Horwitz function up to 100 % ?
The results of the Horwitz function (H, 2H, HORRAT) cannot be used for main components of sugar factory products, such as Pol of raw sugar, Pol of beet or Pol of molasses. The following table demonstrates this for some more concentrations of main components, using % Pol (°Z) as an example:
Analyte concentration
Conventional Fractional Log C H(%) 2H(%) SR(% Pol) R (% Pol)
100 % Pol C = 10244.0011.2
50 % Pol C = 0.5-0.302.24.42.226.2
30 % Pol C = 0.3-0.522.44.81.444.0
17 % Pol C = 0.17-0.772.65.20.892.5


The last line of the table, with Pol-sugar (°Z) of beet, shall be discussed in detail:  From H = 2.6 % and from 2H = 5.2 % (limit of "Relative Reproducibility Standard deviation RSDR") nobody will get satisfying results. The limit of "Reproducibility Standard deviation SR" equals 17 * 5.2 / 100 = 0.89 % Pol-sugar and the "Reproducibility R" equals 17 * 5.2 * 2.8 / 100 = 2.5 % Pol-sugar.

If differences between a beet laboratory and a supervision laboratory with a difference of 2.4 % Pol-sugar have to be regarded as equal, nobody will be satisfied. Horwitz limits are not strict enough in case of main components of sugar factory trade products. The results of collaborative tests must be judged by practical experience and not by "RSDR < 2H" or "HORRAT < 2". What about "Expected = Accepted" for main components?
2004-08-14       G. Pollach
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