Estimating difference in lsmeans

Question

How does one use lsmeans to estimate the difference in two pairwise contrasts? For instance--imagine a continuous dv and two factor predictors

library(lsmeans)
library(tidyverse)

dat <- data.frame(
  y = runif(30),
  x1 = 1:2 %>% 
    factor %>% 
    sample(30, T),
  x2 = letters[1:3] %>%
    factor %>% 
    sample(30, T)
  )

lm1 <- lm(y ~ x1 * x2, data = dat)

This call gets me the estimates of the effect of x1 by x2

lsmeans(lm1, ~ x1 | x2) %>% 
  pairs

returns

x2 = a:
 contrast     estimate        SE df t.ratio p.value
 1 - 2    -0.150437681 0.2688707 24  -0.560  0.5810

x2 = b:
 contrast     estimate        SE df t.ratio p.value
 1 - 2    -0.048950972 0.1928172 24  -0.254  0.8018

x2 = c:
 contrast     estimate        SE df t.ratio p.value
 1 - 2    -0.006819473 0.2125610 24  -0.032  0.9747

This is fine, but I now want the difference of these contrasts, to see if these 1 - 2 differences are themselves different according to x2 levels.

score 2 · Accepted Answer · answered Oct 25 '17 at 12:53

2

Use

lsm = lsmeans(lm1, ~ x1 * x2)
contrast(lsm, interaction = “pairwise”)

answered Oct 25 '17 at 12:53

Russ Lenth

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To my mind, this is not quiet well solution because it will make too much comparisons and this will decrease p-value. – Alex Oct 26 '17 at 06:04
It creates 3 contrasts total. – Russ Lenth Oct 26 '17 at 12:02

score 1 · Answer 2 · answered Oct 25 '17 at 08:30

As I understand your problem right, you can use this solution :

lm1 <- lm(y ~ x1 * x2, data = dat)
means.int <- lsmeans(lm1, ~x1 + x2)
dd <- contrast(means.int, list(`a--b` = c(1,-1,-1,1,0,0), 
                               `a--c`=c(1,-1,0,0,-1,1),
                               `b--c` = c(0,0,1,-1,-1,1)), 
               adjust = 'mvt')

Estimating difference in lsmeans

2 Answers2