MD =
∑D =
188,00 =
4,95 N
38
t =
4,95 = 8,87
437,89 38 38
- 1 -1 = 37 , t
0.9528
= 2,026
-2,03 2,03
8,87
Based on the computation above, I obtained that t
value
was 8.8 and the t
table
was 2.03 so the t
value
was higher than t
table.
It meant that there was significant difference between mean of the pre-test and post-test using CLL method.
4.3.1 Normality
1. Normality test for pre-test
Hipothesis
Ho : The data distributed normality
Ha : The data not distributed normality
The Calculation
Formula :
Ho is accepted if
x
2
x
2 tabel
x
2 k-3
k 1
i i
2 i
i 2
E E
O
Maximum score =
74,00 Panjang Kelas
= 2,3
Minimum Score =
60,00 Mean X
= 66,1
Range =
14,00 S
= 4,8
Class with =
6,0 N
= 38
Class Interval X
pz p
Z Ei
Oi Oi-Ei²
Ei 58,00
- 60,00
57,50 -1,81
0,4646 0,0842
3,201 6
2,448 61,00
- 63,00
60,50 -1,18
0,3803 0,1726
6,557 6
0,047 64,00
66,00 63,50
-0,55 0,2078
0,2408 9,151
11 0,374
67,00 -
69,00 66,50
0,08 0,0330
0,2289 8,700
5 1,574
70,00 -
72,00 69,50
0,71 0,2620
0,1483 5,636
5 0,072
73,00 -
75,00 72,50
1,34 0,4103
0,0654 2,487
5 2,541
75,50 1,97
0,4757 38
x
2
= 7,055
for α = 5, dk = 6 - 3 = 3, x² table = 7,815
7,055 7,81
Because c² 7,81 then the pre test is said to be normallly distributed. 2. Normality test for post-test
Hipothesis
Ho :
The data distributed normality Ha
: The data not distributed normality
The Calculation
Formula :
Ho is accepted if x
2
x
2 tabel
x
2 αk-3
Maximum score =
78,00 Panjang Kelas
= 2,7
Minimum Score =
62,00 Mean X
= 71,1
Range =
16,00 S
= 4,2
Class with =
6,0 N
= 38
Class Interval x
pz p
z Ei
Oi Oi-Ei²
Ei 60,00
- 62,00
59,50 -2,78
0,4973 0,0170
0,646 1
0,193 63,00
- 65,00
62,50 -2,06
0,4803 0,0709
2,693 3
0,035 66,00
68,00 65,50
-1,34 0,4094
0,1788 6,794
6 0,093
69,00 -
71,00 68,50
-0,61 0,2306
0,2735 10,395
9 0,187
72,00 -
74,00 71,50
0,11 0,0429
0,2539 9,648
11 0,189
75,00 -
78,00 74,5
0,83 0,2968
0,1668 6,337
8 0,437
78,5 1,79
0,4636 38
x² =
1,134 for α = 5, dk = 6 - 3 = 3, x² table =
7,815
1,134 7,81
Because c² 7,81 then the post test is said to be normallly distributed.
4.3.2 Homogenity
Hipothesis
Ho :
σ
1 2
= σ
2 2
Ha :
σ
1 2
= σ
2 2
k 1
i i
2 i
i 2
E E
O
The Calculation
Formula :
Ho is accepted if F F
12α nb-1:nk-1
F
12αnb-1:nk-1
Post test Pre test
Sum 2700
2512 N
38 38
X 71,05
66,11 Variance s
2
17,2404 22,6913
Standart deviation s 4,15
4,76 F
= 22,69
= 1,3162 17,24
For α = 5 with: df1
= n1
- 1
= 38 - 1 = 37 df2
= n2
- 1
= 38 - 1 = 37 F
0.02537:37
= 1,92
1,32 1,92
Since F value F table, the pre test and post test of experimental group have the same variance
VK Vb
F
4.4 Description of Students’ Overall Ability before Treatment
