测试跨因素计算的边际效应之间的差异 [英] Testing the difference between marginal effects calculated across factors

问题描述

我正在尝试测试两个边际效应之间的差异.我可以得到R来计算效果，但是找不到任何资源来解释如何测试它们的差异.

I'm trying to test the difference between two marginal effects. I can get R to calculate the effects, but I can't find any resource explaining how to test their difference.

我查看了边距文档和其他边际效果包，但是找不到能够测试差异的东西.

I've looked in the margins documentations and other marginal effects packages but have not been able to find something that tests the difference.

data("mtcars")

mod<-lm(mpg~as.factor(am)*disp,data=mtcars)

(marg<-margins(model = mod,at = list(am = c("0","1"))))
 at(am)     disp    am1
      0 -0.02758 0.4518
      1 -0.05904 0.4518
summary(marg)
 factor     am     AME     SE       z      p   lower   upper
    am1 1.0000  0.4518 1.3915  0.3247 0.7454 -2.2755  3.1791
    am1 2.0000  0.4518 1.3915  0.3247 0.7454 -2.2755  3.1791
   disp 1.0000 -0.0276 0.0062 -4.4354 0.0000 -0.0398 -0.0154
   disp 2.0000 -0.0590 0.0096 -6.1353 0.0000 -0.0779 -0.0402

我想进行一次测试，以确定每行玛格的边际效应是否显着不同；即，边际效应图中的斜率是不同的.这似乎是正确的，因为置信区间不重叠-表示对于am = 0和am = 1，位移的影响是不同的.

I want to produce a test that decides whether or not the marginal effects in each row of marg are significantly different; i.e., that the slopes in the marginal effects plots are different. This appears to be true because the confidence intervals do not overlap -- indicating that the effect of displacement is different for am=0 vs am=1.

我们在下面的评论中讨论，我们可以使用emmeans测试对比，但这是对am = 0和am = 1的平均响应的测试.

We discuss in the comments below that we can test contrasts using emmeans, but that is a test of the average response across am=0 and am=1.

emm<-emmeans(mod,~ as.factor(am)*disp)
emm
 am disp emmean    SE df lower.CL upper.CL
  0  231   18.8 0.763 28     17.2     20.4
  1  231   19.2 1.164 28     16.9     21.6
cont<-contrast(emm,list(`(0-1)`=c(1,-1)))
cont
 contrast estimate   SE df t.ratio p.value
 (0-1)      -0.452 1.39 28 -0.325  0.7479 

此处p值较大，表明am = 0时的平均响应与am = 1时的平均响应没有显着差异.

Here the p-value is large indicating that average response when am=0 is not significantly different than when am=1.

这样做是否合理(例如测试两种方法的差异)?

Is it reasonable to do this (like testing the difference of two means)?

smarg<-summary(marg)
(z=as.numeric((smarg$AME[3]-smarg$AME[4])/sqrt(smarg$SE[3]^2+smarg$SE[4]^2)))
[1] 2.745
2*pnorm(-abs(z))
[1] 0.006044

此p值似乎与非重叠置信区间的分析相符.

This p-value appears to agree with the analysis of non overlapping confidence intervals.

推荐答案

如果我理解您的问题，可以使用emtrends回答:

If I understand your question, it can be answered using emtrends:

library(emmeans)

emt = emtrends(mod, "am", var = "disp")

emt  # display the estimated slopes

## am disp.trend      SE df lower.CL upper.CL
##  0    -0.0276 0.00622 28  -0.0403  -0.0148
##  1    -0.0590 0.00962 28  -0.0787  -0.0393
##
## Confidence level used: 0.95 

pairs(emt) # test the difference of slopes

## contrast estimate     SE df t.ratio p.value
## 0 - 1      0.0315 0.0115 28 2.745   0.0104

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