0
$\begingroup$

This thread provides an excellent explanation of the calculation of the standard deviation of the sample standard deviation. However, I'm not sure if it's appropriate in my case.

I'm attempting to measure the stability of a classification scheme across groups of documents across years. My data contain 10 variables for ~20,000 observations. The observations are groups of documents and the variables are the fraction of each group assigned to one part of a binary classification each year. For instance (using 6 instead of all 10 to save space):

Group 2010 2011 2012 2013 2014 2015
Group1 0.27 0.28 0.27 0.26 0.26 0.28
Group2 0.62 0.64 0.61 0.60 0.66 0.64
Group3 0.00 0.00 0.01 0.00 0.02 0.00

To measure the stability of the fractional classification within a group across years, the population standard deviation seems to make the most sense. Using the sample above, this would be: \begin{equation} \sigma = \sqrt\frac{\sum{(X-\mu)^2}}{N} \end{equation}

Group 2010 2011 2012 2013 2014 2015 SD
Group1 0.27 0.28 0.27 0.26 0.26 0.28 0.0082
Group2 0.62 0.64 0.61 0.60 0.66 0.64 0.0203
Group3 0.00 0.00 0.01 0.00 0.02 0.00 0.0076

Now, I want to find the stability between groups. Would it be appropriate to simply calculate the population SD of the population SD values? It's important to note that the distribution of SDs by group is not normal in my complete data (e.g., Skewness = 1.58, Kurtosis = 5.39)

$\endgroup$
1
  • 1
    $\begingroup$ Use counts instead of fractions, because the fractions have lost almost all information about how much variability to expect. $\endgroup$
    – whuber
    Jun 1, 2022 at 16:09

1 Answer 1

0
$\begingroup$

Based on what I'm reading elsewhere, it seems like the appropriate thing to do here is to calculate the variance for each group then sum these variances and take the standard deviation of that sum. In other words, the square root of the sum of the variances is what I want.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.