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

问题描述

aov 是否适用于不平衡的数据集.根据帮助...为lm提供了一个包装器，用于将线性模型拟合到平衡或不平衡的实验设计.但后来它说aov 是为平衡设计而设计的，如果没有平衡，结果可能很难解释.

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.

我应该如何对 R 中的不平衡数据集执行 2 向方差分析?

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).我记得我们对不平衡数据集使用 III 型平方和.

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.

提前致谢.

推荐答案

Function anova(或 summary.aov)会给你所谓的类型 I(或序列) 平方和.要获得类型 III 的平方和，您可以使用 Anova 函数来自库 car，参数为 type="III".这两种方法在不平衡数据集(以及生成两个表的示例 R 代码)中的区别在此处.

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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