Calculation of Normality Test
B. Calculation of Normality Test
1. Normality Testing of Experimental Group
Table 4.3
Frequency Distribution of Pre Test in Experimental Group
Fi.Xi
Xi2 Fi.Xi 2
Based on the data above, the result of Fi.Xi 2 was 93675 and FiXi was 1575. Then the following was the calculation of mean, variant and
standartdeviation .
a. Mean
Where: = mean of variabel Σ FiXi = total number of score
ΣFi =number of sample
So, ∑
b. Variant
Where : S 2 = Variant n = number of sample
c. Standard Deviation
S= √²
After getting the calculation of mean, variant and deviation standart, the the text step was to found the normality of the test. It means that the test was given to the students was observed by Liliefors test. The calculation of normality writing speaking can be seen in the following table :
Table 4.4
Normality Testing Of Pre Test in Experimental Group
F(Zi) S(Zi)
F(Zi)-S(Zi)
a. Finding Z Score
Formula: Zi 1= Where ;
= value
= mean = standard deviation
b. Finding S(Zi)
From the table above it can be seen that Liliefors observationor Lo = 0.02 with n = 28 and real level = 0.05 from the list of critical value of Liliefors table Lt = 0.161. It was known that the coefficient of Lo (0.02) < Lt (0.161). So, it can be concluded that the data distribution of the student’s achievement in writing was normal.
Table 4.5 Frequency of Distribution Post Test in Experimental
Fi Fi.Xi
Xi2
Fi.Xi2
Based on the data above, the result of Fi.Xi 2 was 148475 and FiXi was 2025. Then the following was the calculation of mean, variant and
standarddeviation .
a. Mean
Where: = mean of variabel
Σ FiXi = total number of score ΣFi
= number of sample So,
b. Variant
Where : S 2 = Variant
n = number of sample 2 . Σ
c. Standard Deviation
S= √²
After getting the calculation of mean, variantand deviation standart, the the text step was to found the normality of the test. It means After getting the calculation of mean, variantand deviation standart, the the text step was to found the normality of the test. It means
Table 4.6 Normality Testing Of Post Test in Experimental
Group
F(Zi)- No
Score
Zi
F(Zi) S(Zi)
S(Zi)
Mean 72.32 Lt
a. Finding Z Score
Zi 1= Where ;
= value = mean = standard deviation
b. Finding S(Zi)
From the table above it can be seen that Liliefors observation or Lo = -0.021 with n = 28 and real level = 0.05 from the list of critical value of Liliefors table Lt = 0.161. It was known that the coefficient of Lo (0.02) < Lt (0.161). So, it can be concluded that the data distribution of the student’s achievement in writing was normal.
2. Normality Testing of Control Group
Tabel 4.7
Frequency Distribution of Pre Test in Control Group No
Xi
Fi
Fi.Xi
Xi2
Fi.Xi2
Based on the data above, the result of Fi.Xi 2 was 81475 and FiXi was 1485. Then the following was the calculation of mean, variant and standart deviation.
a. Mean
Where: = mean of variabel
ΣFiXi = total number of score ΣFi
=number of sample =number of sample
Where : S 2 = Variant
n = number of sample
c. Standard Deviation
S= √²
After getting the calculation of mean, variantand deviation standart, the next step was to found the normality of the test. It means that the test was given to the students was observed by Liliefors test. The calculation of normality writing can be seen in the following table :
Table 4.8 Normality Testing Of Pre Test in Control Group
No Score
Zi
F(Zi) S(Zi) F(Zi)-S(Zi)
1.00 -0.045 Total 1485
a. Finding Z Score
Formula: Zi 1= Where ;
= value = mean = standard deviation
b. Finding S(Zi)
From the table above it can be seen that Liliefors observation or Lo = -0.009 with n = 28 and real level = 0.05 from the list of critical value of Liliefors table Lt = 0.161. It was known that the coefficient of Lo (0.02) < Lt (0.161). So, it can be concluded that the data distribution of the student’s achievement in writing was normal.
Tabel 4.9 Frequency Distribution of Post Test in Control
Fi.Xi
Xi2
Fi.Xi2
Based on the data above, the result of Fi.Xi 2 was 112775 and FiXi was 1765. Then the following was the calculation of mean, variant and
standarddeviation .
a. Mean
Where: = mean of variabel ΣFiXi = total number of score
ΣFi =number of sample So,
b. Variant
Where : S 2 = Variant
n = number of sample . Σ
c. Standard Deviation
S= √²
After getting the calculation of mean, variant and deviation standard, the the text step was to found the normality of the test. It means that the test was After getting the calculation of mean, variant and deviation standard, the the text step was to found the normality of the test. It means that the test was
Table 4.10 Normality Testing Of Post Test in Control Group
No Score
Zi
F(Zi)
S(Zi)
F(Zi)-S(Zi)
a. Finding Z Score
Formula: Zi 1= Where ;
= value = mean = standard deviation
b. Finding S(Zi)
From the table above it can be seen that Liliefors observation or Lo = -0.030 with n = 28 and real level = 0.05 from the list of critical value of Liliefors table Lt = 0.161. It was known that the coefficient of Lo (0.02) < Lt (0.161). So, it can be concluded that the data distribution of the student’s achievement in writing was normal.