4
$\begingroup$

I have three Pearson correlation coefficients (.8978, .5676 and .7865) for three age groups (i.e. 21 to 30 years, 31 to 40 years and 41 to 50 years) whose behavior I am studying in regard to their shopping habits versus weight gain.

Can I say that .8978 is the strongest relationship between shopping habits and weight gain?

Based on the difference in the coefficients, can I say that there is a difference in the shopping habits and weight gain of the three age groups?

Finally, can I just add the the three coefficients and divide by three to come up with an average?

All these are 'face value' interpretations. Are they acceptable or do I need to perform some sort of statistical analysis? If it is the latter (heaven forbid), can SPSS do it?

$\endgroup$
  • $\begingroup$ Just to make sure, the correlation coefficients are between shopping habits and weight gain for three subpopulations? $\endgroup$ – mpiktas Aug 31 '11 at 13:07
  • $\begingroup$ Yes, the correlation coefficients are between shopping habits and weight gain as follows: 21 to 30 years (.8978, sample size 105), 31 to 40 years (.5678, sample size 95), and 41 to 50 years (.7865, sample size 120). $\endgroup$ – Adhesh Josh Aug 31 '11 at 20:15
  • 1
    $\begingroup$ You need to read "Averaging correlation values". A discussion there (since deleted) indicated some people do use the Fisher transformation for averaging while others raise doubts about its meaningfulness. One thing is clear: any mathematical combination of your three correlation coefficients does not necessarily have any predictable relationship to the correlation between shopping and weight for all ages 21-50 years. $\endgroup$ – whuber Sep 2 '11 at 13:44
  • $\begingroup$ The average correlation may have no relationship to the correlation in the entire dataset. See: Simpson's Paradox $\endgroup$ – Lizzie Silver Nov 6 '17 at 3:52
11
$\begingroup$

Can I say .8978 is the strongest relationship between shopping habits and weight gain?

Descriptively, you can say that it is the strongest relationship. Whether it is significantly stronger than the other two depends on your sample size. There's an online calculator for that.

Based on the diffference in the coefficients, can I say that there is a difference in the shopping habits and weight gain of the three age groups?

That's the same statistical question as above. Test each pair of correlations for the significance of the difference. As you perform three tests, you might want to think about a correction of the $\alpha$ level. Another possibility elaborated here would be to add age group as a dummy coded variable into a regression analysis.

Finally, can I just add the the three coefficients and divide by three to come up with an average?

No. To get an average correlation you have to do an $r$-to-$Z$ transformation (Fisher's $Z$), average these transformed values, and backtransform the average $Z$ to an $r$ again. For the transformation, there are several online calculators.

$\endgroup$
  • $\begingroup$ Welcome to our site, Felix! $\endgroup$ – whuber Sep 1 '11 at 14:21
  • $\begingroup$ In what sense is the method using Fisher's transformation a legitimate average of correlation coefficients? What would this average mean? Note that with this approach, if one of the correlations were 1.0 and all the others were greater than -1.0, this "average" would equal 1.0. $\endgroup$ – whuber Sep 2 '11 at 13:39
  • 1
    $\begingroup$ But only for the boundary case of a correlation of 1.0, which leads to an Z of Ìnf`. In the empirically more probable case of correlations < |1|, the backtransformed average of the Zs is less biased than the average of the original rs (see e.g. here or here). $\endgroup$ – Felix S Sep 3 '11 at 16:12
6
$\begingroup$

Averaging correlation coefficients is a meaningless operation. Correlation is $$\rho = \frac{\mbox{Cov}[X,Y]}{\sqrt{\mbox{Var}[X]\mbox{Var}[Y]}}.$$ You cannot even average the components of it (the covariance and two variances), unless the means of all groups on both variables are the same. If they are not, your population variance/covariance will be larger than/different from the (weighted) sum of variances/covariances due to between-group differences.

$\endgroup$
  • 4
    $\begingroup$ Actually, averaging correlations is done in many statistical contexts, e.g. for the calculation of Cronbach's alpha. I cannot really understand your argument ... Of course common sense is needed for this averaging. If you average coefficients from different variables or from very different samples, the resulting mean may not make any sense at all. But, IMO, this rather a question of content and not of statistical procedures. $\endgroup$ – Felix S Sep 2 '11 at 7:56
  • 1
    $\begingroup$ @Felix The concern may be about the potential for misinterpretation. "The mean correlation was 0.75" tells us something about the age groups separately, but it tells us almost nothing about the overall correlation for all age groups, which in principle could be almost as small as -1 and almost as great as +1 (and still give an average of 0.75). $\endgroup$ – whuber Sep 2 '11 at 13:50
  • 1
    $\begingroup$ @StasK: I totally agree with you that Cronbach's alpha is another scenario than the one posted in the original question. My example was rather meant as a reply to "Averaging correlation coefficients is a meaningless operation". Doing this averaging definitely has many pitfalls and misuses; but AFAIK there are scenarios, where it is not a meaningless operation. Otherwise, a lot of statistical textbooks and a lot of psychological research would be meaningless (... but maybe it is ...). $\endgroup$ – Felix S Sep 3 '11 at 15:52
  • 3
    $\begingroup$ @whuber: That's true. Especially in the case of hierarchical data sets, the correlation within subgroups could go into the opposite direction compared to the correlation of all data points (which leads to an ecological fallacy). Computing a mean in general can be meaningless in many situations (e.g., with bimodal distributions), and calculating a mean of aggregate measures can be even more problematic. I only want to argue that if one wants to calculate the mean, Fisher's Z is way to do (see e.g. here). $\endgroup$ – Felix S Sep 3 '11 at 16:06
  • 1
    $\begingroup$ @StasK You need a square root in the denominator of the correlation coefficient. $\endgroup$ – Michael R. Chernick Jan 23 '17 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.