6
$\begingroup$

I'm working on a dataset of 2x2 crossover study. I have 10 subjects, each of them underwent both A and B treatment but in a different sequence. (This is a balanced study.)

enter image description here

I want to see how A and B treatments improve lipid levels. My thought process was to create a linear mixed model with subjects as a random effect; treatment, sequence, and period as the fixed effects; finally, sex and age as covariates.

My data:

#Reproducible data
id <- rep(1:10,3)
age <- rep(c("59","59","70","67","66","70","70","68","71","57"),3)
sex <- rep(c("F","M","F","M","F","F","F","M","F","M"),3)
sequence <- rep(c("1","2","1","2","1","2","1","2","2","2"),3)
period <- c(rep(0,10),rep(1,10),rep(2,10))
Treatment <- c(rep("C",10), rep(c("A","B"),4), "B","B",rep(c("B","A"),4), "A","A") #C is baseline
lipid <- c(18,6,30,12,14,19,10,22,22,27,13,28,14,23,12,27,9,10,13,22,13,22,29,12,16,24,15,13,17,11)
DF <- data.frame(id,age,sex,sequence,period,Treatment,lipid)

 > head(DF)
      id age sex sequence period Treatment lipid
    1  1  59   F        1      0         C    18
    2  2  59   M        2      0         C     6
    3  3  70   F        1      0         C    30
    4  4  67   M        2      0         C    12
    5  5  66   F        1      0         C    14
    6  6  70   F        2      0         C    19

My linear mixed model:

library(lmerTest)
lm1 <- lmer(lipid~Treatment + sequence + period + sex + age + (1|id), data = DF, REML = F)

> summary(lm1)

Random effects:
 Groups   Name        Variance Std.Dev.
 id       (Intercept)  1.344   1.159   
 Residual             34.986   5.915   
Number of obs: 30, groups:  id, 10

Fixed effects:
            Estimate Std. Error      df t value Pr(>|t|)
(Intercept)  23.7890    18.2664 10.1410   1.302    0.222
TreatmentA   -2.8500     2.8572 20.0000  -0.997    0.330
TreatmentB    2.2750     3.1018 20.0000   0.733    0.472
sequence2     4.7080     3.3324 10.0000   1.413    0.188
period1      -1.1250     2.6998 20.0000  -0.417    0.681
sexM         -3.8351     3.7742 10.0000  -1.016    0.334
age          -0.1078     0.2734 10.0000  -0.394    0.702

After building a linear mixed model, I wanted to do post-hoc test to compare treatment A and B. I tried both emmeans and multcomp but they are giving me different results.

Emmeans:

library(emmeans)
emm <- emmeans(lm1,"Treatment")
pairs(emm, adjust = "fdr")

> pairs(emm, adjust = "fdr")
 contrast estimate   SE   df t.ratio p.value
 C - A      nonEst   NA   NA     NA      NA 
 C - B      nonEst   NA   NA     NA      NA 
 A - B       -5.12 2.93 23.5 -1.750  0.2794 

Multicomp:

library(multcomp)
summary(glht(lm1, linfct = mcp(Treatment = "Tukey")), test = adjusted("fdr"))

> summary(glht(lm1, linfct = mcp(Treatment = "Tukey")), test = adjusted("fdr"))
Linear Hypotheses:
           Estimate Std. Error z value Pr(>|z|)
A - C == 0   -2.850      2.857  -0.997    0.463
B - C == 0    2.275      3.102   0.733    0.463
B - A == 0    5.125      2.700   1.898    0.173
(Adjusted p values reported -- fdr method)

I guess question would be,

1) Based on the study design, does my linear mixed model lm1 <- lmer(lipid~Treatment + sequence + period + sex + age + (1|id), data = DF, REML = F) look ok? Or should I account for other interaction terms (ex. Treatment*sequence)?

2) Why does emmeans give me NAs in C-A and C-B when multcomp gives me values? Which one would you recommend to conduct the post-hoc test on lmer model since the results are different?

Any thought is appreciated, thank you!

