I'm completely confused about how to calculate R-squared for given lists of predicted and actual values.

As an example, assume that my predicted values are: [3, 8, 10, 17, 24, 27] and my actual values are [2, 8, 10, 13, 18, 20].

According to wikipedia, I do the following:

  1. get the mean of the actual values (y-bar) = 14.8333
  2. compute the residual sum of squares (RSS). For each pair of values, I'm getting the difference, squaring it, and summing the results. i.e. (3-2)^2 + (8-8)^2 + (10-10)^2 and so on. For my data this is 102.
  3. Compute the total sum of squares (TSS). For each actual value, subtract it from the mean of the actual values, square the result, and sum all of these. i.e. (2-14.833)^2 + (8-14.833)^2 and so on. So TSS = 220.83333.
  4. R^2 = 1 - RSS/TSS = .53811

Contrast this method with one described here, which says I also need to be using the average of the predicted values, as well as what Excel gives using the RSQ formula (.9729).

Am I doing something wrong above? Which is the correct formula/method to use?

  • 1
    $\begingroup$ That applet and Excel are fitting a least squares line to these data, which is totally different: they aren't treating the first set as "predicted" values--they treat them as predictors. Because these are lousy predicted values, the least squares fit is a huge improvement, as reflected in the much larger value of $R^2$ it yields. $\endgroup$
    – whuber
    Nov 30, 2022 at 23:22

1 Answer 1


There are multiple ways of calculating $R^2$ that are equivalent for ordinary least squares linear regression but are not equal in other circumstances.

I like $R^2=1-\frac{ SSRes }{ SSTotal }$, for reasons I discuss here. (Fair warning: this is not an introductory topic.)

However, some people like $R^2=(corr(y,\hat y))^2$. That appears to be the formula used in Excel.

If you fit by ordinary least squares, these two equations will be equal. If you do not fit by ordinary least squares, the two need not be equal.

The disagreement comes from the fact that your predictions are not great predictions of the true values, hence the low value of $R^2$ calculated the way I prefer. However, there is a solid linear relationship between the two, hence the high correlation.

As for what equation you should use to calculate your quantity of interest, that depends on what you find interesting! Both of these equations can give useful information, depending on what you want to know.

  • $\begingroup$ I feel like I'm in over my head already, but are you saying that how the predicted values were obtained matters? Can I not just take any two lists of number and compute the R-squared? $\endgroup$
    – Jer
    Nov 30, 2022 at 23:28
  • 1
    $\begingroup$ Also, you lost me at "I like..." vs. "some other people like...". Isn't there just a definition of what this is, with no opinions involved? I mean, we don't have different opinions of how to calculate the sum of a list of numbers, right? $\endgroup$
    – Jer
    Nov 30, 2022 at 23:31
  • 3
    $\begingroup$ The trouble is that there are multiple calculations that yield the same value in a simple setting, and each can be claimed to be the true $R^2$. When we then extend those ideas to more complicated work, that’s where the disagreements start. I do not see anything analogous for sums. $\endgroup$
    – Dave
    Nov 30, 2022 at 23:44
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    $\begingroup$ To add to Dave's last comment, $R^2$ has a standard meaning within the framework of ordinary least squares regression. You are working outside that framework. There appears to be one useful generalization that would work--it corresponds to your manual calculation--but by trying to apply software that assumes you are doing OLS, you are not getting the calculations you need. Indeed, even sums suffer from a similar problem: they can be defined for far more than real numbers, as in modular arithmetic; but if you ask Excel to perform sums in a finite field, it will get the wrong results. $\endgroup$
    – whuber
    Dec 1, 2022 at 0:33
  • 1
    $\begingroup$ @whuber - got it, thanks - I can appreciate that analogy (as I'm a little more well-versed in algebra than stats). Addition means one thing over the integers (4 + 4 = 8), but a different thing over the integers modulo 5 (4 + 4 = 3). This is kind of like that I guess. Thanks to both you and Dave for humoring me a bit. $\endgroup$
    – Jer
    Dec 1, 2022 at 0:38

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