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I was working through the example here calculating least squares means by doBy. In order to make sure I understand what is going on, I would like to replicate the analysis manually. I can do it when the number of untreated and treated are the same, but failing to do it when they are not the same.

I am copying the example here:

require(doBy)
zz <- "treat year y
1 t1 1 0.5
2 t1 1 1.0
3 t1 1 1.5
4 t2 1 3.0
5 t1 2 3.0
6 t2 2 4.5
7 t2 2 5.0
8 t2 2 5.5"

simdat <- read.table(text = zz, header =TRUE)
msim <- lm(y ~ treat + year, data=simdat)
LSmeans( msim, effect="treat")

It gives the following:

 estimate        se df    t.stat      p.value treat year
1        2 0.2415229  5  8.280787 4.191542e-04    t1  1.5
2        4 0.2415229  5 16.561573 1.465478e-05    t2  1.5

The estimates can be manually calcualted as follows:

t1_y1 <- mean(simdat$y[simdat$treat == "t1" & simdat$year == 1])
t1_y2 <- mean(simdat$y[simdat$treat == "t1" & simdat$year == 2])
(t1_y1 + t1_y2)/2
[1] 2

t2_y1 <- mean(simdat$y[simdat$treat == "t2" & simdat$year == 1])
t2_y2 <- mean(simdat$y[simdat$treat == "t2" & simdat$year == 2])
(t2_y1 + t2_y2)/2
[1] 4

However, when I make it unbalanced, I don't know how to calculate the estimates. For example, as below:

zz <- "treat year y
1 t1 1 0.5
2 t1 1 1.0
3 t1 1 1.5
4 t2 1 3.0
5 t1 2 3.0
6 t2 2 4.5
7 t2 2 5.0
8 t1 2 5.5"

simdat <- read.table(text = zz, header =TRUE)
msim <- lm(y ~ treat + year, data=simdat)
LSmeans( msim, effect="treat")
  estimate        se df   t.stat     p.value treat year
1 2.571429 0.4400255  5 5.843817 0.002076749    t1  1.5
2 3.714286 0.5729889  5 6.482299 0.001302694    t2  1.5

#unweighted means
t1_y1 <- mean(simdat$y[simdat$treat == "t1" & simdat$year == 1])
t1_y2 <- mean(simdat$y[simdat$treat == "t1" & simdat$year == 2])
(t1_y1 + t1_y2)/2
[1] 2.625

t2_y1 <- mean(simdat$y[simdat$treat == "t2" & simdat$year == 1])
t2_y2 <- mean(simdat$y[simdat$treat == "t2" & simdat$year == 2])
(t2_y1 + t2_y2)/2
[1] 3.875

#maybe weighted means?
t1_y1 <- mean(simdat$y[simdat$treat == "t1" & simdat$year == 1])
t1_y2 <- mean(simdat$y[simdat$treat == "t1" & simdat$year == 2])
(t1_y1*(5/8) + t1_y2 * (3/8))
[1] 2.21875

t2_y1 <- mean(simdat$y[simdat$treat == "t2" & simdat$year == 1])
t2_y2 <- mean(simdat$y[simdat$treat == "t2" & simdat$year == 2])
(t2_y1*(5/8) + t2_y2 * (3/8))
[1] 3.65625

How can I manually calculate the least square means?

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  • $\begingroup$ You fitted a model with year as a covariate rather than a 2-level factor. Refit the model with factor(year) in place of year. $\endgroup$ – rvl Dec 30 '16 at 4:06
  • $\begingroup$ @rvl I made it a factor, and got the exact same estimates for the means (2.571, 3.714). How can I calculate these numbers manually? $\endgroup$ – user310374 Dec 30 '16 at 4:19
  • $\begingroup$ The means should be unweighted. Looking at the data you list, I get a manual lsmean of 2.0 for treatment 1. (t1y1=1.0, t1y2=3.0). So I'm not even convinced your model and calculations are based on the same data. $\endgroup$ – rvl Dec 30 '16 at 4:41
  • $\begingroup$ @rvl there are two "datasets" in my question. The first, where the number of samples treated with t1 and t2 are equal (which it seems you are using your calculation), I can also replicate manually. The second, where they are unequal, I cannot. My question is regarding the second. $\endgroup$ – user310374 Dec 30 '16 at 4:54
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    $\begingroup$ @mdewey I suppose I will do so. "Least square means" is a standard statistical calculation though, I'm just surprised it is not what I thought it was based on many explanations I read. And it's not specific to this package... I get the exact same result if I use another package called LSmeans. So clearly, I just don't understand what it is doing. $\endgroup$ – user310374 Dec 30 '16 at 12:57
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The issue, which I overlooked initially, is that you fitted an additive model (no interaction).

