Question

5. A farmer is interested in finding the oat variety that will produce the greatest yield on his farm. To do this he divides his field into 6 distinct blocks and plants each of three oat varieties at random locations in each block. Below are the yield results from the farmers experiment. Block ii iv vi Vict. 61 68 74 53 Gold 117 70 60 64 80 89 Marv. 105 96 89 70 63 97 -- v 111 62 The total sum of squares for the yield data is 6394.2, SSblocks=3858.9 and SSvariety=693.4. Complete an ANOVA table and test at the a = .05 level whether yield varies by variety controlling for the variation observed in the blocks.

Answer 1

Let ss variety SS blocks Xij = itoba (variety of 7th block, i = 1 (03, J = 1(1) 6. By A NOVA, we have ²6 (3; -700) & 3 (Xo Too? SS Error = SS Total - SS variety - SSblocks, = Žj (xij - 800 i=1;=1 Now, df (variety) = 34 = 2, of (blocks) = 6-1 = 5 df (total) 18-1 = 17 df Cerror) = 17-(5+2) = 10 = ss variety/2 SS blocks/5, SS Total 36 Ms variety MS blocks MS Error = SS Enor per 10 :F-statistic = MS variety MS Error F-critical Fa; 2, 10. - Fo.05; 2,10 , ford = 0.05. Here, given that, SS variety 693-4, SS blocks = 3858-9, Stotal 6394.2. :SSEooor 6394:2-693.4-3858.9.= 1841-9 - MS variety = 693-4/2 = 346.7 SBlocks 3858.9/5 771078, 1841.9/10 = 184.19, 346.7 F-statistic 1.882, 184*19 F-critical F0:05; 2,10 = 4.1. MS MS Error

ANOVA table df SS source of MS E-stat F-critical variation din Variety 693-4 34607 1.882 Blocks 3858.9 7778 Error 1841.9 184.19 Total 17 6394-2 F-stat 1•88 2 (4./ = F-critical, → We accept Ho, at level 0.05. So, the different variety gives seme yield on his farm.