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I am trying to grasp the concept of magnitude-of-effect estimations, as opposed to standard null hypothesis testing procedures.

According to a paper by (Snyder & Lawson, 1993) the magnitude-of-effect indices can be grouped into effect size (such as Cohen's d) and measures of association strength. I wish to calculate the Pearson r strength of association, which would be part of the second group in this classification.

Now my question is, is the correlation coefficient obtained by taking the square root of the correlation coefficient $r^2$ provided by standard least-squares linear regression software output (such that generated by a basic Excel linear regression) the same $r$ as the Pearson's $r$ associated with magnitude-of-effect calculations?

I am confused, because the standard output of a normal regression analysis delivers the Pearson product-moment correlation coefficient, if I am not mistaken. However, I wish to calculate the Pearson's $r$ strength of association, as described in the linked paper.

Reference
Snyder & Lawson, J Exp Edu (1993); 61(4): 334-49

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One line summary: Pearson correlation $r$ is a measure of strength of (linear) association and indeed numerically and necessarily equal to the square root of $r^2$ as reported with regressions for one response and one predictor variable.

$r^2$ as reported for a regression of one $y$ and one $x$ is indeed the square of the Pearson correlation coefficient $r$. That is the reason for the notation.

However, the usual name for $r^2$ is the coefficient of determination. $r^2$ is definitely not the correlation coefficient. But coefficient of determination is a wordy name, and it cannot be claimed to be transparent either. I couldn't give you a really convincing story on why we should be talking about "determination" in this context. The idea is presumably that some fraction of the variation, precisely $r^2$ of it, is "determined" by the relationship and the rest, the remaining fraction 1 $- r^2$, is not so determined. But to my taste that terminology gets us too close for comfort to arguments about determinism, cause and effect, and so forth, although some researchers in some moods are keen to discuss all that.

So statistical people usually get into the habit of talking about "r-square" or "r-squared" (in my book, either is fine). In reports, it is better to write $r^2$ unless tribal habits or instructions from elders indicate otherwise.

Let's visit an example. Here are regression and correlation results (which happen to be from Stata, but any good statistics program will do something similar) for gallons per 100 miles predicted from car weight in pounds:

. regress gp100m weight

      Source |       SS       df       MS              Number of obs =      74
-------------+------------------------------           F(  1,    72) =  194.71
       Model |  87.2964969     1  87.2964969           Prob > F      =  0.0000
    Residual |  32.2797639    72  .448330054           R-squared     =  0.7300
-------------+------------------------------           Adj R-squared =  0.7263
       Total |  119.576261    73  1.63803097           Root MSE      =  .66957

------------------------------------------------------------------------------
      gp100m |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
      weight |    .001407   .0001008    13.95   0.000      .001206    .0016081
       _cons |   .7707669   .3142571     2.45   0.017     .1443069    1.397227
------------------------------------------------------------------------------

. corr gp100m weight
(obs=74)

             |   gp100m   weight
-------------+------------------
      gp100m |   1.0000
      weight |   0.8544   1.0000


. display sqrt(0.7300)
.85440037

So, $r^2$ is reported as 0.7300 which is the square of the correlation coefficient 0.8544, as an independent calculator-like display does show.

Side notes:

  1. A little confusingly, $R^2$ is also common notation in reporting regression results, while $R$ is only rarely used for a correlation between two variables. In fact the Stata output I just showed did precisely this. The history and logic to this appear to be that regression can use several predictors or $x$ variables (historically called multiple regression) and in that more general context $r^2$ is less common notation than $R^2$. There are very small issues here; watching what your teacher uses and being consistent about using one symbol or the other is the best first advice.

  2. Exceptionally if $r=0$ or $r=1$ then numerically the values of $r$ and $r^2$ will be identical, but the concepts remain distinct.

  3. 4 decimal places for $r^2$ or $r$ as above is overkill practically, and rounding a bit is usually a good idea. The programmer's implication and inclination is to give more decimal places than you need and to let you choose what rounding you want: indeed, good programs let you specify explicitly how much rounding you want.

  4. You may often see $r^2$ (or $R^2$) reported in percent terms: in this example, any of 85%, 85.4%, 85.44% could be a researcher's choices. The idea here behind using percent presentation is that $r^2$ is a fraction. However, the correlation $r$ itself is not to be thought of a fraction of anything, not least because it can be negative too.

  5. $R$ as the positive square root of $R^2$ from (multiple) regression is also quite often seen and has a simple, concrete and useful interpretation as the correlation between observed response $y$ and predicted response $\hat y$.

  6. Despite widespread use of these conventions in notation, other uses of $r$ and $R$ are certainly possible within statistical science, so watching out for exact definitions remains imperative.

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  • $\begingroup$ Many many thanks. One question on r versus R - I understand they are different things entirely? I have to dig up the article I was reading, but it mentioned that r, R, eta squared and so on were different measures. Anyway, that's a side story I guess, my question is perfectly answered, thanks! $\endgroup$ – AliceD Jun 17 '15 at 10:55
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    $\begingroup$ Thanks for the positive comments. Please see edits, points 5 and 6. $\endgroup$ – Nick Cox Jun 17 '15 at 11:15
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    $\begingroup$ What a nice, thorough, and clear answer. And while in principle one could have answered this question with maybe a couple sentences, this is a nice example of how one can take a relatively straightforward (and also nicely/clearly formulated) question and make something even more educational and instructive out it by adding context, an example, noting potential other sources of confusion, clarifying notation, and so on. This couldn't be upvoted often enough. $\endgroup$ – Wolfgang Jun 17 '15 at 11:50

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