Covariate slopes in Proc Mixed: an interaction problem

[DiegoGil]

Hi there,

I'm out of my depth here and I wonder if someone has a solution. I have this analysis in which I test the effect of temperature in nestling growth. I have several chicks per nest, thus the nest is a random factor and I run this in Proc Mixed. The effect of temperature is different for two treatment groups, and this is reflected in a significant interaction of temperature*group. Fine, so the two slopes are different frome each other. Cool, this is expected. The problem is that I need to know whether these two slopes are significantly different from 0 (in the graph it looks as if one of them is negative, but the other one could be positive or simply not different from 0). My referees want to know that (and so do I!). The SAS estimate shows that one of the slopes is set to 0 by default, so I only have an estimate for the other slope. How can I obtain the two "real" slope estimates? If I split the dataset in two, and I calculate the slopes separately, the effect disappears, so this is not good. Any ideas?? Thanks millions!

Best wishes

Diego Gil

[SteveDenham]

Without seeing your model statement I can't be completely sure, but is it something like:

model dependent_var=group temperature group*temperature/solution;

If this is close, then there should be an estimate for temperature and an estimate for the first group*temperature, with a zero for the last group*temperature.

There are two ways to go from here.  The first is to realize that the slope for temperature actually applies to the last group (adding the estimates together), while the slope for the first group*temperature is the difference between the two slopes.  You can get the slope for this group by adding the estimates.  That's great for the estimates, but the standard errors are a problem.

Try:

model dependent_var= group temperature group*temperature/solution noint;

This should give separate estimates of the slope, with correct standard errors, for each of the slopes.

Steve Denham

[DiegoGil]

Dear Steve,

Thanks for your mail. You were right about the model, it's a bit more complex but I reduced it for this particular problem. So, then if we use /solution we obtain the following ("puesta" is group; and "t014" is temperature):

You said that the slope for group 2 is actually that of temperature, thus: -1.85; while that of the first group would be then (group 2 minus group 1) = 0.2, is this correct?

Then, for the SE, if I use noint I get this, but there is no different SE for the slopes as you said, what am I doing wrong?

Thanks a lot for your help, this is most useful!!

Diego

[SteveDenham · Accepted solution]

You are not doing anything wrong.  I gave you some code that doesn't do what you need.  For the no intercept model, you also need to remove the main effect of t014, fitting only the interactions.  This shouldl give a two degree of freedom F test of the equality of slopes, and a two degree of freedom F test for the equality of intercepts.

model dependent_var= puesta puesta*t014/solution noint;

Steve Denham

[DiegoGil]

Spot on Steve! Thanks a lot!

One final thing, I assumed that one should never put terms in an interaction that have not appear on their own in the model, so I guess this method you gave me is not good for fixed effects, just for estimates of interactions, is this right?

Best wishes

Diego

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Diego Gil

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  Museo Nacional de Ciencias Naturales (CSIC)

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  28006 Madrid

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[SteveDenham]

To continue it, this is for interactions with continuous covariates.  I would consider there to be one fixed classification variable, puesta, which is included.

If this were a factorial or split-plot where there were more classification variables, I would certainly not fit just the interactions, unless I needed to fit a means model, where only the highest order interaction would be in the model.

Also, if there multiple continuous covariates, this gets really, really messy.

Steve Denham