我有一个数据集,想比较物种和栖息地对家域大小的影响-同时使用类型III错误和物种和栖息地内的成对比较。
以下是数据的子集:
species<- c("a","b","c","c","b","c","b","b","a","b","c","c","a","a","b","b","a","a","b","c")
habitat<- c("x","x","x","y","y","y","x","x","y","z","y","y","z","z","x","x","y","y","z","z")
homerange<-c(6,5,7,8,9,4,3,5,6,9,3,6,6,7,8,9,5,6,7,8)
data1<-data.frame(cbind(species, habitat, homerange))
data1$homerange<-as.numeric(as.character(data1$homerange))
目前我正在将数据分成三个物种,然后针对每个物种单独运行ANOVA,但我认为通过一个ANOVA同时询问物种和栖息地更有意义。以下是我为一个物种运行的ANOVA示例:
data.species.a<-subset(data1, species=="a")
fit<-aov(homerange ~ habitat, data=data.species.a)
summary(fit)
TukeyHSD(fit)
aov()似乎使用I型错误……我认为这不合适;此外,我认为Tukey的测试可能过于保守,用于成对比较。有人能帮助我找到一种方法,使我能够运行一个考虑物种和栖息地对家域影响的ANOVA分析,同时考虑III型误差,并允许更少保守的物种和栖息地的成对比较吗?
aov
结果与@DWin的答案以及他的答案的一个小变体进行比较,其中您将使用Anova(...,type="II")
)。您上面是不是想测试aov(homerange ~ habitat*species)
?我赞同CrossValidated的建议。 - Ben Bolker