# Calculate standard deviation when subtracting and dividing means

I have a series of independent uptake rate (u) measurements at different times (2h, 4h, 6h, 24h) (n=3) and want to find the difference between the mean values for each time with standard deviation. Then I want to see that value per hour.

For the means this is easy: (mean u (4h)-mean u(2h))/2 = mean u/h (then the same for 6h-4h and so on)

I do not know how to make these calculations for the SD, I just know it is more complicated than this.

I would really appreciate some help as this is very important for my research, thank you!

• I think in this case you are assuming mean is changing linearly with time, is your standard deviation also changing linearly with time? Commented Mar 2, 2016 at 5:01
• No, my mean uptake rate is actually highest between 2 and 4h, after that it depends on the treatment, some have steep declines first, others more linear, some rise again. Very variable over all. The SD is not proportional to the mean in any way. Sometimes rather high, sometimes low. Commented Mar 2, 2016 at 5:07

## 1 Answer

Assume two measures $(A,B)$, $C$ is constant and that $Cov\left[A,B\right]=0$

$$\sigma_{\frac{A-B}{C}}=\sqrt{\left(\frac{\sigma_A}{C}\right)^{2}+\left(\frac{\sigma_B}{C}\right)^{2}} = \frac{1}{C}\sqrt{\sigma_{A}^{2}+\sigma_{B}^{2}}$$

In your case $(A,B)$ are the consecutive measures. $C$ is the time elapsed between the measures.

If you are interested check the derivation and sources in the Wiki article on Propagation of uncertainty.