Linearity of an analytical method

[sofiane]

Hello,
I have validated an analytical method using HPLC. The method is specific then I test the linearity. My probleme is the interpretation of the obtained result.

In general, we have the equation with an intercept y=ax+b. There are some tests like Student test and Fischer test.

I use OriginPro 8, my results are:

R Value : 0,99923

Intercept:

value: -381754,98699
Standard Error: 98292,45204
t-Value : -3,88387
Prob>|t|: 0,00147
LCL: -591260,38653
UCL : -172249,58745
Dependency : --
CI Half-Width: 209505,39954

Slope:

value: 44552,43809
Standard Error: 4507,6373
t-Value : 98,86531
Prob>|t|: 0
LCL: 43591,92653
UCL : 45512,94966
Dependency : --
CI Half-Width: 960,51157

Is the intercept significative or not? why?

Thank you

[Rndirk]

Can you post a plot of the calibration curve with your standards marked?

What detection do you use and what is the analyte?

It looks very significant but i think we could use the info above.

[HPLC chemist]

The y-intercept is significant. If it is more than 5% (generally) then you would have to run a linearity series of standards for your sample quantitation. You cannot use a single point calibration otherwise your quantitation will be off by >5%.

[sofiane]

Thank you,

C(µg/ml) area
99 4,14534E6
99 4,1095E6
99 4,11171E6
154 6,46189E6
154 6,47718E6
154 6,48218E6
196 8,12518E6
196 8,1273E6
196 8,14697E6
251 1,07338E7
251 1,07337E7
251 1,0748E7
313 1,33669E7
313 1,33945E7
313 1,34524E7


I use UV detection at 270 nm.
I have calculated the biais and it was 1.10

What you think?
Thank you

[HPLC chemist]

While your (correlation coefficient) coefficient of linearity looks good (~0.999). Your DL and QL (was LOQ) suck! The DL was 18216 and QL was 60720 ug/mL. In order to improve your accuracy I would perform a 1/10 dilution and increase the injection volume (maybe 100 uL for a 10 ppm solution).

[HPLC chemist]

Any analytical method MUST have at least 2 acceptance criteria;

1. Correlation coefficient >0.99 (which you have and thus demonstrated the method is linear)

2. %RSD of peaks (height or area) < or = 2.0% in order to demonstrate the method is accurate and can meet the drug product's specifications.

[sofiane]

Any analytical method MUST have at least 2 acceptance criteria;

1. Correlation coefficient >0.99 (which you have and thus demonstrated the method is linear)

2. %RSD of peaks (height or area) < or = 2.0% in order to demonstrate the method is accurate and can meet the drug product's specifications.

What about the y-intercept when it differs from 0 ?

[rb6banjo]

Perhaps you are bound by a regulatory agency of some sort. In that case, you must yield to what they say to do. My argument is a philosphical one so here goes.

The correlation coefficient tells you some things but it's not the end-all. I took your data and I calculated the 1-parameter (y = mx) least-squares slope for it. It came out 42439. Your 2-parameter fit (y = mx + b) yielded 43446. Those are within 2% of each other, which is very good agreement in my opinion. To me, this is the first indicator that the intercept is not terribly large compared to the instrument responses used to calculate it. Know that the slope and intercept in the 2-parameter fit are not independent of each other AND the intercept is always going to be the dumping ground for the majority of the error in your calibration data. The intercept is actually dictated by the midpoint of your calibration data (average of x and average of y).

I had to do some fooling around to get your data to work in the US version of Excel so I could have some roundoff error going on. Please try to ignore the fact that I didn't get the exact values for your slope and intercept.

I also created some data to overlay with yours. Series 1 in the graphic is your data. Series 2 is:

y = 42439*x^0.8 (this data slightly concave down at higher concentrations)

and Series 3 is:

y = 42439*x^1.2 (this data is slightly concave up at higher concentrations)

In effect, these treatments add nonlinearity to the data. Here's the plot:

https://s20.postimg.org/c2oc3r2xp/Linea ... mple_I.jpg

The R^2 values are still "good" yet there's a fair amount of curvature in the data.

If your samples are running in the middle of your calibrated range, you are far from zero where the intercept would could/would have a big effect on your calculated concentration. Far from the origin:

(y - b)/m ~ y/m

Take an instrument response of 1x10^7 cts:

1x10^7/43446 = 230

Accounting for the nonzero intercept you get (my 2 parameter fit says the intercept is -231607):

(1x10^7 + 231607)/43446 = 236

1x10^7/42439 = 236 (1 parameter least-squares response factor)

How big of a deal is 230 vs. 236 in your application? Obviously, as you get closer to the origin, the intercept becomes a bigger factor in your calculation. Your relative standard deviations on your triplicate analyses of your standards are wonderful (0.5% or less). Perhaps being off by 2% on the "sample" I described above is a problem?

I fit your data to y = a1*x + a2*x^2 and the a2 term is small but positive. This is probably why your intercept is negative (your data is slightly concaved upward).

[Peter Apps]

Sometimes it is better to ask why the calibration looks like it does, and try to fix any technical issues with the analysis, rather than argue the fine points of data analysis.

In each set of triplicates there is a clear upward trend. If you injected level 1, level 1, level 1, level 2, level 2, level 2, level 3 ........ this is a symptom of carryover. If you injected level 1, level 2, level 3 ........, level 1, 2, 3 ......... it is a sign of long-term drift.

Peter

[sofiane]

I am very grateful for you
thank you