I have a query about the output statistics gained from linear mixed models (using the lmer function) relative to the output statistics taken from the estimated marginal means gained from this model
Essentially, I am running an LMM comparing the within-subjects effect of different contexts (with "Negative" coded as the baseline) on enjoyment ratings. The LMM output suggests that the difference between negative and polite contexts is not significant, with a p-value of .35. See the screenshot below with the relevant line highlighted:
LMM output
However, when I then run the lsmeans function on the same model (with the Holm correction), the p-value for the comparison between Negative and Polite context categories is now .05, and all of the other statistics have changed too. Again, see the screenshot below with the relevant line highlighted:
LSMeans output
I'm probably being dense because my understanding of LMMs isn't hugely advanced, but I've tried to Google the reason for this and yet I can't seem to find out why? I don't think it has anything to do with the corrections because the smaller p-value is observed when the Holm correction is used. Therefore, I was wondering why this is the case, and which value I should report/stick with and why?
Thank you for your help!
Regression coefficients and marginal means are not one and the same. Once you learn these concepts it'll be easier to figure out which one is more informative and therefore which one you should report.
After we fit a regression by estimating its coefficients, we can predict the outcome yi given the m input variables Xi = (Xi1, ..., Xim). If the inputs are informative about the outcome, the predicted yi is different for different Xi. If we average the predictions yi for examples with Xij = xj, we get the marginal effect of the jth feature at the value xj. It's crucial to keep track of which inputs are kept fixed (and at what values) and which inputs are averaged over (aka marginalized out).
In your case, contextCatPolite in the coefficients summary is the difference between Polite and Negative when smileType is set to its reference level (no reward, I'd guess). In the emmeans contrasts, Polite - Negative is the average difference over all smileTypes.
Interactions have a way of making interpretation more challenging and your model includes an interaction between smileType and contextCat. See Interaction analysis in emmeans.
To add to #dipetkov's answer, the coefficients in your LMM are based on treatment coding (sometimes called 'dummy' coding). With the interactions in the model, these coefficients are no longer "main-effects" in the traditional sense of factorial ANOVA. For instance, if you have:
y = b_0 + b_1(X_1) + b_2(X_2) + b_3 (X_1 * X_2)
...b_1 is "the effect of X_1" only when X_2 = 0:
y = b_0 + b_1(X_1) + b_2(0) + b_3 (X_1 * 0)
y = b_0 + b_1(X_1)
Thus, as #dipetkov points out, 1.625 is not the difference between Negative and Polite on average across all other factors (which you get from emmeans). Instead, this coefficient is the difference between Negative and Polite specifically when smileType = 0.
If you use contrast coding instead of treatment coding, then the coefficients from the regression output would match the estimated marginal means, because smileType = 0 would now be on average across smile types. The coding scheme thus has a huge effect on the estimated values and statistical significance of regression coefficients, but it should not effect F-tests based on the reduction in deviance/variance (because no matter how you code it, a given variable explains the same amount of variance).
https://stats.oarc.ucla.edu/spss/faq/coding-systems-for-categorical-variables-in-regression-analysis/
Related
Still fairly new to GLM and a bit confused about how to establish my model.
About my project:
I sampled the microbiome (and measured a diversity index value = Shannon) from the root system of a sample of 9 trees (=tree1_cat).
In each tree I sampled fine and thick roots (=rootpart) and each tree was sampled four times (=days) over the course of one season. Thus I have a nested design but have to keep time in mind for autocorrelation. Also not all values are present, thus I have a few missing values). So far I have tried and tested the following:
Model <- gls(Shannon ~ tree1_cat/rootpart + tree1_cat + days,
na.action = na.omit, data = psL.meta,
correlation = corAR1(form =~ 1|days),
weights = varIdent(form= ~ 1|days))
Furthermore I've tried to get more insight and used anova(Model) to get the p-values of those factors. Am I allowed to use those p-values? Also I've used emmeans(Model, specs = pairwise ~ rootpart)for pairwise comparisons but since rootpart was entered as nested factor it only gives me the paired interactions.
It all works, but I am not sure, whether this is the right model! Any help would be highly appreciated!
It would be helpful to know your scientific question, but let's suppose you're interested in differences in Shannon diversity between fine and thick roots and in time trends. A model you could use would be:
library(lmerTest)
lmer(Shannon ~ rootpart*days + (rootpart*days|tree1_cat), data = ...)
The fixed-effect component rootpart*days can be expanded into 1 + rootpart + days + rootpart:days (where 1 signifies the intercept)
intercept: SD in fine roots on day 0 (hopefully that's the beginning of the season)
rootpart: difference between fine and thick roots on day 0
days: change per day in SD in fine roots (slope)
rootpart:days difference in slope between thick roots and fine roots
The random-effect component (rootpart*days|tree1_cat) measures how all four of these effects vary across trees, and their correlations (e.g. do trees with a larger-than-average difference between fine and thick roots on day 0 also have a larger-than-average change over time in fine root SD?)
