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Why do anova(lm( )) and summary(lm( )) (in R) yield different p-values for a factor f with two levels?

The model is y ~ f + c, f has two levels (unequal sample size) and c is numeric.

set.seed(123)

df <- data.frame(
  y = rnorm(1000),
  f = sample(c("a","b"), 1000, replace=TRUE, prob=c(.2,.8)),
  c = rnorm(1000)
)


fit <- lm(y ~ f + c, data = df);


anova(fit)                 # p = 0.54553
summary(fit)               # p = 0.5191


library(emmeans)
emmeans(fit, pairwise ~ f) # p = 0.5191


edit: in addition...


library(car)
Anova(fit)                 # p = 0.5191
```
arb
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  • If you use library(car); Anova(fit) instead of anova(fit), you'll get the same p-value as in the other methods. – Sal Mangiafico Dec 18 '22 at 19:07
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    @SalMangiafico But why is that the case? – Dave Dec 18 '22 at 19:24
  • Try fit2 <- lm(y ~ c + f, data = df); you'll get the same results for the f p-value as with car::Anova() or emmeans() or summary(). That for c is now different. With unbalanced designs the Type I ANOVA done by anova() isn't to be trusted. See this page. – EdM Dec 18 '22 at 20:31
  • Oh my, I thought it defaulted to Type II – arb Dec 18 '22 at 22:01
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    You're clearly not the only one who thought that! Type II tests are default for car::Anova(). – EdM Dec 18 '22 at 22:07

0 Answers0