8
H 2.4
-0.29 0.9
9 I
3 0.31
0.09 10
J 3
0.31 0.09
11 K
2.8 0.11
0.01 12
L 2.8
0.11 0.01
13 M
2.2 -0.49
0.24 14
N 2
-0.69 0.48
15 O
3 0.31
0.09
N= 15 40.4
2.87 =∑
= 2.69
S = √
∑
= 0.45
As mention inthe table above, it can be clarified that few students from MA reach the rate 3 of GPA, about half of them only reach the rate under 3 with the
lowest GPA 1.8. The result of mean was 2.69 lower than the mean of SMA, so was the result of deviation standard, it was 0.45.
The next table concluded the result of the calculation above.
Table 4.7 The Statistic Descriptive of the Research
Statistic SMA
MA
The Highest Score 3.6
3.4 The Lowest Score
1.4 1.8
Mean 2.88
2.69 Standard Deviation
0.60 0.45
The tables above described that the mean scores of students from SMA was 2.88, while the mean scores of students from MA was 2.69 and the deviation standard
of SMA was 0.60, while the deviation standard of MA was 0.45. Looking on the table above, there were differences both the result of the mean and deviation
standard. SMA was fine superior 0.19 on mean score and 0.15 on deviation standard from MA.
From the data above, the writer analyzed the score from both educational backgrounds by integrating through following formula based on the statistical
procedure written by Sugiyono:
Mean of variable X
1
∑ =
= 2.88
Mean of variable X
2
∑
=
= 2.69
Deviation StandardvariableX
1
S =
√
∑
= √
= √
= √ = 0.60
Deviation StandardvariableX
2
S =
√
∑
= √
= √
= √ = 0.45
After analyzing the data and counting the formula, it has been found the result of the means, the deviation standard of students learning achievement from both
schools, and finally gave interpretation of ‘t
o
’.
2. Statistical Test T-Test
In analyzing the data from the result above, the writer used statistical calculation of the t-test formula written by Sugiyono. For determining the
formula, he suggested to measure the homogeneity varian of both samples. It is the biggest varian divided by the smallest varian the varian is taken from the
deviation standard, and the result is compared to F table based on the result of the degree of freedomDF from both samples.
For DF of this research is N
1
– 1 and N
2
-
1 = 13- 1 and15 - 1= 12 and 14 The homogeneity varian is F=
= 1. 33 The F tabel of the degree freedom 5 of 12 and 14 is 2.53
Based on the calculation presented above the result of F is smaller than F table 1. 332.53,thus, it can be interpreted that the varian is homogenous.
