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Aug 14, 2021 at 12:48 history edited ttnphns CC BY-SA 4.0
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Sep 23, 2017 at 10:33 history edited ttnphns CC BY-SA 3.0
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Sep 23, 2017 at 10:15 comment added Silverfish I think this answer would be improved (potentially less confusing to future readers) if it showed the formula for $b$ as well as $\beta$
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Jan 6, 2017 at 16:42 history edited ttnphns CC BY-SA 3.0
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Jan 6, 2017 at 16:35 history edited ttnphns CC BY-SA 3.0
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Sep 25, 2015 at 9:32 history edited ttnphns CC BY-SA 3.0
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Apr 3, 2015 at 18:54 history edited ttnphns CC BY-SA 3.0
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Apr 3, 2015 at 18:48 history edited ttnphns CC BY-SA 3.0
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Apr 3, 2015 at 18:20 history edited ttnphns CC BY-SA 3.0
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Mar 31, 2015 at 17:36 history edited ttnphns CC BY-SA 3.0
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Mar 31, 2015 at 17:28 comment added ttnphns @amoeba, thank you very much for noticing that lapse!
Mar 31, 2015 at 17:27 history edited ttnphns CC BY-SA 3.0
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Mar 31, 2015 at 14:51 comment added amoeba Why is there a square root in the denominator of the formula for $\beta_{x_1}$? I tried to obtain this formula by applying the standard equation $\beta = (X^\top X)^{-1} X^\top y$ for the case when $X$ has two standardized columns, and I get the same formula but without a square root. Have I made a mistake, or have you made a mistake? :)
Nov 19, 2013 at 15:56 comment added ttnphns This may happen in suppression stats.stackexchange.com/q/73869/3277
Nov 19, 2013 at 15:49 comment added user34927 Individually the Rsquared of X1 and X2 with Y is <0.1 and in a multiple regression of X1, X2 and Y it is 0.4. I might be misinterpreting what this means so it is probably a good idea to let this go. Thanks for your help.
Nov 19, 2013 at 15:34 comment added ttnphns 1) significantly you shouldn't rely on p-value in such problems as yours; the very size of the correlation matters more. 2) You have to define your understanding of "combination" before you can post a question (when you do it, please post a new, another question. Let this old one be as is).
Nov 19, 2013 at 15:28 comment added user34927 All I know is that individually neither X1 nor X2 are significantly correlated with Y. But I know that Y is determined by the combination of X1 and X2. I was now looking for the correct way to statistically analyze that relationship. But I do not understand if multiple regression or partial correlation is appropriate for that purpose and what the difference between those is in that case.
Nov 19, 2013 at 15:02 comment added ttnphns Wait a bit, @user34927. to prove that the DV (Y) is significantly correlated with one of two IVs (X1) if the effect of the other IV (X2) is removed The effect removed from where? If you "remove" X2 from both Y and X1 then the corr. between Y and X1 is the partial correlation. If you "remove" X2 from X1 only then the corr. between Y and X1 is called the part (or semi-partial) correlation. Were you really asking about it?
Nov 19, 2013 at 14:16 comment added user34927 Okay, obviously I do not get it. Thanks anyway.
Nov 19, 2013 at 14:12 comment added ttnphns Once again - please be attentive to reread the answer. Multiple linear regression coefficient b (or its standardized form beta) associated with a predictor has the same meaning and the same p-value as the partial r between Y and the predictor. Now, if you got it, why are speaking of Rsquared of the regression? It has nothing to do with partial r of a specific predictor.
Nov 19, 2013 at 13:50 comment added user34927 So, just to make sure I understood you correctly. If I want to prove that the DV is significantly correlated with one of two IVs if the effect of the other IV is removed, I can use either multiple regression or partial correlation and both lead me to the same conclusion (i.e. DV is low for given IV2 when IV1 is low) and I can show that by giving the partial r OR the Rsquared of the multiple regression?
Nov 17, 2013 at 23:00 comment added ttnphns Fixing X1 ("x1 given") = removing (controlling) the effect of X1. There is no such thing as "combined effect" in multiple regression (unless you add the interaction X1*X2). Effects in multuple regression are competitive. Linear regression effects are actually partial correlations.
Nov 17, 2013 at 22:44 comment added user34927 I tried to clarify my question..
Nov 17, 2013 at 22:00 comment added ttnphns Obviously, you are free to choose: the numerators are the same, so they convey the same information. As for your (not fully clarified) question, it seems to be about topics "can regr. coef. be 0 when r isn't 0"; "can regr. coef. be not 0 when r is 0". There's a lot questions about that on the site. Just for example, you might read stats.stackexchange.com/q/14234/3277; stats.stackexchange.com/q/44279/3277.
Nov 17, 2013 at 21:38 comment added user34927 Thank you. But how do I decide which one to go with, e.g. for the purpose described in my question?
Nov 17, 2013 at 19:53 history edited ttnphns CC BY-SA 3.0
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Nov 17, 2013 at 19:36 history answered ttnphns CC BY-SA 3.0