This 'maximal' random effects model is almost certainly too complex for your data; a rough rule of thumb says you should have 10-20 data points per parameter estimated, the fixed-effect model takes 4 parameters. A full model with 4 random effects requires the estimate of a 4×4 covariance matrix, which has (4*5)/2 = 10 parameters all by itself. I might just try (1+days|tree1_cat) (random slopes) or (rootpart|tree_cat) (among-tree difference in fine vs. thick differences), with a bias towards allowing for the variation in the effect that is your primary interest (e.g. if your primary question is about fine vs. thick then go with (rootpart|tree_cat).
I probably wouldn't worry about autocorrelation at all, nor heteroscedasticity by day (your varIdent(~1|days) term) unless those patterns are very strongly evident in the data.
If you want to allow for autocorrelation you'll need to fit the model with nlme::lme or glmmTMB (lmer still doesn't have machinery for autocorrelation models); something like
library(nlme)
lme(Shannon ~ rootpart*days,
random = ~days|tree1_cat,
data = ...,
correlation = corCAR1(form = ~days|tree1_cat)
)
You need to use corCAR1 (continuous-time autoregressive order-1) rather than the more common corAR1 for unevenly sampled data. Be aware that lme is more finicky/worse at dealing with singular models, so you may discover you need to simplify your model before you can actually get this model to run.
I have a response Y that is a percentage ranging between 0-1. My data is nested by taxonomy or evolutionary relationship say phylum/genus/family/species and I have one continuous covariate temp and one categorial covariate fac with levels fac1 & fac2.
I am interested in estimating:
is there a difference in Y between fac1 and fac2 (intercept) and how much variance is explained by that
does each level of fac responds differently in regard to temp (linearly so slope)
is there a difference in Y for each level of my taxonomy and how much variance is explained by those (see varcomp)
does each level of my taxonomy responds differently in regard to temp (linearly so slope)
A brute force idea would be to split my data into the lowest taxonomy here species, do a linear beta regression for each species i as betareg(Y(i)~temp) . Then extract slope and intercepts for each speies and group them to a higher taxonomic level per fac and compare the distribution of slopes (intercepts) say, via Kullback-Leibler divergence to a distribution that I get when bootstrapping my Y values. Or compare the distribution of slopes (or interepts) just between taxonomic levels or my factor fac respectively.Or just compare mean slopes and intercepts between taxonomy levels or my factor levels.
Not sure is this is a good idea. And also not sure of how to answer the question of how many variance is explained by my taxonomy level, like in nested random mixed effect models.
Another option may be just those mixed models, but how can I include all the aspects I want to test in one model
say I could use the "gamlss" package to do:
library(gamlss)
model<-gamlss(Y~temp*fac+re(random=~1|phylum/genus/family/species),family=BE)
But here I see no way to incorporate a random slope or can I do:
model<-gamlss(Y~re(random=~temp*fac|phylum/genus/family/species),family=BE)
but the internal call to lme has some trouble with that and guess this is not the right notation anyways.
Is there any way to achive what I want to test, not necessarily with gamlss but any other package that inlcuded nested structures and beta regressions?
Thanks!
In glmmTMB, if you have no exact 0 or 1 values in your response, something like this should work:
library(glmmTMB)
glmmTMB(Y ~ temp*fac + (1 + temp | phylum/genus/family/species),
data = ...,
family = beta_family)
if you have zero values, you will need to do something . For example, you can add a zero-inflation term in glmmTMB; brms can handle zero-one-inflated Beta responses; you can "squeeze" the 0/1 values in a little bit (see the appendix of Smithson and Verkuilen's paper on Beta regression). If you have only a few 0/1 values it won't matter very much what you do. If you have a lot, you'll need to spend some serious time thinking about what they mean, which will influence how you handle them. Do they represent censoring (i.e. values that aren't exactly 0/1 but are too close to the borders to measure the difference)? Are they a qualitatively different response? etc. ...)
As I said in my comment, computing variance components for GLMMs is pretty tricky - there's not necessarily an easy decomposition, e.g. see here. However, you can compute the variances of intercept and slope at each taxonomic level and compare them (and you can use the standard deviations to compare with the magnitudes of the fixed effects ...)
The model given here might be pretty demanding, depending on the size of your phylogeny - for example, you might not have enough replication at the phylum level (in which case you could fit the model ~ temp*(fac + phylum) + (1 + temp | phylum:(genus/family/species)), i.e. pull out the phylum effects as fixed effects).
This is assuming that you're willing to assume that the effects of fac, and its interaction with temp, do not vary across the phylogeny ...
I have a question similar to the one here: Testing the difference between marginal effects calculated across factors. I used the same code to generate average marginal effects for two groups. The difference is that I am running a logistic rather than linear regression model. My average marginal effects are on the probability scale, so emmeans will not provide the correct contrast. Does anyone have any suggestions for how to test whether there is a significant difference in the average marginal effects between group 1 and group 2?
