R: Standard Errors for Contrasts in Model Terms

se.contrast {base} | 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, ...)
    se.contrast.aov(object, contrast.obj,
      coef = contr.helmert(ncol(contrast))[, 1], data = NULL, ...)
    se.contrast.aovlist(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 as returned by `contrasts`.
`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 directly with 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.

### Value

A vector giving the standard errors for each contrast.

### Author(s)

B.D. Ripley

### See Also

`contrasts`, `model.tables`

### Examples

    ## From Venables and Ripley (1997) p.210.
    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)

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