Question

For the given model \(x_{ijk} = \mu + \alpha_i + \beta_j + \gamma_{ij} + e_{ijk}; i=1, \dots,p; j=1, \dots,q; k=1, \dots,m\), the degrees of freedom corresponding to sum of squares due to error is:

• A. \( (p-1)(q-1) \)
• B. \( pq(m-1) \)
• C. \( (pqm - p - q) \)
• D. \( (pqm - 1) \)

Hint:

For ANOVA models with replications, the error degrees of freedom are calculated based on the variation within the cells. If there are \(N_{cell}\) observations in each of the \(C\) cells, the error df is \(C \times (N_{cell} - 1)\).

Answer 1

The Correct Option is B

Step 1: Problem Definition:
The problem asks us to determine the degrees of freedom for the error term (SSE) in a two-way ANOVA, considering interaction effects and \(m\) replicates per cell.

Step 2: Core Concepts:
The error degrees of freedom can be calculated using two approaches: 1. Subtraction Method: df(Error) = df(Total) - df(A) - df(B) - df(AB). 2. Direct Method: Error degrees of freedom reflect within-cell variation. With \(pq\) cells and \(m\) observations per cell, each cell contributes \(m-1\) degrees of freedom. Summing across all cells gives the total error df.

Step 3: Detailed Solution:
Using the direct method: We have \(p \times q\) factor level combinations (cells). Each cell \((i, j)\) has \(m\) observations (\(x_{ij1}, x_{ij2}, \dots, x_{ijm}\)). The within-cell variation is quantified by \( \sum_{k=1}^m (x_{ijk} - \bar{x}_{ij.})^2 \). This within-cell variation has \(m-1\) degrees of freedom. Since we assume independent errors across the \(pq\) cells, we sum the degrees of freedom from each cell to get the total error degrees of freedom: \[ \text{df(Error)} = \sum_{i=1}^p \sum_{j=1}^q (m-1) = pq(m-1) \] Alternatively, using the subtraction method: - Total observations = \(pqm\), so df(Total) = \(pqm-1\). - df for factor A (\(\alpha_i\)) = \(p-1\). - df for factor B (\(\beta_j\)) = \(q-1\). - df for interaction (\(\gamma_{ij}\)) = \((p-1)(q-1) = pq - p - q + 1\). - df(Error) = df(Total) - df(A) - df(B) - df(AB) \[ = (pqm-1) - (p-1) - (q-1) - (pq - p - q + 1) \] \[ = pqm - 1 - p + 1 - q + 1 - pq + p + q - 1 \] \[ = pqm - pq = pq(m-1) \] Both approaches lead to the same conclusion.
Step 4: Conclusion:
The degrees of freedom for the error sum of squares is \( pq(m-1) \).