Regstats : Multiple regression/comparing regression slopes

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laurie
laurie 2012-3-16
Hi,
I want to compare the regression slopes of 3 data sets. Apparently you have to set some dummy variables =0/1 and regress against them to do this.
So I tried to do it using regstats but I can't get regstats to work with several regressors.
I tried the following :
X =
% the "real" regressor (1 2 3) repeated 3 times (one for each set)
1
2
3
1
2
3
1
2
3
Z12 =
% the first dummy regressor (1 if data is from set number 2)
0
0
0
1
1
1
0
0
0
Z20 =
% second dummy regressor (1 if data is from group 3)
0
0
0
0
0
0
1
1
1
V=[X,Z12,Z20]
V =
% all three regressors put together in one matrix
1 0 0
2 0 0
3 0 0
1 1 0
2 1 0
3 1 0
1 0 1
2 0 1
3 0 1
Y=2+3*X+4*Z12+5*Z20*X'
Y =
% fake dataset to see if regstats would give me the right coefficients
35
38
41
39
42
45
35
38
41
then i tried
s=regstats(Y,V,'interactions','all')
and got the following warnings
Warning: Matrix is singular to working precision.
> In regstats at 180
Warning: Matrix is singular to working precision.
> In regstats at 195
and none of the numbers is correct i get a lot of NaNs and Inf for example
>> s.beta
ans =
NaN
NaN
NaN
NaN
NaN
NaN
Inf
I don't have any clue on what is wrong with the code I tried. I have never used regstats before. Any idea on what I could do ?
Thank you very much for your help !
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laurie
laurie 2012-3-17
thank you for your answer
sorry about the stray character. the output version is the right one. i don't know what a lagged value is.
i am doing this because i have 3 sets of data y1,y2, and y3, and i want to compare the slopes of their respective regressions against x. (are the slopes equal or not?).
i think my problem is very similar to this one here
http://www.mathworks.com/matlabcentral/newsreader/view_thread/270278
but i am struggling to understand the method.
Oleg Komarov
Oleg Komarov 2012-3-17
The thread you found is a good one, follow it thoroughly and pay attention to details, for example .* is different from *

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采纳的回答

Oleg Komarov
Oleg Komarov 2012-3-16
Your data (in copy-pastable form):
X = [1;2;3;1;2;3;1;2;3];
Z12 = [0;0;0;1;1;1;0;0;0];
Z20 = [0;0;0;0;0;0;1;1;1];
V = [X,Z12,Z20];
Y = 2 + 3*X + 4*Z12 + 5*Z20'*X;
Setting the model to 'interaction' gives this matrix:
x2fx(V,'interaction')
1 1 0 0 0 0 0
1 2 0 0 0 0 0
1 3 0 0 0 0 0
1 1 1 0 1 0 0
1 2 1 0 2 0 0
1 3 1 0 3 0 0
1 1 0 1 0 1 0
1 2 0 1 0 2 0
1 3 0 1 0 3 0
Is this what you're looking for? (Not likely). To understand what's going on with the choice of the model read the documentation of x2fx, which is used by regstats.

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laurie
laurie 2012-3-17
Thanks you very much, it worked fine !! :-)
Just checking : if regstats gives a p-value > criterium for one regressor, it means that the coefficient for this regressor is probably 0 right ? Also, should'nt the criterium be adjusted to the number of regressors ?
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Oleg Komarov
Oleg Komarov 2012-3-17
You usually read things like, the p-values are too high, thus the coefficients are not statistically significant at any conentional level (where the conventional levels are 5% or 1%).
A p-value is the (smallest) probability to observe a value that is different from the null just by chance.
Rephrasing, a p-value is the percentage of values you would erroneously accept (when they should be considered as the null).
In regressions, the p-value of single coefficients is ususally the t-test against the null of 0.
So, say you have a coefficient of 0.32 and the question you want to ask is: "is it truly 0.32 or should I consider it as 0?"
Suppose you get a p-value of 9.81%. It says that the probability of 0.32 being a 0 is 9.81%.
Now, are you willing to accept such a risk? Usually, the threshold is 5%, thus observing a p-value lower than the threshold is good sign.
DISCLAIMER: I am abusing the words truly and accept.
Truly is with respect to the assumed distribution for your parameters and you NEVER accept the NULL but you fail to reject it.

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