The table below represents the yield of 4 x 4 experiment. Carry out the ANOVA

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Suresh Chand · Accepted Answer

Solution: 1) Null Hypothesis(H0):There is no significance difference between treatment. 2) Alternative Hypothesis(H1): There is significance difference between treatment. 3) Test statistics under H0 Treatment: Treatment Amount Σxij2 ΣTi A 18 15 18 23 1402 74 B 23 16 18 18 1433 74 C 12 12 23 20 1217 67 D 17 19 19 19 1372 74 Σxij2 = 5424 ΣTi = 290 Now, CF = (frac{G^2}{N} = frac{(2190)^2}{16}) = 5256.25 TSS = Σxij2 - CF = 5424 - 5256.25 = 167.75 SST = (frac{sum T_{i}^{2}}{n_i} - CF) = (frac{74^2}{4} + frac{75^2}{4} + frac{67^2}{4} + frac{74^2}{4} - 5256.25) = 5266.5 - 5256.25 = 10.25 SSE = TSS - SST = 167.75 - 10.25 = 157.5 ANOVA Table   Source of variation S.S. D.F. M.SS F-ratio Due to treatment 10.25 3 3.4167 0.282 Due to error 157.5 13 12.1154 Total 16 - - Here, Fratio = 0.282    Ftab (3, 13) = 3.4105 Decision: Since Fratio <  Ftab So, we accept H0. There may be error in calculation. Please follow the steps only. If you find any error then please report to us.

A18	B23	A15	C12
B16	C12	D17	D19
C23	A18	B18	A23
D19	D19	C20	B18

The table below represents the yield of 4 x 4 experiment. Carry out the ANOVA

A18 B23 A15 C12

B16 C12 D17 D19

C23 A18 B18 A23

D19 D19 C20 B18

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The table below represents the yield of 4 x 4 experiment. Carry out the ANOVA A18 B23 A15 C12 B16 C12 D17 D19 C23 A18 B18 A23 D19 D19 C20 B18

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The table below represents the yield of 4 x 4 experiment. Carry out the ANOVA

A18 B23 A15 C12

B16 C12 D17 D19

C23 A18 B18 A23

D19 D19 C20 B18