1
$\begingroup$

I have a monthly average for two values and a standard error corresponding to that average. The two values are the mean abundance of a spp of different sections of a plant. I need now a total abundance for the whole plant, so I am adding the monthly averages to have a value of the total abundance of spp. How can I "sum" the se, (stdev/sqrt(n))?

For example:

            mean 1  se 1    mean 2  se 2    **Total**   **se**
January     0,299   0,023   0,359   0,156   **0,658**   **?**
February    0,634   0,186   0,566   0,2     **1,2**     **?**
March       4,806   0,871   0,416   0,12    **5,222**   **?**
April       1,131   0,148   0,989   0,151   **2,12**    **?**
$\endgroup$
  • $\begingroup$ You might look into the delta method. The Wikipedia page isn't very helpful, but here it is: en.wikipedia.org/wiki/Delta_method $\endgroup$ – Jeremy Miles Jan 8 '15 at 21:58
  • 1
    $\begingroup$ @Jeremy The delta method gives exact results in this situation! In fact, that machinery is not needed at all. $\endgroup$ – whuber Jan 8 '15 at 22:01
  • $\begingroup$ Natalia, are those two means statistically independent or might they be correlated? $\endgroup$ – whuber Jan 8 '15 at 22:03
  • $\begingroup$ For example, 6 plants were sample at each month,there are means of the abundance of ants in the leaves and in the roots for all six plants. so all 6 roots are averaged for the monthly vakue and the same for 6 leaves are average. I think they are independent $\endgroup$ – Natalia Jan 8 '15 at 22:09
  • $\begingroup$ Since each set of roots and leaves is associated with the same plant, you have good reason to suspect that the ant abundances in the leaves and roots are not independent. This is something you must investigate if you want to justify formulas for the standard errors of the totals. The simple way is to return to your raw data, compute the totals for each plant each month, and directly compute their standard errors. $\endgroup$ – whuber Jan 8 '15 at 23:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.