How to Compare Variances? I have a dataset which contains height and weight variables. The mean and variance for height are 165 cm and 25 cm. The mean and variance for weight are 70 kg and 16 kg. How to compare variances of height and weight to determine which one has a higher variance? The naive way is to rescale height so that height and weight has the same mean 70 so that scaled variance for height becomes 25*(70/165)^2 = 4.49. Since 16 > 4.49, so we conclude weight's variance is higher that that of height. Is this a correct approach?
 A: It may be futile to try to compare variances in height with variances in weight.

*

*For adults, height stays fairly constant for years and measurements can be
pretty accurate to nearest cm or half inch.

*For a given individual, weights can vary by time of day, by season, and over the years. Also, even medical scales in doctors'
offices and clinics are not very accurate. (And rules whether to remove
coats and shoes vary.)

If you really want to make this comparison the coefficient of variation (CV)
may be what you want. CV is defined for variables that are always positive.
Population CV is $\sigma/\mu$ and sample CV is $S/\bar X.$
(The CV for ants is greater than the CV for elephants.)
Distributions of CVs depend on the distribution of the population. For example,
in an exponential distribution CV $\approx 1.$ For most normal data in common
use, considerably smaller. If you can't find useful formulas for your data,
then maybe consider bootstrap CIs for CVs.
One exponential sample:
set.seed(626)
x = rexp(100, 1/20)
cv = sd(x)/mean(x);  cv
[1] 0.905036

stripchart(x, pch="|")


Crude quantile 95% bootstrap CI: $(0.78, 1.03)$
set.seed(2020)
m = 5000;  cv.re=numeric(m)
for(i in 1:m) {
 x.re = sample(x, 100, rep=T)
 cv.re[i] = sd(x.re)/mean(x.re) }
quantile(cv.re, c(.025,.975))
       2.5%     97.5% 
  0.7761134 1.0291901 