Thank you so much,
Ilana
It is a bit unclear what the issue really is, but I'll try. I'm supposing your logistic regression model was fitted using, say, glm:
mod <- glm(cbind(heads, tails) ~ treat, data = mydata, family = binomial())
If you then do
emm <- emmeans(mod, "treat")
emm ### marginal means
pairs(emm) ### differences
Your results will be presented on the logit scale.
If you want them on the probability scale, you can do
summary(emm, type = "response")
summary(pairs(emm), type = "response")
However, the latter will back-transform the differences of logits, thereby producing odds ratios.
If you actually want differences of probabilities rather than ratios of odds, use regrid(), which will construct a new grid of values after back-transforming (and hence it will forget the log transformation):
pairs(regrid(emm))
It seems possible that two or more factors are present and you want contrasts of contrasts on the probability scale. In that case, extend this idea by calling regrid() on the table of EMMs to put everything on the probability scale, then follow the analogous procedure used in the linked article.
I want to determine the marginal effects of each dependent variable in a probit regression as follows:
predict the (base) probability with the mean of each variable
for each variable, predict the change in probability compared to the base probability if the variable takes the value of mean + 1x standard deviation of the variable
In one of my regressions, I have a multiplicative variable, as follows:
my_probit <- glm(a ~ b + c + I(b*c), family = binomial(link = "probit"), data=data)
Two questions:
When I determine the marginal effects using the approach above, will the value of the multiplicative term reflect the value of b or c taking the value mean + 1x standard deviation of the variable?
Same question, but with an interaction term (* and no I()) instead of a multiplicative term.
Many thanks
When interpreting the results of models involving interaction terms, the general rule is DO NOT interpret coefficients. The very presence of interactions means that the meaning of coefficients for terms will vary depending on the other variate values being used for prediction. The right way to go about looking at the results is to construct a "prediction grid", i.e. a set of values that are spaced across the range of interest (hopefully within the domain of data support). The two essential functions for this process are expand.grid and predict.
dgrid <- expand.grid(b=fivenum(data$b)[2:4], c=fivenum(data$c)[2:4]
# A grid with the upper and lower hinges and the medians for `a` and `b`.
predict(my_probit, newdata=dgrid)
You may want to have the predictions on a scale other than the default (which is to return the linear predictor), so perhaps this would be easier to interpret if it were:
predict(my_probit, newdata=dgrid, type ="response")
Be sure to read ?predict and ?predict.glm and work with some simple examples to make sure you are getting what you intended.
Predictions from models containing interactions (at least those involving 2 covariates) should be thought of as being surfaces or 2-d manifolds in three dimensions. (And for 3-covariate interactions as being iso-value envelopes.) The reason that non-interaction models can be decomposed into separate term "effects" is that the slopes of the planar prediction surfaces remain constant across all levels of input. Such is not the case with interactions, especially those with multiplicative and non-linear model structures. The graphical tools and insights that one picks up in a differential equations course can be productively applied here.
I ran the same probit regression in SAS and R and while my coefficient estimates are (essentially) equivalent, the reported test statistics are different. Specifically, SAS reports test statistics as t-statistics whereas R reports test statistics as z-statistics.
I checked my econometrics text and found (with little elaboration) that it reports probit results in terms of t statistics.
Which statistic is appropriate? And why does R differ from SAS?
Here's my SAS code:
proc qlim data=DavesData;
model y = x1 x2 x3/ discrete(d=probit);
run;
quit;
And here's my R code:
> model.1 <- glm(y ~ x1 + x2 + x3, family=binomial(link="probit"))
> summary(model.1)
Just to answer a little bit - it's seriously off topic, question should be closed in fact - but neither the t-statistic nor the z-statistic are meaningful. They're both related though, as Z is just the standard normal distribution and T is an adapted "close-to-normal" distribution that takes into account the fact that your sample is limited to n cases.
Now, both the z and the t statistic provide the significance for the null hypothesis that the respective coefficient is equal to zero. The standard error on the coefficients, used for that test, is based on the residual error. Using the link function, you practically transform your response in such a way that the residuals become normal again, whereas in fact the residuals represent the difference between the observed and the estimated proportion. Due to this transformation, calculation of the degrees of freedom for the T-statistic isn't useful anymore and hence R assumes the standard normal distribution for the test statistic.
Both results are completely equivalent, R will just give slightly sharper p-values. It's a matter of debate, but if you look at proportion difference tests, they're also always done using the standard normal approximation (Z-test).
Which brings me back to the point that neither of these values actually has any meaning. If you want to know whether or not a variable has a significant contribution with a p-value that actually says something, you use a Chi-squared test like the Likelihood Ratio test (LR), Score test or Wald test. R just gives you the standard likelihood ratio, SAS also gives you the other two. But all three tests are essentially equivalent, if they differ seriously it's time to look again at your data.
eg in R :
anova(model.1,test="Chisq")
For SAS : see the examples here for use of contrasts, getting the LR, Score or Wald test