My data look like this
Study NDF ADF CP Eeff
1 35.8 24.4 18.6 34.83181476
1 35.8 24.4 18.6 33.76824264
1 35.8 24.4 18.6 32.67390287
1 35.8 24.4 18.6 33.05520666
2 39.7 23.4 16.1 33.19730252
2 39.4 22.9 16.3 34.04709188
3 28.9 20.6 18.7 33.22501606
3 27.1 18.9 17.9 33.80766289
Of course, I have 80 lines like this.
I used lme function to run a mixed model (Study as random effect), as following:
fm1<-lme(Eeff~NDF+ADF+CP,random=~1|Study, data=na.omit(phuong))
I got this result:
Fixed effects: Ratio ~ ADF + CP + FCM + DMI + DIM
Value Std.Error DF t-value p-value
(Intercept) 3.1199808 0.16237303 158 19.214896 0.0000
ADF -0.0265626 0.00406990 158 -6.526603 0.0000
CP -0.0534021 0.00539108 158 -9.905636 0.0000
FCM -0.0149314 0.00353524 158 -4.223598 0.0000
DMI 0.0072318 0.00498779 158 1.449894 0.1491
DIM -0.0008994 0.00019408 158 -4.634076 0.0000
Correlation:
(Intr) ADF CP FCM DMI
ADF -0.628
CP -0.515 0.089
FCM -0.299 0.269 -0.203
DMI -0.229 -0.145 0.083 -0.624
DIM -0.113 0.127 -0.061 0.010 -0.047
These results show the case where intercept is random but slope is fixed. How can I see my 80 intercept, for example, like below when I used study as fixed effect:
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.0021083 0.0102536 -0.206 0.837351
ADF 0.0005248 0.0002962 1.772 0.078313 .
CP 0.0021131 0.0003277 6.448 1.26e-09 ***
factor(Study)2 0.0057274 0.0038709 1.480 0.140933
factor(Study)3 0.0117722 0.0035262 3.338 0.001046 **
factor(Study)4 0.0091049 0.0043227 2.106 0.036730 *
factor(Study)6 0.0149733 0.0045345 3.302 0.001182 **
factor(Study)7 0.0065518 0.0036837 1.779 0.077196 .
factor(Study)8 0.0066134 0.0035371 1.870 0.063337 .
factor(Study)9 0.0086758 0.0036641 2.368 0.019083 *
factor(Study)10 0.0105657 0.0041296 2.559 0.011434 *
factor(Study)11 0.0083694 0.0040194 2.082 0.038900 *
factor(Study)16 0.0171258 0.0028962 5.913 1.95e-08 ***
factor(Study)18 0.0019277 0.0042300 0.456 0.649209
factor(Study)20 0.0172469 0.0040412 4.268 3.36e-05 ***
factor(Study)23 0.0132676 0.0031658 4.191 4.57e-05 ***
factor(Study)24 0.0063313 0.0031519 2.009 0.046236 *
factor(Study)25 0.0050929 0.0039135 1.301 0.194989
Thank you very much,
Phuong
You didn't give us a reproducible question, but the answer is to use coef, for example:
> library(nlme)
> fm1 <- lme(distance~age,random=~1|Subject,data=Orthodont)
> coef(fm1)
(Intercept) age
M16 15.84314 0.6601852
M05 15.84314 0.6601852
M02 16.17959 0.6601852
M11 16.40389 0.6601852
M07 16.51604 0.6601852
M08 16.62819 0.6601852
M03 16.96464 0.6601852
[snip]
use fixef() to get just the fixed effect coefficients
use ranef() to get just the random effects (i.e. deviations of each individual from the fixed coefficients
the Orthodont example in lme actually uses a random-slope(+intercept) model; here I have fitted a random-intercept model, so the estimated slope (age parameter) is the same for every individual
it looks like individuals are sorted in increasing order of estimated random effect
Related
I really need help with this. I want to make a predict model for my glm quasipoisson. I have a problems since i wrongly make a glm model with my dataset.
I used to make a predict model based on my glm quasipoisson for all my parameters, but I ended up predicting for each parameter, and the result is different from the glm quasipoisson data.
Here is my dataset. I use a csv file for all my dataset. Idk how to upload this csv data in this post, pardon me for this.
