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In making scatterplot for correlations between two continuous variables, can we use the choice cubic instead of linear choice in "create a fit line at total", as shown in the copied Figure, please? does this sort of Figure acceptable scientifically to show negative correlation?

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    $\begingroup$ It seems that your SPSS plot shows a regression, not a correlation. As you know, (ordinary) regression assumes (virtually) no error in the predictor, X. In correlation, no such assumption is made, both variables (X1 and X2, if you like) are uncertain. The correlation line must hence minimise distance to observed values perpendicular to the line itself. Model 2 regressions (as a special case of "error-in-variable" models) can fit such models, but I have only ever seen linear ones. See also stats.stackexchange.com/questions/159261/… $\endgroup$ – Carsten Jan 26 at 19:52
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Yes and no, mostly no. Correlation asks for how strong the linear relationship is between two variables. Although a cubic fit to data may be more explanatory than a linear one, that does not mean that we can extract a better correlation from a cubic, not directly anyway. For example, for a cubic under the conditions of relative monotonicity the linear term (proportional to $x$) and cubic term (proportional to $x^3$) are more related to correlation in some transformed coordinate system, and the squared term (proportional to $x^2$) often less so. However, that argument would likely be rejected if it were, for example, included in an article for publication because such methodology has not been characterized, and more direct approaches for obtaining improved correlation are available.

More typically, to extract a better correlation from data one would either 1) do significance testing of a Spearman's rank correlation or related non-parametric test (like Kendall's $\tau$). Such tests, since they are rank tests, and since the data shown are approximately monotonic, would give a test of significance under the assumption of monotonicity. However, more appropriate would likely be to 2) transform the coordinate system so that the plot is more linear, which in the case of the data shown could be log-log transformation or reciprocal plotting. However, this requires trying a number of transforms to see which is best at rendering the plot more linear. Even after transformation of coordinate systems, Spearman's rank may be more appropriate than Pearson's correlation, such that both could be tried.

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    $\begingroup$ Could you please explain what you mean by a term in a regression being "related to correlation"? $\endgroup$ – whuber Jan 26 at 20:23
  • $\begingroup$ I would be happy to, however, I do not see exactly what you are asking. I will take a guess, if the assumption of relative monotonicity is exact, then odd terms of a poly have that property, and even powers do not. As I mentioned, that is not an approach that I, or in fact anyone I know, has ever suggested as being a method worth considering, but that is what the question asked, and any other response would not actually address the question, i.e., "Why can't I do this?" $\endgroup$ – Carl Jan 26 at 20:39
  • $\begingroup$ @whuber.... for example, if the linear term cannot be discarded, and is negative, but the cubic term is positive and cannot be discarded, then any statement regarding correlation is discounted. To put it another way, "We do not do that because it is at best poorly characterized if not totally incorrect." $\endgroup$ – Carl Jan 26 at 20:46
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    $\begingroup$ Because your replies make no sense to me, let me try again. I understand "correlation" to refer to the Pearson correlation coefficient. You appear to make statements about some kind of relationship between including or removing variables from a multiple regression, the regression results, and correlation. What do they mean? $\endgroup$ – whuber Jan 26 at 20:49
  • $\begingroup$ @whuber I did not even consider adding or removing terms from a poly regression, that is yet another reason to NOT use a poly to explore correlation. I added some text to try to help. The answer is because it hasn't been characterized so no one knows how to do that at present. Moreover, we have better methods; so why bother? $\endgroup$ – Carl Jan 26 at 21:01

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