= Solution
For a normal sample of size $m$,
$$
\bar X=\frac1m\sum_{i=1}^mX_i,\qquad S_X^2=\frac1{m-1}\sum_{i=1}^m(X_i-\bar X)^2.
$$
Then $\bar X\sim N(\mu_X,\sigma^2/m)$ and $cS_X^2\sim\chi^2_{m-1}$ with $c=(m-1)/\sigma^2$. These two statistics are independent.
Combining the difference distribution and pooled <variance> calculation gives, with $\nu=m+n-2$,
$$
T_*=\frac{\bar X-\bar Y-(\mu_X-\mu_Y)}{S\sqrt{1/m+1/n}}\sim t_\nu.
$$
\b[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 $\alpha$ when
$$
\left|\frac{\bar X-\bar Y}{S\sqrt{1/m+1/n}}\right|>t_{\nu,1-\alpha/2}.
$$
Here $\nu=8$ and $|T|=2.19<2.31$, so \b[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 The independent, identically distributed two-sample model no longer represents this design: random pair effects induce within-pair dependence, and fixed differing ages can instead produce different individual means.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 \b[<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>.
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