se.contrast {stats}R Documentation

Standard Errors for Contrasts in Model Terms

Description

Returns the standard errors for one or more contrasts in an aov object.

Usage

se.contrast(object, ...)
## S3 method for class 'aov':
se.contrast(object, contrast.obj,
           coef = contr.helmert(ncol(contrast))[, 1],
           data = NULL, ...)

Arguments

object A suitable fit, usually from aov.
contrast.obj The contrasts for which standard errors are requested. This can be specified via a list or via a matrix. A single contrast can be specified by a list of logical vectors giving the cells to be contrasted. Multiple contrasts should be specified by a matrix, each column of which is a numerical contrast vector (summing to zero).
coef used when contrast.obj is a list; it should be a vector of the same length as the list with zero sum. The default value is the first Helmert contrast, which contrasts the first and second cell means specified by the list.
data The data frame used to evaluate contrast.obj.
... further arguments passed to or from other methods.

Details

Contrasts are usually used to test if certain means are significantly different; it can be easier to use se.contrast than compute them directly from the coefficients.

In multistratum models, the contrasts can appear in more than one stratum; the contrast and standard error are computed in the lowest stratum and adjusted for efficiencies and comparisons between strata.

Suitable matrices for use with coef can be found by calling contrasts and indexing the columns by a factor.

Value

A vector giving the standard errors for each contrast.

See Also

contrasts, model.tables

Examples

## From Venables and Ripley (2002) p.165.
N <- c(0,1,0,1,1,1,0,0,0,1,1,0,1,1,0,0,1,0,1,0,1,1,0,0)
P <- c(1,1,0,0,0,1,0,1,1,1,0,0,0,1,0,1,1,0,0,1,0,1,1,0)
K <- c(1,0,0,1,0,1,1,0,0,1,0,1,0,1,1,0,0,0,1,1,1,0,1,0)
yield <- c(49.5,62.8,46.8,57.0,59.8,58.5,55.5,56.0,62.8,55.8,69.5,
55.0, 62.0,48.8,45.5,44.2,52.0,51.5,49.8,48.8,57.2,59.0,53.2,56.0)

npk <- data.frame(block = gl(6,4), N = factor(N), P = factor(P),
                  K = factor(K), yield = yield)
options(contrasts=c("contr.treatment", "contr.poly"))
npk.aov1 <- aov(yield ~ block + N + K, npk)
se.contrast(npk.aov1, list(N=="0", N=="1"), data=npk)
# or via a matrix
cont <- matrix(c(-1,1), 2, 1, dimnames=list(NULL, "N"))
se.contrast(npk.aov1, cont[N, , drop=FALSE]/12, data=npk)

## test a multi-stratum model
npk.aov2 <- aov(yield ~ N + K + Error(block/(N + K)), npk)
se.contrast(npk.aov2, list(N == "0", N == "1"))
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