$\endgroup$
4
  • 1
    $\begingroup$ Your Treatment C and period=0 data points all seem to coincide, so when you include period as a predictor the model can’t properly distinguish their effects. Try repeating the modeling without the period predictor (not clear to me what that accomplishes) and one part of your question might solve itself. $\endgroup$
    – EdM
    Jun 7, 2020 at 22:45
  • 1
    $\begingroup$ @EdM is right - the figure you provided indicates that two treatments were used in your study: A and B. Where does a third treatment, C, come from? Also, the figure indicates that there are only 2 treament periods: Period 1 and Period 2. Where does Period 0 come from? If Period 0 refers to baseline, then you can include the baseline value of your response variable (if you have it) as a predictor variable in your model. Your R model formulation needs to reflect your study design - currently, it doesn't seem to do that. $\endgroup$ Jun 8, 2020 at 2:10
  • $\begingroup$ @IsabellaGhement thank you for replying! So Treatment C means baseline and that's why I put it as period 0. Should I just take out Treatment C and create a column (variable) for baseline? If so, how should I do the comparison of A vs. baseline and B vs. baseline since there will only be A and B in the Treatment group and it wouldn't make sense for me to do a post-hoc test then? $\endgroup$ Jun 8, 2020 at 2:59
  • $\begingroup$ @EdM thank you for replying! I removed period from my model and it worked! Treatment C actually means baseline and that's why I put it down as period 0, I'm sorry I didn't make that part clear. If I wanted to include the possible effect of period, what would be the best way to do it? $\endgroup$ Jun 8, 2020 at 2:59

3 Answers 3

3
$\begingroup$

In modeling you have to be careful not to include the exact same situation in different ways. For example, you already found that the design with all the period = 0 cases having Treatment C made it impossible to get useful results. In the summary(lm1) output, that led to reporting only 1 coefficient for period when the 3 levels meant there should have been 2 coefficients (that's how I figured out the problem) and a correct refusal by emmeans to provide contrasts involving Treatment C when that couldn't be disentangled from period = 0.

Similarly, you can't include both sequence and period in the model, as they represent the same thing. The sequence = 1 value means getting Treatment A then Treatment B; sequence = 2 means the reverse. So at period = 1 all measurements for those having sequence = 1 will be taken following Treatment A, measurements for those having sequence = 2 will be taken following Treatment B, with the opposite at period = 2.

If you only have 10 individuals with 3 measurements each (baseline, after first treatment, after second treatment) you face a risk of overfitting if you try to evaluate more than 1 or 2 predictors. The usual rule of thumb for biomedical situations with continuous outcomes is about 15 observations per predictor evaluated.

But if this is a randomized trial then you can take advantage of randomization itself, which ideally should wash out any contributions to outcome except for those included in the design: in your case, the two Treatments and the order of their application. It would be good to demonstrate that the randomization did a reasonable job of balancing the things you know about (baseline lipid values, sex, age), but you might not have to include sex or age in the model as covariates. Correction for covariates can be helpful in larger studies but you don't have enough cases to do that adequately with only 10 participants.

I like the recommendation in a comment by @Isabella Ghement to incorporate the baseline lipid values as covariates rather than modeling them as outcomes. There seems to have been no actual Treatment C, or if there was there doesn't seem to be a record of lipid values before that Treatment, so there's really nothing to model. Following that recommendation might remove the need to include a random-effects intercept term, which would impose a Gaussian structure on the intercepts that might not be appropriate.

Then your model comes down to an evaluation of Treatment A versus Treatment B and the 2 orders in which the treatments were given. That would seem to require an interaction term (lipid ~ baselineLipid + Treatment + sequence + Treatment:sequence), where Treatment is 0 for TreatmentA and 1 for TreatmentB. Your study still might not be large enough to handle even this many predictors, but I think that this model has a better chance of representing your results fairly.

$\endgroup$
1
  • $\begingroup$ thank you so much for your answer! This is really helpful and makes me think about a proper way to approach model building. Really appreciated! $\endgroup$ Jun 8, 2020 at 22:24
2
$\begingroup$

I will attempt to answer (2).