year as a factor

I will illustrate, but for reasons to be explained later, I'm using fyear, a factor version of year:

> simdat = transform(simdat, fyear = factor(year))

> msim <- lm(y ~ treat + fyear, data = simdat)
> doBy::LSmeans(msim, "treat")
  estimate        se df   t.stat     p.value treat
1 2.571429 0.4400255  5 5.843817 0.002076749    t1
2 3.714286 0.5729889  5 6.482299 0.001302694    t2

> msimi <- lm(y ~ treat * fyear, data = simdat)
> doBy::LSmeans(msimi, "treat")
  estimate        se df   t.stat     p.value treat
1    2.625 0.4419417  4 5.939697 0.004028851    t1
2    3.875 0.5929271  4 6.535374 0.002832366    t2

Your calculations are thus correct for the interaction model msimi. That's because the fitted values for that model are the cell means. Least-squares means are averages of predictions, equally weighted. Here they are obtained by hand for the additive model, msim:

First, obtain the predictions:

> grid = expand.grid(treat = c("t1","t2"), fyear = factor(c(1,2)))
> pred = predict(msim, newdata = grid)
> cbind(grid, pred=pred)
  treat fyear     pred
1    t1     1 1.214286
2    t2     1 2.357143
3    t1     2 3.928571
4    t2     2 5.071429

Now, average those predictions:

> (pred[1] + pred[3])/2
       1 
2.571429 

> (pred[2] + pred[4])/2
       2 
3.714286

These are the same as the results that LSmeans obtained.

year as numeric

There is a subtle difference when we use the model with year as a numeric predictor:

> msimq = lm(y ~ treat + year, data = simdat)

Least-squares means are obtained from a "reference grid" defined by the model. The lsmeans package allows obtaining that reference grid explicitly:

> library(lsmeans)

> summary(ref.grid(msim))
 treat fyear prediction        SE df
 t1    1       1.214286 0.5190258  5
 t2    1       2.357143 0.7340133  5
 t1    2       3.928571 0.6086117  5
 t2    2       5.071429 0.6086117  5

> summary(ref.grid(msimq))
 treat year prediction        SE df
 t1     1.5   2.571429 0.4400255  5
 t2     1.5   3.714286 0.5729889  5

As you see, the reference grid for msim consists of the four combinations of the two factors, whereas the reference grid for msimq has only two points, with year set to its average. The least-squares means for treat are the same for both models, because the linear effect of year is used, which implies that the mean at the average is the average of the means of years 1 and 2.

Summary

To understand least-squares means correctly, focus on the fact that they are based on predictions from a model -- not directly on data without a model context.

You might want to take a look at the documentation and vignettes in the lsmeans package, which has more comprehensive support for obtaining least-squares means from various models.

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  • $\begingroup$ Thanks for the great answer @rvl. Tangentially, why are the calculations based on the predictions from the model, and not the data itself? In the general case, what do I gain from assuming this model (which may be a very poor fit...), predicting the y for each combination, and then averaging over the combinations...when I could just average over the true y for each combination? $\endgroup$ – user310374 Dec 30 '16 at 18:16
  • $\begingroup$ This is not tangential at all. Technically, the calculations do depend on the data because we need data before we can fit a model. But least-squares means are definitely a summary of a fitted model, not of data, and perhaps that's why Walt Harvey used "least squares" in coining the term. (continued in next comment) $\endgroup$ – rvl Dec 30 '16 at 19:46
  • $\begingroup$ Now, I want to emphasize that all statistical inferences are based on models. There are models for how the mean response depends on the predictors, but also models for the distribution of the errors around those means. And in all statistical inferences, if the model is a poor fit (or a poor distributional assumption), the statistical inference you have made is nonsense. I would encourage you to always think about statistics from the perspective of first ensuring the model you are using is reasonable. (continued next comment) $\endgroup$ – rvl Dec 30 '16 at 19:47
  • $\begingroup$ In this particular example, just calculating treatment means from unweighted averages of cell means could be very misleading, because if interaction is present, those marginal means don't make much sense. (Apologies for abuse of comments. But I was answering a question that was stated in a comment.) $\endgroup$ – rvl Dec 30 '16 at 19:48

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