The Result Score of Experimental Class
The Result Score of Experimental Class
From the table above, the students’ speaking ability was taught by applying Bamboo Dancing method showed the minimum score of pre-test was
40, the maximum score of pre-test was 80 and the mean of pre-test was 56,85. On the other hand the minimum score of post-test was 60, the maximum score of post-test was 90, and the mean of pre-test was 73,33.
Table VI
The Result Score Control Class No.
Students’ Initial
Pre-Test Post-Test
1 ASS
2 AI 40 60
3 AO
4 AAP
5 AZ
6 FR
7 FA 40 55
8 GM
9 MA
10 MHAM
11 MH
12 MHA
13 MAA
14 MP
15 MS
16 NP
17 NA
18 N
19 PA
20 RP
21 RP
22 SA
23 TA
24 T
25 WL
26 ZM
27 ZA
Total
Mean
From the table above, the students’ speaking ability was taught without applying Bamboo Dancing method showed the minimum score of pre-test was
40, the maximum score of pre-test was 70 and the mean of pre-test was 52,96. On the other hand the minimum score of post-test was 50, the maximum score of post-test was 75, and the mean of pre-test was 62,59.
2. The Calculation of Normality Test
a. Normality Testing of Experimental Group
Table VII
Frequency Distribution of Pre Test in Experimental Group
2 No. 2 Xi Fi Fi.Xi Xi Fi.Xi
Based on the data above, the result of Fi.Xi 2 is 90825 and FiXi is 1535. Then the following is the calculation of mean, variant and standard deviation.
a. Mean x = ∑
∑ Fi
Where: x
= Mean of variable x ΣFiXi = Total number of score
ΣFi = Number of sample
So, x = ∑
∑ Fi x = = 56,85 ∑ Fi x = = 56,85
S 2 = Variant N = Number of sample
c. Standard Deviation
S= √ = √ 136,82 = 11,69
After getting the calculation of mean, variant and deviation standard, then the next step is to found out the normality of the test. It means that the test was given to the students is observed by Liliefors test. The calculation of normality speaking can be seen in the following table:
Table VIII
Normality Testing of Pre Test in Experimental Group
No. Score
Zi
F(Zi)
S(Zi)
F(Zi) - S(Zi)
1 40 -1,44
2 40 -1,44
3 40 -1,44
4 40 -1,44
5 40 -1,44
Normality Testing of Pre Test in Experimental Group
1 -0,024 Total
1535 Lo = 0,014 Mean
56,85 Lt = 0,173
a. Finding Z score Formula: Zi 1 = Where,
= value = mean
S = standard deviation
Zi 1 = , = -1,44
b. Finding S(Zi) S(Zi) =
S(Zi) = = 0,22 S(Zi) = = 0,29 S(Zi) = = 0,55 S(Zi) = = 0,66 S(Zi) = = 0, 74 S(Zi) = = 0,96 S(Zi) = = 1
From the table above, it can be seen that Liliefors observation or Lo = 0,014 with n = 27 and at real level = 0.05 from the list of critical value of Liliefors table Lt =.0,173. It is known that the coefficient of Lo (0,014 ) < Lt (0,173 ). So it can be concluded that the data distribution of the student’s ability in speaking is normal.
Table IX
Frequency Distribution of Post Test in Experimental Group
2 No 2 Xi Fi Fi.Xi Xi Fi.Xi
Based on the data above, the result of Fi.Xi 2 is 147050 and FiXi is 1980. Then the following is the calculation of mean, variant and standard deviation.
a. Mean x = ∑
∑ Fi
Where: x
= Mean of variable x ΣFiXi = Total number of score ΣFi = Number of sample
So, x = ∑
∑ Fi
b. Variant Where:
S 2 = Variant N = Number of sample
c. Standard Deviation
S= √ = 71,15 = 8,43
After getting the calculation of mean, variant and deviation standard, then the next step is to found out the normality of the test. It means that the test was given to the students is observed by Liliefors test. The calculation of normality speaking can be seen in the following table:
Table X Normality Testing of Post Test in Experimental Group
No Score
Zi
F(Zi)
S(Zi)
F(Zi) - S(Zi)
1 60 -1,58
2 60 -1,58
3 60 -1,58
4 65 -0,98
5 65 -0,98
6 65 -0,98
7 70 -0,39
8 70 -0,39
9 70 -0,39
10 70 -0,39
Normality Testing of Post Test in Experimental Group
1 -0,025 Total
a. Finding Z score Formula: Zi 1 =
Where, = value
= mean s = standard deviation
Zi 1 = ,
= -1,58
Zi 2 = , = -0,98
Zi 3 = ,
= -0,39
Zi 4 = , = 0,19 Zi 4 = , = 0,19
S(Zi) = = 0,11 S(Zi) = = 0,22 S(Zi) = = 0,48 S(Zi) = = 0,74 S(Zi) = = 0, 88 S(Zi) = = 1
From the table above, it can be seen that Liliefors observation or Lo = - 0,025 with n = 27 and at real level = 0.05 from the list of critical value of Liliefors table Lt = 0,173. It is known that the coefficient of Lo (-0,025) < Lt ( 0,173). So it can be concluded that the data distribution of the student’s ability in speaking is normal.
