How to extract the value (minimum difference) the function multcomp::glht is using to calculate the-CodePudding

I have this example where I'd like to have multiple comparisons across treatments. Here is the data:

data.1 <-read.csv(text = "
location,treat,response
loc1,T1,120
loc1,T2,60
loc1,T3,59
loc1,T4,10
loc2,T1,129
loc2,T2,55
loc2,T3,59
loc2,T4,8
loc3,T1,134
loc3,T2,60
loc3,T3,58
")

And this is what I did:

library(lme4)
library(lmerTest)
library(emmeans)
library(multcomp)

model.fit <- lmer(response ~ treat   (1|location), data = data.1)

model.fit.emmeans <- emmeans(model.fit, ~ treat, 
                             options = list(estName = "response"))

pairs.comp.glht<-glht(model.fit, linfct=mcp(treat="Tukey"))

pairs.comp.glht.cld <-cld(pairs.comp.glht)

Running this pairs.comp.glht.cld gave me the output I needed.

I am looking for the value of the minimum difference to call a difference and display a different letter. I am assuming the value should be in this object: pairs.comp.glht or here pairs.comp.glht.cld, but I cannot extract the value.

CodePudding user response：

You can get a more detailed summary of the results with

summary(pairs.comp.glht)
# 
#    Simultaneous Tests for General Linear Hypotheses
# 
# Multiple Comparisons of Means: Tukey Contrasts
# 
# 
# Fit: lmer(formula = response ~ treat   (1 | location), data = data.1)
# 
# Linear Hypotheses:
#               Estimate Std. Error z value Pr(>|z|)    
# T2 - T1 == 0  -69.3333     3.3806 -20.509   <1e-08 ***
# T3 - T1 == 0  -69.0000     3.3806 -20.410   <1e-08 ***
# T4 - T1 == 0 -118.6667     3.7796 -31.396   <1e-08 ***
# T3 - T2 == 0    0.3333     3.3806   0.099        1    
# T4 - T2 == 0  -49.3333     3.7796 -13.052   <1e-08 ***
# T4 - T3 == 0  -49.6667     3.7796 -13.141   <1e-08 ***
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
# (Adjusted p values reported -- single-step method)

Note that the standard errors vary so the minimum difference will also vary. For the first comparison, to get a two-tailed p-value of .05 you would need a difference of 1.96 * 3.3806 = /-6.625976.

CodePudding user response：

This question presumes there is a single value that can serve the stated purpose. If there is such a value, it can be found via

confint(pairs(model.fit.emmeans))

The half-width of each confidence interval is the critical value for that comparison. If all, those half-widths are equal, that's your answer. (And they'll be equal if the df are all equal and the SE are all equal.)

But with a mixed model, or unbalanced data, etc., it is often the case that the comparisons all have different critical values.