15 15
60 72
12
16 16
60 70
10
17 17
52 57
5
18 18
62 68
6
19 19
51 66
15
20 20
59 65
6
21 21
52 59
7
22 22
52 55
3
23 23
48 57
9
24 24
50 61
11
25 25
59 64
5
Σ
1455 1659
204
X
58.2 66.36
8.16
The table presents the scores of students in the control class included pre-test and post-test scores. It also mentions the result for minimum score,
maximum score, and total score of both pre-test and post-test. The minimum score of pre-test is 43, the maximum score is 72, and the total
score is 1455. The results for post-test are 50 as the minimum, 76 as the maximum, and 1659 as the total score.
Besides that, the total score of gained score is 204, the mean of pre-test is 58.2, the mean of post-test is 66.36, and the mean of gained score is
8.16.
B. The Analysis of Data
After collecting and describing data, the researcher analyzed the data and presents the result in three points; normality test, homogeneity test,
and hypothesis test.
1. Normality Test
The normality test is aimed to know whether the data is normally distributed or not. To do the normality test, the researcher uses
Liliefors. After finishing the normality test, he got two kinds of value; L
max
and L
table.
The both values are going to be used to see the normality of the data.
The researcher uses this criterion to see the normality of data: H
1
: L L
table
H : L ≤ L
table
Note: H
1
= Data is not normally distributed H
= Data is normally distributed
a. The Normality Test of Experiment Class
Table 4.3 Calculation of Pre-test Normality in Experiment Class
No. X
f fX
fX
2
p=fn ∑P
z = Xi- Xs
ф T=ф-∑p
1 48
5 240
11520 0.2000
0.2000 -1.56454
0.0594 0.1406
2 52
1 52
2704 0.0400
0.2400 -1.04302
0.1492 0.0908
3 56
3 168
9408 0.1200
0.3600 -0.52151
0.3015 0.0585
4 60
4 240
14400 0.1600
0.5200 0.5
0.02 5
64 5
320 20480
0.2000 0.7200
0.521512 0.6982
0.0218 6
65 2
130 8450
0.0800 0.8000
0.65189 0.7422
0.0578 7
68 1
68 4624
0.0400 0.8400
1.043025 0.8508
0.0108 8
70 3
210 14700
0.1200 0.9600
1.303781 0.9032
0.0568
9
72 1
72 5184
0.0400 1.0000
1.564537 0.9406
0.0594
Total
∑X=
∑
F=
∑
fX=
∑f
X
2
=
555 25
1500 91470
S= 7.67
S
2
= 58.8
rata2x= 60
Lmax= 0.0908
Ltable= 0.173
∑ [
∑ ]
[ ]
√
The Liliefors shows that the significant degree of 0.05 in L0.0525 = 0.173
H
1
: L 0.173 H
: L ≤ 0.173 In the table 4.3, the L
max
value is 0.0908. Therefore, H is accepted
because the result shows that L
max
is lower than L
table
0.1406 0.173. It means that the data in experiment class pre-test is normally
distributed.
