I'm doing an experiment with a cryostat to determine the critical temperature for lead. To avoid asymmetries, I determine the critical temperature both through heating (going from 2 K to 10 K) and cooling (10 K -> 2 K).

Now I have two values, that differ slighty and I average them. So a measurement of (6.942 $\pm$ 0.020) K and (6.959 $\pm$ 0.019) K gives me an average of 6.951 K.

Now the question is: what is the error of that average?

One way to do it would be to calculate the variance of this sample (containing two points), take the square root and divide by $\sqrt{2}$. This gives me an SEM of 0.0085 K, which is too low for my opinion (where does this precision come from?)

The other way is to say the the mean is a function of two variables, $\bar{T} = \frac{T_1 + T_2}{2}$, therefore by error propagation the error is $\Delta T = \frac12\sqrt{(\Delta T_1)^2+(\Delta T_2)^2}$, and that gives me a much more rational value of 0.014.

I see how those values differ in terms of numbers, but which one is correct when talking about the correct estimate for the standard deviation?

  • $\begingroup$ @COOLSerdash That's actually another point I have thought about: The numbers after the $\pm$ denote the error of the thermometer, as given by the manufacturer. My interpretation of that was always that the manufacturer did a lot of measurements with a calibrated source and calculated the 'descriptive' variance of those values, therefore saving me the fuss of doing several measurements. $\endgroup$ – Wojciech Morawiec Sep 29 '13 at 21:46
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    $\begingroup$ Understanding systematic errors is the key. Let $\mu$ be the critical temperature (CT). Let's posit that the expected CT measured through heating equals $\mu-\delta_h$ and measured through cooling equals $\mu+\delta_c$. I presume a value like $6942\pm 20$ represents the mean and standard error of some heating measurements; $6959\pm 19$ are the mean and SE of some cooling measurements. It will be hard to estimate $\mu$ because you have little information about $\delta_h$ or $\delta_c$. You can estimate $(\mu-\delta_h)+(\mu+\delta_c)/2$ = $\mu+(\delta_c-\delta_h)/2$. $\endgroup$ – whuber Sep 29 '13 at 21:48
  • $\begingroup$ @whuber That is an excellent comment, I never would have thought of it that way! But I have to admit that I have the feeling it doesn't completely answer my question: What if I had done the two measurements one after another through heating or I wouldn't mind systematic errors at all? How would I then correctly estimate the error of the average? $\endgroup$ – Wojciech Morawiec Sep 29 '13 at 22:17
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    $\begingroup$ Even if you don't mind systematic errors, if you agree that they are present, then no amount of averaging your data will estimate them for you. You have several options, ranging from making simplifying assumptions (such as $\delta_h = \delta_c$) to just reporting the likely interval containing $\mu$ and providing error estimates for its endpoints. What approach is best depends on how you are thinking about this problem and what, precisely, you are attempting to estimate. (That's why I am writing comments here instead of answers!) $\endgroup$ – whuber Sep 30 '13 at 0:16
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    $\begingroup$ Hmm. Can you confirm the calibration of your system? This will allow you to quantify the likely window within which your bias lives. Can you confirm there is no systemic error by repeated melt/freeze/melt/freeze cycles? If you can quantify uncertainty associated with your process independent of calibration then you can account for that source of variability within your measurement. itl.nist.gov/div898/handbook/mpc/mpc.htm $\endgroup$ – EngrStudent Sep 30 '13 at 0:49

If you take the standard deviation of your two values (first way you suggested), this does not include the error within each measurement, only the error between the two methods (heating vs. cooling). That is why the first way shows surprisingly low standard deviation- it's only incorporating one of the sources of error.

Therefore, the correct way is to propagate error (the second way you suggested). This incorporates both the measurement error of each value and the error between your two methods.

(Also a note based on the question comments: the manufacturer error listed is the +/- range guaranteed by the company based on all the devices they make. So yes, it does save you time if you use it as the equipment's error, but if you take many measurements of the same thing yourself you can calculate the error of your specific equipment and that can be more precise than the range guaranteed by the manufacturer. I don't know about thermometers, but this is what analytical chemists do with their glassware.)

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  • $\begingroup$ and please correct if I mixed up when to say "deviation" vs "error" $\endgroup$ – rrr Apr 22 '18 at 21:30

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