Box-Cox back transformation with emmeans not working properly - r

Question

I am doing a linear regression with data that needs transformation, for it, I am using a Box-Cox power transformation, followed by back-transformation to write a report using the original scale. I've been trying to do this with the emmeans packages, and I followed the steps described in the emmeans package vignette, however, I find that the summary results for the estimated means are not at all similar to the untransformed data. In fact, the output is not transformed at all.
Here is a reproducible example using the examples from the emmeans package:
require(emmeans)
# Fit a model using an oddball transformation:
bctran <- make.tran("boxcox", 0.368)
warp.bc <- with(bctran,
lm(linkfun(breaks) ~ wool * tension, data = warpbreaks))
# Obtain back-transformed LS means:
emmeans(warp.bc, ~ tension | wool, type = "response")
# Fit a model without transformation:
warp <- lm(breaks ~ wool * tension, data = warpbreaks)
# Obtain LS means:
emmeans(warp, ~ tension | wool)
which returns:
> emmeans(warp.bc, ~ tension | wool, type = "response")
wool = A:
tension emmean SE df lower.CL upper.CL
L 8.07 0.419 48 7.23 8.92
M 5.91 0.419 48 5.07 6.75
H 5.94 0.419 48 5.10 6.79
wool = B:
tension emmean SE df lower.CL upper.CL
L 6.45 0.419 48 5.61 7.29
M 6.53 0.419 48 5.69 7.37
H 5.22 0.419 48 4.38 6.07
Confidence level used: 0.95
> emmeans(warp, ~ tension | wool)
wool = A:
tension emmean SE df lower.CL upper.CL
L 44.6 3.65 48 37.2 51.9
M 24.0 3.65 48 16.7 31.3
H 24.6 3.65 48 17.2 31.9
wool = B:
tension emmean SE df lower.CL upper.CL
L 28.2 3.65 48 20.9 35.6
M 28.8 3.65 48 21.4 36.1
H 18.8 3.65 48 11.4 26.1
Confidence level used: 0.95
when in fact the estimated mean for tension:L should be 42.37, as calculated using the formula:
> origin + (1 + param * pmax(eta))^(1/param)
> 0 + (1 + 0.368 * pmax(8.07))^(1/0.368)
[1] 42.37179
Is there something I am missing or not understanding properly?

Answer 1

Hmmmm. I reproduced this problem. I'm not sure what's wrong, but so far I can tell that bctran itself is in order:
> emm = as.data.frame(emmeans(warp.bc, ~tension|wool))
> emm
tension wool emmean SE df lower.CL upper.CL
1 L A 8.074761 0.4192815 48 7.231739 8.917783
2 M A 5.911710 0.4192815 48 5.068688 6.754732
3 H A 5.942335 0.4192815 48 5.099313 6.785357
4 L B 6.449869 0.4192815 48 5.606847 7.292891
5 M B 6.531085 0.4192815 48 5.688063 7.374107
6 H B 5.224939 0.4192815 48 4.381917 6.067961
> bctran$linkinv(emm$emmean)
[1] 42.42263 23.10060 23.32407 27.22827 27.88877 18.43951
So these back-transformed EMMs are in-order. I'll trace the code and see why the results aren't back-transformed.
Update
I found a logic error from a revision a few months ago whereby if a transformation is character (e.g., "log") it works fine, but if it is a list (e.g., your bctran) it is ignored.
I fixed that error in the next version to push to the github site (version >= 1.3.3.0999902), and the fix will be in the next CRAN update (version > 1.3.3).
> emmeans(warp.bc, ~ tension | wool)
wool = A:
tension emmean SE df lower.CL upper.CL
L 8.07 0.419 48 7.23 8.92
M 5.91 0.419 48 5.07 6.75
H 5.94 0.419 48 5.10 6.79
wool = B:
tension emmean SE df lower.CL upper.CL
L 6.45 0.419 48 5.61 7.29
M 6.53 0.419 48 5.69 7.37
H 5.22 0.419 48 4.38 6.07
Results are given on the Box-Cox (lambda = 0.368) (not the response) scale.
Confidence level used: 0.95
> emmeans(warp.bc, ~ tension | wool, type = "response")
wool = A:
tension response SE df lower.CL upper.CL
L 42.4 4.48 48 34.0 52.0
M 23.1 3.05 48 17.5 29.8
H 23.3 3.07 48 17.7 30.0
wool = B:
tension response SE df lower.CL upper.CL
L 27.2 3.38 48 20.9 34.6
M 27.9 3.44 48 21.5 35.3
H 18.4 2.65 48 13.6 24.3
Confidence level used: 0.95
Intervals are back-transformed from the Box-Cox (lambda = 0.368) scale
Notice that even without back-transforming, there is an annotation of that fact. The fact that no annotation at all is present in your results was a tip-off.