Richness = as.matrix(dat1[,14])
Richness
8
3
3
4
3
5
4
3
7
8
Parameter = as.matrix(dat1[,15:22])
Parameter
JE Temp Hmdt Sond HE WE L MH
1 31.3 93 63.3 3.89 4.32 80 7.82
2 26.9 92 63.5 9.48 8.85 60 8.32
1 27.3 93 67.4 1.23 2.37 60 10.10
3 31.6 99 108.0 1.90 3.32 80 4.60
1 29.3 99 86.8 2.42 7.83 460 12.20
2 29.4 85 86.1 4.71 15.04 200 10.10
1 29.4 87 93.5 3.65 14.70 200 12.20
1 29.5 97 87.5 1.42 3.17 80 4.07
1 25.9 95 62.3 5.23 16.89 140 10.03
1 29.5 95 63.5 1.85 6.50 120 6.97
Rich = glm(Richness ~ Parameter, family=quasipoisson, data = dat1)
summary(Rich)
Call:
glm(formula = Richness ~ Parameter, family = quasipoisson, data = dat1)
Deviance Residuals:
1 2 3 4 5
-0.017139 0.016769 -0.008652 0.002194 -0.003153
6 7 8 9 10
-0.016828 0.022914 -0.013823 -0.012597 0.030219
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -7.4197959 0.5061733 -14.659 0.0434 *
ParameterJE 0.1833651 0.0224198 8.179 0.0775 .
ParameterTemp 0.2441301 0.0073380 33.269 0.0191 *
ParameterHmdt 0.0393258 0.0032176 12.222 0.0520 .
ParameterSond -0.0319313 0.0009662 -33.050 0.0193 *
ParameterHE -0.0982213 0.0060587 -16.212 0.0392 *
ParameterWE 0.1001758 0.0027575 36.329 0.0175 *
ParameterL -0.0014170 0.0001554 -9.117 0.0695 .
ParameterMH 0.0137196 0.0073704 1.861 0.3138
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for quasipoisson family taken to be 0.002739787)
Null deviance: 7.8395271 on 9 degrees of freedom
Residual deviance: 0.0027358 on 1 degrees of freedom
AIC: NA
Number of Fisher Scoring iterations: 3
This is the model that i tried make with ggplot
ggplot(dat1, aes(Temp, Richness))+
geom_point() +
geom_smooth(method = "glm", method.args = list(family = quasipoisson),
fill = "grey", color = "black", linetype = 2)``
and this is the result.
I make for each parameters, but i just know this result turn wrong because it used a quasipoisson data for each parameter, what i want is the predict model based on quasipoisson data like in the summary above.
I tried to used the code from plot the results glm with multiple explanatories with 95% CIs, but i really confuse to set my data like the example there. But the result in that example is nearly like what i want.
Can anyone help me with this? How can I put the glm predict model for all parameters in one frame with ggplot?
Hope anyone can help me to fix this. Thank you so much!
Have you tried the plot_model function from sjplot package?
I'm writing from my phone, but the code is something Like this.
library(sjPlot)
plot_model(glm_model)
More info:
http://www.strengejacke.de/sjPlot/reference/plot_model.html
code:
data("mtcars")
glm_model<-glm(am~.,data = mtcars)
glm_model
library(sjPlot)
plot_model(glm_model, vline.color = "red")
plot_model(glm_model, show.values = TRUE, value.offset = .3)
I would like to assign a variable with a custom factor from an ANOVA model to the emmeans() statement. Here I use the oranges dataset from R to make the code reproducible. This is my model and how I would usually calculate the emmmeans of the factor store:
library(emmeans)
oranges$store<-as.factor(oranges$store)
model <- lm (sales1 ~ 1 + price1 + store ,data=oranges)
means<-emmeans(model, pairwise ~ store, adjust="tukey")
Now I would like to assign a variable (lsmeanfact) defining the factor for which the lsmeans are calculated.
lsmeanfact<-"store"
However, when I want to evaluate this variable in the emmeans() function it returns an error, it basically does not find the variable lsmeanfact, so it does not evaluate this variable.
means<-emmeans(model, pairwise ~ eval(parse(lsmeanfact)), adjust="tukey")
Error in emmeans(model, pairwise ~ eval(parse(lsmeanfact)), adjust = "tukey") :
No variable named lsmeanfact in the reference grid
How should I change my code to be able to evaluate the variable lsmeanfact so that the lsmeans for "plantcode" are correctly calculated?
You can make use of reformulate function.
library(emmeans)
lsmeanfact<-"store"
means <- emmeans(model, reformulate(lsmeanfact, 'pairwise'), adjust="tukey")
Or construct a formula with formula/as.formula.
means <- emmeans(model, formula(paste('pairwise', lsmeanfact, sep = '~')), adjust="tukey")
Here both reformulate(lsmeanfact, 'pairwise') and formula(paste('pairwise', lsmeanfact, sep = '~')) return pairwise ~ store.
