Solved – in r, lognormal, glm, transformations, what should I do

generalized linear model

(Updated)

I have biomass (grams) as my response variable, and weather data (wind, air temperature, relative humidity, precipitation) as well as vegetation measurements (basal area, canopy closure, stem counts) as explanatory variables. I have some zeros in the data, like no wind speed (it just wasn't windy) or there was no precipitation for that day so it's 0.

I've also got different survey locations that I surveyed at different times.

I want to see what factors influence my response variable. Hypothetically, there should be a model with something like this:

biomass~wind+airtemp+rH+precip+ba+closure+stems+(1|location)

I looked at the biomass data to see if they fit normal distributions:

library(car)
library(MASS)
qqp(biomass,"norm")

doesn't fit as well as qqp(biomass,"lnorm") so I think that a lognormal distribution fits it better right?

Also, following advice given in answers and comments (below) I graphed the residuals and I get a cone-shaped residuals vs fitted graph (= non constant variance) and a curved normal q-q plot.

Should my biomass should be logged? :

log(biomass)~wind+airtemp+rH+precip...

Or else should I transform my biomass data before adding them to the model? (Although shouldn't log(variable) be the same as using a previously logged variable?

From previous answers and comments (see below), my equation has evolved into something like this:

library(lme4)

fit3=lmer(log(biomass)~wind+temp...+(1|location),data=mydata)

Plus, my adviser asked me to add treatment type (categorical variable) to the equation, so it really looks like this:

fit4=lmer(log(biomass)~treatmenttype+wind+temp....+(1|location),data=mydata)

When I try that though, I get some warning and error messages:

The variance-covariance matrix is not symmetric, returning NA matrix
There's an error in evaluating the argument 'x' in selecting a method for function

The R comments look like this: 

Warning message:
In vcov.merMod(object, use.hessian = use.hessian) :
Computed variance-covariance matrix problem: matrix is not symmetric [1,2];
  returning NA matrix
Error in diag(vcov(object, use.hessian = use.hessian)) : 
error in evaluating the argument 'x' in selecting a method for function 'diag': Error          in rr@factors$correlation <- if (!is.na(sigm)) as(rr, "corMatrix") else rr : 
  trying to get slot "factors" from an object of a basic class ("matrix") with no    slots

I'm not sure what this means. Is this an error on my part in the regression or something I have wrong in the R code?

Best Answer

There are a variety of confusions here: most have probably been dealt with at one time or another on CrossValidated ...

  • generally, the distributional assumptions of regression modeling (whether linear, generalized linear, or mixed) refer to the conditional distribution of the response variable: that is, the assumption is that $y \sim \textrm{Dist}(...)$, where the $...$ contains the information from the input variables (approximately the same as "predictor variables", "covariates", or "independent variables").
  • sometimes people also transform the predictor variables, but this is to improve the linearity of the relationship between the input variables and the response. There are very few cases where any important assumptions are made about the distribution of the input variables.
  • if you have a continuous response variable, you can probably get away with a linear model (implemented via lm() in base R) or, if you want to include a random effect of site, lmer() from the lme4 package (or lme from the nlme package).
  • first you should plot your data. You should probably start by looking at univariate relationships (plot(biomass~wind,data=mydata)), even though they can miss a lot of higher-order structure.

I would probably try 

  fit1 <- lm(biomass~.-location,data=mydata) 
  ## dot in the formula stands for "everything but the response"; 
  ##    -location takes out the location
  plot(fit1)   ## diagnostic plots
  ## maybe, if the Q-Q plot and scale-location plot look funny ...
  library("MASS")
  boxcox(fit1)

first, and then try the equivalent

library("lme4")
fit2 <- lmer(biomass~...,data=mydata)
## here you need to fill in the ... yourself since lme4 doesn't have the same
##  shortcuts

This is just scratching the surface, but might get you started.