R语言对回归模型进行协方差分析
原文链接:http://tecdat.cn/?p=9529
目录
怎么做测试
协方差分析
拟合线的简单图解
模型的p值和R平方
检查模型的假设
具有三类和II型平方和的协方差示例分析
协方差分析
拟合线的简单图解
组合模型的p值和R平方
检查模型的假设
怎么做测试
具有两个类别和II型平方和的协方差示例的分析
本示例使用II型平方和 。参数估计值在R中的计算方式不同,
Data = read.table(textConnection(Input),header=TRUE)
plot(x = Data$Temp,y = Data$Pulse,col = Data$Species,pch = 16,xlab = "Temperature",ylab = "Pulse")legend('bottomright',legend = levels(Data$Species),col = 1:2,cex = 1,pch = 16)
协方差分析
Anova Table (Type II tests)Sum Sq Df F value Pr(>F)Temp 4376.1 1 1388.839 < 2.2e-16 ***Species 598.0 1 189.789 9.907e-14 ***Temp:Species 4.3 1 1.357 0.2542### Interaction is not significant, so the slope across groups### is not different.model.2 = lm (Pulse ~ Temp + Species,data = Data)library(car)Anova(model.2, type="II")Anova Table (Type II tests)Sum Sq Df F value Pr(>F)Temp 4376.1 1 1371.4 < 2.2e-16 ***Species 598.0 1 187.4 6.272e-14 ***### The category variable (Species) is significant,### so the intercepts among groups are differentCoefficients:Estimate Std. Error t value Pr(>|t|)(Intercept) -7.21091 2.55094 -2.827 0.00858 **Temp 3.60275 0.09729 37.032 < 2e-16 ***Speciesniv -10.06529 0.73526 -13.689 6.27e-14 ***### but the calculated results will be identical.### The slope estimate is the same.### The intercept for species 1 (ex) is (intercept).### The intercept for species 2 (niv) is (intercept) + Speciesniv.### This is determined from the contrast coding of the Species### variable shown below, and the fact that Speciesniv is shown in### coefficient table above.nivex 0niv 1
拟合线的简单图解
plot(x = Data$Temp,y = Data$Pulse,col = Data$Species,pch = 16,xlab = "Temperature",ylab = "Pulse")

模型的p值和R平方
Multiple R-squared: 0.9896, Adjusted R-squared: 0.9888F-statistic: 1331 on 2 and 28 DF, p-value: < 2.2e-16
检查模型的假设

线性模型中残差的直方图。这些残差的分布应近似正态。

残差与预测值的关系图。残差应无偏且均等。
### additional model checking plots with: plot(model.2)### alternative: library(FSA); residPlot(model.2)
具有三类和II型平方和的协方差示例分析
本示例使用II型平方和,并考虑具有三个组的情况。
### --------------------------------------------------------------### Analysis of covariance, hypothetical data### --------------------------------------------------------------Data = read.table(textConnection(Input),header=TRUE)
plot(x = Data$Temp,y = Data$Pulse,col = Data$Species,pch = 16,xlab = "Temperature",ylab = "Pulse")legend('bottomright',legend = levels(Data$Species),col = 1:3,cex = 1,pch = 16)
协方差分析
options(contrasts = c("contr.treatment", "contr.poly"))### These are the default contrasts in RAnova(model.1, type="II")Sum Sq Df F value Pr(>F)Temp 7026.0 1 2452.4187 <2e-16 ***Species 7835.7 2 1367.5377 <2e-16 ***Temp:Species 5.2 2 0.9126 0.4093### Interaction is not significant, so the slope among groups### is not different.Anova(model.2, type="II")Sum Sq Df F value Pr(>F)Temp 7026.0 1 2462.2 < 2.2e-16 ***Species 7835.7 2 1373.0 < 2.2e-16 ***Residuals 125.6 44### The category variable (Species) is significant,### so the intercepts among groups are differentsummary(model.2)Coefficients:Estimate Std. Error t value Pr(>|t|)(Intercept) -6.35729 1.90713 -3.333 0.00175 **Temp 3.56961 0.07194 49.621 < 2e-16 ***Speciesfake 19.81429 0.66333 29.871 < 2e-16 ***Speciesniv -10.18571 0.66333 -15.355 < 2e-16 ***### The slope estimate is the Temp coefficient.### The intercept for species 1 (ex) is (intercept).### The intercept for species 2 (fake) is (intercept) + Speciesfake.### The intercept for species 3 (niv) is (intercept) + Speciesniv.### This is determined from the contrast coding of the Species### variable shown below.contrasts(Data$Species)fake nivex 0 0fake 1 0niv 0 1
拟合线的简单图解

组合模型的p值和R平方
Multiple R-squared: 0.9919, Adjusted R-squared: 0.9913F-statistic: 1791 on 3 and 44 DF, p-value: < 2.2e-16
检查模型的假设
hist(residuals(model.2),col="darkgray")

线性模型中残差的直方图。这些残差的分布应近似正态。
plot(fitted(model.2),residuals(model.2))

残差与预测值的关系图。残差应无偏且均等。
### additional model checking plots with: plot(model.2)### alternative: library(FSA); residPlot(model.2)

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