You do not need to do anything special at all. The specs argument to emmeans() can be a character value. You can get the pairwise comparisons in a separate call, which is actually a better way to go anyway.
library(emmeans)
model <- lm(sales1 ~ price1 + store, data = oranges)
lsmeanfact <- "store"
( EMM <- emmeans(model, lsmeanfact) )
## store emmean SE df lower.CL upper.CL
## 1 8.01 2.61 29 2.67 13.3
## 2 9.60 2.30 29 4.89 14.3
## 3 7.84 2.30 29 3.13 12.6
## 4 10.44 2.35 29 5.63 15.2
## 5 10.19 2.28 29 5.53 14.9
## 6 15.22 2.28 29 10.56 19.9
##
## Confidence level used: 0.95
pairs(EMM)
## contrast estimate SE df t.ratio p.value
## 1 - 2 -1.595 3.60 29 -0.443 0.9976
## 1 - 3 0.165 3.60 29 0.046 1.0000
## 1 - 4 -2.428 3.72 29 -0.653 0.9856
## 1 - 5 -2.185 3.50 29 -0.625 0.9882
## 1 - 6 -7.209 3.45 29 -2.089 0.3206
## 2 - 3 1.761 3.22 29 0.546 0.9936
## 2 - 4 -0.833 3.23 29 -0.258 0.9998
## 2 - 5 -0.590 3.23 29 -0.182 1.0000
## 2 - 6 -5.614 3.24 29 -1.730 0.5239
## 3 - 4 -2.593 3.23 29 -0.802 0.9648
## 3 - 5 -2.350 3.23 29 -0.727 0.9769
## 3 - 6 -7.375 3.24 29 -2.273 0.2373
## 4 - 5 0.243 3.26 29 0.075 1.0000
## 4 - 6 -4.781 3.28 29 -1.457 0.6930
## 5 - 6 -5.024 3.23 29 -1.558 0.6314
##
## P value adjustment: tukey method for comparing a family of 6 estimates
Created on 2021-06-29 by the reprex package (v2.0.0)
Moreover, in any case what is needed in specs are the name(s) of the factors involved, not the factors themselves. Note also that it was unnecessary to convert store to a factor before fitting the model
Hoping that you can clear some confusion in my head.
Linear mixed model is constructed with lmerTest:
MODEL <- lmer(Ca content ~ SYSTEM +(1 | YEAR/replicate) +
(1 | YEAR:SYSTEM), data = IOSDV1)
Fun starts happening when I'm trying to get the confidence intervals for the specific levels of the main effect.
Commands emmeans and lsmeans produce the same intervals (example; SYSTEM A3: 23.9-128.9, mean 76.4, SE:8.96).
However, the command as.data.frame(effect("SYSTEM", MODEL)) produces different, narrower confidence intervals (example; SYSTEM A3: 58.0-94.9, mean 76.4, SE:8.96).
What am I missing and what number should I report?
To summarize, for the content of Ca, i have 6 total measurements per treatment (three per year, each from different replication). I will leave the names in the code in my language, as used. Idea is to test if certain production practices affect the content of specific minerals in the grains. Random effects without residual variance were left in the model for this example.
Linear mixed model fit by REML. t-tests use Satterthwaite's method ['lmerModLmerTest']
Formula: CA ~ SISTEM + (1 | LETO/ponovitev) + (1 | LETO:SISTEM)
Data: IOSDV1
REML criterion at convergence: 202.1
Scaled residuals:
Min 1Q Median 3Q Max
-1.60767 -0.74339 0.04665 0.73152 1.50519
Random effects:
Groups Name Variance Std.Dev.
LETO:SISTEM (Intercept) 0.0 0.0
ponovitev:LETO (Intercept) 0.0 0.0
LETO (Intercept) 120.9 11.0
Residual 118.7 10.9
Number of obs: 30, groups: LETO:SISTEM, 10; ponovitev:LETO, 8; LETO, 2
Fixed effects:
Estimate Std. Error df t value Pr(>|t|)
(Intercept) 76.417 8.959 1.548 8.530 0.0276 *
SISTEM[T.C0] -5.183 6.291 24.000 -0.824 0.4181
SISTEM[T.C110] -13.433 6.291 24.000 -2.135 0.0431 *
SISTEM[T.C165] -7.617 6.291 24.000 -1.211 0.2378
SISTEM[T.C55] -10.883 6.291 24.000 -1.730 0.0965 .
