Marginal Means accounting for the random effect uncertainty - mixed-models

Question

When we have repeated measurements on an experimental unit, typically these units cannot be considered 'independent' and need to be modeled in a way that we get valid estimates for our standard errors.
When I compare the intervals obtained by computing the marginal means for the treatment using a mixed model (treating the unit as a random effect) and in the other case, first averaging over the unit and THEN runnning a simple linear model on the averaged responses, I get the exact same uncertainty intervals.
How do we incorporate the uncertainty of the measurements of the unit, into the uncertainty of what we think our treatments look like?
In order to really propogate all the uncertainty, shouldn't we see what the treatment looks like, averaged over "all possible measurements" on a unit?
``` r
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
library(emmeans)
library(lme4)
#> Loading required package: Matrix
library(ggplot2)
tmp <- structure(list(treatment = c("A", "A", "A", "A", "A", "A", "A",
"A", "A", "A", "A", "A", "B", "B", "B", "B", "B", "B", "B", "B",
"B", "B", "B", "B"), response = c(151.27333548, 162.3933313,
159.2199999, 159.16666725, 210.82, 204.18666667, 196.97333333,
194.54666667, 154.18666667, 194.99333333, 193.48, 191.71333333,
124.1, 109.32666667, 105.32, 102.22, 110.83333333, 114.66666667,
110.54, 107.82, 105.62000069, 79.79999821, 77.58666557, 75.78666928
), experimental_unit = c("A-1", "A-1", "A-1", "A-1", "A-2", "A-2",
"A-2", "A-2", "A-3", "A-3", "A-3", "A-3", "B-1", "B-1", "B-1",
"B-1", "B-2", "B-2", "B-2", "B-2", "B-3", "B-3", "B-3", "B-3"
)), row.names = c(NA, -24L), class = c("tbl_df", "tbl", "data.frame"
))
### Option 1 - Treat the experimental unit as a random effect since there are
### 4 repeat observations for the same unit
lme4::lmer(response ~ treatment + (1 | experimental_unit), data = tmp) %>%
emmeans::emmeans(., ~ treatment) %>%
as.data.frame()
#> treatment emmean SE df lower.CL upper.CL
#> 1 A 181.0794 10.83359 4 151.00058 211.1583
#> 2 B 101.9683 10.83359 4 71.88947 132.0472
#ggplot(.,aes(treatment, emmean)) +
#geom_pointrange(aes(ymin = lower.CL, ymax = upper.CL))
### Option 2 - instead of treating the unit as random effect, we average over the
### 4 repeat observations, and run a simple linear model
tmp %>%
group_by(experimental_unit) %>%
summarise(mean_response = mean(response)) %>%
mutate(treatment = c(rep("A", 3), rep("B", 3))) %>%
lm(mean_response ~ treatment, data = .) %>%
emmeans::emmeans(., ~ treatment) %>%
as.data.frame()
#> treatment emmean SE df lower.CL upper.CL
#> 1 A 181.0794 10.83359 4 151.00058 211.1583
#> 2 B 101.9683 10.83359 4 71.88947 132.0472
#ggplot(., aes(treatment, emmean)) +
#geom_pointrange(aes(ymin = lower.CL, ymax = upper.CL))
### Whether we include a random effect for the unit, or average over it and THEN model it, we find no difference in the
### marginal means for the treatments
### How do we incoporate the variation of the repeat measurments to the marginal means of the treatments?
### Do we then ignore the variation in the 'subsamples' and simply average over them PRIOR to modeling?
<sup>Created on 2021-07-31 by the [reprex package](https://reprex.tidyverse.org) (v2.0.0)</sup>

Answer 1

emmeans() does take into account the errors of random effects. This is what I get when I remove the complex sequences of pipes:
> mmod = lme4::lmer(response ~ treatment + (1 | experimental_unit), data = tmp)
> emmeans(mmod, "treatment")
treatment emmean SE df lower.CL upper.CL
A 181 10.8 4 151.0 211
B 102 10.8 4 71.9 132
Degrees-of-freedom method: kenward-roger
Confidence level used: 0.95
This is as shown. If I fit a fixed-effects model that accounts for experimental units as a fixed effect, I get:
> fmod = lm(response ~ treatment + experimental_unit, data = tmp)
> emmeans(fmod, "treatment")
NOTE: A nesting structure was detected in the fitted model:
experimental_unit %in% treatment
treatment emmean SE df lower.CL upper.CL
A 181 3.25 18 174.2 188
B 102 3.25 18 95.1 109
Results are averaged over the levels of: experimental_unit
Confidence level used: 0.95
The SEs of the latter results are considerably lower, and that is because the random variations in experimental_unit are modeled as fixed variations.
Apparently the piping you did accounts for the variation of the random effects and includes those in the EMMs. I think that is because you did things separately for each experimental unit and somehow combined those results. I'm not very comfortable with a sequence of pipes that is 7 steps long, and I don't understand why that results in just one set of means.
I recommend against the as.data.frame() at the end. That zaps out annotations that can be helpful in understanding what you have. If you are doing that to get more digits precision, I'll claim that those are digits you don't need, it just exaggerates the precision you are entitled to claim.
Notes on some follow-up comments
Subsequently, I am convinced that what we see in the piped operations in the second part of the OP doe indeed comprise computing the mean of each EU, then analyzing those.
Let's look at that in the context of the formal model. We have (sorry MathJax doesn't work on stackoverflow, but I'll leave the markup there anyway)
$$ Y_{ijk} = \mu + \tau_i + U_{ij} + E_{ijk} $$
where $Y_{ijk}$ is the kth response measurement on the ith treatment and jth EU in the ith treatment, and the rhs terms represent respectively the overall mean, the (fixed) treatment effects, the (random) EU effects, and the (random) error effects. We assume the random effects are all mutually independent. With a balanced design, the EMMs are just the marginal means:
$$ \bar Y_{i..} = \mu + \tau_i + \bar U_{i.} + \bar E_{i..} $$
where a '.' subscript means we averaged over that subscript. If there are n EUs per treatment and m measurements on each EU, we get that
$$ Var(\bar Y_{i..} = \sigma^2_U / n + \sigma^2_E / mn $$
Now, if we aggregate the data on EUs ahead of time, we are starting with
$$ \bar Y_{ij.} = \mu + U_{ij} + \bar E_{ij.} $$
However, if we then compute marginal means by averaging over j, we get exactly the same thing as we did before with $\bar Y_{i..}$, and the variance is exactly as already shown. That is why it doesn't matter if we aggregated first or not.