1. **State the problem:** Calculate Spearman's rank correlation coefficient for the given ranks of 6 students in Math and Physics, and determine if there is a significant positive correlation at the 0.01 level. 2. **Formula:** Spearman's rank correlation coefficient $r_s$ is given by: $$r_s = 1 - \frac{6 \sum d_i^2}{n(n^2 - 1)}$$ where $d_i$ is the difference between the ranks of each pair and $n$ is the number of pairs. 3. **Calculate differences and squared differences:** | Student | Math Rank ($x_i$) | Physics Rank ($y_i$) | $d_i = x_i - y_i$ | $d_i^2$ | |---------|------------------|---------------------|------------------|---------| | A | 1 | 2 | $1 - 2 = -1$ | 1 | | B | 2 | 1 | $2 - 1 = 1$ | 1 | | C | 3 | 4 | $3 - 4 = -1$ | 1 | | D | 4 | 3 | $4 - 3 = 1$ | 1 | | E | 5 | 6 | $5 - 6 = -1$ | 1 | | F | 6 | 5 | $6 - 5 = 1$ | 1 | Sum of squared differences: $$\sum d_i^2 = 1 + 1 + 1 + 1 + 1 + 1 = 6$$ 4. **Apply formula:** $$r_s = 1 - \frac{6 \times 6}{6(6^2 - 1)} = 1 - \frac{36}{6(36 - 1)} = 1 - \frac{36}{6 \times 35}$$ 5. **Simplify denominator:** $$6 \times 35 = 210$$ 6. **Calculate fraction:** $$\frac{36}{210} = \frac{\cancel{36}}{\cancel{210}} = \frac{6}{35} \approx 0.171$$ 7. **Calculate $r_s$:** $$r_s = 1 - 0.171 = 0.829$$ 8. **Interpretation:** For $n=6$, the critical value of Spearman's $r_s$ at the 0.01 significance level (two-tailed) is approximately 0.886. Since $0.829 < 0.886$, the correlation is not significant at the 0.01 level. **Final answer:** Spearman's rank correlation coefficient is $\boxed{0.829}$. There is no significant positive correlation at the 0.01 level.