Hi Ondřej,
if you want to calculate the stddev of a sample of values where the
exact mean is not known, but only the mean of the values in the sample,
you have to divide by (n-1) instead of n.
https://en.wikipedia.org/wiki/Standard_deviation#Corrected_sample_standard_deviationSo you should have
...
return sqrt(SUM/(n-1));
A hand-waving argument for this:
The denominator is zero if you have only one value (n=1): For a sample
with only one value you cannot determine the standard deviation. For
very few values, the mean value of these values will on average still be
quite off from the mean value that you would get from an infinitely
large sample. So the deviations (x-mean) from this mean will be too
small and the equation with dividing by n will give you a stddev that is
too small. The (n-1) corrects for this fact.
Michael
________________________________________________________________
On 2016-06-30 15:30, Ondřej Šebesta wrote:
> Hello all.
>
> I would like to know why I get a different number with custom written
> functions to get standard deviation (according to
>
http://www.investopedia.com/terms/s/standarddeviation.asp) compare to
> Array.getStatistics build-in macro function? Is there a bug or am I missing
> something?
>
> Thank you very much.
> Ondrej
>
> minimal macro:
>
> volumeArray=newArray(3, 5, 6, 4, 4, 1, 2, 7);
> volume=Mean(volumeArray);
> volStDev=StDev(volumeArray);
> print("custom functions send "+volume+"+-"+volStDev);
> Array.getStatistics(volumeArray, min, max, volume, volStDev);
> print("Array.getStatistics send "+volume+"+-"+volStDev);
>
> //******************** functions **********************
> function StDev(pole) {
> SUM=0;
> n=pole.length;
> mean=Mean(pole);
> for (x=0; x<n; x++) {
> SUM = SUM+square(pole[x]-mean);
> }
> return sqrt(SUM/n);
> }
>
> function Mean(pole) {
> mean=0;
> n=pole.length;
> for (x=0; x<n; x++) {
> mean+=pole[x];
> }
> return mean/n;
> }
> function square(a) {
> return a*a;
> }
>
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