I'm working my way through a Linear Regression Textbook and am trying to replicate the results from a section on the Test of the General Linear Hypothesis, but I need a little bit of help on how to do so in R.
I've already taken a look at a number of other posts, but am hoping someone can give me some example code. I have data on twenty-six subjects which has the following form:
Group, Weight (lb), HDL Cholesterol mg/decaliters
1,163.5,75
1,180,72.5
1,178.5,62
2,106,57.5
2,134,49
2,216.5,74
3,163.5,76
3,154,55.5
3,139,68
Given this data I am trying to test to see if the regression lines fit to the three groups of subjects have a common slope. The models postulated are:
y=βo + β1⋅x + ϵ
y=γ0 + γ1⋅xi + ϵ
y= δ0 + δ1⋅xi + ϵ
So the hypothesis of interest is H0: β1 = γ1 = δ1
I have been trying to do this using the linearHypothesis function in the car library, but have been having trouble knowing what the model object should be, and am not confident that this is the correct approach (or package) to be using.
Any help would be much appreciated – Thanks!
Tim, your question doesn't seem so much to be about R code. Instead, it appears that you have questions about how to test the interaction of your Group and Weight (lb) variables on the outcome HDL Cholesterol mg/decaliters. You don't state this specifically, but I'm taking a guess that these are your predictors and outcome, respectively.
So essentially, you're trying to see if the predictor Weight (lb) has differential effects depending on the level of the variable Group. This can be done in a number of ways using the linear model. A simple regression approach would be lm(hdl ~ 1 + group + weight + group*weight). And then the coefficient for the interaction term group*weight would tell you whether or not there is a significant interaction (i.e., moderation) effect.
However, I think we would have a major concern. In particular, we ought to worry that our hypothesized effect is that the group variable and the hdl variable do not interact. That is, you're essentially predicting the null. Furthermore, you're predicting the null despite having a small sample size. Therefore, it would be rather unlikely that we have sufficient statistical power to detect an effect, even if there were one to be observed.
Related
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/
I am relatively new to both R and Stack overflow so please bear with me. I am currently using GLMs to model ecological count data under a negative binomial distribution in brms. Here is my general model structure, which I have chosen based on fit, convergence, low LOOIC when compared to other models, etc:
My goal is to characterize population trends of study organisms over the study period. I have created marginal effects plots by using the model to predict on a new dataset where all covariates are constant except year (shaded areas are 80% and 95% credible intervals for posterior predicted means):
I am now hoping to extract trend magnitudes that I can report and compare across species (i.e. say a certain species declined or increased by x% (+/- y%) per year). Because I use poly() in the model, my understanding is that R uses orthogonal polynomials, and the resulting polynomial coefficients are not easily interpretable. I have tried generating raw polynomials (setting raw=TRUE in poly()), which I thought would produce the same fit and have directly interpretable coefficients. However, the resulting models don't really run (after 5 hours neither chain gets through even a single iteration, whereas the same model with raw=FALSE only takes a few minutes to run). Very simplified versions of the model (e.g. count ~ poly(year, 2, raw=TRUE)) do run, but take several orders of magnitude longer than setting raw=FALSE, and the resulting model also predicts different counts than the model with orthogonal polynomials. My questions are (1) what is going on here? and (2) more broadly, how can I feasibly extract the linear term of the quartic polynomial describing response to year, or otherwise get at a value corresponding to population trend?
I feel like this should be relatively simple and I apologize if I'm overlooking something obvious. Please let me know if there is further code that I should share for more clarity–I didn't want to make the initial post crazy long, but happy to show specific predictions from different models or anything else. Thank you for any help.
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 ...
Seems like a very basic question but I just wanted to confirm. I'm running a multivariable linear regression model adjusted for different types of covariates (some numeric, some categorical, etc.). A sample of the model is shown below:
fit <- ols(outcome ~ exposure + age + zbmi + income + sex + ethnicity)
Both the "outcome" and "exposure" are continuous numerical variables.
My question is, if say I run the model and the beta estimate, 95% CI, and p-value looks something like the one below:
B = -0.20 // 95%CI: [-0.50, -0.001] // p = 0.04
Would it be appropriate to interpret this as: "For every 1 unit increase of the exposure is a 0.20 decrease in the outcome"?
What I want to know is how did it determine the order of "per 1 unit increase"? Is that just the default order of how R sorts continuous variables when running it in a regression model? Also, since both my outcome and exposure are continuous variables, does this mean that it automatically sorted these variables in ascending order (by default?) when I ran the model?
Just a bit confused on whether this sorting order matters before I run any regression model using continuous variables. Any tips / help would be appreciated!
Under OLS, there is no ordering or sorting of the predictors. The right hand side of the equation is summed before subtracting it from the left hand side. Then the square of this difference is minimized. So with this technique, the predictors do not have to be sorted in any way.
For interpretation of your betas, the predictors are supposed to be independent, so it doesn't matter in which order you take them.
Side note: In reality, you might get some dependence among the predictors, and this will be reflected in the standard errors being slightly larger.
I would like to run repeated measure anova in R using regression models instead an 'Analysis of Variance' (AOV) function.
Here is an example of my AOV code for 3 within-subject factors:
m.aov<-aov(measure~(task*region*actiontype) + Error(subject/(task*region*actiontype)),data)
Can someone give me the exact syntax to run the same analysis using regression models? I want to make sure to respect the independence of residuals, i.e. use specific error terms as with AOV.
In a previous post I read an answer of the type:
lmer(DV ~ 1 + IV1*IV2*IV3 + (IV1*IV2*IV3|Subject), dataset))
I am really not sure about this solution since it still treats variables as between subjects, and I don't understand how adding random factors would change this.
Does someone know how to run repeated measure anova with lm/lmer taking into account residual independence?
Many thanks,
Solene
I have some worked examples with more detail here: https://keithlohse.github.io/mixed_effects_models/lohse_MER_chapter_02.html
But if you want to get a mixed model that is homologous to your ANOVA, you can include random intercepts for your each subject:factor with your within-subject factors. E.g.,
aov(DV~W1*W2*W3 + Error(SUBJECT/(W1*W2*W3)),data)
has a mixed-model equivalent of:
lmer(speed ~
# Fixed Effects
W1*W2*W3 +
# Random Effects
(1|SUBJECT) + (1|W1:SUBJECT) + (1|W2:SUBJECT) + (1|W3:SUBJECT),
data = DATA,
REML = TRUE)
With REML set to TRUE and a balanced design, you should get degrees of freedom and f-values that are identical to your ANOVA. ML tends to underestimate variance components, so if you are comparing nested models and need to use ML your results will not match precisely. If you are not comparing nested models and can use REML, then the ANOVA and mixed-model should match (again, in a balanced design).
To #skan's earlier answer and other ideas people might have, I am not saying this is THE random-effects structure (as it might be more appropriate to include random slopes for W1 compared to random-intercepts), but if you have one observation per subject:condition, then these random-effects produce an equivalent result.
If your aov example is right (maybe you don't want to nest things) you want this:
lmer(measure~(task*region*actiontype) + 1(1|subject/(task:region:actiontype))
If residual independence means intercept and slope independently calculated you need to specify them separately:
+(1|yourfactors)+(0+variable|yourfactors)
or use the symbol:
+(1||yourfactors)
Anyway if you read the help files you can find that lme4 can't deal with the most general problems.