Use of F-table to find a confidence interval for a ratio of variances

Question

What are other ways to write the CI for the ratio of the variances $\sigma_1^2/\sigma_2^2$?

One way is $((s_1^2/s_2^2)(1/f_{1-\alpha/2,n_1-1,n_2-1}),(s_1^2/s_2^2)(f_{\alpha/2,n_2-1,n_1-1}))$

Could you provide alternative ways of the confidence interval?

Thank you

Answer 1

Suppose you have $n_1 = 10$ observations $X_{1i}$ from $\mathsf{Norm}(\mu_1=20, \sigma_1=2)$ and $n_2 = 15$ observations $X_{1i}$ from $\mathsf{Norm}(\mu_1=30, \sigma_1=5)$. Then $\varphi = 4/25 = 0.16.$ Perhaps (rounded):

x1 = round(rnorm(10, 20, 2),2);  x2 = round(rnorm(15, 30, 5),2)
x1
[1] 22.70 21.39 19.49 23.15 22.31 18.80 19.65 21.09 19.18 20.55
x2
 [1] 29.74 31.69 31.84 24.65 32.46 30.24 28.52 32.03 38.30 29.98 24.76 28.11
[13] 45.31 24.00 21.16

Sample variances are $S_1^2 = 2.3956,\, S_2^2 = 35.3782,$ so the observed variance ratio is $W = S_1^2/S_2^2 = 0.6771.$

var(x1); var(x2)
[1] 2.395632
[1] 35.37823
w = var(x1)/var(x2);  w
[1] 0.06771488

The relevant F-distribution is $\mathsf{F}(9, 14).$ In R statistical software the values $L =0.2633$ and $U = 3.209$ of my Comment can be found as shown below. Please notice that the 'percentage points' (often denoted by subscripts) refer to upper tails, whereas the R quantile function (inverse CDF) looks at probabilities in lower tails.

qf(c(.025, .975), 9, 14)
[1] 0.2632998 3.2093003

From most printed tables of F-distributions, you can get $U = 3.209$ directly, but $L$ only indirectly. Please look now at a such a table so you know exactly what I mean.

From a printed table that gives only upper-tail cut-off values, you can obtain $L = 0.2633$ by taking the reciprocal of an upper-tail cut-off value, as shown below. Notice that the degrees of freedom are interchanged. Please look now at a printed table to find the value 3.798 that cuts probability 2.5% from the upper tail of $\mathsf{F}(14, 9).$

1/qf(.975, 14, 9)
[1] 0.2632998
qf(.975, 14, 9)
[1] 3.797952     # rounded value available in printed table

Finally, a 95% confidence interval for $\varphi,$ in my example with fake data, is given as $(W/3.209, W/0.2633) = (0.0211, 0.2572),$ which does include the value $\varphi = 0.16,$ used to generate my fake data.

w/qf(c(.975, .025), 9, 14)
[1] 0.02109958 0.25717789
w/qf(.975, 9, 14);  w*qf(.975, 14, 9)  
[1] 0.02109958
[1] 0.2571779   

Note: I have used the notation of my Comment throughout, partly because it avoids confusion between upper and lower tails and partly because it is compatible with the conventions used in most statistical software. Someday soon I suppose textbooks will stop filling a dozen pages at the end with printed tables of various distributions used in inference. Software is increasingly available; it is more flexible to use and provides more accurate values.