Compute the ANOVA table using the following figures obtained from three (3) sample areas: Group A [3, 1, 3, 2, 4, 3], Group B [4, 3, 8, 5, 4, -], Group C [9, 7, 8, 11, 9, -]. Test the hypothesis at 95% probability level.

Question 1

Answer · 2026-03-18T10:05:27.450Z

Step 1: Record the data and compute group totals, means, and grand mean. Data: c|ccc & A & B & C \\ Rep 1 & 1 & 3 & 8 \\ Rep 2 & 2 & 4 & 7 \\ Rep 3 & 4 & 5 & 11 \\ Sum & 7 & 12 & 26 \\ Mean & (7)/(3) & 4 & (26)/(3) \\ Total sum T = 7 + 12 + 26 = 45. Grand mean X = (45)/(9) = 5. k=3 groups, n_i=3 per group, N=9. Step 2: Sum of Squares Between Groups (SSB). Formula: SSB = _i=1^k n_i (X_i - X)^2. X_A - 5 = (7)/(3) - 5 = -(8)/(3), (-(8)/(3))^2 = (64)/(9). X_B - 5 = 4 - 5 = -1, (-1)^2 = 1 = (9)/(9). X_C - 5 = (26)/(3) - 5 = (11)/(3), ((11)/(3))^2 = (121)/(9). (X_i - X)^2 = (64 + 9 + 121)/(9) = (194)/(9). SSB = 3 × (194)/(9) = (194)/(3). Step 3: Sum of Squares Within Groups (SSW). Formula: SSW = _i=1^k _j=1^n_i (X_ij - X_i)^2. Group A: 1 - (7)/(3) = -(4)/(3), (-(4)/(3))^2 = (16)/(9); 2 - (7)/(3) = -(1)/(3), (1)/(9); 4 - (7)/(3) = (5)/(3), (25)/(9). SS_A = (16+1+25)/(9) = (42)/(9) = (14)/(3). Group B: 3-4 = -1, 1; 4-4=0, 0; 5-4=1, 1. SS_B = 1 + 0 + 1 = 2. Group C: 8 - (26)/(3) = -(2)/(3), (4)/(9); 7 - (26)/(3) = -(5)/(3), (25)/(9); 11 - (26)/(3) = (7)/(3), (49)/(9). SS_C = (4+25+49)/(9) = (78)/(9) = (26)/(3). SSW = (14)/(3) + 2 + (26)/(3) = (14 + 6 + 26)/(3) = (46)/(3). Step 4: Total Sum of Squares (SST). SST = SSB + SSW = (194)/(3) + (46)/(3) = (240)/(3) = 80. Step 5: Degrees of freedom. Between: df_B = k-1 = 2. Within: df_W = N - k = 6. Total: df_T = N-1 = 8. Step 6: Mean squares. MSB = (SSB)/(df_B) = (194/3)/(2) = (194)/(6) = (97)/(3). MSW = (SSW)/(df_W) = (46/3)/(6) = (46)/(18) = (23)/(9). Step 7: F-statistic. F = (MSB)/(MSW) = (97/3)/(23/9) = (97)/(3) × (9)/(23) = (97 × 3)/(23) = (291)/(23). ANOVA \ table: c|cccc Source & df & SS & MS & F \\ Between & 2 & (194)/(3) & (97)/(3) & (291)/(23) \\ Within & 6 & (46)/(3) & (23)/(9) & \\ Total & 8 & 80 & & \\ Step 8: Hypothesis test. H_0: _A = _B = _C (means equal). H_a: At least one mean differs. Significance level: 99% probability = 0.01. Critical value: F_0.01, 2,6 = 18.16. Computed F = (291)/(23) ≈ 12.65 < 18.16. Fail to reject H_0. There is not sufficient evidence at the 1% significance level to conclude that the group means differ. ANOVA table as above; F = 291/23 ≈ 12.65 < F_crit=18.16; fail to reject H_0

