原文連結：http://tecdat.cn/?p=9529
目錄
怎麼做測試
協方差分析
拟合線的簡單圖解
模型的p值和R平方
檢查模型的假設
具有三類和II型平方和的協方差示例分析
協方差分析
拟合線的簡單圖解
組合模型的p值和R平方
檢查模型的假設
怎麼做測試
具有兩個類别和II型平方和的協方差示例的分析
本示例使用II型平方和 。參數估計值在R中的計算方式不同，
Data = read.table(textConnection(Input),header=TRUE) 複制
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) 複制
協方差分析
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 複制
拟合線的簡單圖解
plot(x = Data$Temp, y = Data$Pulse, col = Data$Species, pch = 16, xlab = "Temperature", ylab = "Pulse") 複制
模型的p值和R平方
Multiple R-squared: 0.9896, Adjusted R-squared: 0.9888F-statistic: 1331 on 2 and 28 DF, p-value: < 2.2e-16 複制
檢查模型的假設
線性模型中殘差的直方圖。這些殘差的分布應近似正态。
殘差與預測值的關系圖。殘差應無偏且均等。
### additional model checking plots with: plot(model.2)### alternative: library(FSA); residPlot(model.2) 複制
具有三類和II型平方和的協方差示例分析
本示例使用II型平方和，并考慮具有三個組的情況。
### --------------------------------------------------------------### Analysis of covariance, hypothetical data### --------------------------------------------------------------Data = read.table(textConnection(Input),header=TRUE) 複制
plot(x = Data$Temp, y = Data$Pulse, col = Data$Species, pch = 16, xlab = "Temperature", ylab = "Pulse")legend('bottomright', legend = levels(Data$Species), col = 1:3, cex = 1, pch = 16) 複制
協方差分析
options(contrasts = c("contr.treatment", "contr.poly")) ### These are the default contrasts in RAnova(model.1, type="II") Sum Sq Df F value Pr(>F) Temp 7026.0 1 2452.4187 <2e-16 ***Species 7835.7 2 1367.5377 <2e-16 ***Temp:Species 5.2 2 0.9126 0.4093 ### Interaction is not significant, so the slope among groups### is not different. Anova(model.2, type="II") Sum Sq Df F value Pr(>F) Temp 7026.0 1 2462.2 < 2.2e-16 ***Species 7835.7 2 1373.0 < 2.2e-16 ***Residuals 125.6 44 ### The category variable (Species) is significant,### so the intercepts among groups are differentsummary(model.2)Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -6.35729 1.90713 -3.333 0.00175 **Temp 3.56961 0.07194 49.621 < 2e-16 ***Speciesfake 19.81429 0.66333 29.871 < 2e-16 ***Speciesniv -10.18571 0.66333 -15.355 < 2e-16 ***### The slope estimate is the Temp coefficient.### The intercept for species 1 (ex) is (intercept).### The intercept for species 2 (fake) is (intercept) + Speciesfake.### The intercept for species 3 (niv) is (intercept) + Speciesniv.### This is determined from the contrast coding of the Species### variable shown below.contrasts(Data$Species) fake nivex 0 0fake 1 0niv 0 1 複制
拟合線的簡單圖解
組合模型的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 複制
檢查模型的假設
hist(residuals(model.2), col="darkgray") 複制
線性模型中殘差的直方圖。這些殘差的分布應近似正态。
plot(fitted(model.2), residuals(model.2)) 複制
殘差與預測值的關系圖。殘差應無偏且均等。
### additional model checking plots with: plot(model.2)### alternative: library(FSA); residPlot(model.2) 複制