First, the code given does not produce the results shown. In order to obtain those results, I needed the following manipulations of the data before fitting the model:

DF$Treatment = relevel(factor(DF$Treatment), ref = "C")
DF$age = as.numeric(DF$age)
DF$period = factor(DF$period)

In future postings, please actually test the code that is presented to ensure you are accurately representing what happened.

Now, on to the question. The emmeans package, unlike many (most) others such as multcomp, tests for estimability. Estimability has to do with ambiguities arising from rank-deficient models. The fact that the model is rank deficient is an important omission from what is shown in the question. Note the warning message:

> lm1 <- lmer(lipid~Treatment + sequence + period + sex + age + (1|id), data = DF, REML = F)
fixed-effect model matrix is rank deficient so dropping 1 column / coefficient

When there is rank deficiency, that means that certain model predictions are not unique; they differ depending on how the model is parameterized. In this case, the model lm1 was fitted with the default contrast coding, contr.treatment, in which 0--1 indicators are used for each factor, with the one for the reference level omitted. With that contrast coding, we obtain the results shown in the OP: (some output is omitted)

> library(emmeans)
> pairs(emmeans(lm1, "Treatment"))

 contrast estimate   SE   df t.ratio p.value
 C - A      nonEst   NA   NA     NA      NA 
 C - B      nonEst   NA   NA     NA      NA 
 A - B       -5.12 2.93 23.5 -1.750  0.2078 

> library(multcomp)
> summary(glht(lm1, linfct = mcp(Treatment = "Tukey")), test = adjusted("fdr"))

           Estimate Std. Error z value Pr(>|z|)
A - C == 0   -2.850      2.857  -0.997    0.463
B - C == 0    2.275      3.102   0.733    0.463
B - A == 0    5.125      2.700   1.898    0.173
(Adjusted p values reported -- fdr method)

Now, let's fit the same model using contr.sum codings:

> lm2 = update(lm1, contrasts = list(Treatment = "contr.sum", 
+   sequence = "contr.sum", period = "contr.sum", sex = "contr.sum"))
fixed-effect model matrix is rank deficient so dropping 1 column / coefficient

> fixef(lm2)
(Intercept)  Treatment1  Treatment2   sequence1     period1        sex1         age 
 23.6587644   1.6916667  -3.4083333  -2.3539871  -1.1250000   1.9175647  -0.1077586 

Note that most of the regression coeffiicients are different from those in lm1, because of the different contrast coding. Now do the same analyses with this model:

> pairs(emmeans(lm2, "Treatment"))

 contrast estimate   SE   df t.ratio p.value
 C - A      nonEst   NA   NA     NA      NA 
 C - B      nonEst   NA   NA     NA      NA 
 A - B       -5.12 2.93 23.5 -1.750  0.2078 

> summary(glht(lm2, linfct = mcp(Treatment = "Tukey")), test = adjusted("fdr"))

            Estimate Std. Error z value Pr(>|z|)
A - C == 0   -5.100      5.065  -1.007    0.471
B - C == 0    0.025      4.613   0.005    0.996
B - A == 0    5.125      2.700   1.898    0.173
(Adjusted p values reported -- fdr method)

The emmeans results are identical for the two models. However, the multcomp results are different, albeit the same for the B - A contrast. That contrast is the one that is uniquely estimable. I hope this explains why emmeans does not show two of the comparisons, and why multcomp really should test estimability also.

I cannot answer why multcomp produces a different standard error for the one estimable comparison.

$\endgroup$
1
  • $\begingroup$ Thank you for the reminder, I will double-check my codes in the future. And thank you for clarifying the error message and estimability of emmeans. I think this is super interesting how the result changes based on different codings. Definitely something to keep in mind $\endgroup$ Jun 10, 2020 at 17:23
1
$\begingroup$

Following the current advice of removing sequence, I suggest also including period as nested within ID and removing it from fixed effects i.e. lmer(lipid~Treatment + sex + age + (1|id/period), data = DF, REML = F)

I have found this guide to be quite helpful: https://bbolker.github.io/mixedmodels-misc/glmmFAQ.html#model-specification

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.