I have a couple of measurements for an environmental variable on different locations (7 measurement stations) as time series with averages for every year.

For the single locations I have calculated the deviations of single years from the long term average. The annual deviations are not normally distributed as there are trends in the data. I cannot describe these trends as I have not enough knowledge about the influences. Further more the data for the single locations show different trends because of local influences. Also the absolute hight of the measurement value is different for the single locations (by more than 10% in the long term average).

To get an idea on how the locations compare to each other I calculated the relative root mean square deviation of the single years from the long term average and the 5th and the 95th percentile of the relative deviations to get an symmetric coverage interval for the uncertainty at the 90% level.

My results look as follows:

location  RMSD[%] P5[%] P95[%]

1         3.8      -5.6   -1.3  
2         3.1      -5.1    5.2
3         5.1      -0.6    8.6 
4         3.3      -6.2    3.6
5         3.8      -6.7    1.8
6         2.8       0.2    4.0
7         3.7      -6.1    3.9

I want to get an estimate of the expected deviation of a single year from the long term average and the uncertainty of the deviation for an arbitrary location in the region the measurements are taken from.

What would be a statistically valid way to do this? I am thinking about using the average over all measurement locations I have, but as the locations are a lot different from another, I am not sure if this is ok.


1 Answer 1


Compute the sample estimate of mean absolute deviation. Obtain an estimate of the variance for the absolute deviation divide it by n to get the estimate for the average absolute deviation.

  • $\begingroup$ do you mean the mean absolute deviation over all measurement values for all locations? I wanted to avoid using absolute values, as the absolute values are different from location to location (I have updated the question to make this more clear). $\endgroup$
    – bmu
    Commented Jun 6, 2012 at 11:41
  • $\begingroup$ Averaging deviations is a bad idea because positive and negative deviations cancel destroying any picture they might give you of the variation. $\endgroup$ Commented Jun 6, 2012 at 13:07
  • $\begingroup$ Thanks a lot for your answer and your comment. I understand now that averaging is a bad idea, but I don't really understand your answer than. Can you please give me some more details. $\endgroup$
    – bmu
    Commented Jun 6, 2012 at 15:08
  • $\begingroup$ Well I didn't spell out how you get the estimate of the variance of the mean absolute deviation. This is the difficult part. I will see if I can work that out. $\endgroup$ Commented Jun 6, 2012 at 15:29

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