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Correlation of Fixed Effects:
(Intr) SISTEM[T.C0 SISTEM[T.C11 SISTEM[T.C16
SISTEM[T.C0 -0.351
SISTEM[T.C11 -0.351 0.500
SISTEM[T.C16 -0.351 0.500 0.500
SISTEM[T.C5 -0.351 0.500 0.500 0.500
optimizer (nloptwrap) convergence code: 0 (OK)
boundary (singular) fit: see ?isSingular
> ls_means(MODEL, ddf="Kenward-Roger")
Least Squares Means table:
Estimate Std. Error df t value lower upper Pr(>|t|)
SISTEMA3 76.4167 8.9586 1.5 8.5299 23.9091 128.9243 0.02853 *
SISTEMC0 71.2333 8.9586 1.5 7.9514 18.7257 123.7409 0.03171 *
SISTEMC110 62.9833 8.9586 1.5 7.0305 10.4757 115.4909 0.03813 *
SISTEMC165 68.8000 8.9586 1.5 7.6797 16.2924 121.3076 0.03341 *
SISTEMC55 65.5333 8.9586 1.5 7.3151 13.0257 118.0409 0.03594 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Confidence level: 95%
Degrees of freedom method: Kenward-Roger
> emmeans(MODEL, spec = c("SISTEM"))
SISTEM emmean SE df lower.CL upper.CL
A3 76.4 8.96 1.53 23.9 129
C0 71.2 8.96 1.53 18.7 124
C110 63.0 8.96 1.53 10.5 115
C165 68.8 8.96 1.53 16.3 121
C55 65.5 8.96 1.53 13.0 118
Degrees-of-freedom method: kenward-roger
Confidence level used: 0.95
> as.data.frame(effect("SISTEM", MODEL))
SISTEM fit se lower upper
1 A3 76.41667 8.958643 57.96600 94.86734
2 C0 71.23333 8.958643 52.78266 89.68400
3 C110 62.98333 8.958643 44.53266 81.43400
4 C165 68.80000 8.958643 50.34933 87.25067
5 C55 65.53333 8.958643 47.08266 83.98400
Many thanks.
I'm pretty sure this has to do with the dreaded "denominator degrees of freedom" question, i.e. what kind (if any) of finite-sample correction is being employed. tl;dr emmeans is using a Kenward-Roger correction, which is more or less the most accurate available option — the only reason not to use K-R is if you have a large data set for which it becomes unbearably slow.
load packages, simulate data, fit model
library(lmerTest)
library(emmeans)
library(effects)
dd <- expand.grid(f=factor(letters[1:3]),g=factor(1:20),rep=1:10)
set.seed(101)
dd$y <- simulate(~f+(1|g), newdata=dd, newparams=list(beta=rep(1,3),theta=1,sigma=1))[[1]]
m <- lmer(y~f+(1|g), data=dd)
compare default emmeans with effects
emmeans(m, ~f)
## f emmean SE df lower.CL upper.CL
## a 0.848 0.212 21.9 0.409 1.29
## b 1.853 0.212 21.9 1.414 2.29
## c 1.863 0.212 21.9 1.424 2.30
## Degrees-of-freedom method: kenward-roger
## Confidence level used: 0.95
as.data.frame(effect("f",m))
## f fit se lower upper
## 1 a 0.8480161 0.2117093 0.4322306 1.263802
## 2 b 1.8531805 0.2117093 1.4373950 2.268966
## 3 c 1.8632228 0.2117093 1.4474373 2.279008
effects doesn't explicitly tell us what/whether it's using a finite-sample correction: we could dig around in the documentation or the code to try to find out. Alternatively, we can tell emmeans not to use finite-sample correction:
emmeans(m, ~f, lmer.df="asymptotic")
## f emmean SE df asymp.LCL asymp.UCL
## a 0.848 0.212 Inf 0.433 1.26
## b 1.853 0.212 Inf 1.438 2.27
## c 1.863 0.212 Inf 1.448 2.28
## Degrees-of-freedom method: asymptotic
## Confidence level used: 0.95
Testing shows that these are equivalent to about a tolerance of 0.001 (probably close enough). In principle we should be able to specify KR=TRUE to get effects to use Kenward-Roger correction, but I haven't been able to get that to work yet.
