R语言对回归模型进行协方差分析

Posted 2021-03-22 tecdat

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 different

 

 

Coefficients:

             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.

 

 

    niv

ex    0

niv   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:  0.9888

F-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 R

 
Anova(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 different

 

 

summary(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 niv

ex      0   0

fake    1   0

niv     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:  0.9913

F-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) 

 

 

如果您有任何疑问，请在下面发表评论。 

 

 

大数据部落 -中国专业的第三方数据服务提供商，提供定制化的一站式数据挖掘和统计分析咨询服务

统计分析和数据挖掘咨询服务：y0.cn/teradat（咨询服务请联系官网客服）

?QQ：3025393450

 

?QQ交流群：186388004 

【服务场景】  

科研项目; 公司项目外包;线上线下一对一培训;数据爬虫采集;学术研究;报告撰写;市场调查。

【大数据部落】提供定制化的一站式数据挖掘和统计分析咨询

 

欢迎关注微信公众号，了解更多数据干货资讯！

 

 

欢迎选修我们的R语言数据分析挖掘必知必会课程！