非平衡数据集上的2向方差分析 [英] 2-way anova on unbalanced dataset

问题描述

aov适用于不平衡的数据集.根据帮助...provides a wrapper to lm for fitting linear models to balanced or unbalanced experimental designs.但稍后会说aov is designed for balanced designs, and the results can be hard to interpret without balance.

Is aov appropriate for unbalanced datasets. According to help ...provides a wrapper to lm for fitting linear models to balanced or unbalanced experimental designs. But later on it says aov is designed for balanced designs, and the results can be hard to interpret without balance.

我应该如何在R中的不平衡数据集上执行2次方差分析?

How should I perform a 2-way anova on an unbalanced dataset in R?

我想重现SAS输出的I型和III型平方和的不同结果(使用proc glm时).我记得我们使用type III sum of squares表示不平衡的数据集.

I would like to reproduce the different results for type I and type III sum of squares of SAS output (when using proc glm). I remember we were using type III sum of squares for unbalanced datasets.

谢谢.

推荐答案

函数anova(或summary.aov)将为您提供所谓的I型(或顺序)平方和.要获取III型平方和，您可以使用 Anova <库car中的/a>函数，其参数为type="III".

Function anova (or summary.aov) will give you the so called type I (or sequential) sum of squares. To get type III sum of squares, you can use the Anova function from library car with parameter type="III". The difference between these two approaches in unbalanced datasets (and also sample R code to produce both tables) is presented in detail here.

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