much additional
information about
recount text. 3
To know the students’
perception about
peer feedback through FB.
a. Student 1 thought that peer feedback was
really good because it told the mistakes that they had made and the excellence of
their writing. Student 2 and 3 said that they did not really like to give comment to
others. b.
Peer feedback developed their writing skill because they could correct their
writing into the right form stated student 1 and 2 while student 3 stated it did not
influence his writing skill. c.
They all said it was beneficial in developing
critical thinking,
learner autonomy and social interaction among
students even though the student 3 did not feel the effect.
From the result of interview above, the highest and middle score student student 1 and 2 gave the positive response to the treatment but the lowest score
student student 3 gave the negative response although he admitted the benefit of peer feedback through Facebook. It can be concluded that the advantages of using
Facebook as a media in writing and doing peer feedback through Facebook was beneficial for them.
b The Data of Controlled Class
In the following table is the score of pre-test and post-test in controlled class.
Table 4.3 The Test Result of Controlled Class Students
Pre-test Post-test
Gained Score
1 63
66 3
2 60
70 10
3 50
63 13
4 70
76 6
5 50
76 26
6 76
93 17
7 56
70 14
8 86
96 10
9 46
73 27
10 73
86 13
11 83
90 7
12 73
76 3
13 86
96 10
14 83
96 13
15 66
76 10
16 40
56 16
17 80
80 18
80 83
3 19
73 83
10 20
46 70
24 21
60 70
10 22
60 76
16 23
66 80
14 24
73 76
3 25
53 66
13
Ʃ 1652
1943 291
Mean 66
77.7 11.6
The table 4.2 above informs that the mean score of pre-test in control class was 66, while the mean score of post-test here is 77.7. The total gained score was
291.
2. Data Analysis
Based on pre-test and post-test result scores of both experimental and controlled classes, the writer used statistic calculation of the t-test formula with the
degree of significance 5 in analyzing the data to know the significant effectiveness of using Social Networking Site to students’ recount writing.
First, the writer determined the mean of X experimental class and Y controlled class. To determine the mean of X by using this formula:
M
1
=
=
= 16.68 Meanwhile, the writer also determined the mean of Y or controlled class by
using the following formula:
M
2
= =
= 11.64 To make it clearer, the writer provided the table to show the result of mean of
experimental and controlled class as following:
Table 4.4 Standard Deviation Table Students
X Y
x X-M
1
y Y-M
2
x
2
y
2
1 33
3 16.32
-8.64 266.34
74.64 2
30 10
13.32 -1.64
177.42 2.68
3 6
13 -10.68
1.36 114.06
1.84 4
20 6
3.32 -5.64
11.02 31.80
5 36
26 19.32
14.36 373.26
206.20 6
6 17
-10.68 5.36
114.06 28.72
7 7
14 -9.68
2.36 93.70
5.56 8
17 10
0.32 -1.64
0.10 2.68
9 20
27 3.32
15.36 11.02
235.92 10
33 13
16.32 1.36
266.34 1.84
11 20
7 3.32
-4.64 11.02
21.52 12
16 3
-0.68 -8.64
0.46 74.64
13 20
10 3.32
-1.64 11.02
2.68 14
7 13
-9.68 1.36
93.70 1.84
15 13
10 -3.68
-1.64 13.54
2.68 16
13 16
-3.68 4.36
13.54 19.00
17 4
-12.68 -11.64
160.78 135.48
18 16
3 -0.68
-8.64 0.46
74.64 19
7 10
-9.68 -1.64
93.70 2.68
20 3
24 -13.68
12.36 187.14
152.76 21
13 10
-3.68 -1.64
13.54 2.68
22 7
16 -9.68
4.36 93.70
19.00 23
16 14
-0.68 2.36
0.46 5.56
24 24
3 7.32
-8.64 53.58
74.64 25
30 13
13.32 1.36
177.42 1.84
ƩX= 417 ƩY= 291 Ʃx= 0
Ʃy= 0 Ʃx
2
= 2351.38
Ʃy
2
= 1183.52
After determining the mean of both experimental X and controlled class Y, the writer determined the Standard Deviation of experimental class by using a
formula as follows: SD
1
=
√
=
√
= √
= 9.69 To determine the Standard Deviation of controlled class Y used this
formula: SD
2
=
√
=
√
= √
= 6.88 Next, after knowing the result of both Standard Deviation X and Y, the
writer calculated the Standard Error Mean of experimental class X by using the formula below:
SE
=
√
=
√
= 1.97 Meanwhile, the formulation below was used to determine the Standard Error
Mean of controlled class Y: SE
=
√
=
√
= 1.40 Then, the writer calculated the difference of standard error between mean of
experimental class X and mean of controlled class Y by using this formula:
SE
- =
√ =
√ =
√ = 1.83
The next step is determining t
observed:
t
o
=
–
= =
= 2.75 Finally, the writer calculated t
table
t
t
in significance level of 5 with degree of freedom df:
df = N
1
+ N
2
- 2 = 25+25
– 2 = 48