However, I will also say that there's something a little bit funky about your example. If we compute the distance between the mean and the lower CI in units of standard error, for emmeans we get (76.4-23.9)/8.96 = 5.86, which implies a very small effect degrees of freedom (e.g. about 1.55). That seems questionable to me unless your data set is extremely small ...
From your updated post, it appears that Kenward-Roger is indeed estimating only 1.5 denominator df.
In general it is dicey/not recommended to try fitting random effects where the grouping variable has a small number of levels (although see here for a counterargument). I would try treating LETO (which has only two levels) as a fixed effect, i.e.
CA ~ SISTEM + LETO + (1 | LETO:ponovitev) + (1 | LETO:SISTEM)
and see if that helps. (I would expect you would then get on the order of 7 df, which would make your CIs ± 2.4 SE instead of ± 6 SE ...)
I have a mixed effects model (with lme4) with a 2-way interaction term, each term having multiple levels (each 4) and I would like to investigate their effects in reference to their grand mean. I present this example here from the car data set and omit the error term since it is not neccessary for this example:
## shorten data frame for simplicity
df=Cars93[c(1:15),]
df=Cars93[is.element(Cars93$Make,c('Acura Integra', 'Audi 90','BMW 535i','Subaru Legacy')),]
df$Make=drop.levels(df$Make)
df$Model=drop.levels(df$Model)
## define contrasts (every factor has 4 levels)
contrasts(df$Make) = contr.treatment(4)
contrasts(df$Model) = contr.treatment(4)
## model
m1 <- lm(Price ~ Model*Make,data=df)
summary(m1)
as you can see, the first levels are omitted in the interaction term. And I would like to have all 4 levels in the output, referenced to the grand mean (often referred to deviant coding). These are the sources I looked at: https://marissabarlaz.github.io/portfolio/contrastcoding/#coding-schemes and How to change contrasts to compare with mean of all levels rather than reference level (R, lmer)?. The last reference does not report interactions though.
The simple answer is that what you want is not possible directly. You have to use a slightly different approach.
In a model with interactions, you want to use contrasts in which the mean is zero and not a specific level. Otherwise, the lower-order effects (i.e., main effects) are not main effects but simple effects (evaluated when the other factor level is at its reference level). This is explained in more details in my chapter on mixed models:
http://singmann.org/download/publications/singmann_kellen-introduction-mixed-models.pdf
To get what you want, you have to fit the model in a reasonable manner and then pass it to emmeans to compare against the intercept (i.e., the unweighted grand mean). This works also for interactions as shown below (as your code did not work, I use warpbreaks).
afex::set_sum_contrasts() ## uses contr.sum globally
library("emmeans")
## model
m1 <- lm(breaks ~ wool * tension,data=warpbreaks)
car::Anova(m1, type = 3)
coef(m1)[1]
# (Intercept)
# 28.14815
## both CIs include grand mean:
emmeans(m1, "wool")
# wool emmean SE df lower.CL upper.CL
# A 31.0 2.11 48 26.8 35.3
# B 25.3 2.11 48 21.0 29.5
#
# Results are averaged over the levels of: tension
# Confidence level used: 0.95
## same using test
emmeans(m1, "wool", null = coef(m1)[1], infer = TRUE)
# wool emmean SE df lower.CL upper.CL null t.ratio p.value
# A 31.0 2.11 48 26.8 35.3 28.1 1.372 0.1764
# B 25.3 2.11 48 21.0 29.5 28.1 -1.372 0.1764
#
# Results are averaged over the levels of: tension
# Confidence level used: 0.95
emmeans(m1, "tension", null = coef(m1)[1], infer = TRUE)
# tension emmean SE df lower.CL upper.CL null t.ratio p.value
# L 36.4 2.58 48 31.2 41.6 28.1 3.196 0.0025
# M 26.4 2.58 48 21.2 31.6 28.1 -0.682 0.4984
# H 21.7 2.58 48 16.5 26.9 28.1 -2.514 0.0154
#
# Results are averaged over the levels of: wool
# Confidence level used: 0.95
emmeans(m1, c("tension", "wool"), null = coef(m1)[1], infer = TRUE)
# tension wool emmean SE df lower.CL upper.CL null t.ratio p.value
# L A 44.6 3.65 48 37.2 51.9 28.1 4.499 <.0001
# M A 24.0 3.65 48 16.7 31.3 28.1 -1.137 0.2610
# H A 24.6 3.65 48 17.2 31.9 28.1 -0.985 0.3295
# L B 28.2 3.65 48 20.9 35.6 28.1 0.020 0.9839
# M B 28.8 3.65 48 21.4 36.1 28.1 0.173 0.8636
# H B 18.8 3.65 48 11.4 26.1 28.1 -2.570 0.0133
#
# Confidence level used: 0.95
Note that for coef() you probably want to use fixef() for lme4 models.
