A. Description of the Data 1. The result of Pre-Test of the Experiment Class - The Effect Of Peer Correction And Teacher Written Feedback On The Paragraph Writing Ability At The Third Semester Students Of English Study Program Of STAIN Palangka Raya - Dig

59

CHAPTER IV
RESULT OF THE STUDY
In this chapter, the researcher presented the data which had been
collected from the research in the field of study which consists of description of
the data, result of data analysis, and interpretations.
A. Description of the Data
1. The result of Pre-Test of the Experiment Class
The Pre-Test at the experiment class had been conducted in class C
with the number of student was 16 students. The Pre-test scores of class
were presented in table 4.1.
Table 4.1 Pre-Test scores of Experimental class
NO Students’ Initial Names
1
A
2
BL
3
DR
4

DSW
5
ES
6
FF
7
LAR
8
MR
9
N
10
N
11
RH
12
RAA
13
SP
14

SW
15
UJ
16
YSS
Highest Score
Lowest Score

59

Pre-Test
46
54
53
40
54
31
47
67
54

46
55
49
28
38
36
32
74
29

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Based on the calculation result scores of pre test the highest score
was 67 and the lowest one was 28. To determine the range of score, the
class interval, and interval of temporary, the writer calculated using
formula as follows:
The Highest Score (H)

= 67

The Lowest Score (L)

= 28

The Range of Score I

=H–L+1
= 67 – 28 + 1
= 40

The Class Interval (K)

= 1 + (3.3) x Log n
= 1 + (3.3) x Log 16
= 1 + (3.3) x 1.2041199827
= 1 + 3.9735959429
= 4.9735959429
=5

Interval of Temporary (I)

=

R 40

K
5

=8
So, the range of score was 40, the class interval was 5, and interval
of temporary was 8. Then it was presented using frequency distribution in
the following table:

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Table 4.2 The Frequency Distribution of the Pre Test Scores of the
Experiment Class
Class
(k)

Interval
(I)

Frequency
(F)

Midpoint
(X)

Class
Boundaries

Relative
Frequency
(%)
1
60-67
1
63.5
59.5 – 67.5

18.75
2
52-59
5
55.5
51.5 – 60.5
18.75
3
44-51
4
47.5
43.5 – 52.5
25
4
36-43
3
39.5
35.5 – 44.5
31.25
5

28-35
3
31.5
27.5 – 36.5
6.25
∑F = 16
∑P = 100
Figure 4.1 The Frequency Distribution of the Pre test Score of the
Experiment Class
The Frequency Distribution of the Pre test Score of the
Experiment Class

6

F
R
E
Q
U
E

N
C
Y

5
4
3

5

2

4
3

3

1

1

0

27.5 – 36.5

35.5 – 44.5

43.5 – 52.5

51.5 – 60.5

59.5 – 67.5

Class Boundaries

2. The result of Post-Test score of the Experiment Class
The Post-Test at the experiment class had been conducted in class C
with the number of student was 16 students. The Post-Test of class were
presented in table 4.3.

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Table 4.3Post-Test scores of Experimental class
NO
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

Students’ Initial Names
A
BL
DR
DSW
ES
FF
LAR
MR
N
N
RH
RAA
SP
SW
UJ
YSS
Highest Score
Lowest Score

Post-Test
64
73
80
70
69
69
73
78
69
71
62
72
63
69
78
66
80
62

Based on the calculation result scores of post test, the highest score
was 80 and the lowest score was 62. To determine the range of score, the
class interval, and interval of temporary, the writer calculated using
formula as follows:
The Highest Score (H)

= 80

The Lowest Score (L)

= 62

The Range of Score I

=H–L+1
= 80 – 62 + 1
= 19

The Class Interval (K)

= 1 + (3.3) x Log n

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= 1 + (3.3) x Log 16
= 1 + (3.3) x 1.2041199827
= 1 + 3.9735959429
= 4.9735959429
=5
Interval of Temporary (I)

=

R 19
 = 3.8
K 5

=4
So, the range of score was 18, the class interval was 5, and interval
of temporary was 4. Then it was presented using frequency distribution in
the following table:
Table 4.4 The Frequency Distribution of the Post Test Scores of the
Experiment Class
Class
(k)

Interval
(I)

Frequency
(F)

Midpoint
(X)

Class
Boundaries

1
2
3
4

74-78
70-73
66-69
62-65

3
5
5
3
∑F = 16

75.5
71.5
67.5
63.5

73.5 – 77.5
69.5 – 72.5
65.5 – 68.5
61.5 – 64.5

Relative
Frequency
(%)
18.75
31.25
31.25
18.75
∑P = 100

Figure 4.2 The Frequency Distribution of the Post test Score of the
Experiment Class

64

8

The Frequency Distribution of the Post test Score of the
Experiment Class

7
F
R
E
Q
U
E
N
C
Y

6
5
4
3

5

5

2

3

3

1
0
61.5 – 64.5

65.5 – 68.5

69.5 – 72.5

73.5 – 77.5

Class Boundaries
3. Scores of students who are taught using Peer Correction Techniques
NO
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

Students’ Initial Names
A
BL
DR
DSW
ES
FF
LAR
MR
N
N
RH
RAA
SP
SW
UJ
YSS
Highest Score
Lowest Score
Total

