我正在尝试运行Box的M-检验协方差矩阵的同质性,用于双向MANOVA。
从昨天下午开始,我一直在寻找一个例子。我看到了在单向MANOVA中使用boxM的许多例子。在每一种情况下,如果源还包括双向MANOVA,则不包括演示在双向情况下运行boxM测试。我只需要一个有用的例子。一旦我有了语法,我就可以让它工作了。
生物工具包中的boxM函数表示它是一个分类因子(单向MANOVA)。
https://www.rdocumentation.org/packages/biotools/versions/3.1/topics/boxM
海图包中的boxM函数表示它与一个或多个分类因素一起工作--
https://www.rdocumentation.org/packages/heplots/versions/1.3-5/topics/boxM
--然而,当我试图使用它时,我遇到了一个错误:“模型必须完全交叉公式。”
下面,我展示了在单独使用任何一个因素时,我不会得到一个错误,但是任何交叉因素的安排都会导致这个错误。注意:我没有得到这个错误运行Levene的测试与交叉变量。
Response1、Response2和Response3是连续的。
Factor1有2级。Factor2有5级。
library(heplots)
> Model2 <- lm(cbind(Response1, Response2, Response3) ~ Factor1, data=Data40)
> boxM(Model2)
Box's M-test for Homogeneity of Covariance Matrices
data: Y
Chi-Sq (approx.) = 3.5562, df = 6, p-value = 0.7365
> Model2 <- lm(cbind(Response1, Response2, Response3) ~ Factor2, data=Data40)
> boxM(Model2)
Box's M-test for Homogeneity of Covariance Matrices
data: Y
Chi-Sq (approx.) = 35.079, df = 24, p-value = 0.06724
> Model2 <- lm(cbind(Response1, Response2, Response3) ~ Factor1 * Factor2, data=Data40)
> boxM(Model2)
Error in boxM.formula(formula(Y), data = eval(data, envir = environment(formula(Y))), :
Model must be completely crossed formula only.
> Model2 <- lm(cbind(Response1, Response2, Response3) ~ Factor1 + Factor2 + Factor1 * Factor2, data=Data40)
> boxM(Model2)
Error in boxM.formula(formula(Y), data = eval(data, envir = environment(formula(Y))), :
Model must be completely crossed formula only.
> Model2 <- lm(cbind(Response1, Response2, Response3) ~ Factor1 + Factor2 + Factor1:Factor2, data=Data40)
> boxM(Model2)
Error in boxM.formula(formula(Y), data = eval(data, envir = environment(formula(Y))), :
Model must be completely crossed formula only.发布于 2020-10-05 18:54:11
不知道这个包从未使用过它,但在几分钟的时间内,您可能会以它不喜欢的方式指定公式.使用iris,因为包作者这样做了,而您没有提供任何数据。
library(heplots)
# adding a bogus second factor to iris
iris$nonsense <- rep(1:2)
iris$nonsense <- factor(iris$nonsense)
# one factor
boxM( cbind(Sepal.Length, Sepal.Width, Petal.Length, Petal.Width) ~ nonsense, data=iris)
#>
#> Box's M-test for Homogeneity of Covariance Matrices
#>
#> data: Y
#> Chi-Sq (approx.) = 16.389, df = 10, p-value = 0.08904
# second factor
boxM( cbind(Sepal.Length, Sepal.Width, Petal.Length, Petal.Width) ~ Species, data=iris)
#>
#> Box's M-test for Homogeneity of Covariance Matrices
#>
#> data: Y
#> Chi-Sq (approx.) = 140.94, df = 20, p-value < 2.2e-16
# crossed note not including the `lm`
boxM( cbind(Sepal.Length, Sepal.Width, Petal.Length, Petal.Width) ~ Species * nonsense, data=iris)
#>
#> Box's M-test for Homogeneity of Covariance Matrices
#>
#> data: Y
#> Chi-Sq (approx.) = 169.1, df = 50, p-value = 7.609e-15https://stackoverflow.com/questions/64212658
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