Euij2 - 872 - 369 - 1312 also 2560 - how did you get these numbers? A = color, B

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Euij2 - 872 - 369 - 1312 also 2560 - how did you get these numbers?

A = color, B = Black & white, C = combination to test hypothesis, H_0: mu_A = mu_B = mu_c vs H_1: Atleast one mean is different T_i = sigma_j uij sigma_j uij^2 A = 120 872 B = 77 369 C = 149 1312 G = sigma_j sigma_j uij = 346 sigma sigma uij^2 = 2560 = RSS correction factor (C.F) = G^2/N = (346)^2/51 = 2347.3725 Therefore Total S.S = RSS - CF = 2560 - 2347.3725 TSS = 212.62 Between S.S = sigma_i = 1^3 T_i^2/n_i - C.F = [(120)^2/17 + (77)^2/17 + (149)^2/17] - 2347.3725 BSS = 154.39 WSS = TSS - BSS = 212.62 - 154.39 WSS = 58.24 Degrees of freedom rightarrow Total SS = N_i - 1 = 51 - 1 = 50 BSS = 3 - 1 = 2 WSS = 48 ANOVA S.V d.f S.S. MSS. F value P value Between S.S 2 154.39 154.39/2 = 77.196 63.63 0.0001 Within S.S 48 58.24 58.24/48 = 1.213 Total 50 212.63 F_0.05, 2, 50 = 3.18 P value = 0.0001

Explanation / Answer

Suppose the value under A block is 10, 13, 7 , 9 etc (like this 17 values)

Sum uij = 10+13+ 7 + 9+......= 120 ( since question of s nopt given i have assumed values just for understanding)

Sum uij 2 = 102 + 132 + 72 + .......= 872

Similarly for B and C block we get 369 , 1312

Now double sum uij 2 = 872+ 369+ 1312 = 2560

Note the values 10,13, 7,9 are just for understanding

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