PC_1
69
66
67
61
61
35
78
66
50
54
62
58
69
71
70
59

PC_2
72
71
75
65
61
39
78
76
52
64
67
67
78
71
71
66

78
35
996

78
39
1007

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Mean

62.25

67.0625

Based on the calculation result of scores of the students who are taught
using peer correction, the highest score achieved by students in the first
meeting was 78 and the lowest score was 35. The range was 44, from the
student’s number (N) 16. From the calculation result of statistic, the mean
score achieved by students was 62.25. In the last meeting of peer
correction, the highest score achieved by students was 78 and the lowest
one was 39. The range was 40, from the student’s number (N) 16. From
the calculation result of statistic, the mean score achieved by students was
67.0625.
4. Scores of students who are taught using Teacher Written Feedback
Techniques
NO
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

Students’ Initial Names
A
BL
DR
DSW
ES
FF
LAR
MR
N
N
RH
RAA
SP
SW
UJ
YSS
Highest Score
Lowest Score

TWF_1
68
67
75
63
61
55
70
77
60
60
57
63
73
74
71
50
77
50

TWF_2
74
70
75
66
66
55
76
78
63
65
61
68
76
76
80
77
80
55

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Total
Mean

994
62.25

1049
70.375

Based on the calculation result of scores of the students who are taught
using teacher written feedback, the highest score achieved by students in the
first meeting was 77 and the lowest score was 50. The range was 28, from the
student’s number (N) 16. From the calculation result of statistic, the mean
score achieved by students was 62.25. In the last meeting of teacher written
feedback, the highest score achieved by students was 80 and the lowest one
was 55. The range was 26, from the student’s number (N) 16. From the
calculation result of statistic, the mean score achieved by students was
70.375.
B. Result of Data Analysis
1. The difference between the result of pre-test and post-test

NO
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

Students’Initial
Names
A
BL
DR
DSW
ES
FF
LAR
MR
N
N
RH
RAA
SP
SW
UJ

PreTest
46
54
53
40
54
31
47
67
54
46
55
49
28
38
36

PostTest
64
73
80
70
69
69
73
78
69
71
62
72
63
69
78

Gain
(d)
18
19
27
30
15
38
26
11
15
25
7
23
35
31
42

xd
-9,9375
-8,9375
-0,9375
2,0625
-12,9375
10,0625
-1,9375
-16,9375
-12,9375
-2,9375
-20,9375
-4,9375
7,0625
3,0625
14,0625

X2d
98,75390625
79,87890625
0,87890625
4,25390625
167,3789063
101,2539063
3,75390625
286,8789063
167,3789063
8,62890625
438,3789063
24,37890625
49,87890625
9,37890625
197,7539063

67

16

t=

YSS

32
∑X1=
730

66
∑X2=
1126

34
∑d=
396

6,0625

36,75390625
∑ X2d = 1513

√

N (N-1)
Where:
Md

: mean of the difference pre-test and post-test (post test – pre test)

xd

: deviation of each subject (d-Md)

∑ X2d

: sum of squared deviations

N

: total the subject on sample

d.b

: N-1

Md =

=

t=

√

= 24. 75

=
N (N-1)

√

=

16 x 15

tvalue =5.6911591786
Based on the manual calculating, the tvalue was 5.6911591786. Then it
was consulted with ttable with df= (N-1) and level of significance ttable 0.05 =
2.13. Because tvalue = 5.6911591786was higher than ttable= 2.13 It meant
that both of pre-test and post-test were significant.
2. Normality and Homogeneity Test
The researcher calculated the result of pre-test and post-test score of
experiment by using SPSS 17.0 programs. It was done to know the
normality of the data that is going to be analyzed having normal

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distribution or not.
a. Normality test of Pre-Test
One-Sample Kolmogorov-Smirnov Test
Pre_Test
N
16
a,,b
Normal Parameters
Mean
45.63
Std. Deviation
10.695
Most Extreme
Absolute
.139
Differences
Positive
.128
Negative
-.139
Kolmogorov-Smirnov Z
.556
Asymp. Sig. (2-tailed)
.917

Based on the calculation used SPSS program, the dvalue of pre-test
was 0.139. From the table of critical value of Kolmogrov-Smirnov test
with the student’s number (N) = 16 at the significance level α= 0.05,
the score of dtable was 0.327. Because dvalue was lower than dtable (0.139
< 0.327), it could be concluded that the data was in normal
distribution.
b. Normality test of Post- Test
One-Sample Kolmogorov-Smirnov Test
Post_Test
N
16
a,,b
Normal Parameters
Mean
70.38
Std. Deviation
5.277
Most Extreme
Absolute
.147
Differences
Positive
.122
Negative
-.147
Kolmogorov-Smirnov Z
.589
Asymp. Sig. (2-tailed)
.879

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Based on the calculation used SPSS program, the dvalue of post-test
was 0.147. From the table of critical value of Kolmogrov-Smirnov test
with the student’s number (N) = 16 at the significance level α= 0.05,
the score of dtable was 0.327. Because dvalue was lower than dtable
(0.147< 0.327), it could be concluded that the data was in normal
distribution.
c. Homogeneity Test
Homogeneity test was conducted to know whether data are
homogeneous or not.
Levene's Test of Equality of Error Variances
Dependent Variable:Post_Test
F
df1
df2
Sig.
.979
12
3
.583
Tests the null hypothesis that the error variance of the dependent variable
is equal across groups.

Based on the result of homogeneity test, the fvalue was 0.979 and the
significant
value

value

was 0.583. The data are homogeneous if the significant

is higher than significance level α= 0.05. Because thesignificant

value

(0.583) was higher than significance level α= 0.05, it could be concluded
that the data were homogeneous.
3. Testing Hypotheses
The researcher used SPSS 17.0 Program calculation to test the
hypotheses used Two-Ways Repeated Measures ANOVA. The criteria of
Ha was accepted when Fvalue>Ftable, and Ha was refused when Fvalue