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I have two sets of data

Set 1: 0.051, 0.047, 0.044, 0.049, 0.043, 0.048, and 0.042. Using Excel, the average is 0.046286 and the Std Dev is 0.003352

Set 2: 0.051, 0.046, 0.053, 0.047, 0.047, 0.048, and 0.049. Using Excel, the average is 0.048714 and the Std Dev is 0.002498

If I average all of the data in Set 1 and Set 2 together, I get an average of 0.0475. If I average the average from set 1 (0.046286) and the average from set 2 (0.048714), I get the same thing of 0.0475. Okay, that seems reasonable to me.

If I want the SD of all of the data in Set 1 and Set 2 together, I get a Std Dev of 0.003107. If I average the Std Dev from set 1 (0.003352) and the Std Dev from set 2 (0.002498) I get 0.0029250, which is not the same thing. Ugh, that does NOT seem reasonable to me.

Why does my average std dev from the two sets not equal the total std dev of the two sets together? Why does it work for average, but not std dev?

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    $\begingroup$ Variances add, not standard deviations (assuming independence). $\endgroup$
    – Adrian Keister
    Oct 15, 2021 at 18:54
  • $\begingroup$ @AdrianKeister But this question is not about adding $\endgroup$
    – Henry
    Oct 15, 2021 at 21:38
  • $\begingroup$ @Henry But if they add, they average, right? $\endgroup$
    – Adrian Keister
    Oct 15, 2021 at 21:54
  • $\begingroup$ No, not in this question (try it on the data by squaring the standard deviations given). Variances might average in this question if (A) the sample sizes were the same, (B) the sample means were the same, and (C) the method of calculating sample variances used $\frac1n$ and not $\frac1{n-1}$. Neither B nor C apply to this question $\endgroup$
    – Henry
    Oct 15, 2021 at 21:58
  • $\begingroup$ @Henry I'd say it does relate to adding (a property of sums is why it works for means of equal sized groups but not means of groups of different sizes). The same property also tells us how to modify that second stage mean so it does 'work' $\endgroup$
    – Glen_b
    Oct 15, 2021 at 22:46

3 Answers 3

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There's a simple principle that exlains this:

$E(x)+E(y) = E(x+y)$

$Var(x+y)=Var(x) + Var(y) +2*Cov(x,y) $

When working with means (expected values) they are just linearly related. So adding first or averaging first gives the same results. However, with variance (and, by extension, standard deviation) you have to take into account the covariance. So the order really matters.

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  • $\begingroup$ I think you meant $\operatorname{Cov}(x, y)$ instead of $\operatorname{Cov}(x + y)$. $\endgroup$
    – COOLSerdash
    Oct 15, 2021 at 19:35
  • $\begingroup$ Edited, thank you! $\endgroup$
    – Tanner Phillips
    Oct 15, 2021 at 19:43
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    $\begingroup$ But the question is for the standard deviation of the combined data, not the sum $\endgroup$
    – Henry
    Oct 15, 2021 at 20:24
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Another answer... from an intuitive perspective...

The standard deviation is a measurement of the dispersion around the mean. Suppose you have two data sets distributed as below, set 1 is on the left and set 2 is on the right.

enter image description here

The red points represents the mean of each set.

In each set, the data points are close together. The standard deviation for set 1 is 3.74 (with a mean 8.7) and the one for set 2 is 4.72 (with a mean of 92). In both cases, the data points are "gathered" around their respective means, so their standard deviations are "similar" in terms of scale.

If you average them, you get a standard deviation of 4.23.

Now if we consider the entire data set (set 1 + set 2), the average has moved to the middle (green point)... In this configuration, the data points are quite scattered around the mean. The resulting standard deviation is 43! Very far from the averaged standard deviation of the two separate sets.

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The sample standard deviation is the root mean square deviation from the mean, with a correction to make the sample variance unbiased. So it is a much more complicated calculation (for example the means are different and then change) than taking the average for two different sets of the same size (you have $7$ elements of each)

What you have done with the mean or average is really via the sums of the data and back

= (0.046286 * 7 + 0.048714 * 7) / (7 + 7)

though you took a shortcut and did = (0.046286 + 0.048714) / 2

The equivalent combining calculation with the standard deviation is much more complicated because there are so many steps to undo and redo, in effect going to the variances and sum of squares of the original data and back; indeed I cannot fit it on one line:

=SQRT(((0.003352^2 * (7-1) + 0.046286^2 * 7 + 0.002498^2 * (7-1) + 0.048714^2 * 7) - 
(0.046286 * 7 + 0.048714 * 7)^2 / (7+7)) / (7+7-1))

Even if you have both samples equal size and if they had the same mean (let's suppose we leave the first sample the same and changed the last observation of the second sample to 0.032 so the second sample now also has mean 0.046286 and has standard deviation 0.0067753, and you used both shortcuts from having the same mean and having the same sample sizes, it would still not be as simple as with the means, since you would use =SQRT((0.003352^2+0.0067753^2)*6/13) to get the new combined standard deviation of about 0.005135

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