11 60
70 10
12 50
67 17
13 50
57 7
14 33
60 27
15 57
70 13
16 43
47 4
17 57
60 3
18 60
60 19
20 37
17 20
33 57
24 21
50 60
10 22
43 60
17 23
33 73
40 24
40 43
3 25
30 33
3 26
37 70
33 27
50 53
3 28
60 70
10 29
43 57
14
30
73 77
4
Total
∑ 1418 ∑ 1721
∑ 303
Based on the table 4.2 above, the highest score in the controlled class was 77 obtained by one student. Meanwhile, the lowest score in the controlled
class was 33 obtained by one student.
B. The Analysis of the Data
Table 4.3 The Comparison of Students’ Score in the Experiment Class and
Controlled Class Students
Experiment Class X
Controlled Class Y
x X-MX y Y-MX
X-MX
2
Y-MX
2
1 34
4.13 -10.1
17.08 102.01
2 16
7 -13.86
-3.1 192.28
9.61 3
40 7
10.13 -3.1
102.68 9.61
4 24
-14 -5.86
-24.1 34.41
580.81 5
57 7
27.13 -3.1
736.21 9.61
6 23
20 -6.86
9.9 47.15
98.01 7
30 13
0.13 2.9
0.01 8.41
8 13
-16.86 -10.1
284.48 102.01
9 30
7 0.13
-3.1 0.01
9.61 10
43 -3
13.13 -13.1
172.48 171.61
11 40
10 10.13
-0.1 102.68
0.01 12
37 17
7.13 6.9
50.88 47.61
13 26
7 -3.86
-3.1 14.95
9.61 14
23 27
-6.86 16.9
47.15 285.61
15 23
13 -6.86
2.9 47.15
8.41 16
23 4
-6.86 -6.1
47.15 37.21
17 43
3 13.13
-7.1 172.48
50.41 18
37 7.13
-10.1 50.88
102.01 19
37 17
7.13 6.9
50.88 47.61
20 47
24 17.13
13.9 293.55
193.21 21
33 10
3.13 -0.1
9.81 0.01
22 37
17 7.13
6.9 50.88
47.61 23
13 40
-16.86 29.9
284.48 894.01
24 27
3 -2.86
-7.1 8.21
50.41
25 40
3 10.13
-7.1 102.68
50.41 26
7 33
-22.86 22.9
522.88 524.41
27 23
3 -6.86
-7.1 47.15
50.41 28
33 10
3.13 -0.1
9.81 0.01
29 27
14 -2.86
3.9 8.21
15.21 30
10 4
-19.86 -6.1
394.68 37.21
Total
∑
896
∑
303
∑ ∑
∑
2
= 3903.46
∑
2
= 3552.7 Mean
∑ 29.86 ∑
=10.1
The procedures of caluculation were as follow: 1. Determining mean of variable X the formulation is:
=
∑
=
29.86
2. Determining men of variable Y the formulation is: =
∑
= =
10.1
3. Determining Standard Deviation X SDx, with formula: √
∑
√
√
4. Determining Standard Deviation Y SDy, with formula: √
∑
√ √
5. Determining Standard Errors Mean Variable X SEmx, with formula: √
√ √
2.11 6. Determining Standard Errors Mean Variable X SEmx, with formula:
√ √
√
7. Determining the difficulties of Mean Variable X and Mean Variable Y, with formula:
√ √
√ √
8. Determining with formula:
6.76 9. Determining
in significance level 5 with degree of freedom df: df = N1+N2 -2
df = 30+30-2 df= 60-2
df= 58
Because the score of 58 was not mentioned in the table, the writer used the closest to 58 which was 60 degree of freedom. Thus, the writer gained
: Degree of significance 5 = 2. 00
10. The comparison between and
From the calculation above, it was found that the result of was 6.76
and the degree of freedom df was 60. In this research, the writer used the degree of significance 5. It can be seen that degree of freedom df and the degree of
significance 5=2.00. By comparing the result of
and on the degree of
significance 5 2.00, thus the writer summarized that was higher than
. It means that
alternative hypothesis was accepted and null
hypothesis was rejected because . Thus, there was significant difference
after teaching reading comprehension by using reciprocal teaching technique at
