Timeline for Negative adjusted coefficients of determination - why?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Oct 14, 2015 at 14:32 | vote | accept | Sos | ||
| Oct 6, 2015 at 15:14 | answer | added | JimB | timeline score: 2 | |
| Oct 6, 2015 at 9:11 | comment | added | Sos | @JimBaldwin - regarding your last point, using only 9 points is just to illustrate my main question. There is no point in including more data here in order to grasp my doubts | |
| Oct 5, 2015 at 23:21 | comment | added | JimB | And fitting a model with 8 parameters (1 intercept, 6 covariates, and a variance) with 9 data points is, well, kinda silly. | |
| Oct 5, 2015 at 23:06 | comment | added | JimB | This is a statistical question rather than a Mathematica question as negative adjusted R2's are also possible when one does include an intercept. This question would be best asked at CrossValidated: stats.stackexchange.com. | |
| S Oct 5, 2015 at 18:43 | history | suggested | gwr | CC BY-SA 3.0 |
improved formatting
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| Oct 5, 2015 at 18:39 | comment | added | gwr | I just removed the 5th and 6th column of your data and the reduced model without $a$ and $t$ gets achieves an $R^2_{adj}$ of $-0.42$. | |
| Oct 5, 2015 at 18:36 | comment | added | Sos |
By opposition, you are suggesting that terms a and t are simply worsening the ability of the model to explain the data?
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| Oct 5, 2015 at 18:35 | comment | added | Sos |
To reply to your 1st comment: You are correct! Indeed, when I look at lmf["ParameterTable"] not only is their value close to 0 but they are statistically not different than 0 as indicated by their p-values. To answer your 2nd comment: In practical terms you are correct, an R2<0 is no different than R2=0, but I was suspecting that a R2<0 could indicate that something fundamentally wrong was occurring in the fitting algorithm.
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| Oct 5, 2015 at 18:27 | comment | added | gwr | Why is getting $R^2_{adj} = 0$ a goal of yours, even if it is comparatively better than a negative value? | |
| Oct 5, 2015 at 18:24 | review | Suggested edits | |||
| S Oct 5, 2015 at 18:43 | |||||
| Oct 5, 2015 at 18:18 | comment | added | gwr | Adjusted $R^2$ might be negative if there are terms in your model that do not help to predict the response, as this page about goodness of fit tells us. If I look at the proposed linear model equation, then $a$ and $t$ have coefficients of approximately zero. | |
| Oct 5, 2015 at 18:11 | history | edited | Sos | CC BY-SA 3.0 |
added 9 characters in body
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| Oct 5, 2015 at 18:10 | comment | added | Sos |
@gwr - I meant the other way around. I apologize and will correct that right away. Indeed, I notice that the documentation adds an intercept by default, but having that {1} changes nothing. The main issue remains
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| Oct 5, 2015 at 18:06 | comment | added | gwr | Are you really trying to fit "data to a model (polynomial)" - an intriguing endeavor? I do not quite understand, why you joining the {1} to the list of functions; it does not change a thing and by definition an intercept is automatically added (e.g. beta0)? | |
| Oct 5, 2015 at 17:45 | comment | added | Sos | ehm, now that I look at both questions at the bottom, I am wondering if the title is the most appropriate | |
| Oct 5, 2015 at 17:40 | history | asked | Sos | CC BY-SA 3.0 |