Is my LOQ correct ?

[aceto_81]

Ok guys

i have done it but why most of my result after the log appeared negative

Do you know someone in the litreature who did it , come on help me more


( Thanks god that i am not working in the industry otherwise i will be fired )

Why don't you show us your results, give us something to work with, maybe give your calculations, so we can see where you are doing wrong?

Ace

[tom jupille]

i have done it but why most of my result after the log appeared negative

If you express your CV as a decimal fraction, the log *will* be negative. For example, log(0.01) = -1. If you express it as a percentage, it will be positive. It doesn't matter, because you will convert back to CV in any case.

Using fictitious data, a CV vs. A plot looks something like this:



If you need to quantitate with a CV <= 3%, then the LOQ is 0.1 ppm (that's where the plot crosses 3% on the vertical axis). If you can tolerate a CV <= 5%, then the LOQ is 0.04 ppm.

In this case, I didn't bother calculating the logs, because Excel lets you use a logarithmic axis.

There is a brief description of using this approach to set LOQ in Snyder, Glach, & Kirklands "Practical HPLC Method Development (2nd ed)" book on pages 695 & 696. Here's a link to the book on Amazon.

[Ronaldo]

Tom ,

You are unbelievable great human being ( Knowledge + patient + generous)

[tom jupille]

Awwww.

[joelyue]

Hi,

lets say i have this set of data. How do i calculate CV and A?
FYI, i am using standard addition for calibration.

Peak Area Concentration (ppb)
0.014 1.4176
0.013 1.2549
0.013 1.2398
0.066 6.4636
0.064 6.2981
0.063 6.2256
0.113 11.0976
0.118 11.5792
0.115 11.329

[tom jupille]

From those data, you can't calculate CVs

"A" is the amount of analyte, so your A values would be 1.4176, 1.2549, . . . etc.

"CV" is the coefficient of variation (otherwise known as the % relative standard deviation). It is the standard deviation divided by the mean, expressed as a percentage. So, if your mean peak area at a particular concentration were 1000, and the standard deviation were 50, then CV = 50/1000 = 0.05 = 5%.

You need a *minimum* of 3 replicates at each sample concentration in order to calculate standard deviations (5 would be better).

[aceto_81]

From those data, you can't calculate CVs

"A" is the amount of analyte, so your A values would be 1.4176, 1.2549, . . . etc.

"CV" is the coefficient of variation (otherwise known as the % relative standard deviation). It is the standard deviation divided by the mean, expressed as a percentage. So, if your mean peak area at a particular concentration were 1000, and the standard deviation were 50, then CV = 50/1000 = 0.05 = 5%.

You need a *minimum* of 3 replicates at each sample concentration in order to calculate standard deviations (5 would be better).

What one can do is divide each area by its concentration, so that you get an normalised area. Then calculate CV on this normalised area. (about 1.64 for your data)

Ace

[tom jupille]

What one can do is divide each area by its concentration, so that you get an normalised area. Then calculate CV on this normalised area. (about 1.64 for your data)

Which would not help with a CV vs A plot. What you need to determine is the CV at each analyte level.

[aceto_81]

What one can do is divide each area by its concentration, so that you get an normalised area. Then calculate CV on this normalised area. (about 1.64 for your data)

Which would not help with a CV vs A plot. What you need to determine is the CV at each analyte level.

Yep, Tom, you're absolutely right, but he asked for a CV, not for a CV vs A.
I thought he was hijacking this thread by asking another question.

For LOD/LOQ determination you really need more points, IF you want to use the approach of a CV vs A plot.
But on the other hand, if you use the approach of standard error /slope, you get an LOD of 0.09.

Ace

[tom jupille]

There is no question that the CV vs. A plot requires a *lot* more work (which may be why it's not used very much! ). It *does* provide a major advantage in attaching some context to the value and making very clear that there is no such thing as "*the* LOQ".

By the way, there is an excellent discussion of the related topic of detection limits in the Coleman & Vanetta series of articles on Statistics (sections 28 -30; Nov, 2007; February, 2008; & June, 2008 issues of American Laboratory).

[rindtr]

hi all,
I have read all of your comments on how to calculate LOQ and i have been impressed a lot. but there are some points that should be enlightened for me to understand what is being done correctly. first of all, I'd like to get why you are not using S/N ratio. it is simple especially if LOQ is calculated for analyst to make sure that he/she is giving result within a safe range. so there is not an exact value for LOQ. the second way to calculate it, of course it is more precise, drawing a calibration curve and decreasing the concentration as much as you get meaningful peaks. at the lowest value of your calibration curve make multiple injections, i prefer 10, calculate the standart deviation of those 10 injections and add the lowest concentration to 3 times of that calculated standart deviation. it is LOD 3 times of LOD is the LOQ.

[tom jupille]

LOQ can be estimated in a number of ways. As I have indicated, I like the CV vs A plot because it relates the LOQ to the desired level of precision in the analysis (both the S/N ratio and the "standard error of the intercept" method implicitly assume about 5% RSD). On the downside, it *is* more work, and it is not truly algorithmic, because it depends on visual interpretation of a graph.

The S/N method has a number of pitfalls: how is noise measured? over how long a period of time? where? from where is peak height measured (top of the baseline noise? middle?). Is noise taken into account in identifying the top of the peak? To give you an idea of the complications, read this long thread: viewtopic.php?t=11618&highlight=

Statistical methods avoid the S/N pitfalls, but, as mentioned above, ignore the fact that the there is no such thing as "the" LOQ; the value will depend on the desired level of repeatability.

In any case, all of the above are only *estimates*, and must be confirmed by running replicates at the estimated LOQ to document that the quantitation *does* in fact provide the requisite reproducibility.

[rindtr]

with my all respect Tom, it is very professional way to calculate LOQ by using CV vs "A" graph. I made my mind to do so in my works from now on. but would you please give a reference (literature) in which this method of calculation is described clearly? I used to calculate LOQ by injecting the lowest concentration of calibration curve as much as i can then making statistics as i explained before, it is true that in this method the actual LOQ can never be reached but the analyst make sure not to report values under the LOQ. I would really appreciate if you could give some details

Rind

[tom jupille]

There may be other references to the technique, but the one I'm familiar with is the brief description in Snyder, Glach, & Kirkland's "Practical HPLC Method Development (2nd ed)" book on pages 695 & 696. Here's a link to the description of the book on Amazon: http://tinyurl.com/2e63ksn

[lmh]

See also Epshtein, Pharmaceutical Chemistry Journal 38:212-225.

Just to back up what everyone has said about sufficient replicate points: in most circumstances we plot a mean +/- s.d., so all we want is a reliable mean. This time you are plotting the error, and the error-on-the-error is even more wobbly than the error itself. Therefore you need even more replicate points so the error-on-the-error isn't so huge that the points you plot become too wobbly for sensible interpolation.