Question 2

Step 1: Record the data and compute group totals, means, and grand mean. Data: c|ccc & A & B & C \\ Rep 1 & 1 & 3 & 8 \\ Rep 2 & 2 & 4 & 7 \\ Rep 3 & 4 & 5 & 11 \\ Sum & 7 & 12 & 26 \\ Mean & (7)/(3) & 4 & (26)/(3) \\ Total sum T = 7 + 12 + 26 = 45. Grand mean X = (45)/(9) = 5. k=3 groups, n_i=3 per group, N=9. Step 2: Sum of Squares Between Groups (SSB). Formula: SSB = _i=1^k n_i (X_i - X)^2. X_A - 5 = (7)/(3) - 5 = -(8)/(3), (-(8)/(3))^2 = (64)/(9). X_B - 5 = 4 - 5 = -1, (-1)^2 = 1 = (9)/(9). X_C - 5 = (26)/(3) - 5 = (11)/(3), ((11)/(3))^2 = (121)/(9). (X_i - X)^2 = (64 + 9 + 121)/(9) = (194)/(9). SSB = 3 × (194)/(9) = (194)/(3). Step 3: Sum of Squares Within Groups (SSW). Formula: SSW = _i=1^k _j=1^n_i (X_ij - X_i)^2. Group A: 1 - (7)/(3) = -(4)/(3), (-(4)/(3))^2 = (16)/(9); 2 - (7)/(3) = -(1)/(3), (1)/(9); 4 - (7)/(3) = (5)/(3), (25)/(9). SS_A = (16+1+25)/(9) = (42)/(9) = (14)/(3). Group B: 3-4 = -1, 1; 4-4=0, 0; 5-4=1, 1. SS_B = 1 + 0 + 1 = 2. Group C: 8 - (26)/(3) = -(2)/(3), (4)/(9); 7 - (26)/(3) = -(5)/(3), (25)/(9); 11 - (26)/(3) = (7)/(3), (49)/(9). SS_C = (4+25+49)/(9) = (78)/(9) = (26)/(3). SSW = (14)/(3) + 2 + (26)/(3) = (14 + 6 + 26)/(3) = (46)/(3). Step 4: Total Sum of Squares (SST). SST = SSB + SSW = (194)/(3) + (46)/(3) = (240)/(3) = 80. Step 5: Degrees of freedom. Between: df_B = k-1 = 2. Within: df_W = N - k = 6. Total: df_T = N-1 = 8. Step 6: Mean squares. MSB = (SSB)/(df_B) = (194/3)/(2) = (194)/(6) = (97)/(3). MSW = (SSW)/(df_W) = (46/3)/(6) = (46)/(18) = (23)/(9). Step 7: F-statistic. F = (MSB)/(MSW) = (97/3)/(23/9) = (97)/(3) × (9)/(23) = (97 × 3)/(23) = (291)/(23). ANOVA \ table: c|cccc Source & df & SS & MS & F \\ Between & 2 & (194)/(3) & (97)/(3) & (291)/(23) \\ Within & 6 & (46)/(3) & (23)/(9) & \\ Total & 8 & 80 & & \\ Step 8: Hypothesis test. H_0: _A = _B = _C (means equal). H_a: At least one mean differs. Significance level: 99% probability = 0.01. Critical value: F_0.01, 2,6 = 18.16. Computed F = (291)/(23) ≈ 12.65 < 18.16. Fail to reject H_0. There is not sufficient evidence at the 1% significance level to conclude that the group means differ. ANOVA table as above; F = 291/23 ≈ 12.65 < F_crit=18.16; fail to reject H_0

Compute the ANOVA table using the following figures obtained from three (3) sample areas: Group A [3, 1, 3, 2, 4, 3], Group B [4, 3, 8, 5, 4, -], Group C [9, 7, 8, 11, 9, -]. Test the hypothesis at 95% probability level.

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