Question:
Error sum of squares in RBD as compared to CRD using the same material is:
Error sum of squares in RBD as compared to CRD using the same material is:
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RBD removes block variation from the residual before computing error SS, so it can only reduce, not increase, the error sum of squares compared to CRD.
Updated On: Jul 4, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Write the total sum of squares for the same set of observations under a completely randomized design (CRD) and under a randomized block design (RBD) analysis.
Step 2: In CRD, \(TSS = SS_{Treatment} + SSE_{CRD}\), since the CRD analysis only separates out variation due to treatments.
Step 3: In RBD, the same total variation is split three ways: \(TSS = SS_{Treatment} + SS_{Block} + SSE_{RBD}\), because RBD additionally isolates the variation due to blocks.
Step 4: Equating the two expressions for \(TSS\) (same data, same treatment effect being removed in both): \(SSE_{RBD} = SSE_{CRD} - SS_{Block}\).
Step 5: Since a sum of squares can never be negative, \(SS_{Block} \ge 0\), which gives \(SSE_{RBD} \le SSE_{CRD}\). Whenever the blocking variable genuinely accounts for real variation among the units (which is the entire reason RBD is preferred over CRD), \(SS_{Block} > 0\) and the error sum of squares in RBD is strictly less than in CRD.
Final answer: Less (Option B).
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