Question for the statisticians

[Noser222]

Does anyone ever use a log-log plot , (ie. log of integrated area vs. log of concentration) when the standard curve covers several orders of magnitudes? I did for a sample where I used concentrations of 10,000, 1,000, 100 and 10 ppb and got linearity that was as good as if I simply plotted int. area vs. conc.

Obviously the residuals would be much greater as a percentage on the lower end standards than the higher oneswhen you get to the very low conc. samples, but on the log-log the points are evenly spread and residuals would be treated about equally.

So anyway, I was just looking for thoughts on whether this is as good, better, or worse, than a weighted regression.

[Uwe Neue]

I am not a statistician, so somebody who knows better may correct me.

However, I have frequently used log-log plots for calibration curves in LC that stretch over several orders of magnitude. In this case you can use without prejudice the standard curvefitting procedures.

Note that the standard assumption for a linear regression without weighted correlation is that the errors are the same throughout the range of measurement, independent of the value of the Y-parameter. In standard linear regression in chromatography, this is never the case, since the errors are proportional to the injection volume, but not to the amount injected at constant volume. Therefore you will always be better off doing a log-log correlation, since now your errors are constant over the entire range.

With other words, your procedure is better than the "standard" procedure.

One thing to watch out for though is that in your case, the slope is expected to be 1.000000000. Significant deviations from a slope of 1 indicate complications with your detection scheme.

[Noser222]

Thanks Uwe. In my case, for the standards I mentioned, I was getting a slope of 0.98. What time of problems with the detection scheme might this indicate?

[Uwe Neue]

Your detector is not completely linear. If you are using a MS system, this might be normal. For a UV detector, this should not be the case.

[Noser222]

Yes it is MS, and I know that over several orders of magnitude you often have to use a quadratic fit.