If \(X_1,X_2,\ldots,X_n \sim N(\mu_X,\sigma^2_X)\) and \(Y_1,Y_2,\ldots,Y_m \sim N(\mu_Y,\sigma^2_Y)\) are independent random samples, and:

1. \(c=F_{1-\alpha/2}(m-1,n-1)=\dfrac{1}{F_{\alpha/2}(n-1,m-1)}\) and

2. \(d=F_{\alpha/2}(m-1,n-1)\),

then a \((1−\alpha) 100\%\) confidence interval for \(\sigma^2_X/\sigma^2_Y\) is:

\(\left(\dfrac{1}{F_{\alpha/2}(n-1,m-1)} \dfrac{s^2_X}{s^2_Y} \leq \dfrac{\sigma^2_X}{\sigma^2_Y}\leq F_{\alpha/2}(m-1,n-1)\dfrac{s^2_X}{s^2_Y}\right)\)

Proof

Because \(X_1,X_2,\ldots,X_n \sim N(\mu_X,\sigma^2_X)\) and \(Y_1,Y_2,\ldots,Y_m \sim N(\mu_Y,\sigma^2_Y)\) , it tells us that:

\(\dfrac{(n-1)S^2_X}{\sigma^2_X}\sim \chi^2_{n-1}\) and \(\dfrac{(m-1)S^2_Y}{\sigma^2_Y}\sim \chi^2_{m-1}\)

Then, by the independence of the two samples, we well as the definition of an F random variable, we know that:

\(F=\dfrac{\dfrac{(m-1)S^2_Y}{\sigma^2_Y}/(m-1)}{\dfrac{(n-1)S^2_X}{\sigma^2_X}/(n-1)}=\dfrac{\sigma^2_X}{\sigma^2_Y}\cdot \dfrac{S^2_Y}{S^2_X} \sim F(m-1,n-1)\)

Therefore, the following probability statement holds:

\(P\left[F_{1-\frac{\alpha}{2}}(m-1,n-1) \leq \dfrac{\sigma^2_X}{\sigma^2_Y}\cdot \dfrac{S^2_Y}{S^2_X} \leq F_{\frac{\alpha}{2}}(m-1,n-1)\right]=1-\alpha\)

Finding the \((1-\alpha)100\%\) confidence interval for the ratio of the two population variances then reduces, as always, to manipulating the quantity in parentheses. Multiplying through the inequality by:

\(\dfrac{S^2_X}{S^2_Y}\)

and recalling the fact that:

\(F_{1-\frac{\alpha}{2}}(m-1,n-1)=\dfrac{1}{F_{\frac{\alpha}{2}}(n-1,m-1)}\)

the \((1-\alpha)100\%\) confidence interval for the ratio of the two population variances reduces to:

\(\dfrac{1}{F_{\frac{\alpha}{2}}(n-1,m-1)}\dfrac{S^2_X}{S^2_Y}\leq \dfrac{\sigma^2_X}{\sigma^2_Y} \leq F_{\frac{\alpha}{2}}(m-1,n-1)\dfrac{S^2_X}{S^2_Y}\)

as was to be proved.