1. **State the problem:** We want to determine if there is a significant difference in mean sodium amounts among three groups: Condiments, Cereals, and Desserts, using the between-group variance at a 0.05 significance level. 2. **Data given:** Condiments: 270, 130, 230, 180, 80, 70, 217 Cereals: 260, 220, 290, 290, 200, 320, 195 Desserts: 100, 180, 250, 250, 300, 360, 300, 174 3. **Formula for between-group variance (Mean Square Between, MSB):** $$MSB = \frac{\sum_{i=1}^k n_i (\bar{x}_i - \bar{x})^2}{k-1}$$ where: - $k$ = number of groups - $n_i$ = number of observations in group $i$ - $\bar{x}_i$ = mean of group $i$ - $\bar{x}$ = overall mean 4. **Calculate group means:** - Condiments mean $\bar{x}_1 = \frac{270 + 130 + 230 + 180 + 80 + 70 + 217}{7} = \frac{1177}{7} = 168.14$ - Cereals mean $\bar{x}_2 = \frac{260 + 220 + 290 + 290 + 200 + 320 + 195}{7} = \frac{1775}{7} = 253.57$ - Desserts mean $\bar{x}_3 = \frac{100 + 180 + 250 + 250 + 300 + 360 + 300 + 174}{8} = \frac{1914}{8} = 239.25$ 5. **Calculate overall mean $\bar{x}$:** Total sum = 1177 + 1775 + 1914 = 4866 Total observations = 7 + 7 + 8 = 22 $$\bar{x} = \frac{4866}{22} = 221.18$$ 6. **Calculate each term $n_i (\bar{x}_i - \bar{x})^2$:** - Condiments: $7 \times (168.14 - 221.18)^2 = 7 \times (-53.04)^2 = 7 \times 2812.24 = 19685.68$ - Cereals: $7 \times (253.57 - 221.18)^2 = 7 \times 32.39^2 = 7 \times 1049.06 = 7343.42$ - Desserts: $8 \times (239.25 - 221.18)^2 = 8 \times 18.07^2 = 8 \times 326.52 = 2612.16$ 7. **Sum these values:** $$19685.68 + 7343.42 + 2612.16 = 29641.26$$ 8. **Calculate between-group variance:** Number of groups $k=3$ $$MSB = \frac{29641.26}{3-1} = \frac{29641.26}{2} = 14820.63$$ 9. **Final answer rounded to two decimal places:** $$\boxed{14820.63}$$ This is the between-group variance for the sodium amounts among the three food groups.