I have a set of around 160 samples from different sediment bodies, or stratigraphic units, which I am attempting to correlate using geochemical analysis of 25 elements for each sample, and a similarity coefficient often applied in correlating volcanic ashes. The literature gives no mention of assumptions of normality for the coefficient. It is assumed that each sample is representative of the variability in that stratigraphic unit, as representative sections were sampled while in the field.

To do these analyses, the sample is split to a smaller fraction that is loaded in to a plastic capsule, which contains the sample for the XRF irradiation. Some workers in my department who have done XRF on solid objects such as larger rock fragments, insist that 25-30 sub-samples must be analyzed per sample, So in other words, for each of my 160 samples, I would prepare 25-30 splits of the sample for XRF analysis. These ideas have propagated through the department. They are however using MDFA with its assumption of multivariate normality. Also, I am measuring sands, silts and clays, plus smaller granule gravels, of which hundreds if not thousands are visible to the XRF beam while in the capsules.

Being that there does not seem to be any assumption of normality with the Correlation Coefficient being used, is there any basis for this idea that I would need to prepare 25-30 splits of each sample in my case?

I apologize if this has been explained poorly.


I have had a chance to find some more sources on the subject, and the similarity coefficient of Borchardt and others (1972) that I am using is not a statistically rigorous method, thus no assumptions of normality. Its results however have been compared in a few studies favorably with the results of rigorous methods.

Borchardt, G.A., P.J. Aruscavage and H.T. Millard, Jr. 1972. Correlation of the Bishop Ash, A Pleistocene Marker Bed, Using Instrumental Neutron Activation Analysis. Journal of Sedimentary Petrology 42(2), 301-306.

| cite | improve this answer | |

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.