Appendix 5
Homogeneity Test by Using Bartlett Test
Subject df 1/df Si2 Log Si2 (df) Log Si2
1
2
3 41
44
39 0.0244
0.0227
0.0256 63.9907
78.4727
76.6769 1.8061
1.8947
1.8847 74.0501
83.3668
73.5033
Total 124 0.0727 - - 230.9202
Find total variance of all subject
Find value of B by using formula:
B = (Log S2) ∑ (ni-1) = (Log 73.1195)124 = 1.8640 x 124 = 231.1360
Bartlett test by using Chi-Square statistic
2 = (Ln 10) (B – ∑{(ni-1)Log Si2})
= 2.3026 (231.1360 – 230.9202) = 2.3026 x 0.2158 = 0.4969
ratio2 = 0.4969
By requiring significance level () at the 0.05 level and dk = k – 1 = 3 – 1 = 2, is gotten 2(1 - ) (dk) = 2(0.95) (2) = 2.92. So, ratio2 < table2 ; 0.4969 < 2.92; the result is: all of population variance is homogeny.
Appendix 6
Analysis of Variance
Calculate the mean of squares
= (2264 + 2418 + 2152)2 = 68342 = 367744.5354
42 + 45 + 40 127
Calculate a between-group sum of square
= (2264)2 + (2418)2 + (2152)2 - 367744.5354
42 45 40
= ( 122040.381 + 129927.2 + 115777.6) – 367744.5354
= 367745.181 – 367744.5354
= 0.6456
Calculate the sum of squares from all of groups
∑ X2 = X12 + X22 + X32 = 22642 + 24182 + 21522
= 5125696 + 5846724 + 4631104 = 15603524
Calculate a within-group sum of square
Dy = ∑ X2 – Ry – Ay = 15603524 - 367744.5354 - 0.6456 = 15235778.82
Make a table of analysis of variance (ANOVA)
Analysis of Variance (ANOVA)
Source of Variance Df Jk KT F
Means
Between Groups
Within Groups 1
2
124 367744.5354
0.6456
15235778.82 367744.5354
0.3228
122869.184 0.000002627
Total 131 - - -
From list of F distribution, with df for numerator = 2, df for denominator = 124 and significance at the level 0.05 (), is gotten Ftable = 3.0708. So, Fratio < Ftable It means that the English ability of the populations is not different.
