An Introduction to Probability and Statistical Inference, 2nd Edition

14.3.2 Testing the Hypothesis of No Row Or No Column Effects

First, consider the null hypothesis of no row effect; namely

$H_{0,A}:{\alpha }_1=\cdots ={\alpha }_I=0.$

  (30)

Under H_0,A, the likelihood function in (24), to be denoted for convenience by L_A(y;μ, β, σ²), becomes:

$L_A\left(\mathbf{y};\mu ,\mathbf{\beta },{\sigma }^2\right)={\left(\frac{1}{\sqrt{2\pi {\sigma }^2}}\right)}^{IJ}\text{?exp}\left[-\frac{1}{2{\sigma }^2}{\displaystyle \sum _i{\displaystyle \sum _j{\left(y_{ij}-\mu -{\beta }_j\right)}^2}}\right].$

  (31)

Maximization of this likelihood with respect to β_j's and μ, for each fixed σ², is equivalent to minimization, with respect to β_j's and μ of the expression:

$\mathcal{S}\left(\mu ,{\beta }_1,\dots ,{\beta }_J\right)=\mathcal{S}\left(\mu ,\mathbf{\beta }\right)={\displaystyle \sum _i{\displaystyle \sum _j{\left(y_{ij}-\mu -{\beta }_j\right)}^2.}}$

  (32)

Working exactly as in (25), we obtain ...