Score f
j
P
j
100P
j
P
j
– 100 P
j
P
j
– 100P
j 2
100P
j
90-97 1
5 2.037
2.963 4.308
Jumlah 20 100
36.61 Table 4.5
X
2
Table
j j
j
P P
P n
o 100
100 100
2 2
61 .
36 100
20
=
7.322 X
2 t
with degree of freedom 3 at significance 5= 7.815. Comparing X
2
observation and X
2
table X
2 t
, we know that X
2 o
is lower than X
2 t
. The result is as follow:
7.815 7.322 Because of X
2 o
is lower than X
2 t,
so the H is accepted and H
1
is rejected. It means that the distribution data from this population is normal.
2. The Controlled Class
a. Frequency Distribution
The steps are: 1
Determining the Range R = X
maks
- X
min
= 73.7 – 40.2
= 33.5 2
Determing the Size Class Class
= 1 + 3.3 log n = 1 + 3.3 log 20
= 1 + 3.3 1.31 = 1 + 4.29
= 5.29
3 Determining the Interval Class
3 .
6 29
. 5
5 .
33
i i
C R
i
So, the table of frequency distribution of controlled class as seen in the table 4.6 below.
Table 4.6 The Table of Frequency Distribution of Controlled Class
Score B
b
B
a Frequency
X
i
X
i 2
f
i
X
i
f
i
X
i 2
f
i
f
k
40-46 39.5
46.5 5
5 43
1849 215
9245 47-53
46.5 53.5
2 7
50 2500
100 5000
54-60 53.5
60.5 9
16 57
3249 513
29241 61-67
60.5 67.5
3 19
64 4096
192 12288
68-74 67.5
74.5 1
20 71
5041 71
5041
Total 20
1091 60815
55 .
54 20
1091
i i
i
f X
f X
28 .
8 19
95 .
1300 19
05 .
59514 60815
1 20
20 1091
60815 1
2 2
2
n n
X f
X f
s
i i
i i
From the data above, mean score of experiment class is 54.55 and standard deviation is 8.28.
b. Normality Test of Controlled Class
The steps and formula that used in this class is same with experiment class.
a. Determining Hypothesis
H : Distribution of population is normal.
H
1
: Distribution of population is abnormal. So.
If
2 obs
≤
2 table.
H
o
is accepted and H
1
is rejected If
2 obs
2 table
. H
o
is rejected and H
1
is accepted b.
Determining X
2
table From Chi square table with df = 5
– 3 = 2, significant level 5 is 5.991. c.
Determining Proportion to j Pj Pj =
d. Determining 100Pj
Score Bb
Ba Zbb
Zba Za
tab
Zb
tab
Wide 100pj
40-46 39.5
46.5 -1.82
-0.97 0.0346 0.1655 0.1309 13.09
47-53 46.5
53.5 -0.97
-0.13 0.1655 0.4495 0.2841 28.41
54-60 53.5
60.5 -0.13
0.72 0.4495 0.7638 0.3143 31.43
61-67 60.5
67.5 0.72
1.56 0.7638 0.9411 0.1773 17.73
68-74 67.5
74.5 1.56
2.41 0.9411 0.922 0.0509 5.092
= 54.5 SD =8.28
Table 4.7 100 Pj Table
e. Determining X
2 o
with the formula Table 4.8
X
2
Table
Score f
j
P
j
100P
j
P
j
– 100 P
j
P
j
– 100P
j 2
100P
j
40-46 5
25 13.09
11.91 10.834
47-53 2
10 28.1
-18.4 11.928
54-60 9
45 31.43
13.57 5.863
61-67 3
15 17.73
-2.73 0.42
Score f
j
P
j
100P
j
P
j
– 100 P
j
P
j
– 100P
j 2
100P
j
68-74 1
5 5.092
-0.09 0.0017
Jumlah 20 100
29.046
j j
j
P P
P n
o 100
100 100
2 2
046 .
29 100
20
= 5.089 From the result above, X
2 o
is lower than X
2 t
with degree of freedom 2 at significance 5= 5.991, so H
o
is accepted and H
1
is rejected. It means that the distribution data in controlled class is normal.
3. Homogeneity Test
After calculating the distribution data of this population is normal, next step is homogeneity test. This step is to know whether this sample is homogeny or
not. Fisher formula was used with the degree of significance 5 F
table
= 2.12.
73 .
1 47
. 68
57 .
118 Varians
smallest The
Varians biggest
The F
obs
F
obs
F
table
1.73 2.12 So, from the calculation of homogeneity test with Fisher formula, F
obs
F
table.
It means that H is accepted and H
1
is rejected; both of group sample have same variants or homogeny.
4. Statistical Test T-Test
As mentioned before, in analyzing the data from the result of pre-test and post-test, the writer used statistic calculation of the t-test formula with the degree
of significance 5.
