I am piggybacking on my question from yesterday
Suppose we have three treatments, and want to get all pairwise differences. The default in R is to use contrasts and only display 2. I.E 2 vs 1, and 3 vs 1. In order to get 3 vs 2, we need to subtract the beta coefficients (3vs 1 - 2 vs 1).
So we have the estimate now, but is there a way to get the variance of that estimate? Or do we have to run the regression again with a different reference group to get it?
We can use the glht function from the multcomp package to do post-hoc testing, with options to adjust for multiple testing. (see http://www.ats.ucla.edu/stat/r/faq/testing_contrasts.htm for more examples)
A small example
library(multcomp)
data(mtcars)
mtcars$cyl <- factor(mtcars$cyl, levels=c(4, 6, 8))
# run model
m <- lm(mpg ~ cyl, data=mtcars)
coef(summary(m))
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) 26.663636 0.9718008 27.437347 2.688358e-22
# cyl6 -6.920779 1.5583482 -4.441099 1.194696e-04
# cyl8 -11.563636 1.2986235 -8.904534 8.568209e-10
coef(m)[2] - coef(m)[3]
# cyl6
# 4.642857
# use glht function to estimate other contrasts
# define the linfct matrix to test specific contrast
mf <- glht(m, linfct = matrix(c(0, 1, -1), 1))
summary(mf)
# Simultaneous Tests for General Linear Hypotheses
#
# Fit: lm(formula = mpg ~ cyl, data = mtcars)
#
# Linear Hypotheses:
# Estimate Std. Error t value Pr(>|t|)
# 1 == 0 4.643 1.492 3.112 0.00415 **
# ---
# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
# (Adjusted p values reported -- single-step method)
# look at estimate by changing the reference level manually
mtcars$cyl2 <- factor(mtcars$cyl, levels=c(6, 4, 8))
m2 <- lm(mpg ~ cyl2, data=mtcars)
coef(summary(m2))
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) 19.742857 1.218217 16.206357 4.493289e-16
# cyl24 6.920779 1.558348 4.441099 1.194696e-04
# cyl28 -4.642857 1.492005 -3.111825 4.152209e-03
We can also get the answer by knowing the theory behind it. Perhaps this answer is better suited for CV, but I figured I would post it here.
The answer comes down to knowing the statistics behind it. We have a difference of two random variables, that we want to get the variance of.
In mathematical terms, we would like var(beta_{2vs1} -beta_{3vs1}). Knowing our elementary statistics, we know that this is equal to var(beta_{2vs1}) +var(beta_{3vs1})-2*cov(beta_{2vs1},beta_{3vs1})
v <- vcov(m)
sqrt(v[2,2] + v[3,3] - 2*v[2,3])
# [1] 1.492005
