# Quantiles of the Chi-square over degrees of freedom distribution [duplicate]

I've been studying the EPA's procedures for calculating Method Detection Limits (MDLs) based on 40 CFR 136 (Appendix B).

After calculating the MDL, they give the constants for calculating a confidence interval for the MDL

according to the following equations derived from percentiles of the chi square over degrees of freedom distribution ($\chi^2$/df).

LCL = 0.64 MDL

UCL = 2.20 MDL

Based on an example at http://www.chemiasoft.com/chemd/node/58, I found that $\sqrt \frac{df}{\chi^2(\alpha/2, df)}$ and $\sqrt \frac{df}{\chi^2(1 -\alpha/2, df)}$ give the constants for the MDL limits. I confirmed this using R with

# The EPA document anticipates a sample size of 7, or 6 df
sqrt(6 / qchisq(.975, 6)) # 0.6443934
sqrt(6 / qchisq(.025, 6)) # 2.202066


I'm afraid I don't understand the derivation, however. It seems to be a chi-squared variable divided by it's degrees of freedom would be the same as a chi-square variable divided by a constant, which has a Gamma distribution. So I'm fairly certain I'm missing something obvious and would appreciate any help in understanding the derivation of the confidence interval for the MDL.

• It's a confidence interval for the standard deviation, written "$S$" in the regs. Since $S^2$ is the variance and $S$ must be positive, the CI is obtained by taking the square roots of the CI for the variance. – whuber May 26 '17 at 12:58
• @whuber Oh. So if $MDL = T * S$, then $LCL = 0.64 * MDL = 0.64 * T * S$. In other words, the confidence interval for the MDL is $T$ multiplied by the confidence limits for $S$? – Benjamin May 26 '17 at 13:11