What explains the differences in p-values in the following aov and lm calls ?
Is the difference only due to different types of sums-of-squares calculations ?
set.seed(10)
data=rnorm(12)
f1=rep(c(1,2),6)
f2=c(rep(1,6),rep(2,6))
summary(aov(data~f1*f2))
summary(lm(data~f1*f2))$coeff
summary(aov) uses so called Type I (sequential) sums of squares. summary(lm) uses so called Type III sums of squares, which is not sequential. See gung's answer for details.
Note that you need to call lm(data ~ factor(f1) * factor(2)) (aov()automatically converts the RHS of the formula to factors). Then note the denominator for the general $t$-statistic in linear regression (see this answer for further explanations):
$$t = \frac{\hat{\psi} - \psi_{0}}{\hat{\sigma} \sqrt{\bf{c}' (\bf{X}'\bf{X})^{-1} \bf{c}}}$$
$\bf{c}' (\bf{X}'\bf{X})^{-1} \bf{c} $ differs for each tested $\beta$ coefficient because the vector $\bf{c}$ changes. In contrast, the denominator in the ANOVA $F$-test is always MSE.
set.seed(10)
data=rnorm(12)
f1=rep(c(1,2),6)
f2=c(rep(1,6),rep(2,6))
summary(aov(data~f1*f2))
Df Sum Sq Mean Sq F value Pr(>F)
f1 1 0.535 0.5347 0.597 0.462
f2 1 0.002 0.0018 0.002 0.966
f1:f2 1 0.121 0.1208 0.135 0.723
Residuals 8 7.169 0.8962
summary(lm(data~f1*f2))$coeff
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.05222024 2.732756 0.0191090 0.9852221
f1 -0.17992329 1.728346 -0.1041014 0.9196514
f2 -0.62637109 1.728346 -0.3624106 0.7264325
f1:f2 0.40139439 1.093102 0.3672066 0.7229887
These are two different codes. from the Lm model you need the coefficients. while from the aov model you are just tabulating the sources of variation. Try the code
anova(lm(data~f1*f2))
Analysis of Variance Table
Response: data
Df Sum Sq Mean Sq F value Pr(>F)
f1 1 0.5347 0.53468 0.5966 0.4621
f2 1 0.0018 0.00177 0.0020 0.9657
f1:f2 1 0.1208 0.12084 0.1348 0.7230
Residuals 8 7.1692 0.89615
This gives the tabulation of the sources of variation leading to the same results.
