逻辑回归：方差卡方检验与系数的显着性（R中的anova（）vs summary（））

我有一个8个变量的逻辑GLM模型。我在R中进行了卡方检验，anova(glm.model,test='Chisq')当在测试的顶部进行排序时，其中两个变量可预测，而在底部的排序中则没有那么多。在summary(glm.model)表明它们的系数不显着（高p值）。在这种情况下，变量似乎并不重要。

我想问问哪个是变量显着性更好的检验-模型摘要中的系数显着性或来自的卡方检验anova()。还有-什么时候一个比另一个更好？

我想这是一个广泛的问题，但是任何有关考虑因素的建议将不胜感激。

除了@gung的答案外，我还将尝试提供该anova函数实际测试内容的示例。我希望这使您能够决定哪些测试适合您对测试感兴趣的假设。

假设您有一个结果和3个预测变量：x 1，x 2和x 3。现在，如果您的逻辑回归模型为。运行时，该函数按顺序比较以下模型：$y$$x_{1}$$x_{2}$$x_{3}$my.mod <- glm(y~x1+x2+x3, family="binomial")anova(my.mod, test="Chisq")

1. glm(y~1, family="binomial") 与 glm(y~x1, family="binomial")
2. glm(y~x1, family="binomial") 与 glm(y~x1+x2, family="binomial")
3. glm(y~x1+x2, family="binomial") 与 glm(y~x1+x2+x3, family="binomial")

因此，通过在每个步骤中添加一个变量，它依次将较小的模型与下一个更复杂的模型进行比较。每个比较都是通过似然比测试（LR测试；请参见下面的示例）完成的。据我所知，这些假设很少引起人们的兴趣，但这必须由您决定。

这是一个示例R：

mydata      <- read.csv("https://stats.idre.ucla.edu/stat/data/binary.csv")
mydata$rank <- factor(mydata$rank)

my.mod <- glm(admit ~ gre + gpa + rank, data = mydata, family = "binomial")
summary(my.mod)

Coefficients:
             Estimate Std. Error z value Pr(>|z|)    
(Intercept) -3.989979   1.139951  -3.500 0.000465 ***
gre          0.002264   0.001094   2.070 0.038465 *  
gpa          0.804038   0.331819   2.423 0.015388 *  
rank2       -0.675443   0.316490  -2.134 0.032829 *  
rank3       -1.340204   0.345306  -3.881 0.000104 ***
rank4       -1.551464   0.417832  -3.713 0.000205 ***
   ---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

# The sequential analysis
anova(my.mod, test="Chisq")

Terms added sequentially (first to last)    

     Df Deviance Resid. Df Resid. Dev  Pr(>Chi)    
NULL                   399     499.98              
gre   1  13.9204       398     486.06 0.0001907 ***
gpa   1   5.7122       397     480.34 0.0168478 *  
rank  3  21.8265       394     458.52 7.088e-05 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

# We can make the comparisons by hand (adding a variable in each step)

  # model only the intercept
mod1 <- glm(admit ~ 1,                data = mydata, family = "binomial") 
  # model with intercept + gre
mod2 <- glm(admit ~ gre,              data = mydata, family = "binomial") 
  # model with intercept + gre + gpa
mod3 <- glm(admit ~ gre + gpa,        data = mydata, family = "binomial") 
  # model containing all variables (full model)
mod4 <- glm(admit ~ gre + gpa + rank, data = mydata, family = "binomial") 

anova(mod1, mod2, test="LRT")

Model 1: admit ~ 1
Model 2: admit ~ gre
  Resid. Df Resid. Dev Df Deviance  Pr(>Chi)    
1       399     499.98                          
2       398     486.06  1    13.92 0.0001907 ***

anova(mod2, mod3, test="LRT")

Model 1: admit ~ gre
Model 2: admit ~ gre + gpa
  Resid. Df Resid. Dev Df Deviance Pr(>Chi)  
1       398     486.06                       
2       397     480.34  1   5.7122  0.01685 *

anova(mod3, mod4, test="LRT")

Model 1: admit ~ gre + gpa
Model 2: admit ~ gre + gpa + rank
  Resid. Df Resid. Dev Df Deviance  Pr(>Chi)    
1       397     480.34                          
2       394     458.52  3   21.826 7.088e-05 ***

$p$summary(my.mod)

• 对于系数x1：glm(y~x2+x3, family="binomial")与 glm(y~x1+x2+x3, family="binomial")
• 对于系数x2：glm(y~x1+x3, family="binomial")与glm(y~x1+x2+x3, family="binomial")
• 对于系数x3：glm(y~x1+x2, family="binomial")与glm(y~x1+x2+x3, family="binomial")

因此，每个系数相对于包含所有系数的完整模型。Wald检验是似然比检验的近似值。我们还可以进行似然比检验（LR检验）。方法如下：

mod1.2 <- glm(admit ~ gre + gpa,  data = mydata, family = "binomial")
mod2.2 <- glm(admit ~ gre + rank, data = mydata, family = "binomial")
mod3.2 <- glm(admit ~ gpa + rank, data = mydata, family = "binomial")

anova(mod1.2, my.mod, test="LRT") # joint LR test for rank

Model 1: admit ~ gre + gpa
Model 2: admit ~ gre + gpa + rank
  Resid. Df Resid. Dev Df Deviance  Pr(>Chi)    
1       397     480.34                          
2       394     458.52  3   21.826 7.088e-05 ***

anova(mod2.2, my.mod, test="LRT") # LR test for gpa

Model 1: admit ~ gre + rank
Model 2: admit ~ gre + gpa + rank
  Resid. Df Resid. Dev Df Deviance Pr(>Chi)  
1       395     464.53                       
2       394     458.52  1   6.0143  0.01419 *

anova(mod3.2, my.mod, test="LRT") # LR test for gre

Model 1: admit ~ gpa + rank
Model 2: admit ~ gre + gpa + rank
  Resid. Df Resid. Dev Df Deviance Pr(>Chi)  
1       395     462.88                       
2       394     458.52  1   4.3578  0.03684 *

$p$summary(my.mod)

rankanova(my.mod, test="Chisq")rankanova(mod1.2, my.mod, test="Chisq")$p$$7.088\cdot 10^{-5}$rank