You are looking for linear hypothesis test by check p-value of some linear combination of regression coefficients. Based on my answer: How to conduct linear hypothesis test on regression coefficients with a clustered covariance matrix?, where we only considered sum of coefficients, I will extend the function LinearCombTest
to handle more general cases, supposing alpha
as some combination coefficients of variables in vars
:
LinearCombTest <- function (lmObject, vars, alpha, .vcov = NULL) {
## if `.vcov` missing, use the one returned by `lm`
if (is.null(.vcov)) .vcov <- vcov(lmObject)
## estimated coefficients
beta <- coef(lmObject)
## linear combination of `vars` with combination coefficients `alpha`
LinearComb <- sum(beta[vars] * alpha)
## get standard errors for sum of `LinearComb`
LinearComb_se <- sum(alpha * crossprod(.vcov[vars, vars], alpha)) ^ 0.5
## perform t-test on `sumvars`
tscore <- LinearComb / LinearComb_se
pvalue <- 2 * pt(abs(tscore), lmObject$df.residual, lower.tail = FALSE)
## return a matrix
form <- paste0("(", paste(alpha, vars, sep = " * "), ")")
form <- paste0(paste0(form, collapse = " + "), " = 0")
matrix(c(LinearComb, LinearComb_se, tscore, pvalue), nrow = 1L,
dimnames = list(form, c("Estimate", "Std. Error", "t value", "Pr(>|t|)")))
}
Consider a simple example, where we have a balanced design for three groups A
, B
and C
, with group mean 0, 1, 2, respectively.
x <- gl(3,100,labels = LETTERS[1:3])
set.seed(0)
y <- c(rnorm(100, 0), rnorm(100, 1), rnorm(100, 2)) + 0.1
fit <- lm(y ~ x)
coef(summary(fit))
# Estimate Std. Error t value Pr(>|t|)
#(Intercept) 0.1226684 0.09692277 1.265631 2.066372e-01
#xB 0.9317800 0.13706949 6.797866 5.823987e-11
#xC 2.0445528 0.13706949 14.916177 6.141008e-38
Since A
is the reference level, xB
is giving B - A
while xC
is giving C - A
. Suppose we are now interested in the difference between group B
and C
, i.e., C - B
, we can use
LinearCombTest(fit, c("xC", "xB"), c(1, -1))
# Estimate Std. Error t value Pr(>|t|)
#(1 * xC) + (-1 * xB) = 0 1.112773 0.1370695 8.118312 1.270686e-14
Note, this function is also handy to work out the group mean of B
and C
, that is (Intercept) + xB
and (Intercept) + xC
:
LinearCombTest(fit, c("(Intercept)", "xB"), c(1, 1))
# Estimate Std. Error t value Pr(>|t|)
#(1 * (Intercept)) + (1 * xB) = 0 1.054448 0.09692277 10.87926 2.007956e-23
LinearCombTest(fit, c("(Intercept)", "xC"), c(1, 1))
# Estimate Std. Error t value Pr(>|t|)
#(1 * (Intercept)) + (1 * xC) = 0 2.167221 0.09692277 22.36029 1.272811e-65
Alternative solution with lsmeans
Consider the above toy example again:
library(lsmeans)
lsmeans(fit, spec = "x", contr = "revpairwise")
#$lsmeans
# x lsmean SE df lower.CL upper.CL
# A 0.1226684 0.09692277 297 -0.06807396 0.3134109
# B 1.0544484 0.09692277 297 0.86370603 1.2451909
# C 2.1672213 0.09692277 297 1.97647888 2.3579637
#
#Confidence level used: 0.95
#
#$contrasts
# contrast estimate SE df t.ratio p.value
# B - A 0.931780 0.1370695 297 6.798 <.0001
# C - A 2.044553 0.1370695 297 14.916 <.0001
# C - B 1.112773 0.1370695 297 8.118 <.0001
#
#P value adjustment: tukey method for comparing a family of 3 estimates
The $lsmeans
domain returns the marginal group mean, while $contrasts
returns pairwise group mean difference, since we have used "revpairwise" contrast. Read p.32 of lsmeans
for difference between "pairwise"
and "revpairwise"
.
Well this is certainly interesting, as we can compare with the result from LinearCombTest
. We see that LinearCombTest
is doing correctly.
lsmeans
besides reading through your answer. How do you do you import the console result? – Underside