variance of the variance for finite population for normal distribution

Question

I am interested in calculating the variance of the sample variance for a finite population...of course in the limit of sampling the entire population, this will be zero. There seems to be one author who is doing this, and he has many similar versions of the expressions within. It seems easy enough to implement, however when I looked at the limit, it disagrees with my derivation (which agrees with mathworld). For a normal distribution, in the limit as the population goes to infinity, I get the following expressions

http://www.ijpam.eu/contents/2005-21-3/10/10.pdf
$\frac{2S^2(2n - 3)}{n(n - 1)}$...substituting in $\mu_2$ and $\mu_4$ for those of a standard normal with $\mu_1$=0

http://mathworld.wolfram.com/SampleVarianceDistribution.html
$\frac{2S^2}{n - 1}$...converting from population variance to sampling variance

where $S$ is the sampling variance

Am I making a mistake in interpreting the authors expressions. Ultimately, I am after confidence intervals for the variance when sampling a finite population

Answer 1

You might also be interested in the following paper which addresses variance of sample variance for a finite population, and confidence interval of population variance in a finite population problem:

O'Neill, B. (2014) Some useful moment results in sampling problems. The American Statistician 68(4), pp. 282-296.