= Solution
\b[Prefer Student 2's paired analysis], because the observational units are matched pairs: two shoes exposed to the same child's activity. A <paired t-test> uses $Y_i=A_i-B_i$, so shared child-to-child variation is removed before estimating the <standard error>. The independent units are the ten pairs, giving $9$ <statistical degrees of freedom> under the model of independent normally distributed pair differences.
The role of <covariance> is explicit:
$$
\operatorname{Var}(A_i-B_i)=\operatorname{Var}(A_i)+\operatorname{Var}(B_i)-2\operatorname{Cov}(A_i,B_i).
$$
With common marginal <variance> $\sigma^2$ and within-pair <correlation coefficient> $\rho$, the true <variance> of the average difference is $2\sigma^2(1-\rho)/10$. The independent-groups calculation ignores the covariance term and uses $2\sigma^2/10$. Positive within-pair association can therefore make the paired <standard error> much smaller, explaining how the reported <p-values> can differ without arithmetic mistakes. The preference follows from the paired design, not from selecting the smaller <p-value>. Marginal normality alone need not imply normally distributed differences; exact paired inference uses the normal-difference assumption, for example justified by a joint <multivariate normal distribution>.
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