I have 3 to 5 measures of a trait per individual in two different conditions (A and B).

I'm plotting the average for each individual in each condition and I use the standard error (i.e., $SD/\sqrt{N}$, with $N$ = number of measurements) as error bars.

Now I want to plot the difference between the average measure per individual in condition A and condition B. I know I can determine the propagated error doing:

$$SD=\sqrt{SD_A^2+SD_B^2}$$ but how can I propagate standard errors (since I'm dealing with averages of measurements) instead of standard deviations? Does this make sense at all?


2 Answers 2


You should simply treat your SE as SD, and use exactly the same error propagation formulas. Indeed, standard error of the mean is nothing else than standard deviation of your estimate of the mean, so the math does not change. In your particular case when you estimate SE of $C=A-B$ and you know $\sigma^2_A$, $\sigma^2_B$, $N_A$, and $N_B$, then $$\mathrm{SE}_C=\sqrt{\frac{\sigma^2_A}{N_A}+\frac{\sigma^2_B}{N_B}}.$$

Please note that another option that could potentially sound reasonable is incorrect: $$\mathrm{SE}_C \ne \sqrt{\frac{\sigma^2_A\sigma^2_B}{N_A+N_B}}.$$

To see why, imagine that $\sigma^2_A=\sigma^2_B=1$, but in one case you have a lot of observations and another case only one: $N_A=100, N_B=1$. The standard error of the mean of the first group is 0.1, and of the second it is 1. Now if you use the second (incorrect) formula, you would get approximately 0.14 as the joint standard error, which is far too small given that you second measurement is known $\pm 1$. The correct formula gives $\approx 1$, which makes sense.

  • $\begingroup$ +1 This is the basis for the unequal-variance, unequal-sample sizes formula for the Student t statistic. $\endgroup$
    – whuber
    Feb 22, 2014 at 14:53

Since you know the number of measurements, my first instinct would be to just calculate the propagated SD and then calculate the SE from the propagated SD by dividing it by the square root of N, as per your equation above.

  • 1
    $\begingroup$ I believe this is incorrect. Please see my answer for the explanation why. $\endgroup$
    – amoeba
    Feb 22, 2014 at 14:19
  • $\begingroup$ Ah, I see. I did not take the unequal sample sizes into account. Thank you for the explanation, @amoeba. If you have the time to help me get my thoughts straight; in a situation where the sample sizes had been equal, my proposed method above would have been correct, right? $\endgroup$
    – Mattias
    Feb 24, 2014 at 14:30
  • $\begingroup$ Yes, absolutely. $\endgroup$
    – amoeba
    Feb 24, 2014 at 15:23

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