b. Normality Testing of Control Group
Table XI
Frequency Distribution of Pre Test in Control Group
2 No 2 Xi Fi Fi.Xi Xi Fi.Xi
14700 Total
Based on the data above, the result of Fi.Xi 2 is 78450 and FiXi is 1430. Then the following is the calculation of mean, variant and standard deviation.
a. Mean x = ∑
∑ Fi
Where: x
= Mean of variable x ΣFiXi = Total number of score ΣFi = Number of sample
b. Variant Where:
S 2 = Variant N = Number of sample
c. Standard Deviation
S= 2 √ = √104 , 34 = 10,21
After getting the calculation of mean, variant and deviation standard, then the next step is to found out the normality of the test. It means that the test was given to the students is observed by Liliefors test. The calculation of normality speaking can be seen in the following table:
Table XII Normality Testing of Pre Test in Control Group
No Score
Zi
F(Zi)
S(Zi)
F(Zi) - S(Zi)
1 40 -1,26
2 40 -1,26
3 40 -1,26
4 40 -1,26
5 40 -1,26
6 40 -1,26
7 45 -0,77
8 45 -0,77
9 45 -0,77
10 45 -0,77
11 50 -0,28
12 50 -0,28
13 50 -0,28
Normality Testing of Pre Test in Control Group
1 -0,049 Total
a. Finding Z score Formula: Zi 1 = Where,
= value = mean
s = standard deviation
b. Finding S(Zi) S(Zi) = S(Zi) = 6
S(Zi) = 10
S(Zi) = 13
S(Zi) = 17
S(Zi) = 22
27 = 0,88 S(Zi) = = 1 From the table above, it can be seen that Liliefors observation or Lo = -
S(Zi) = 24
0,002 with n = 27 and at real level = 0.05 from the list of critical value of Liliefors table Lt =.0,173. It is known that the coefficient of Lo (-0,002) < Lt (0,173). So it can be concluded that the data distribution of the student’s ability in speaking is normal.
Table XIII
Frequency Distribution of Post Test in Control Group
2 No 2 Xi Fi Fi.Xi Xi Fi.Xi
Based on the data above, the result of Fi.Xi 2 is 107150 and FiXi is 1690. Then the following is the calculation of mean, variant and standard deviation.
a. Mean
∑ Fi
Where: x
= Mean of variable x ΣFiXi = Total number of score
ΣFi = Number of sample
So, So,
Where: S2 = Variant N = Number of sample
c. Standard Deviation
S= 2 √ = √52 , 63 = 7,25
After getting the calculation of mean, variant and deviation standard, then the next step is to found out the normality of the test. It means that the test was given to the students is observed by Liliefors test. The calculation of normality speaking can be seen in the following table.
Table XIV
Normality Testing of Post Test in Control Group
No Score
Zi
F(Zi)
S(Zi) F(Zi) - S(Zi)
1 -0,044 Total
-0,029 Mean 62,59259
Lo
Lt
= value = mean
S = standard deviation
a. Finding S(Zi) S(Zi) = S(Zi) = 2
S(Zi) = 7
S(Zi) = 14
S(Zi) = 20
S(Zi) = 24
27 = 0,88 S(Zi) = = 1
From the table above, it can be seen that Liliefors observation or Lo = - 0,029 with n = 27 and at real level = 0.05 from the list of critical value of Liliefors table Lt = 0,173. It is known that the coefficient of Lo (- 0,029 ) < Lt ( 0,173). So it can be concluded that the data distribution of the student’s ability in speaking is normal.
3. Calculation of Homogeneity Test
a. Homogeneity Testing of Pre Test
F obs =
Where : S 1 2 = the biggest variant S 2 2 = the smallest variant
Based on the variants of both samples of pre-test found that:
S ex = 136,82
N = 27
S co = 104,34
N = 27
So,
Fobs =
= 1,31 Then the coefficient of F obs = 1.31 is compared with F table , where F table is
determined at real level = 0.05 and the same numerator dk = N - 1 = 27 - 1 = 26 that was exist dk numerator 26, the denominator dk = n - 1 (27 - 1 = 26). Then
F table can be calculated F 0.05(26,26) = 1,95. So F obs <F table atau ( 1.31 < 1,95 ) so it can be concluded that the variant is homogenous.
b. Homogeneity Testing of Post Test
F obs =
Where : S 1 2 = the biggest variant S 2 2 = the smallest variant
Based on the variants of both samples of post-test found that:
S ex = 71,15
N = 27
S co = 52,63
N = 27
So,
Fobs =
= 1,35 Then the coefficient of F obs = 1,35 is compared with F table , where F table is
determined at real level = 0.05 and the same numerator dk = N - 1 = 27 - 1 = 26 that was exist dk numerator 26, the denominator dk = n - 1 (27 - 1 = 26).Then
F table can be calculated F 0.05(26,26) = 1,95. So F obs <F table atau ( 1.35 < 1,95 ) so it can be concluded that the variant is homogenous.