除了@gung 的回答，我将尝试提供该anova函数实际测试的示例。我希望这使您能够决定哪些测试适合您有兴趣测试的假设。

假设您有一个结果和 3 个预测变量：、和。现在，如果您的逻辑回归模型是. 运行时，该函数按顺序比较以下模型。这种类型也称为 I 型方差分析或 I 型平方和（有关不同类型的比较，请参阅这篇文章）：$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")对比glm(y~x1, family="binomial")
2. glm(y~x1, family="binomial")对比glm(y~x1+x2, family="binomial")
3. glm(y~x1+x2, family="binomial")对比glm(y~x1+x2+x3, family="binomial")

因此，它通过在每个步骤中添加一个变量来依次将较小的模型与下一个更复杂的模型进行比较。这些比较中的每一个都是通过似然比检验（LR 检验；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 ***

这$p$- 输出中的值summary(my.mod)是Wald 测试，用于测试以下假设（请注意，它们是可互换的，测试的顺序无关紧要）：

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

所以每个系数都针对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 *

这$p$- 来自似然比检验的值与上述 Wald 检验获得的值非常相似summary(my.mod)。

注意： 的第三个模型比较与下面示例中rank的anova(my.mod, test="Chisq")比较rank（anova(mod1.2, my.mod, test="Chisq")）相同。每次，$p$-值相同，$7.088\cdot 10^{-5}$. 每次都是没有模型rank与包含它的模型之间的比较。