Bioequivalence and Bioavailability Forum • Standard calibration curve

Ken Peh
★

Malaysia,
2014-01-13 18:20
(4150 d 09:20 ago)

Posting: # 12162
Views: 12,017

Standard calibration curve [Bioanalytics]

Dear Members,

In EMEA Guideline on bioanalytical method validation, under 4.1.4 Standard Calibration Curve, it is written that "The back calculated concentrations of the calibration standards should be within ±15% of the nominal value, except for the LLOQ for which it should be within ±20%"

My questions are as below:

Besides the accuracy values, is regression coefficient value important, eg. R square >0.99 ? Should both be considered as acceptance criteria? The accuracy and R square values.
If we have 8 calibrators, from back calculated concentration of the calibration standards, we found 2 calibration standards failed accuracy. One of calibration standard was LLOQ. To keep the LLOQ, instead of dropping the failed LLOQ, reject another calibration standard which passes the accuracy and reevaluate the calibration curve. As a result, 7 calibration standards pass. Is this the right approach ??
On what occasion weighting method is allowed ? Weighting method is not mentioned in EMEA guideline but in FDA new draft guideline.

Highly appreciate your kind comments.

Thank you.

Regards,
Ken

Ohlbe
★★★

France,
2014-01-13 22:50
(4150 d 04:50 ago)

@ Ken Peh
Posting: # 12164
Views: 10,690

Standard calibration curve

Post reply

Dear Ken,

❝ I. Besides the accuracy values, is regression coefficient value important, eg. R square >0.99 ? Should both be considered as acceptance criteria ? The accuracy and R square values.

There is nothing in the guidelines on R or R square. They are not good indications of linearity, and give little information on goodness of fit. Have a look at Anscombe's quartet.

❝ II. If we have 8 calibrators, from back calculated concentration of the calibration standards, we found 2 calibration standards failed accuracy. One of calibration standard was LLOQ. To keep the LLOQ, instead of dropping the failed LLOQ, reject another calibration standard which passes the accuracy and reevaluate the calibration curve. As a result, 7 calibration standards pass. Is this the right approach :confused: ??

Theoretically, no. You're not supposed to drop a calibration standard that passes. Though you may try and justify it if graphically a borderline standard is thought to be the reason for failure of other standards, it may lead to unpleasant discussions with authorities and inspectors.

❝ III. On what occasion weighting method is allowed ? Weighting method is not mentioned in EMEA guideline but in FDA new draft guideline.

With chromatographic methods weighting is common practice. You should be able to justify the weighting factor you are using, but I would not expect to have to justify weighting itself. In any case it is quite easy to show the interest during method validation: in each validation run you can calculate the calibration curve without weighting and with classical factors such as 1/X and 1/X², then check the back-calculated concentrations and residuals.

—
Regards
Ohlbe

Ken Peh
★

Malaysia,
2014-01-14 17:37
(4149 d 10:02 ago)

@ Ohlbe
Posting: # 12167
Views: 10,605

Use of weighting factor

Post reply

Dear Ohlbe and ElMaestro,

Thank you very much for the comments and inputs.

❝ In any case it is quite easy to show the interest during method validation: in each validation run you can calculate the calibration curve without weighting and with classical factors such as 1/X and 1/X², then check the back-calculated concentrations and residuals.

I have read the lengthy discussion on other threads on weighting factor in standard calibration curve. I have gone through the paper on Weighted least square linear regression (Almeida et al., 2002) and example plus calculation given by Helmut. I am confused due to my poor knowledge in stats. I copied the example (given by Helmut in year 2008) as below for my question.

Example:

model: y = 1.0000 + 2.0000 * x + error

 ┌────┬──────┬───────┬────────┬───────┬────────┬───────┬────────┐

 │ x  │  y   │  w=1  │  bias  │ w=1/x │  bias  │ w=1/x²│  bias  │

 ├────┼──────┼───────┼────────┼───────┼────────┼───────┼────────┤

 │  1 │  3.7 │  1.13 │ 13.0%  │  1.05 │  5.00% │  1.02 │  2.00% │

 │  2 │  5.6 │  2.07 │  3.50% │  2.01 │  0.50% │  1.99 │ -0.50% │

 │  4 │  9.4 │  3.95 │ -1.25% │  3.92 │ -2.00% │  3.93 │ -1.75% │

 │  8 │ 17   │  7.72 │ -3.50% │  7.74 │ -3.25% │  7.82 │ -2.25% │

 │ 16 │ 34   │ 16.1  │  0.63% │ 16.3  │  1.88% │ 16.5  │  3.13% │

 ├────┴──────┼───────┴────────┼───────┴────────┼───────┴────────┤

 │ intercept │ 1.4250 (+42.5%)│ 1.6109 (+61.1%)│ 1.7133 (+71.3%)│

 │     slope │ 2.0185 (+0.93%)│ 1.9876 (-0.62%)│ 1.9537 (-2.31%)│

 └───────────┴────────────────┴────────────────┴────────────────┘

I would like to know how the values of 1/x and 1/x² were calculated. The x is the input as given in the first column (1,2,4,8,16) :confused:

?? I could not get the data as presented in the table.

Kindly let me know the program in R-software that can be used to perform this calculation.

I am really sorry to bring up the old topic again. Also sorry to Helmut for using his example without permission.

Thank you.

Regards,
Ken

Edit: Table BBcoded. Please use the before posting. [Helmut]

Helmut
★★★

Vienna, Austria,
2014-01-14 18:32
(4149 d 09:07 ago)

@ Ken Peh
Posting: # 12168
Views: 10,664

Confusing table

Post reply

Hi Ken,

❝ […] I have gone through the paper on Weighted least square linear regression (Almeida et al., 2002)…

That’s an excellent paper, IMHO.

❝ … and example plus calculation given by Helmut.

❝

❝ I would like to know how the values of 1/x and 1/x² were calculated. The x is the input as given in the first column (1,2,4,8,16) :confused: ?? I could not get the data as presented in the table.

x is the independent variable (in bioanalytics concentration) and y the dependent (observed) one – in BA the response (analyte/IS-ratio, peak height/area, whatsoever). The weights are not given in the table, they are simply 1/x or 1/x². Sorry for the confusion caused. The table entries are the back-calculated x-values based on the intercept/slope of the respective models given below. All calculations were done in full precision, therefore you will see some slight differences.
One example: With a response of 3.7 back-calculated values are (3.7-1.4250)/2.0185=1.13 (unweighted), (3.7-1.6109)/1.9876=1.05 (weight=1/x), and (3.7-1.7133)/1.9537=1.02 (w=1/x²). Bias is given as % relative error or 100(estimated-true)/true, e.g., 100×(1.13-1)/1=+13.0%.

❝ Kindly let me know the program in R-software that can be used to perform this calculation.

I’m not in the office right now; I will post some tomorrow.

❝ […] Also sorry to Helmut for using his example without permission.

No problem. All my posts are open-source .

—
Dif-tor heh smusma 🖖🏼 Довге життя Україна!
Helmut Schütz

The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes

Ken Peh
★

Malaysia,
2014-01-15 05:06
(4148 d 22:34 ago)

@ Helmut
Posting: # 12169
Views: 10,524

Confusing table

Post reply

Dear Helmut,

Thank you very much for your explanation.

May I know how the values of 1/x and 1/x² are generated?

❝ x is the independent variable (in bioanalytics concentration) and y the dependent (observed) one – in BA the response (analyte/IS-ratio, peak height/area, whatsoever). The weights are not given in the table, they are simply 1/x or 1/².

Please kindly show how the weights are generated.
We use the weight to multiply the independent values. Then, plot standard curve. From the curve, we can estimate the slope and intercept. We can then compute the back calculated value. The flow is right :confused:

??

Appreciate your input.

Thank you.

Regards,
Ken

Helmut
★★★

Vienna, Austria,
2014-01-15 14:46
(4148 d 12:54 ago)

@ Ken Peh
Posting: # 12174
Views: 10,605

Weights

Post reply

Hi Ken,

❝ May I know how the values of 1/x and 1/x² are generated?

Hey, that’s a formula. If x=4, 1/x=0.25 and 1/x²=0.0625. :-D

❝ We use the weight to multiply the independent values. Then, plot standard curve. From the curve, we can estimate the slope and intercept. We can then compute the back calculated value. The flow is right?

No, this does not work at all! For some R-code see this post. I’m busy right now, but will post code for Almeida’s paper later.

—
Dif-tor heh smusma 🖖🏼 Довге життя Україна!
Helmut Schütz

The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes

Ken Peh
★

Malaysia,
2014-01-18 06:46
(4145 d 20:54 ago)

@ Helmut
Posting: # 12192
Views: 10,429

Weights

Post reply

Dear Helmut,

❝ Hey, that’s a formula. If x=4, 1/x=0.25 and 1/x²=0.0625.

