1. **Stating the problem:** We are given a contingency table showing joint frequencies of two categorical variables: "Poste" (position) and "But" (goal count categories). The table also provides marginal totals and a value $ \Phi^2 = 0.0987 $. 2. **Goal:** We want to understand the association between the two categorical variables using the $ \Phi^2 $ coefficient and possibly calculate related statistics like $ V^2 $ (Cramér's V squared) and $ \chi^2_n $ (chi-square statistic normalized). 3. **Formula and explanation:** - The $ \Phi^2 $ coefficient is defined as: $$\Phi^2 = \frac{\chi^2}{n}$$ where $ \chi^2 $ is the chi-square statistic and $ n $ is the total sample size. - Cramér's V squared is: $$V^2 = \frac{\Phi^2}{\min(k-1, r-1)}$$ where $ k $ is the number of columns and $ r $ is the number of rows. - $ \chi^2_n $ is the chi-square statistic normalized by sample size, which equals $ \Phi^2 $. 4. **Calculate total sample size $ n $:** From the table, the total sum of frequencies is 1 (since marginals sum to 1), so assume frequencies are proportions. 5. **Determine dimensions:** - Rows (Poste): 3 (Attaquant, Milieu, Défenseur) - Columns (But): 3 (Aucun, 1 à 3, 4 Plus) 6. **Calculate Cramér's V squared:** $$V^2 = \frac{\Phi^2}{\min(3-1,3-1)} = \frac{0.0987}{2} = 0.04935$$ 7. **Interpretation:** - $ \Phi^2 = 0.0987 $ indicates some association between "Poste" and "But". - $ V^2 = 0.04935 $ is a normalized measure of association accounting for table size. **Final answers:** - $ \Phi^2 = 0.0987 $ - $ V^2 = 0.04935 $ - $ \chi^2_n = 0.0987 $ (since $ \chi^2_n = \Phi^2 $)