Hi everybody,
regarding one of the causal relationships analyzed in my project PLS results dislay a very low path coefficient (just below 0,16) but a rather high R² (0,5) of the dependent variable. I am struggling to interpret these results. Do you have any advice?
Regards
Juliane
interdependence path coefficient and R²
Hi Bido,
thanks a lot for your swift answer.
My problem is, that PLS displays the following: R2 = 0.51; B1 = 0.159
Following your calculation r_y1 = 3.208, which is >1 but generally should range between -1 and 1. I don't know how to interpret these results, have you got any idea?
Kind regards
Juliane
thanks a lot for your swift answer.
My problem is, that PLS displays the following: R2 = 0.51; B1 = 0.159
Following your calculation r_y1 = 3.208, which is >1 but generally should range between -1 and 1. I don't know how to interpret these results, have you got any idea?
Kind regards
Juliane
- Diogenes
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Hi,
I understood that one of them has low effect (.16), but the others should have greater effect (path coefficient).
How many arrows arrive in the dependent LV?
What are the values of correlations and path coefficients?
Do you run the PLS algorithm with standardized values (Data metric = Mean 0, Var 1)?
Best regards,
Bido
I understood that one of them has low effect (.16), but the others should have greater effect (path coefficient).
How many arrows arrive in the dependent LV?
What are the values of correlations and path coefficients?
Do you run the PLS algorithm with standardized values (Data metric = Mean 0, Var 1)?
Best regards,
Bido
- Diogenes
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- Joined: Sat Oct 15, 2005 5:13 pm
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Hi,
With just one arrow between the LVs, the structural model is a simple regression, then the path coefficient (standardized) is equal to the correlation, and R2 = path * r = (path)^2 = r ^2
If the R2 = 0.51, the correlation is 0.71, and this should be the value of the standardized path coefficient.
The only explanation that I could to think is that you are not using the (Data metric = Mean o, Var 1) in the PLS algorithm…
Best regards,
Bido
With just one arrow between the LVs, the structural model is a simple regression, then the path coefficient (standardized) is equal to the correlation, and R2 = path * r = (path)^2 = r ^2
If the R2 = 0.51, the correlation is 0.71, and this should be the value of the standardized path coefficient.
The only explanation that I could to think is that you are not using the (Data metric = Mean o, Var 1) in the PLS algorithm…
Best regards,
Bido
Hi Bido,
thank you very much indeed. I can follow your argumentation.
As I am using (Data metric = Mean o, Var 1), I however don't understand why SmartPLS displays this one path coefficient as .159 instead of .715 - especially since all of the other path coefficents and R2 of the model are displayed correctly... .
Again, thanks a lot for your help!
Best regards
Juliane
thank you very much indeed. I can follow your argumentation.
As I am using (Data metric = Mean o, Var 1), I however don't understand why SmartPLS displays this one path coefficient as .159 instead of .715 - especially since all of the other path coefficents and R2 of the model are displayed correctly... .
Again, thanks a lot for your help!
Best regards
Juliane
- Diogenes
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- Posts: 899
- Joined: Sat Oct 15, 2005 5:13 pm
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Could you send me the illustration (print screen) of your model (with results)?
These results don't make sense...
diogenesbido@yahoo.com.br
These results don't make sense...
diogenesbido@yahoo.com.br