Thank you. I understand now.

❝ No, this does not work at all! For some R-code see this post. I’m busy right now, but will post code for Almeida’s paper later.

Sorry for posting at the wrong timing. Yes, we have referred to the post. Tried R code and got the answer. Thought need to use special program in R. Should have tried out first before posting.

Thanks again for your guidance.

Regards,
Ken

ElMaestro
★★★

Denmark,
2014-01-14 11:14
(4149 d 16:25 ago)

@ Ken Peh
Posting: # 12165
Views: 10,726

Standard calibration curve

Post reply

Hi Ken Peh,

I agree with Ohlbe, and I have some add'l comments.

❝ I. Besides the accuracy values, is regression coefficient value important, eg. R square >0.99 ? Should both be considered as acceptance criteria ? The accuracy and R square values.

When we talk R squared then we usually interpret it as the quantification of explanatory variation of the model. If we have an R sq. of 0.99 then it means our model explains 99% of the scatter in the data (at least in comparison to a non-model - a straight line through the mean).
Now, if scatter per se is a problem then we have precision to deal with that so we likely do not need R sq. also*. But I'd say, I see no issue in using R sq. internally at your CRO as a quality criterion (it will help you pass precision), just like column age, grid uptime etc.

Dropping a passing calibrator to make another one pass might not be a good idea. In a sense someone might say you should care for scientific quality rather than for the LLOQ itself.

Bob is 175 cm tall, Jim is 180 cm tall and Matt is 185 cm tall. What is their average height**?

Best regards,
EM.

* Until someone at a Crystal City meeting comes up with the idea to also require a R sq. of 0.99 or better for whatever reason, then suddenly everyone wants it.
** 177.5 cm, because we do not consider Matt. :-D

Roberto
☆

Italy,
2014-02-18 14:27
(4114 d 13:13 ago)

@ ElMaestro
Posting: # 12438
Views: 10,282

Standard calibration curve

Post reply

Hi at all Members,

Anscombe's quartet, is a very interesting example, but i think that it could be true for value of R-square that are outside the normal range accepted in bioanalytics. If we set the limit of R-square at > 0.98, it's very difficult, or impossible, to meet the situation described by Anscombe; for this limit, the line is well defined with a model that fit very well the data. So, with limit like this, i think that the R-square value could be an helpful parameter, obviously not alone but with the Others.

Best Regards
Roberto

Ohlbe
★★★

France,
2014-02-18 15:53
(4114 d 11:46 ago)

@ Roberto
Posting: # 12439
Views: 10,309

Standard calibration curve

Post reply

Dear Roberto,

I have here very nice examples of linear calibrations with 1/x² weighting, with R² > 0.99, where visually I would say that a quadratic fit would be much more appropriate than linear. The fit is not too bad at the HQC level, but I'm not too sure what the accuracy would look like at the ULOQ.

I agree that R² can be of use, combined with other criteria. Just don't expect more information from it than it can give.

—
Regards
Ohlbe

Roberto
☆

Italy,
2014-02-18 18:00
(4114 d 09:39 ago)

@ Ohlbe
Posting: # 12440
Views: 10,279

Standard calibration curve

Post reply

❝ I have here very nice examples of linear calibrations with 1/x² weighting, with R² > 0.99, where visually I would say that a quadratic fit would be much more appropriate than linear. […]

Tanks Ohlbe for your replay,
It would be interesting to have one of your example, in order to better understand.
Did you try to calculate the quadratic fit that you "felt" better represented with the data and which R-square did you obtained?

Thanks a lot
Roberto

Edit: Full quote removed. Please delete everything from the text of the original poster which is not necessary in understanding your answer; see also this post! [Helmut]

AngusMcLean
★★

USA,
2014-03-01 02:27
(4104 d 01:12 ago)

@ Ken Peh
Posting: # 12536
Views: 10,154

Standard calibration curve

Post reply

Ken: you have got a lot of excellent material here. I think you need an SOP, which defines your lab policy. I think you can write one now from your experience and the material given in this forum. In the US the correlation coefficient is not really relevant to the analytical criteria. The SOP I used to use allowed the analyst to drop one aberrant back calculated standard, but criteria were stipulated for that event. The policy with weighted regression was to evaluate the options and if 1/x and 1/x squared are evaluated and give a similar result go with the1/x weighting. Of course if a quadratic option is available then it could be evaluated and preferred.

FDA like to see one general policy in place not one made up on a day to day basis,

Hope this helps,

Angus