The limit of detection ($LOD$) for an analytical method can be derived from its calibration curve using the following equation:
$$ LOD = {\dfrac{(3.3 \cdot {\sigma_ \mathrm{resids}})} {m}} \tag{1} $$ [see reference at bottom]
where:
- $ \sigma_\mathrm{resids} $ is the standard deviation of the residuals of the regression fit
- $m$ is the slope of the regression curve.
Would the above equation still apply if I were to apply it to a calibration curve generated through linear regression of $\ln(y)$ on $\ln(x)$?
Here's an example data set that I generated (artifically - hence apparent rounding errors) to reflect what would be seen in some electrospray ionisation-mass spectrometry data (non-linear and heteroscedastic):
$$\begin{array}{|c|c|c|c|} \hline \text{Calibrant} & x \ (\pu{mg/L}) & y \text{ (peak area, AU)} & \ln(x) & \ln(y) \\ \hline \text{S1} & 1 & 550 & 0 & 6.309918 \\ \hline \text{S2} & 1 & 500 & 0 & 6.214608 \\ \hline \text{S3} & 1 & 450 & 0 & 6.109248 \\ \hline \text{S4} & 2 & 1100 & 0.693147 & 7.003065 \\ \hline \text{S5} & 2 & 1200 & 0.693147 & 7.090077 \\ \hline \text{S6} & 2 & 1300 & 0.693147 & 7.17012 \\ \hline \text{S7} & 10 & 6500 & 2.302585 & 8.779557 \\ \hline \text{S8} & 10 & 7000 & 2.302585 & 8.853665 \\ \hline \text{S9} & 10 & 7500 & 2.302585 & 8.922658 \\ \hline \text{S10} & 50 & 40000 & 3.912023 & 10.59663 \\ \hline \text{S11} & 50 & 45000 & 3.912023 & 10.71442 \\ \hline \text{S12} & 50 & 50000 & 3.912023 & 10.81978 \\ \hline \text{S13} & 125 & 140000 & 4.828314 & 11.8494 \\ \hline \text{S14} & 125 & 150000 & 4.828314 & 11.91839 \\ \hline \text{S15} & 125 & 160000 & 4.828314 & 11.98293 \\ \hline \text{S16} & 250 & 400000 & 5.521461 & 12.89922 \\ \hline \text{S17} & 250 & 420000 & 5.521461 & 12.94801 \\ \hline \text{S18} & 250 & 440000 & 5.521461 & 12.99453 \\ \hline \text{S19} & 500 & 1000000 & 6.214608 & 13.81551 \\ \hline \text{S20} & 500 & 1200000 & 6.214608 & 13.99783 \\ \hline \text{S21} & 500 & 1400000 & 6.214608 & 14.15198 \\ \hline \end{array}$$
Applying equation (1) and exponentiating the result, the $LOD$ would come out as:
$$ LOD = \exp {\left(\dfrac{3.3 \times 0.177}{1.2265} \right) } = \pu{1.61 mg/L} $$
The strange thing for me is that if I multiply the original concentrations ($x$) by 1000, say to convert to from $\pu{mg/L}$ to $\pu{\mu g/L}$,and then generate the calibration curve of $\ln(y)$ on $\ln(x)$, the $LOD$ comes out at exactly the same value, i.e. 1.61. This is because the slope and the standard deviation of the regression fit are unchanged by multiplying the concentration by 1000 - only the intercept value changes. Note, given that concentration axis was in $\pu{\mu g/L}$ before $\ln$-transformation, this would presumably mean an $LOD$ of $\pu{1.61 \mu g/L}$?
My suspicion is that I've missed a crucial step in calculating $LOD$ from the $\ln$-transformed calibration curve model.
Thanks for any pointers you can provide!