The correct answer is A (t-statistic). For a hypothesis test concerning the mean difference between two normally distributed populations with unknown variances, the appropriate test statistic is the t-statistic. This is because the test involves paired observations, and the t-statistic is calculated as:

$t = \frac{d - \mu_{d0}}{s_d / \sqrt{n}}$

where:

• $d$ is the sample mean difference,
• $\mu_{d0}$ is the hypothesized mean difference (often zero),
• $s_d$ is the standard error of the mean difference,
• $n$ is the number of paired observations.

The t-statistic follows a t-distribution with $n - 1$ degrees of freedom.

Option B (F-statistic) is incorrect because the F-test is used for comparing variances, not mean differences. Option C (Chi-square statistic) is also incorrect, as it is used for tests concerning the variance of a single normally distributed population, not mean differences between two populations.