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If you are making a calibration curve for $\ce{CuSO4}$ where you plot absorbance against concentration and the curve deviates slightly from the origin, does it make the calibration curve useless?

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    $\begingroup$ Not necessarily, it depends on the method you are using for collecting the data. For example, the standard addition method basically requires that the curve doesn't go through the origin. Which one are you using? $\endgroup$ – andselisk Mar 3 at 16:25
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    $\begingroup$ The value of the absorbance for a blank is another data point, but a little special. Usually you calibrate the instrument with a blank, so that in essence you take at least two measurements at the origin. The most important question is probably what range of concentrations you want to cover. Whether you include the origin depends on the linearity (not to mention dynamic range) of the instrument response. Also, you should perform measurement and analysis identically for your samples, whose concentration should fall in the range. Also, some analyses force the curve through the origin, bad. $\endgroup$ – Buck Thorn Mar 3 at 16:45
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Good answers, but it is almost never a good idea to force the intercept to zero. Unless there is a clearly justifiable and compelling reason to do so, it is better to let the data set 'vote' for itself and not impose a preference/bias/mandate. Among other things, forcing the intercept to zero changes the standard error of the slope, which is an important statistic.

Below is a screenshot of the result of using Excel's Data Analysis tool on a set of linear calibration data I collected in the past:

OLS statistics

As shown, the intercept is -0.0541 and its standard error is 0.0352, so the t test statistic of the intercept, defined as intercept/(standard error of the intercept) = -1.537. The negative sign is irrelevant for present purposes, but the P-value is 0.264.

The null hypothesis for the intercept is that the true intercept, which is unknown, is zero. The alternative hypothesis (aka research hypothesis) is that the true intercept is either 1) not equal to zero or 2) not equal to zero, but in a known direction. The former is the 2-tailed option and the latter is the 1-tailed option. For intercepts, the 2-tailed option is most common and this would almost certainly be the case for a Beer's law plot.

The P-value of 0.264 is much higher than the usual 0.05 value and this means that it would be quite risky to reject the null hypothesis based on the OLS analysis. In fact, the risk of mistakenly rejecting the null hypothesis, even though it was true, would be 26.4%. The confidence in the alternative hypothesis would be 100% - 26.4% = 73.6%, which is too little confidence for most people. Typically, you want P ≤ 0.05. Otherwise, it is too risky to 'jump ship' for the alternative hypothesis.

What does it mean? Well, the data resulted in a P-value of 26.4%, so the null hypothesis cannot prudently be rejected. But it is also not accepted: it still might be wrong, but the data did not show that. Replacing the intercept by zero would mean that a decision was made that the true intercept was zero, even though the analysis proved no such thing. It is not a good idea to pretend that the data showed what it did not show. Bottom line: the intercept should not be forced to zero unless there is a clear and compelling reasion to do so.

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    $\begingroup$ very nice answer. I'd like to add that there are tons of proper and valid reasons why the signal intensity may have an offset (or, thinking in terms of spectra, that lead to a non-zero baseline). For example, a small amount of some masking agent may lead to a negative offset: in that case also the blank sample should not be included in the calibration. Other, more general thought: using a blank measurement to adjust the instrument to yield zero signal (outside the calculation of the calibration function) will transform noise/random uncertainty on that blank measurement into a systematic term.. $\endgroup$ – cbeleites supports Monica Jul 19 at 17:37
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    $\begingroup$ ...for all following measurements if the calibration function is then forced through the origin. Still, if the calibration allows an offset, that offset corresponds to the fitting function's estimate of the error of that blank measurement, so the calibration function is not harmed as when forcing it through the origin, but the measurement is wasted: Including that blank measurement instead as normal measurement at conc = 0 in the calibration curve (and there wouldn't even be any discussion about forcing the curve through the origin now) allows that measurement to add its information to... $\endgroup$ – cbeleites supports Monica Jul 19 at 17:44
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    $\begingroup$ ... the estimation of the calibration curve. Which is what we typically want. Oh, and of course, the important link is that we cannot measure absorbance, we typically don't even measure transmission but the intensity of light that got through sample and possibly cuvette/window/sample holder and assume that the missing light that was not missing in the blank measurement was absorbed. $\endgroup$ – cbeleites supports Monica Jul 19 at 17:44
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    $\begingroup$ @cbeleites I completely agree on all the points you raise! As an analytical chemist and spectroscopist, I have seen some truly bizarre things people have done to their data. I answered this question to try to stop some of that abuse of data. I do not know whether any good will come of it, but maybe it will. Anyway, thanks for your thoughtful comments! $\endgroup$ – Ed V Jul 19 at 17:56
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    $\begingroup$ @cbeleites Your comment about is right on point! When I used to teach the Instrumental Analysis lab, I used a home-made scanning spectrometer to have students obtain absorbance spectra, from 400 - 700 nm, of aqueous chloride solutions of $Ho^{3+}$, $Er^{3+}$ and $Pr^{3+}$. First they did a dark response scan, with a "black cuvette", then a blank scan (cuvette with water), and then the sample scan. Next they used a spreadsheet template to compute the dark response corrected "pseudo-transmittances", next T($\lambda$) as quotient and then A($\lambda$). Students were a surprised to see that $\endgroup$ – Ed V Jul 19 at 22:39
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You already have an answer above in the comments. However some teachers suggest to force the calibration line zero, because one might think that blank must have a zero value. This is not a correct approach at all. You can have a non-zero intercept as long as your equation is linear (correlation coefficient r2 is closer to 1 e.g. 0.999). MS Excel can easily do this for you. Although you may not need it for your school work, but the rigorous procedure is to do a statistical t-test. This test "decides" whether we should pass the calibration curve through zero or not. Advanced readers may like to read this article http://www.chromatographyonline.com/calibration-curves-part-i-b-or-not-b

Some organizations or protocols may require that we force the calibration line thru zero. It makes some calculations easier.

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