Discussion
The incidence matrix of a block design is given by :
\[ N = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 \end{bmatrix} \]
Hence the design is :
Answer: B
The design is not connected and not orthogonal.
Read the rows as treatments and the columns as blocks. Treatment 2 (row 2) appears only in block 3, and block 3 contains no other treatment. So treatment 2 is never compared with any other treatment within a block, and no chain of blocks links it to the rest: the design is disconnected, and contrasts involving treatment 2 are not estimable.
Orthogonality needs \(n_{ij} = r_{i}k_{j}/n\) for every cell. Here the treatment replications are 2, 1, 2, 2, the block sizes 1, 3, 1, 2, and \(n = 7\); cell (1, 1) has \(n_{11} = 1\) but \(r_{1}k_{1}/n = 2/7\). So it is not orthogonal either. (Reading the rows as blocks leads to the same conclusion, with treatment 3 isolated in block 2.)
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