Calibration curve
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 |
Calibration curve
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 on the chart so yo can get the coiefficents of gthe equaion. "Mauro" wrote: 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 |
Calibration curve
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 a 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" wrote: 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 on the chart so yo can get the coiefficents of gthe equaion. "Mauro" wrote: 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 |
Calibration curve
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" wrote: 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 a 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" wrote: 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 on the chart so yo can get the coiefficents of gthe equaion. "Mauro" wrote: 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 |
Calibration curve
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" wrote: 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" wrote: 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 a 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" wrote: 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 on the chart so yo can get the coiefficents of gthe equaion. "Mauro" wrote: 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 |
Calibration curve
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%. "Mauro" wrote: 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" wrote: 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" wrote: 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 a 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" wrote: 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 on the chart so yo can get the coiefficents of gthe equaion. "Mauro" wrote: 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 |
Calibration curve
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. "Joel" wrote: 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%. "Mauro" wrote: 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" wrote: 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" wrote: 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 a 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" wrote: 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 on the chart so yo can get the coiefficents of gthe equaion. "Mauro" wrote: 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 |
Calibration curve
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 "Mauro" wrote: 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 a 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" wrote: 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 on the chart so yo can get the coiefficents of gthe equaion. "Mauro" wrote: 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 |
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