I read here that if group sizes are equal, ANOVA is robust against the violation of the assumptions of normality and homoscedasticity.

I am wondering if this is the case, and if so why?


1 Answer 1


Welch one-way ANOVA. Maybe so, but I think it is better to use a Welch version of the one-way ANOVA---as shown briefly below in R. This test is similar to the Welch two-sample t test, specifically designed to accommodate unequal variances.

x1 = rnorm(10, 50, 2);   x2 = rnorm(10, 50, 4)
x3 = rnorm(10, 55, 4);   x4 = rnorm(10, 60, 5)
x = c(x1, x2, x3, x4);  gp = rep(1:4, each=10)

These data are simulated to have among-group differences in both means and variances.

boxplot(x ~ gp, col="skyblue2", pch=20)

enter image description here

oneway.test(x ~ gp)

        One-way analysis of means 
     (not assuming equal variances)

data:  x and gp
F = 28.896, num df = 3.000, denom df = 18.466, 
   p-value = 3.463e-07

Note: In a standard on-way ANOVA one would have a denom df of $4(10 - 1) = 36$ degrees of freedom.

References: This web page discussed ad-hoc tests and other relevant details. This excellent Answer by @gung on our own site explores this test and other related issues. Perhaps explore other "Related" links provided in the right-hand margin of this page.

  • $\begingroup$ Thanks, I was not aware of this alternative! $\endgroup$
    – manesioz
    May 29, 2019 at 13:07

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