Hence there are no differences in the mean gains in weight for
the different breeds of cattle.
Solution: Given problem Observation B C D 1 41 45 2 46.5 47.5 50 62 41.5 40 31.5 22 25.5 3 28.5 B с D Row total (4) 1 194.5 46.5 47.5 50 62 41.5 41 22 2 45 31.5 28.5 142.5 3 40 25.5 144 Col total (c) 144 143.5 88.5 105 481
Σ' - = 20729.5 – (Α) Σ (144) + 143.57 + 88.57 + - 10s?) T 1 (20736 - 20592.25 + 7832.25 + 11025) -ξσο1855) = 20061.8333 - (Β) ΣΧ 1 (194.3' + 142.5? + ) + 1443) C 1 (37830.25 + 20306.25 + 20736) 1 (78872.5) 19718.125 - (C) (Σ)? (481)? Η 12 231361 12 19280.0833 – (D)
Sum of squares total (2x) SS, - Ex- = (4) - (D) = 20729.5 - 19280.0833 = 1449.4167 Sum of squares between rows Ex (x)2 SS, = -(C)-(D) C = 19718.125 - 19280.0833 = 438.0417 Sum of squares between columns Σ» (Σx SSC (B)-(D) 7 = 20061.8333 - 19280.0833 = 781.75 Sum of squares Error (residual) SS: = SS7-SSR-SSC = 1449.4167 - 438.0417 - 781.75 = 229.625
ANOVA table Source of Variation Sums of Squares Degrees of freedom SS DF Mean Squares MS F p-value 438.0417 219.0208 Between rows SSR = 438.0417 r. 1 = 2 MSR = 219.0208 = 5.7229 0.0407 2 38.2708 781.75 Between columns SSC = 781.75 C-1 = 3 MSC = 260.5833 260.5833 = 6.8089 0.0286 38.2708 3 229.625 Error (residual) SSE = 229.625 (1 - 1)(c - 1) - 6 MSE 38.2708 6 Total SS, = 1449.4167 rc - 1 = 11 Conclusion: 1. F for between columns F(2.6) at 0.05 level of significance = 5.1433 As calculated FR = 5.7229 > 5.1433 So, H, is rejected, Hence there is significant differentiating between rows 2. F for between columns F(3.6) at 0.05 level of significance = 4.7571 As calculated Fc = 6.8089 > 4.7571 So, H, is rejected, Hence there is significant differentiating between columns