If I have 2 experiments measuring the same thing. For example, I measure the voltage with 2 different tools (2 different voltmeters). I can calculate the final average in 2 different ways. For example, I have 7 measurements with the first tool and 5 measurements with the second tool for whatever reason.

  1. Calculate the average of all 12 values.
  2. Calculate the average of the first tool (7 values), then calculate the average of the second tool (5 values) and then calculate the average of the average.

Specific example:

enter image description here

As we can see, the first way gives the average 1.158 and the second way gives 1.161. The question: What is the correct or better way to calculate the average and why?


As pointed by Andrew in the answer, I updated my example accordingly. Let's assume, that both voltmeters give only 1 digit after the dot. It is not real data, but an artificial example.

I calculated the standard deviation and Voltmeter 1 gives $1.14\pm 0.09$. I understand that I have to put $\pm 2\sigma$ for 95% confidence interval, but let's keep it as is ($\pm \sigma$) for simplicity.

Voltmeter 2 gives $1.18\pm 0.15$. So, the averages from both voltmeters are consistent (values overlap within the confidence interval).

Based on the notation from andrew I calculated it like this

$${\mu}_1 = 1.14$$ $${\mu}_2 = 1.18$$ $${\sigma}_1 = 0.09$$ $${\sigma}_2 = 0.15$$

and based on equations from andrew

$$\bar{\mu}=1.15$$ $$\bar{\sigma}=0.08$$

enter image description here

  • 1
    $\begingroup$ I’m voting to close this question because this is a question about Statistics and is better suited at Cross Validation (Statistics) Stack Exchange. $\endgroup$
    – Farcher
    Sep 22 at 6:45
  • 1
    $\begingroup$ You can use the T-test to see if the two means are significantly different T-Test Calculator for 2 Independent Means $\endgroup$
    – Farcher
    Sep 22 at 6:47
  • $\begingroup$ @Farcher Please keep it here. I want to get an answer from people who are experienced with practical physical experiments, not theoretical ones with long equations where I will get lost. $\endgroup$
    – Zlelik
    Sep 22 at 13:21
  • $\begingroup$ If you have generated the data ten this a very poor example to use. The validity of the third significant figure is suspect if only because you have not stated whether the digital voltmeter truncates a value of 1.16 to give 1.1 or rounds the value to 1.2 The T-test I have suggested would enable you to find out if there was a significant difference between the two sets of data, which there is not, and that would enable you to decide whether or not to lump all the data together to obtain a mean value. $\endgroup$
    – Farcher
    Sep 22 at 14:09

2 Answers 2


First -- do the voltmeters really only give you 2 significant figures? If so, you shouldn't report the final results to more than 2 sig figs, and to do this analysis carefully you should account for quantization error, which would not be negligible. If the voltmeters actually do give you more digits (which I suspect is the case), then you should report them in your data table (and then quantization error isn't an issue).

Second, you haven't reported an uncertainty on your measured values. Do both voltmeters have the same uncertainty? Or is one higher quality? You can estimate this by taking the standard deviation of your measurements for each voltmeter. However, ideally the voltmeters have some information as to their expected precision.

With those two points out of the way, I would recommend doing the following two things with your data:

  1. Report the central value and standard deviation of your measurements for each voltmeter. Eg, Voltmeter 1 gives $x \pm y\ {\rm V}$, while Voltmeter 2 gives $a \pm b\ {\rm V}$. This will let you address whether the two measurements are consistent with each other. If they aren't, I would be very careful about combining them! A better use of your time would be to understand why you get two inconsistent answers.
  2. If the two measurements are consistent, then I would combine them with noise-weighting. If your two measurements gave mean values $\mu_1$ and $\mu_2$ with uncertainties $\sigma_1$ and $\sigma_2$, then your combined mean $\bar{\mu}$ and combined error $\bar{\sigma}$ should be given by \begin{eqnarray} \bar{\mu} &=& \frac{\mu_1/\sigma_1^2 + \mu_2/\sigma_2^2}{\sigma_1^{-2} + \sigma_2^{-2}}\\ \bar{\sigma} &=& \sqrt{\frac{1}{\sigma_1^{-2} + \sigma_2^{-2}}} \end{eqnarray} If the intrinsic uncertainty for each of the two voltmeters are the same, this will reduce to taking the average of all your measurements, weighted equally. (In order to see this, note that the uncertainty in the mean of $N$ measurements scales like $1/\sqrt{N}$).
  • $\begingroup$ This is not real data, just an example. I just made it myself with a random generator :). I just want to understand how to approach this problem in general. I assume that both voltmeters have the same uncertainty. Let me do my homework and apply your approach with the numbers from the original question. $\endgroup$
    – Zlelik
    Sep 21 at 21:45
  • $\begingroup$ I made UPDATE 1 with all the numbers according to your equations. Do you think it is all correct? Strange, that the final $\bar\sigma$ is less then $\sigma_1$ and $\sigma_2$. I am a bit more on practical side and it is easier for me to understand it with specific examples with the specific numbers. $\endgroup$
    – Zlelik
    Sep 22 at 13:22

The amount of data you have is quite small. Anyway, since you changed the tool, I'd do the average value of the averages, since you have different measurements performed with these two tools.

If you evaluate the average of all the measurements you gather, you're weighting the measurements performed with Tool #1 (7 measurements) more than the measurements with Tool #2 (5 measurements). I guess there is no reason to do that, so I'd do the average of the averages.

The difference you get is approximately 0.25%. Are you able to really appreciate this difference?

  • $\begingroup$ This ignores the precision of the two tools. What is one is an order of magnitude more precise than the other? You'd destroy the precision of result if you simply averaged the averages without weighting according to their variance. $\endgroup$
    – Ruslan
    Sep 22 at 12:37

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