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H_0:μ_(1∙)=μ_(2∙)=⋯=μ_(a∙) F=MSA/MSE F_(ν_1,ν_2 ) ν_1=a-1 ,

ν_2=ab(n-1)

H_0:μ_(∙1)=μ_(∙2)=⋯=μ_(∙b) F=MSB/MSE F_(ν_1,ν_2 ) ν_1=b-1 ,

ν_2=ab(n-1)

H_0: there is not interactions between factors F=(MS(AB))/MSE F_(ν_1,ν_2 ) ν_1=(a-1)(b-1),

ν_2=N-ab

Contrast

H_0:c_1 μ_1+c_2 μ_2+⋯c_k μ_k=0 ∑▒c=0 t=(∑_(i=1)^k▒〖c_i x ̅_i 〗)/(s_p √(∑_(i=1)^k▒(c_i^2)/n_i )) Student t(υ) ν=n_1+⋯+n_k-k

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