point that D
table
with the closest Kolmogorov-Smirnov critical points of 45 is 0.198. Therefore the comparison of the D
pre-test
D
table
is 0.148 0.198. In other hand, the absolute difference D of post-test class is
0.109, it is also less than D
table
of critical point 45 = 0.198 or D
post-test
D
table
0.120 0.224. In finding the normality, it can be referred to Z
pre-test
= 0.970 p 0.05 and Z
post-test
= 0.716 p 0.05. Therefore, it can get a conclusion that both data of the test in the experimental class
is normal.
b. Homogeneity of the Test Pre-test and Post-test Table 4.4
Homogeneity of the Pre-test
Test of Homogeneity of Variance
Levene Statistic df1
df2 Sig.
Nilai Based on Mean
.237 1
84 .628
Based on Median .139
1 84
.710 Based on Median and with
adjusted df .139
1 83.689
.710 Based on trimmed mean
.238 1
84 .627
In table 4.4, we know that the degree of significance of the mean of the post-test is 0.628 which bigger than the critical point of
0.05. Based on that finding, in can be summarized that both pre-test and post-test in the experimental class are also homogenous not
heterogeneous.
2. Hypothesis Testing
The writer uses t
test
formula to find the empirical evidence statistically and to make the testing of hypothesis in this research. In the
t
test,
the writer uses manual calculation. After getting the data; the results of students reading comprehension achievement of 8E class, X is the
students’ pre-test score in the experimental class, and Y is the students’
post-test score in the experimental class. Then, the writer analyzes them by using manual statistic calculation of the t
test
formula as follows.
Table 4.5 The Comparison of the Students’ Pre-test and Post-test Score in the
Experimental Class Students N
Pre-test Post-test
Gain Score D D
2
X Y
Y-X 1
30 65
35 1225
2 40
80 40
1600
3 40
70 30
900
4 70
80 10
100
5 50
80 30
900
6 40
50 10
100
7 45
75 30
900
8 30
55 25
625
9 75
100 25
625
10 45
60 15
225
11 45
85 40
1600
12 55
90 35
1225
13 25
50 25
65
14 45
75 30
900
15 15
45 30
900
16 35
70 35
1225
17 65
95 30
900
18 30
45 15
225
19 25
75 50
2500
20 45
55 10
100
21 30
80 50
2500
22 15
55 40
1600
23 35
70 35
1225
24 35
70 35
1225
25 45
60 15
225
26 35
75 40
1600
27 30
65 35
1225
28 35
70 35
1225
29
20 55
35 1225
30 40
60 20
400
31 40
50 10
100
32 40
80 40
1600
33 50
70 20
400
34 25
80 55
3025
35 35
70 35
1225
36 40
50 10
100
37 60
75 15
225
38 35
65 30
900
39 45
50 5
25
40 35
45 10
100
41 65
65
42 55
65 10
100
43 25
75 50
2500
Total ∑ 1720
2900 1180
40150
The writer calculates the data based on the procedure of the calculation. The formulation as follows:
a. Determining mean from differences MD, with formula: =
= 1180
43 = 27.44
b. Determining standard of deviation from difference of variable X and Y SD
D
, with formula: =
D N
D N
= 40150
43 1180
43 =
933.72 27.44
