Question

The formula for estimating one missing value in a randomized block design having \(b\) blocks and \(k\) treatments with usual notations is:

• A. \[\dfrac{bT' + kB' - G'}{(b-1)(k-1)}\]
• B. \[\dfrac{bT' + bT' - G'}{(b-1)(k-1)}\]
• C. \[\dfrac{bT' + kB' - kG'}{(b-1)(k-2)}\]
• D. \[\dfrac{bB' + kT' - G'}{(b-1)(k-1)}\]

Hint:

Minimize the error sum of squares with respect to the missing value: the block total gets weighted by the number of blocks and the treatment total by the number of treatments.

Answer 1

The Correct Option is D

Step 1: Let \(x\) be the missing observation, occurring in the block whose known-values total is \(B'\) and in the treatment whose known-values total is \(T'\), and let \(G'\) be the grand total of the remaining \((bk-1)\) known observations.

Step 2: With \(x\) filled in, write the sums of squares as functions of \(x\): \[TSS(x) = S_{known} + x^2 - \frac{(G'+x)^2}{bk}\] \[SS_{Block}(x) = \frac{(B'+x)^2}{k} + C_1 - \frac{(G'+x)^2}{bk}\] \[SS_{Treat}(x) = \frac{(T'+x)^2}{b} + C_2 - \frac{(G'+x)^2}{bk}\] where \(C_1, C_2\) collect the squared totals of the unaffected blocks and treatments.

Step 3: The error sum of squares is \(SSE(x) = TSS(x) - SS_{Block}(x) - SS_{Treat}(x)\). Substituting and simplifying, the \((G'+x)^2/(bk)\) terms combine to leave: \[SSE(x) = \text{const} + x^2 - \frac{(B'+x)^2}{k} - \frac{(T'+x)^2}{b} + \frac{(G'+x)^2}{bk}\]

Step 4: The value of \(x\) that best fits the additive block+treatment model is the one that minimizes \(SSE(x)\), so differentiate with respect to \(x\) and set it to zero: \[2x - \frac{2(B'+x)}{k} - \frac{2(T'+x)}{b} + \frac{2(G'+x)}{bk} = 0\]

Step 5: Multiply throughout by \(bk\) and collect terms in \(x\): \[bk\,x - b(B'+x) - k(T'+x) + (G'+x) = 0\] \[x\big[bk - b - k + 1\big] = bB' + kT' - G'\] \[x(b-1)(k-1) = bB' + kT' - G'\]

Step 6: Hence \[x = \frac{bB' + kT' - G'}{(b-1)(k-1)}\] This matches Option D.

Final answer: \(\dfrac{bB' + kT' - G'}{(b-1)(k-1)}\) (Option D).