Handout for ANOVA
Use Data Example628Anova.SPSS
Test the mean of pension (pension1) classified by education (edu1)
1. Test normality
Kolmogorov – Smirnov (n > 50)
Shapiro – Wilk (n <=50)
The result show that pension is normality (cannot reject null hypothesis).
2. Test Variance
Equal variance ANOVA
At least one pair unequal Welch Test or Brown’s Test
Test of Homogeneity of Variances
pension1
Levene Statistic / df1 / df2 / Sig.4.926 / 8 / 781 / .000
Reject Null hypothesis (equality of variance) Unequal Brown or Welch
Robust Tests of Equality of Means
pension1
Statistic(a) / df1 / df2 / Sig.Welch / 21.182 / 8 / 38.749 / .000
Brown-Forsythe / 20.319 / 8 / 294.010 / .000
a Asymptotically F distributed.
If we assume that the variances are equal (but we should not in this case) use ANOVA
ANOVA
pension1
Sum of Squares / df / Mean Square / F / Sig.Between Groups / 9454730421.224 / 8 / 1181841302.653 / 10.703 / .000
Within Groups / 86240428570.918 / 781 / 110423083.958
Total / 95695158992.141 / 789
Null hypothesis is rejected unequal mean
Test More than one factor (marital status)
Tests of Between-Subjects Effects
Dependent Variable: pension1
Source / Type III Sum of Squares / df / Mean Square / F / Sig.Corrected Model / 12526232560.329(a) / 31 / 404072018.075 / 3.683 / .000
Intercept / 24653833541.307 / 1 / 24653833541.307 / 224.695 / .000
edu1 / 3298753459.617 / 8 / 412344182.452 / 3.758 / .000
marital1 / 765407863.027 / 3 / 255135954.342 / 2.325 / .074
edu1 * marital1 / 1170006896.356 / 20 / 58500344.818 / .533 / .953
Error / 83168926431.813 / 758 / 109721538.828
Total / 458514256487.802 / 790
Corrected Total / 95695158992.141 / 789
a R Squared = .131 (Adjusted R Squared = .095)