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Negative values in calibration curve

  1. May 19, 2014 #1
    Hi

    I just wondering how to report negative values cuased by non-zero intersection.
    For example, when the absorbance<0.037 in the figure, the concentration would be negative.

    Thank you!
     

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  2. jcsd
  3. May 20, 2014 #2

    Borek

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    Staff: Mentor

    What do you mean by "how to report"?
     
  4. May 20, 2014 #3
    I mean how to explain
     
  5. May 20, 2014 #4

    Borek

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    It is a rather common error. You may have some interference from other substances, you may have some substance present in the reagents used, you may have incorrectly calibrated spectrometer and so on. As long as the calibration curve is nicely linear and all measurements fall within the range covered, there is typically no problem with using it.
     
  6. May 20, 2014 #5
    Can I see the concentration is zero when it becomes slightly negative
     
  7. May 20, 2014 #6

    Borek

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    No, that would mean something went wrong. Note that you will be outside of the calibrated range, and I told you it is OK to use measurements that fall within the range tested.

    Even if you have a plot that looks perfectly linear and looks like it crosses 0,0 point, but you did the calibration for 10..100 range (of whatever unit), extrapolating the calibration outside of the 10..100 and saying "my concentration was measured to be 1" would be generally speaking a bad practice. In some cases it can be acceptable, but it depends on the application.
     
  8. May 20, 2014 #7

    DrDu

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    That's rather a question for the statistics forum than for the chemistry forum. Obviously a reading < 0.037 won't correspond to a negative concentration, simply because there can be no negative concentrations.
    You should also never report an estimator without some confidence interval.
     
  9. May 20, 2014 #8
    Did you subtract out a background reading? Calibration curves almost never behave in a perfect manner where they intersect through 0,0. No instrument is perfect.
     
  10. May 21, 2014 #9

    DrDu

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    Of course he did. If not, there wouldn't be the parameter a for the curve.
    One of the easiest ways to formulate the problem mathematically is the following.
    The absorbance A is distributed (e.g. normally) around ##a+bc##, i.e.
    ##A\sim N(a+bc,\sigma)##. From the maximum likelihood principle we find c as that value which maximizes the probability to find the value of A actually measured under the constraint ##c\ge 0##.
    That is ## \hat{c}=\mathrm{max}(0, (A-a)/b)##, where ##\hat{c} ## is the maximum likelihood estimator for c. If the standard deviation ## \sigma## depends on c, too, or if the distribution is not normal the result may be more complicated but the principle remains the same.
     
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