Timeline for Two negative beta's in a curvilinear regression when mean centered or using standardized values
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 7, 2014 at 0:11 | comment | added | user39659 | orm.sagepub.com/content/15/3/339.abstract To all who come across this in the future, I recommend reading Dalal & Zickar (2012) -- The "solution" of using mean centering is not as straightforward as one would think. | |
| Sep 23, 2012 at 21:35 | answer | added | Peter Flom | timeline score: 5 | |
| Sep 23, 2012 at 21:25 | history | edited | Peter Flom | CC BY-SA 3.0 |
fixed spelling
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| Sep 23, 2012 at 19:56 | comment | added | user14316 | @Fojtasek I also tried to mean centering the squared X, then I have a positive beta for the normal independent variable and a negative beta for the squared one. That is exactly the case when there is an inverted U-shaped relation, but in that case there is a multicolliniearity problem because the VIF value is higher than 10. | |
| Sep 23, 2012 at 19:39 | comment | added | gung - Reinstate Monica | I'm not sure there's a problem here, but if you can provide your data, people may be able to say more. | |
| Sep 23, 2012 at 17:03 | comment | added | Fojtasek | I suspect that you squared the mean-centered x instead of mean centering the squared x. If you mean center before you square, you get a u-shaped predictor. | |
| Sep 23, 2012 at 16:40 | comment | added | Michael R. Chernick | Is y =ax$^2$ with a>0 the onlt way to get a perfect inverted U? I don't think so. If the inverted U corresponds to a different function then there will not be an a that gives a perfect fit to y=ax$^2$. Then it would be possible for some other model to fit better. | |
| Sep 23, 2012 at 16:19 | review | First posts | |||
| Oct 2, 2012 at 13:02 | |||||
| Sep 23, 2012 at 16:16 | history | asked | atsk | CC BY-SA 3.0 |