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: Understanding the Concept:
The question asks for the degrees of freedom for the error term (SSE) in a two-way ANOVA model with interaction and \(m\) replications per cell.

Step 2: Key Formula or Approach:
The degrees of freedom for error can be found in two ways: 1. By subtraction: df(Error) = df(Total) - df(A) - df(B) - df(AB). 2. Directly: The error degrees of freedom represent the variation within each cell. There are \(pq\) cells in total. Within each cell, there are \(m\) observations, so there are \(m-1\) degrees of freedom. Summing over all cells gives the total error df.

Step 3: Detailed Explanation:
Let's use the direct method first as it's more intuitive. The model has \(p \times q\) combinations of factor levels (cells). For each cell \((i, j)\), there are \(m\) observations (\(x_{ij1}, x_{ij2}, \dots, x_{ijm}\)). The variation within this single cell is measured by \( \sum_{k=1}^m (x_{ijk} - \bar{x}_{ij.})^2 \). The degrees of freedom associated with this within-cell variation is \(m-1\). Since there are \(pq\) such cells, and the errors are assumed independent across cells, the total degrees of freedom for error is the sum of the degrees of freedom from each cell: \[ \text{df(Error)} = \sum_{i=1}^p \sum_{j=1}^q (m-1) = pq(m-1) \] Alternatively, using the subtraction method: - Total number of 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 methods yield the same result.
Step 4: Final Answer:
The degrees of freedom corresponding to the sum of squares due to error is \( pq(m-1) \).