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 R2 for multiple linear regression with 0 intercept

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AMartinOrigin Posted - 05/24/2012 : 06:45:50 AM
I wanted to compare how good a fit to data a multiple linear regression was with and without forcing it to go through the origin. The r2 reported for the 2 cases seems to have been calculated in different ways. That for the standard regression uses the expected residual sum of squares divided by usual total sum of squares, the latter calculated with respect to the mean. For the regression forced through the origin however, the r2 calculation uses a total sum of squares that is absolute (i.e. sum y^2) rather than relative to the mean (i.e. sum (y-mean(y))2 ) despite using the usual residual sum of squares. The upshot is that the r2 for regression thorugh origin is lower despite the residual sum of squares being higher simply because a different definition for the total sum of squares is used. I'm guessing there must be a reason for this but it eludes me. Can anyone shed light on this please?
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Echo_Chu Posted - 05/25/2012 : 05:29:37 AM
Hi, Adrian

When intercept is not included in model, R^2 can be negative if SST is corrected. Most statistical package compute R^2 with uncorrected SST. You refer more details in the reference

Anyway, if you want to computer R^2 from corrected SST with no intercept. You can use the nonlinear fit tool in Origin.

But I am not sure whether it is good to compare the model with and without intercept in this way, because according to Hahn, G.J.(1977), "The values of R-sq for the intercept and no-intercept models are thus not comparable"

To compare the model, I would sugggest you use AIC, which is equal to

where rss is Residual Sum of Square, N is number of points, and K is number of Parameters (for example K=2 for linear fit model)

(Sorry that our fit comparison model does ot support multiple regression yet. You need to compute the AIC value manually.)

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