 Solution

For a normal sample of size $m$,Then $\bar{X}\sim N({\mu }_X,{\sigma }^2/m)$ and $c{S}_X^2\sim {\chi }_{m-1}^2$ with $c=(m-1)/{\sigma }^2$. These two statistics are independent.

Combining the difference distribution and pooled variance calculation gives, with $\nu =m+n-2$,The printed statistic has the sign of the centering term reversed. Its numerator subtracts ${\mu }_Y-{\mu }_X$, so it is not generally a central Student statistic; the corrected centering above is necessary. Under the null hypothesis ${\mu }_X={\mu }_Y$, the two expressions coincide, so the requested test is unaffected. Reject the two-sided null at significance level $\alpha$ whenHere $\nu =8$ and $|T|=2.19<2.31$, so do not reject at the 5% level. This does not establish equality of the population means.

Age matching means that the two observations in a pair share an age effect, while ages vary across pairs. The independent, identically distributed two-sample model no longer represents this design: random pair effects induce within-pair dependence, and fixed differing ages can instead produce different individual means. Pairing by itself need not cause stochastic dependence when ages are treated as fixed. The appropriate alternative is a paired t-test on the within-pair differences. Removing shared age variation can reduce the noise and improve power when age strongly affects volume and the members of a pair are positively associated. Improvement is not automatic, since the paired test also uses fewer degrees of freedom.