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.
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
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.
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
Here the p-value is large indicating that average response when am=0 is not significantly different than when am=1.
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
This p-value appears to agree with the analysis of non overlapping confidence intervals.
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