R se.contrast 模型术语对比的标准误差

说明

返回 aov 对象中一个或多个对比的标准错误。

用法

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

参数

object	合适的配合，通常来自aov。
contrast.obj	要求标准误差的对比。这可以通过列表或矩阵来指定。单个对比可以通过给出要对比的单元的逻辑向量列表来指定。多重对比应由矩阵指定，矩阵的每一列都是数值对比向量(总和为零)。
coef	当contrast.obj是列表时使用；它应该是与零和列表长度相同的向量。默认值是第一个 Helmert 对比度，它对比列表指定的第一个和第二个单元格均值。
data	用于评估 contrast.obj 的数据帧。
...	传入或传出其他方法的进一步参数。

细节

对比通常用于测试某些均值是否显著不同；使用 se.contrast 比直接从系数计算它们更容易。

在多层模型中，对比可能出现在多个层中，在这种情况下，标准误差在最低层中计算，并根据效率和各层之间的比较进行调整。 (请参阅 aov 帮助中有关使用正交对比的注释中的注释。)此类标准错误通常是保守的。

通过调用 contrasts 并按因子对列进行索引，可以找到适合与 coef 一起使用的矩阵。

值

给出每个对比的标准误差的向量。

例子

## 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)
## Set suitable contrasts.
options(contrasts = c("contr.helmert", "contr.poly"))
npk.aov1 <- aov(yield ~ block + N + K, data = 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)), data = npk)
se.contrast(npk.aov2, list(N == "0", N == "1"))


## an example looking at an interaction contrast
## Dataset from R.E. Kirk (1995)
## 'Experimental Design: procedures for the behavioral sciences'
score <- c(12, 8,10, 6, 8, 4,10,12, 8, 6,10,14, 9, 7, 9, 5,11,12,
            7,13, 9, 9, 5,11, 8, 7, 3, 8,12,10,13,14,19, 9,16,14)
A <- gl(2, 18, labels = c("a1", "a2"))
B <- rep(gl(3, 6, labels = c("b1", "b2", "b3")), 2)
fit <- aov(score ~ A*B)
cont <- c(1, -1)[A] * c(1, -1, 0)[B]
sum(cont)       # 0
sum(cont*score) # value of the contrast
se.contrast(fit, as.matrix(cont))
(t.stat <- sum(cont*score)/se.contrast(fit, as.matrix(cont)))
summary(fit, split = list(B = 1:2), expand.split = TRUE)
## t.stat^2 is the F value on the A:B: C1 line (with Helmert contrasts)
## Now look at all three interaction contrasts
cont <- c(1, -1)[A] * cbind(c(1, -1, 0), c(1, 0, -1), c(0, 1, -1))[B,]
se.contrast(fit, cont)  # same, due to balance.
rm(A, B, score)


## multi-stratum example where efficiencies play a role
## An example from Yates (1932),
## a 2^3 design in 2 blocks replicated 4 times

Block <- gl(8, 4)
A <- factor(c(0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,
              0,1,0,1,0,1,0,1,0,1,0,1))
B <- factor(c(0,0,1,1,0,0,1,1,0,1,0,1,1,0,1,0,0,0,1,1,
              0,0,1,1,0,0,1,1,0,0,1,1))
C <- factor(c(0,1,1,0,1,0,0,1,0,0,1,1,0,0,1,1,0,1,0,1,
              1,0,1,0,0,0,1,1,1,1,0,0))
Yield <- c(101, 373, 398, 291, 312, 106, 265, 450, 106, 306, 324, 449,
           272, 89, 407, 338, 87, 324, 279, 471, 323, 128, 423, 334,
           131, 103, 445, 437, 324, 361, 302, 272)
aovdat <- data.frame(Block, A, B, C, Yield)
fit <- aov(Yield ~ A + B * C + Error(Block), data = aovdat)
cont1 <- c(-1, 1)[A]/32  # Helmert contrasts
cont2 <- c(-1, 1)[B] * c(-1, 1)[C]/32
cont <- cbind(A = cont1, BC = cont2)
colSums(cont*Yield) # values of the contrasts
se.contrast(fit, as.matrix(cont))
# comparison with lme
library(nlme)
fit2 <- lme(Yield ~ A + B*C, random = ~1 | Block, data = aovdat)
summary(fit2)$tTable # same estimates, similar (but smaller) se's.

也可以看看

contrasts , model.tables