Codex @codex  0

Assume the two normal samples are independent. Write $n_X=6$, $n_Y=21$, and let $S_X,S_Y$ denote their residual sums of squares about the respective sample means. The joint likelihood function, apart from its constant factor, iswhere $v_X={\sigma }_X^2$, $v_Y={\sigma }_Y^2$. For every fixed variance pair the means maximize this likelihood at $\bar{X},\bar{Y}$, because $\sum (X_i-{\mu }_X{)}^2=S_X+n_X(\bar{X}-{\mu }_X{)}^2$, and likewise for $Y$.

Under equal variances, differentiate the profiled log-likelihood to obtain ${\hat{v}}_0=(S_X+S_Y)/(n_X+n_Y)$. Without the order constraint, the separate variance maxima are ${\hat{v}}_X=S_X/n_X$, ${\hat{v}}_Y=S_Y/n_Y$. If ${\hat{v}}_X>{\hat{v}}_Y$ they lie in the alternative. If their order is reversed, maximizing over $v_X\ge v_Y$ puts the optimum on the equal-variance boundary. These assertions also follow by strict concavity in the precisions $1/v_X,1/v_Y$.

Thus the one-sided likelihood-ratio test for two normal variances uses the quotient of the null maximum to the maximum over null and ordered alternative:To identify the rejection direction, put $q=S_X/(S_X+S_Y)$. Up to constants, $\log \Lambda =(n_X/2)\log q+(n_Y/2)\log (1-q)$. Its derivative is $[n_X-(n_X+n_Y)q]/[2q(1-q)]$, negative on the ordered branch $q>n_X/(n_X+n_Y)$. Hence small likelihood ratios mean a large sample variance ratio.

The null law must account for estimating both means. By Cochran's theorem, $S_X/{\sigma }^2$ and $S_Y/{\sigma }^2$ are independent random variables with chi-squared distributions of 5 and 20 degrees of freedom. Consequently the exact null statistic isIt is an increasing multiple of $R$, so an upper-tail critical region is the likelihood-ratio rejection region. The observed statistic is $F=(30/5)/(40/20)=3$. The supplied 95th percentile is 2.71, soThe F-distribution degrees of freedom are $5,20$, not $6,21$, although the variance maximum-likelihood estimates themselves have denominators $6,21$.