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Re: st: Tiao-Goldberger test

From   Michael Stobernack <>
Subject   Re: st: Tiao-Goldberger test
Date   Mon, 16 Feb 2009 18:01:27 +0100

At 16.02.2009 16:47, you wrote:
--- On Mon, 16/2/09, Michael Stobernack wrote:
> Another possibility could be to incorporate dummy variables
> and interaction terms for the different groups of interest.
> Since I have got four different groups I need three dummies
> and many interactions terms, so I loose (to) many degrees of
> freedom (n=50).

By estimating separate models for each group you actually
estimate 4 extra coefficients, i.e. loose 4 degrees of

I estimate the following model 4 times, i.e. once for each of the four groups:

y = b0 + b1*x1 + b2 * x2 + b3*x3 + e, so I have to estimate 4*4 = 16 coefficients.

If I estimate one model for all 4 groups together and D1 is a dummy four group1 etc, I use

y = b0 + b1*x1 + b2*x2 + b3*x3 +
b4*D1 + b5*D2 + b6*D3 + (to see, whether the intercept differs between the groups) b7*D1*x1 + b8*D2*x1 + b9*D3*x1 + (to see, whether the slope of x1 differs between the groups) b10*D1*x2 + b11*D2*x2 + b12*D3*x2 + (to see, whether the slope of x2 differs between the groups) b13*D1*x3 + b14*D2*x3 + b15*D3*x3 + epsilon (to see, whether the slope of x3 differs between the groups)

So I have to estimate 16 coefficients.

What do you mean by 'estimate 4 extra coefficients'? Do you mean 4 extra values of the residual variance?


 compared to a model with all the interaction
effects. The reason is that in the interaction model you
constrain the residual variance to be equal across groups.

-- Maarten

Maarten L. Buis
Department of Social Research Methodology
Vrije Universiteit Amsterdam
Boelelaan 1081
1081 HV Amsterdam
The Netherlands

visiting address:
Buitenveldertselaan 3 (Metropolitan), room N515

+31 20 5986715

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