If your aim is to see if the association (correlation) between weight and RV *depends* on the treatment, you should look at the interactions. These give you the *change* (or difference) in correlation (RV vs. weight) between different treatments.

Short: RV~weight+trt (with weight as a continuous and trt as a categorical variable) will give a model with the same slope but different intercepts for the different treatments. RV~weight*trt will give a model with different slopes (and different intercepts) for the treatments. [RV~weight+weight:trt would force the same intercept but allows different slopes; this is rarely - but might be - sensible]

Just as an illustration:

# construct some data:

# RV depends on weight but the degree depends on the treatment

set.seed(99)

weight = rnorm(3*20, m=60, sd=5)

trt = gl(3,20,labels=letters[1:3])

RV = c(0.5,-0.7,1.2)[as.integer(trt)] * weight + rnorm(length(weight), m=0, sd=2) + c(30,80,-40)[as.integer(trt)]

# plot the values; color indicated the treatment

plot(RV~weight, pch=16, col=as.integer(trt))

# calculate a linear model with interaction

# (try and substitute the * by a + to see what happens)

model = lm(RV ~ weight*trt)

# store the coefficients for plotting

coefs = coef(model)

# add the regression lines based on the coefficients; separately for each treatment

trta = c(coefs["(Intercept)"], coefs["weight"])

abline(trta,col=1)

trtb = c(coefs["(Intercept)"]+coefs["trtb"], coefs["weight"]+coefs["weight:trtb"])

abline(trtb,col=2)

trtc = c(coefs["(Intercept)"]+coefs["trtc"], coefs["weight"]+coefs["weight:trtc"])

abline(trtc,col=3)

legend("bottomleft",legend=levels(trt),title="treatment",pch=16,col=1:3,bty="n",inset=c(0.1,0.1))

I'm not sure if I got the question. However: the + is used if the pair-wise relations between Predictor and RV can be described in an additive way ("main effects"). The * indicates that the change in the value of the RV can not be predicted by one predictor alone without knowing the value of the other predictor ("interaction"). The best way is to have a scientific idea about the possible interactions to be represented in the model. If this is not the case, the analysis of the residuals may help to see if the assumption of interactions (and which) is resonable.

Hi Jochen, thanks for your answer. I am aware that my explanation was not the best, but I am a new R user. As stated I just wanna test if a scaling relationship change with my treatments (1, 2, 3....6). Scaling, is already a linear relationship between a certain variable and animal weight (RV ~ animal weight). Therefore, I am trying to evaluate if this relationship (linear regressions) change in my different treatments (RV ~ Predictor 1 *? or +? Predictor 2). As you pointed out, the best way is to get to the model with a scientific idea of what is going on. In this case, I am expecting to see a change in the scaling relationship with my treatments. Yet, I am not sure what is the proper way to test it: animal weight + treatment? or animal weight * treatment (interaction). I have already run both models, but the post-hoc comparisons at the end are quite different. Both showed nice and sensible results, however, I wanna be sure that I am properly testing it. Thanks again!

If your aim is to see if the association (correlation) between weight and RV *depends* on the treatment, you should look at the interactions. These give you the *change* (or difference) in correlation (RV vs. weight) between different treatments.

Short: RV~weight+trt (with weight as a continuous and trt as a categorical variable) will give a model with the same slope but different intercepts for the different treatments. RV~weight*trt will give a model with different slopes (and different intercepts) for the treatments. [RV~weight+weight:trt would force the same intercept but allows different slopes; this is rarely - but might be - sensible]

Just as an illustration:

# construct some data:

# RV depends on weight but the degree depends on the treatment

set.seed(99)

weight = rnorm(3*20, m=60, sd=5)

trt = gl(3,20,labels=letters[1:3])

RV = c(0.5,-0.7,1.2)[as.integer(trt)] * weight + rnorm(length(weight), m=0, sd=2) + c(30,80,-40)[as.integer(trt)]

# plot the values; color indicated the treatment

plot(RV~weight, pch=16, col=as.integer(trt))

# calculate a linear model with interaction

# (try and substitute the * by a + to see what happens)

model = lm(RV ~ weight*trt)

# store the coefficients for plotting

coefs = coef(model)

# add the regression lines based on the coefficients; separately for each treatment

trta = c(coefs["(Intercept)"], coefs["weight"])

abline(trta,col=1)

trtb = c(coefs["(Intercept)"]+coefs["trtb"], coefs["weight"]+coefs["weight:trtb"])

abline(trtb,col=2)

trtc = c(coefs["(Intercept)"]+coefs["trtc"], coefs["weight"]+coefs["weight:trtc"])

abline(trtc,col=3)

legend("bottomleft",legend=levels(trt),title="treatment",pch=16,col=1:3,bty="n",inset=c(0.1,0.1))

In addition to the clear explaination of Jochen Wilhelm, I would add that if you fitted the two models, with and without interaction, comparing them also answers your question: if the models are (significantly) different, the one with interaction is best, and since you have interaction, the relationship between oxygen consumption and treatment changes with treatment.

If you are new in R, remember not to fall in the classical problem with numeric coding of treatment, leading to linear regression instead of analysis of variance, and do not forget the as.factor( treatment ) --- not a problem if treatment is coded by something else than numbers.