Calibration curve

[Mauro]

I am working with a calibration curve
(http://en.wikipedia.org/wiki/Calibration_curve) and I want to evaluate the
use of 3 point curve instead 5 ponit curve with excel.
I have 5 point and 3 point (3 of the 5 points) curve and I use regression
analysis to obtain the following function Y= mx + b, and when I use 5 or 3
point the parameters m and b are similar. I think the use of the intersection
and slope error to evaluate what its better, but may be there are better ways
of evaluate it
thanks in advance

 

[Joel]

Each curve is different and the best method may not be the same. That is why
there are different methods. It also depends on the accuracy you are trying
to achieve. Do you want to get 3, 5 or 6 sigma. You can calcualte the sigma
by comparing the actual measured results and against curve you obtained.
Calculate the standard deviation of the difference between measured and curve
results to get the sigma. This is the best method to determine which curve
is the best

The should be very little difference between 3 point and 5 point if your
results are linear. If they are not linear then the more points you have the
better the results. If you plot the results on a graph in excel you can add
a trend line to the chart. Then change the parameter of the trendline and
see the differences in the results. You should diplay the trendline equation

 

[Mauro]

Joel,
thanks for your response.

Here I attach an example,

X y
0.352 1.09
0.803 1.78
1.08 2.6
1.38 3.03
1.75 4.01

The linear regresion is
y = 2.0925x + 0.2567
R2 = 0.9877

and slope and interception error function are:
Sm 0.134749235
Sb 0.158317598 (y=mx + b)

But when I use

X y
0.352 1.09
1.08 2.6
1.75 4.01

the linear regresion is
y = 2.0885x + 0.3515
R2 = 1

and the error are
Sm 0.008727309
Sb 0.00527747

So, If I asume that the linear regresion are the same How can evaluate whats
is better ? evaluationg Sm and Sb?
Could you help me?

 

[Joel]

The error is the distance between the linear regression and the actual data
point. You need to caculate the distance from each measured point to the
line. The distance becomes your error. The sigma square is the square root
of the sums of the squares of he error (distance).


You have a line
y = 2.0925x + 0.2567

The slope is 2.095

The slope of the line from the point to the line is perpendiculr so the
slope is 1/2.095.

You know have two equations

y = 2.0925x + 0.2567

and

y = (1/2.0925)x + z
where one point is (0.352,1.09)
so

1.09 = (1/2.095)* .352 + z
z = .921


y = 2.0925x + 0.2567
y = .477x + .921

solve these equation
..477x + .921 = 2.0925x + .2567
..221x = 1.171
x = .188
y = 1.011

The distance between the 2 points is

sqrt((0.352-.188)**2+(1.09-1.011)**2)

Now repeat for each point. Then sum all the distances. Repeat with 2nd
line. The line with the small sum of the distance is the better fit.

 

[Mauro]

Joel, thanks agian:

I did what you siad,

and for the
y = 2.0925x + 0.2567
I found 0.144211147

and for the y = 2.0885x + 0.3515
I found 0.008629727

So, It seems to be the 3-point curve which has the less error, i.e. the
better fit. But If you have to evaluate the use of one and the other way of
using a calibration curve, what criteria did you use? Which of this gave the
most accuracy results when you evlaute your data. I suppossed that the use a
5 point curve give the better results insted the 3 point , but with this
results, 3 point it is better? In the daily, what you recomend?

 

[Joel]

I added a trend line to a scatter chart. The got the equation from the
trendline. I selected the option to display the trendline. I wrote a macro
that extracted the equation from the chart automatically and then extracted
the slope and y intercept (b) from the equation which was a sring. I also
modified the trend line to force the crossing of the y axis to be at .98
instead of 1 to get better results. I was doing some probability
calculations and my results never started at 100% (my best results was 98%).
I also added two additional line to the graph to show confidence of 90% and
95%.

 

[Joel]

I hop eyou used the sum of the square of the distances to get the error.
This is really a standard deviation calculation. If you don't square the
results you end up with the avage which is not what you want. Sigma squared
is the standard deviate which you can use to plot a normal distribution
curve. Signa is the width of the normal distribution curve.

 

[Jerry W. Lewis]

You seem to be approaching this problem backwards. How were the data
collected? How did you come to select which points to potentially omit?

For a potentially curved calibration curve, omitting points at the end of
the range has the potential to improve the linear approximation to the
calibration curve. However you are not omitting points at the end of the
range, the quadratic coefficient is not significant (p=0.42) and the linear
and quadratic fits for the full data have almost identical adjusted R2 values
(0.9837 vs 0.9838)

I would not usually consider omitting interior points from the calibration
curve unless I had reason to suspect that there was a problem with how they
had been obtained. Randomly removing points from the calibration curve would
tend to widen confidence intervals for fitted parameters and increase the
uncertainty of quantitation from the calibration curve.

In your case, I find it interesting that you are enquiring about removing
two points whose removal would most increase R2. If that is in fact the
basis for your considering removing them (in the absence of an external
reason to suspect them), then removing them would likely introduce a small
bias (0.1) into your quantitation, while causing you to underestimate the
uncertainty of quantitation by a factor of nearly 17.

Jerry