I have a mixed effect model with a log(x+1) transformed response variable. The output from emmeans with the type as "response" provides the mean and confidence intervals for both groups that I am comparing. However what I want is the mean and CI of the difference between the groups (i.e. the estimate). emmeans only provides the ratio (with type="response") or the log ratio (with type="link") and I am unsure how to change this into absolute values. If you run the model without the log(x+1) transformation then emmeans provides the estimated difference and CI around this difference, not the ratios. How can I also do this when my response variable is log(x+1) transformed?
bmnameF.lme2 = lme(log(bm+1)~TorC*name, random=~TorC|site,
data=matched.cases3F, method='REML')
emmeans(lme, pairwise~TorC,
type='response')%>%confint(OmeanFHR[[2]])%>%as.data.frame
emmeans.TorC emmeans.emmean emmeans.SE emmeans.df emmeans.lower.CL emmeans.upper.CL contrasts.contrast contrasts.estimate contrasts.SE contrasts.df contrasts.lower.CL contrasts.upper.CL
Managed 376.5484 98.66305 25 219.5120 645.9267 Managed - Open 3.390123 1.068689 217 1.821298 6.310297
Open 111.0722 43.15374 25 49.8994 247.2381 Managed - Open 3.390123 1.068689 217 1.821298 6.310297
Let me show a different example so the results are reproducible to all viewers:
mod = lm(log(breaks+1) ~ wool*tension, data = warpbreaks)
As you see, with a log transformation, comparisons/contrasts are expressed as ratios by default. But this can be changed by specifying transform instead of type in the emmeans() call:
> emmeans(mod, pairwise ~ tension|wool, transform = "response")
$emmeans
wool = A:
tension response SE df lower.CL upper.CL
L 42.3 5.06 48 32.1 52.4
M 23.6 2.83 48 17.9 29.3
H 23.7 2.83 48 18.0 29.4
wool = B:
tension response SE df lower.CL upper.CL
L 27.7 3.32 48 21.0 34.4
M 28.4 3.40 48 21.6 35.3
H 19.3 2.31 48 14.6 23.9
Confidence level used: 0.95
$contrasts
wool = A:
contrast estimate SE df t.ratio p.value
L - M 18.6253 5.80 48 3.213 0.0065
L - H 18.5775 5.80 48 3.204 0.0067
M - H -0.0479 4.01 48 -0.012 0.9999
wool = B:
contrast estimate SE df t.ratio p.value
L - M -0.7180 4.75 48 -0.151 0.9875
L - H 8.4247 4.04 48 2.086 0.1035
M - H 9.1426 4.11 48 2.224 0.0772
P value adjustment: tukey method for comparing a family of 3 estimates
Or, you can do this later via the regrid() function:
emm1 = emmeans(mod, ~ tension | wool)
emm2 = regrid(emm1)
emm2 # estimates
pairs(emm2) # comparisons
regrid() creates a new emmGrid object where everything is already back-transformed, thus side-stepping the behavior that happens with contrasts of log-transformed results. (In the previous illustration, the transform argument just calls regrid after it constructs the reference grid.)
But there is another subtle thing going on: The transformation is auto-detected as log; the +1 part is ignored. Thus, the back-transformed estimates are all too large by 1. To get this right, you need to use the make.tran() function to create this generalization of the log transformation:
> emm3 = update(emmeans(mod, ~ tension | wool), tran = make.tran("genlog", 1))
> str(emm3)
'emmGrid' object with variables:
tension = L, M, H
wool = A, B
Transformation: “log(mu + 1)”
> regrid(emm3)
wool = A:
tension response SE df lower.CL upper.CL
L 41.3 5.06 48 31.1 51.4
M 22.6 2.83 48 16.9 28.3
H 22.7 2.83 48 17.0 28.4
wool = B:
tension response SE df lower.CL upper.CL
L 26.7 3.32 48 20.0 33.4
M 27.4 3.40 48 20.6 34.3
H 18.3 2.31 48 13.6 22.9
Confidence level used: 0.95
The comparisons will come out the same as shown earlier, because offsetting all the means by 1 doesn't affect the pairwise differences.
See vignette("transformations", "emmeans") or https://cran.r-project.org/web/packages/emmeans/vignettes/transformations.html for more details.