Question

In a two-way layout with one observation in each cell, construct a test of the null hypothesis that all the effects of both factor \(A\) and factor \(B\) are\({\bf{0}}\).

Short answer

Reject \({H_0}\)if\(U_{A + B}^2 = \frac{{(I - 1)(J - 1)\left( {S_A^2 + S_B^2} \right)}}{{(I + J - 2)S_{{\rm{Resid }}}^2}} \ge F_{I + J - 2,(I - 1)(J - 1)}^{ - 1}\left( {1 - {\alpha _0}} \right)\).

Step-by-step solution

Definition for probability and statistics

• Probability is concerned with forecasting the possibility of future events, whereas statistics is concerned with analysing the frequency of previous events.
• Probability is largely a theoretical discipline of mathematics that investigates the implications of mathematical notions.

Determine the null hypothesis

• Given: Two-way layout with one observation in each cell.
• \(K = {\rm{ Number of repetitions }} = 1\)
• We are interested in testing whether the effects of both factors (\(A\)and \(B\)) are 0.
• \({H_0}:{\alpha _i} = {\beta _j} = 0{\rm{ for }}i = 1, \ldots ,I,j = 1, \ldots ,J{H_1}:{H_ - }0{\rm{ is not true}}\)
• When\(K = 1\), then we are basically using the two-way layout without replications.
• One of the theorems of this textbook for the two-way layout without replications with the above hypotheses, the statistics\(U_{A + B}^2\)has the \({\rm{F}}\)distribution with \(I + J - 2\)and \((I - 1)(J - 1)\)degrees of freedom if the null hypothesis \({H_0}\)is true.
• \(U_{A + B}^2 = \frac{{S_A^2 + S_B^2}}{{(I + J - 2){\sigma ^{\prime 2}}}} = \frac{{(I - 1)(J - 1)\left( {S_A^2 + S_B^2} \right)}}{{(I + J - 2)S_{{\rm{Resid }}}^2}}\)
• We can then use the test statistic \(U_{A + B}^2\)to test the null hypothesis, where we will reject the null hypothesis \({H_0}\)when \(U_{A + B}^2\)is larger than or equal to the critical value\(F_{I + J - 2,(I - 1)(J - 1)}^{ - 1}\left( {1 - {\alpha _0}} \right)\).
• Reject \({H_ - }0\)if\(U_{A + B}^2 = \frac{{(I - 1)(J - 1)\left( {S_A^2 + S_B^2} \right)}}{{(I + J - 2)S_{{\rm{Resid }}}^2}} \ge F_{I + J - 2,(I - 1)(J - 1)}^{ - 1}\left( {1 - {\alpha _0}} \right)\)