Linear regression - are my models nested or non-nested? I’ve got two main groups, which I am comparing, a) rats and b) rabbits. Each group consists of two subgroups – breed 1 and breed 2.
Using linear regression, I would need to find out, does a model taking into account the subgroups account for more variance than a model with only the main groups.
So I have created two models:
\begin{align}
Y &= a_1+b1_1(\text{rab})+e_1 \tag{1}  \\
Y &= a_2+b1_2(\text{rat2})+b2_2(\text{rab1})+b3_2(\text{rab2})+e_2\ \tag{2}
\end{align}
Where all variables are dummy variables:


*

*$\text{rab}\ \ = 1$ for a rabbit and zero otherwise (e.g., rats)

*$\text{rab1} = 1$ for rabbit breed 1

*$\text{rab2} = 1$ for rabbit breed 2

*$\text{rat2}\, = 1$ for rat breed 2


In order to compare the models statistically, I would need to know, whether my models can be considered nested or not.
If the models are considered nested, I would also need advice in how to compare them statistically - preferably with SPSS. With non-nested models I can handle the rest.
 A: While your data and model don't fit the formal definition of nested, they are in essence nested because you can redefine the variables in such a way that they do fit the definition.  For example change to 2 variables:
rab is 0 for rats and 1 for rabbits
breed is 0 for breed 1 (either species) and 1 for breed 2 (either species)
then fit the model:
$ y = a + b1(rab) + b2(breed) + b3(rab \times breed) + e$
Now you can see that your first model is clearly nested in this model and this model will give the exact same predictions and therefore the exact same sums of squares, likelihood, etc.  so any test using these values will give the same results for your model and my model, so your 2 models can be treated as nested.
A: They are in fact nested models. Model 1 can be considered a special case of Model 2 by imposing the following parameter restrictions:
$$ b1_2=0 $$
$$b2_2=b3_2$$
In effect, this will absorb $rat2$ into the intercept (which already represents $rat1$), and $rab1+rab2$ will be combined into a single vector equivalent to the original $rab$.
