我如何使用glm（）函数？

Question

I'm trying to fit a general linear model (GLM) on my data using R. I have a Y continuous variable and two categorical factors, A and B. Each factor is coded as 0 or 1, for presence or absence.

Even if just looking at the data I see a clear interaction between A and B, the GLM says that p-value>>>0.05. Am I doing something wrong?

First of all I create the data frame including my data for the GLM, which consists on a Y dependent variable and two factors, A and B. These are two level factors (0 and 1). There are 3 replicates per combination.

A<-c(0,0,0,1,1,1,0,0,0,1,1,1)
B<-c(0,0,0,0,0,0,1,1,1,1,1,1)
Y<-c(0.90,0.87,0.93,0.85,0.98,0.96,0.56,0.58,0.59,0.02,0.03,0.04)
my_data<-data.frame(A,B,Y)

Let's see how it looks like:

my_data
##    A B    Y
## 1  0 0 0.90
## 2  0 0 0.87
## 3  0 0 0.93
## 4  1 0 0.85
## 5  1 0 0.98
## 6  1 0 0.96
## 7  0 1 0.56
## 8  0 1 0.58
## 9  0 1 0.59
## 10 1 1 0.02
## 11 1 1 0.03
## 12 1 1 0.04

As we can see just looking on the data, there is a clear interaction between factor A and factor B, as the value of Y dramatically decreases when A and B are present (that is A=1 and B=1). However, using the glm function I get no significant interaction between A and B, as p-value>>>0.05

attach(my_data)
## The following objects are masked _by_ .GlobalEnv:
## 
##     A, B, Y


my_glm<-glm(Y~A+B+A*B,data=my_data,family=binomial)
## Warning: non-integer #successes in a binomial glm!
summary(my_glm)
## 
## Call:
## glm(formula = Y ~ A + B + A * B, family = binomial, data = my_data)
## 
## Deviance Residuals: 
##       Min         1Q     Median         3Q        Max  
## -0.275191  -0.040838   0.003374   0.068165   0.229196  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)
## (Intercept)   2.1972     1.9245   1.142    0.254
## A             0.3895     2.9705   0.131    0.896
## B            -1.8881     2.2515  -0.839    0.402
## A:B          -4.1747     4.6523  -0.897    0.370
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 7.86365  on 11  degrees of freedom
## Residual deviance: 0.17364  on  8  degrees of freedom
## AIC: 12.553
## 
## Number of Fisher Scoring iterations: 6

Answer 1

While you state Y is continuous, the data shows that Y is rather a fraction. Hence, probably the reason you tried to apply GLM in the first place.

To model fractions (ie continuous values bounded by 0 and 1) can be done with logistic regression if certain assumptions are fullfilled. See the following cross-validated post for details: https://stats.stackexchange.com/questions/26762/how-to-do-logistic-regression-in-r-when-outcome-is-fractional . However, from the data description it is not clear that those assumptions are fullfilled.

An alternative to model fractions are beta regression or fractional repsonse models.

See below how to apply those methods to your data. The results of both methods are consistent in terms of signs and significance.

# Beta regression
install.packages("betareg")
library("betareg")
result.betareg <-betareg(Y~A+B+A*B,data=my_data)
summary(result.betareg)

# Call:
#   betareg(formula = Y ~ A + B + A * B, data = my_data)
# 
# Standardized weighted residuals 2:
#   Min      1Q  Median      3Q     Max 
# -2.7073 -0.4227  0.0682  0.5574  2.1586 
# 
# Coefficients (mean model with logit link):
#   Estimate Std. Error z value Pr(>|z|)    
# (Intercept)   2.1666     0.2192   9.885  < 2e-16 ***
#   A             0.6471     0.3541   1.828   0.0676 .  
#   B            -1.8617     0.2583  -7.206 5.76e-13 ***
#   A:B          -4.2632     0.5156  -8.268  < 2e-16 ***
#   
#   Phi coefficients (precision model with identity link):
#   Estimate Std. Error z value Pr(>|z|)  
# (phi)    71.57      29.50   2.426   0.0153 *
#   ---
#   Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 
# 
# Type of estimator: ML (maximum likelihood)
# Log-likelihood: 24.56 on 5 Df
# Pseudo R-squared: 0.9626
# Number of iterations: 62 (BFGS) + 2 (Fisher scoring) 


# ----------------------------------------------------------


# Fractional response model
install.packages("frm")
library("frm")
frm(Y,cbind(A, B, AB=A*B),linkfrac="logit")


*** Fractional logit regression model ***

#   Estimate Std. Error t value Pr(>|t|)    
# INTERCEPT  2.197225   0.157135  13.983    0.000 ***
#   A          0.389465   0.530684   0.734    0.463    
#   B         -1.888120   0.159879 -11.810    0.000 ***
#   AB        -4.174668   0.555642  -7.513    0.000 ***
#   
#   Note: robust standard errors
# 
# Number of observations: 12 
# R-squared: 0.992 

Answer 2

The family=binomial implies Logit (Logistic) Regression, which is itself produces a binary result.

From Quick-R

Logistic Regression

Logistic regression is useful when you are predicting a binary outcome from a set of continuous predictor variables. It is frequently preferred over discriminant function analysis because of its less restrictive assumptions.

Answer 3

The data shows an interaction. Try to fit a different model, logistic is not appropriate.

with(my_data, interaction.plot(A, B, Y, fixed = TRUE, col = 2:3, type = "l"))

An analysis of variance shows clear significance for all factors and interaction.

fit <- aov(Y~(A*B),data=my_data)
summary(fit)
            Df Sum Sq Mean Sq F value   Pr(>F)    
A            1 0.2002  0.2002   130.6 3.11e-06 ***
B            1 1.1224  1.1224   732.0 3.75e-09 ***
A:B          1 0.2494  0.2494   162.7 1.35e-06 ***
Residuals    8 0.0123  0.0015                     
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1