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Weighted Standard deviation
I found the attached 'weightsd.pdf' formula for weighted standard deviation from NIST.
Wolfram shows the variance of the weighted mean in equation 24 in the 'maximum likelihood' attachment.
The link to Wikipedia below states the same formula in hte Wolfram article for determining the weighted standard deviation on the link below:
https://controls.engin.umich.edu/wiki/index.php/Basic_statistics:_mean,_median,_average,_standard_deviation,_z-scores,_and_p-value
My question is, I have yet to be able to reconcile the formula in the NIST attachment with the Wolfram and Wikipedia results, and was wondering which was the more apropriate approach to use (NIST or Wolfram) and why.
Thank you
weightsd.pdf
Maximum-Likelihood----from-Wolfr.pdf
Wolfram shows the variance of the weighted mean in equation 24 in the 'maximum likelihood' attachment.
The link to Wikipedia below states the same formula in hte Wolfram article for determining the weighted standard deviation on the link below:
https://controls.engin.umich.edu/wiki/index.php/Basic_statistics:_mean,_median,_average,_standard_deviation,_z-scores,_and_p-value
My question is, I have yet to be able to reconcile the formula in the NIST attachment with the Wolfram and Wikipedia results, and was wondering which was the more apropriate approach to use (NIST or Wolfram) and why.
Thank you
weightsd.pdf
Maximum-Likelihood----from-Wolfr.pdf
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ASKER
Examples with calculations help. I'll send calculations when they aren't proprietary.
"Did you mean x bar sub w and xi are not weighted in the NIST equation?"
the xi are not weighted but the x (or as I should have said) x bar sub w is.
Use the nist. It is clearer.
the xi are not weighted but the x (or as I should have said) x bar sub w is.
Use the nist. It is clearer.
ASKER
To confirm, If I manage to decipher this mess, both formulas should yield the same result?
Thank you