= Solution
For a sample size $m\ge2$ define $\overline X=m^{-1}\sum_iX_i$ and $S_X^2=(m-1)^{-1}\sum_i(X_i-\overline X)^2$. The normal-sample theorem gives
$$
\boxed{\overline X\sim N(\mu_X,\sigma^2/m),\qquad\frac{(m-1)S_X^2}{\sigma^2}\sim\chi^2_{m-1},\qquad c=\frac{m-1}{\sigma^2}.}
$$
It also states that the mean and <sample variance> are <independent>. Combining (i) and (ii), and the <independence> across samples, the correctly centred pivot is
$$
\boxed{T_0=\frac{\overline X-\overline Y-(\mu_X-\mu_Y)}{S\sqrt{1/m+1/n}}\sim t_{m+n-2}.}
$$
The source reverses the mean difference inside the subtraction. Its literal statistic is generally noncentral when $\mu_X\ne\mu_Y$, rather than having a central Student distribution. Under the <null hypothesis> both differences are zero, so this misprint does not change the requested null test.
For a two-sided level-$\alpha$ test reject when $|T|>t_{m+n-2,1-\alpha/2}$. In the numerical example the pooled <variance> is $0.038$, the standard error is $\sqrt{0.038(2/5)}\approx0.12329$, and $T\approx-2.190$. There are eight <degrees of freedom> and the supplied critical value is $2.31$. Thus \b[do not reject equal mean cell volume at the 5% level]. This does not establish equality of the means; it states that the observed difference does not cross this test's two-sided threshold.
Age <matching> introduces a shared block factor: the two observations in a pair can be associated, and their age-dependent means also undermine the simple unblocked <independent>-samples model. A \b[<paired t-test>] on the within-pair drug-minus-placebo differences accounts for that design, assuming <independent> pairs and the corresponding normal-difference model. Shared age <variation> cancels in those differences; positive within-pair association reduces their <variance> and can increase <power function of a statistical test>[power], although the reduction in <degrees of freedom> means improvement is not automatic. The paired test cannot be computed from the two marginal means and <variances> alone: the within-pair differences or <covariance> are needed.
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