Adding Standard Deviations - Sig figs?

Question

I am working on a question that asks to calculate the answer along with it's standard deviation.

The +- symbol stands for the uncertainty.

(67+-11) + (6+-4) + (21+-3) + (7+-2);

Because this is addition, sig figs only depend on the decimals, of which there aren't any. So I get 101 as the base answer. Then, if I used the formula for getting standard deviation when adding independent deviations (Δz=[(Δx)^2+(Δy)^2]^1/2), I get something like 12.247...

I have a question about what the sig figs for the standard deviation. Do I follow the rules for the addition, and get 12 because there can't be decimals? And thus is is 101+-12? Or do I need to round to three sig figs because 101 is three sig figs? Or even further, I've had some recommend that the answer is 100+-10, because the uncertainty can only be to 1 sig fig, and thus the basis answer must be as well (I don't know). So given these contradictory thoughts, which is the standard and correct way to list it out?

Thanks!

Answer 1

I believe the usual practice is to show one decimal place more for the margin of error than for the estimated quantity. Since your estimated quantity has 0 decimal places, that rule would give you $101 \pm 12.2$

However, practices vary from discipline to discipline. You may want to google 'NIST significant digits' and 'NIST rounding rules' to see sensible comments on this general topic from the U.S. National Institute for Standards and Technology.

However, there is another issue here that no statistician would let pass without further comment:

The practice of quoting plus or minus one standard deviation to indicate margin of error can be quite misleading for most data. For normally distributed data the probability of being within one standard deviation of the mean is only about 68%. A much more realistic indication of the margin of error is to give the base value plus or minus two standard deviations. For normal data this interval turns out to include the true value in about 95% of instances. (Errors for many, but not all, other kinds of data--other than normally distributed--can be realistically be represented by the same rule: mean $\pm$ 2(sd). But you have to know some probability theory or do simulations to judge which kinds of data those are.)

Your method for estimating the standard deviation itself works well for adding independent terms. The R code below simulates your situation a million times, assuming independent normally distributed random variables. The mean of the sum is indeed 101 and the standard deviation is indeed near 12.25. (The simulation itself is accurate to about three significant digits.)

 m = 10^6
 x1 = rnorm(m, 67, 11);  x2 = rnorm(m, 6, 4)
 x3 = rnorm(m, 21, 3); x4 = rnorm(m, 7, 2)
 y = x1 + x2 + x3 + x4
 mean(y)
 ## 101.0192
 sd(y)
 ## 12.24843

Thus, in your instance a reasonable guess is that the sum will lie in the interval $101 \pm 24.5$ about 95% of the time. If you are 'wearing rose colored glasses' and content with an interval that works about 2/3 of the time, then I guess $101 \pm 12.2$ is OK.

The histogram below represents the distribution of sums $Y$ assuming four independent normal terms according to your question. The orange lines represent intervals based on one standard deviation on either side of the mean and the green lines for two standard deviations.