3 Determining the Interval of Class
59 .
7 29
. 5
2 .
40
i i
C R
i
8 So, the table of frequency distribution of experiment class as seen in the
table 4.3 below.
Score B
b
B
a Frequency
X
i
X
i 2
f
i
X
i
f
i
X
i 2
f
i
f
k
50-57 49.5
57.5 5
5 53.5
2862.3 267.5 14311.25
58-65 57.5
65.5 2
7 61.5
3782.3 123
7564.5 66-73
65.5 73.5
7 14
69.5 4830.3
468.5 33811.75 74-81
73.5 81.5
5 19
77.5 6006.3
378.5 30031.25 82-89
81.5 89.5
19 85.5
7310.3 90-97
89.5 97.5
1 20
93.5 8742.3
93.5 8742.25
Total 20
1358 94461
Table 4.3 The Table of Frequency Distribution of Experiment Class
9 .
67 20
1358
i i
i
f X
f X
89 .
10 19
8 .
2252 19
2 .
92208 94461
1 20
20 1358
94461 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 67.9 and standard deviation is 10.89.
b. Normality Test of The Experiment Class
From the data above, analyzing the score from the experiment class was used Chi Square formula. The steps are:
1. 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 2.
Determining X
2
table From Chi square table with df = 6
– 3 = 3, significant level 5 is 7.815. 3.
Determining Proportion to j Pj Pj =
4. Determining 100Pj
Score Bb
Ba Zbb
Zba Za
tab
Zb
tab
Wide 100pj
50-57 49.5
57.5 -1.69
-0.96 0.046 0.17
0.124 12.42
58-65 57.5
65.5 -0.955 -0.22
0.17 0.413 0.243
24.3 66-73
65.5 73.5
-0.22 0.514 0.413 0.696 0.284
28.37 74-81
73.5 81.5
0.514 1.249 0.696 0.894 0.198 19.77
82-89 81.5
89.5 1.249 1.983 0.894 0.976 0.082
8.22 90-97
89.5 97.5
1.983 2.728 0.976 0.997 0.02 2.037
= 67.9 SD =10.89
Table 4.4 100Pj Table
5. Determining X
2 o
with the formula
Score f
j
P
j
100P
j
P
j
– 100 P
j
P
j
– 100P
j 2
100P
j
50-57 5
25 12.42
12.58 12.73
58-65 2
10 24.3
-14.3 8.415
66-73 7
35 28.37
6.633 1.551
74-81 5
25 19.77
5.232 1.384
82-89 8.22
-8.22 8.22
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
