## The confidence interval for the ratio of population variances

Consider two normal distributions $\small{N(\mu_X, \sigma_X^2)}$ and $\small{N(\mu_Y, \sigma_Y^2)}$. We are interested in comparing the population variance $\small{\sigma_X^2}$ and $\small{\sigma_Y^2}$.

Suppose we randomly draw n and m samples from these two distributions X and Y respectively and estimate their sample variances $\small{S_X^2}$ and $\small{S_Y^2}$. Using these estimates, we can derive a confidence interval for the ratio $\small{\dfrac{\sigma_X^2}{\sigma_Y^2}}$ as follows:

According to the definition of the F distribution, the ratio $\small{ \dfrac{S_Y^2/\sigma_Y^2}{S_X^2/\sigma_X^2} }$ follows an F distribution with $\small{r_1 = m-1}$ and $\small{r_2 = n-1 }$ degrees of freedom.
(Note : n is the number of samples in X and m is the number of samples in Y )

For a given significance $\small{\alpha}$, choose two F values F1 and F2 on the F distribution such that the area under the F curve between $\small{F_1}$ and $\small{F_2}$ is $\small{\alpha}$. We therefore let,
$\small{ F_1 = F_{\alpha/2}(m-1,n-1)~~ }$ and $~~\small{F_2 = F_{1-\alpha/2}(m-1, n-1) }$
(The area under the curve from 0 to $\small{F_1}$ is $\small{\alpha/2}$ and from 0 $\small{F_2}$ is $\small{1-\alpha/2}$ )

and can write the inequality

$~~~~~~~~~~~~\small{1 - \alpha~=~ P\left( F_1~\leq~\dfrac{S_Y^2/\sigma_Y^2}{S_X^2/\sigma_X^2}~\leq~F_2 \right) }$

$~~~~~~~~~~~~~~\small{~~~~~~~=~P\left(F_1 \dfrac{S_X^2}{S_Y^2}~\leq~\dfrac{\sigma_X^2}{\sigma_Y^2}~\leq~F_2 \dfrac{S_X^2}{S_Y^2} \right)}$

$~~~~~~~~~~~~~~\small{~~~~~~~=~P\left(F_{\alpha/2}(m-1,n-1) \dfrac{S_X^2}{S_Y^2}~\leq~\dfrac{\sigma_X^2}{\sigma_Y^2}~\leq~F_{1-\alpha/2}(m-1, n-1) \dfrac{S_X^2}{S_Y^2} \right)}$

Therefore, the interval given by

$\small{\left[ F_{\alpha/2}(m-1,n-1) \dfrac{S_X^2}{S_Y^2}, ~~ F_{1-\alpha/2}(m-1, n-1) \dfrac{S_X^2}{S_Y^2} \right] }~~~~~$
is a $\small{100(1-\alpha)\%}$ confidence interval for the ration of proportions $\small{\dfrac{\sigma_X^2}{\sigma_Y^2} }$.

It follows that the interval,

$\small{\left[ \sqrt{F_{\alpha/2}(m-1,n-1)} \dfrac{S_X}{S_Y}, ~~ \sqrt{F_{1-\alpha/2}(m-1, n-1)} \dfrac{S_X}{S_Y} \right] }~~~~~$
is a $\small{100(1-\alpha)\%}$ confidence interval for the ration of proportion $\small{\dfrac{\sigma_X}{\sigma_Y} }$ of population standard deviations.

Important Note : The above mentioned confidence interval for the ratio of variances are fully valid only when the two underlying distributions are perfectly Gaussian. If the distributions are non-Gaussian, the estimate of confidence intervals for the ratio of variances are not accurate and erro prone.

Example-1 : Two data sets, X and Y are assumed to have normal distributions $\small{N(\mu_X, \sigma_X^2)}$ and $\small{N(\mu_Y, \sigma_Y^2)}$. A random draw of 12 and 15 data points respectively from X and Y resulted in the following data:

$~~~~~~~~~~~~\small{ X = \{ 9.1, 12.5, 10.2, 9.5, 7.3, 5.6, 10.1, 13.0, 12.8, 9.0, 7.9, 7.7 \} }$
$~~~~~~~~~~~~\small{ Y = \{11.6, 21.0, 20.9, 7.1, 15.9, 15.6, 17.9, 10.3, 16.5, 17.4, 15.7, 17.1, 13.5, 12.7, 19.0 \} }$
Find a two sided $\small{95\% }$ confidence interval for the ratio of population variances $\small{\dfrac{\sigma_X^2}{\sigma_Y^2 } }$

From the data, n = 12, m = 15, $\small{\S_X = 2.32 }$ and $\small{\S_Y = 3.86~~ }$, $~~\small{\dfrac{S_X}{S_Y} = 0.5998}~~~$, and $\small{\dfrac{S_X^2}{S_Y^2} = 0.3597 }$

Also, $\small{\alpha = 0.05 }$ and hence $\small{\alpha/2 = 0.025 }$

We find $\small{F_{\alpha/2}(m-1, n-1) = F_{0.025}(14, 11) = 0.3231 }$
and, $~~~\small{F_{1 - \alpha/2}(m-1, n-1) = F_{0.925}(14, 11) = 2.4048 }$

With these numbers, we can write a $\small{95\%}$ confidence interval for $\small{\dfrac{\sigma_X^2}{\sigma_Y^2 } }$ as,

$\small{\left[ F_{\alpha/2}(m-1,n-1) \dfrac{S_X^2}{S_Y^2}, ~~ F_{1-\alpha/2}(m-1, n-1) \dfrac{S_X^2}{S_Y^2} \right] ~ = ~[ 0.3231 \times 0.5998, 2.4048 \times 0.5992]~=~[0.194, 1.44 ] }$

We note that the point estimate $\small{\dfrac{S_X^2}{S_Y^2} = 0.3597}$ is outside this confidenc interval.