Question:

For the given model \(x_{ijk} = \mu + \alpha_i + \beta_j + \gamma_{ij} + e_{ijk}; i=1, \dots,p; j=1, \dots,q; k=1, \dots,m\), under the assumption of the normality of the parent population, the p.d.f. of \(y = \frac{S_{AB}^2}{\sigma_e^2}\), is

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Remember the fundamental theorem of ANOVA: For a normal linear model, any Sum of Squares (SS) divided by the error variance \(\sigma^2\) follows a Chi-squared distribution with degrees of freedom equal to the degrees of freedom of that SS. Also, remember the relationship: \(\chi^2_v \equiv \text{Gamma}(v/2, 2)\).
Updated On: Sep 20, 2025
  • Gamma with parameters \((q/2, 2)\)
  • Gamma with parameters \((pq/2, 2)\)
  • Gamma with parameters \(((p-1)(q-1)/2, 2)\)
  • Gamma with parameters \(((p-1)/2, 2)\)
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
The model given is a two-way ANOVA model with interaction (\(\gamma_{ij}\)). The quantity \(S_{AB}^2\) represents the sum of squares for the interaction between factors A (\(\alpha_i\)) and B (\(\beta_j\)). The question asks for the distribution of this sum of squares, scaled by the error variance \(\sigma_e^2\).

Step 2: Key Formula or Approach:
Under the standard assumptions of ANOVA (normality, independence, homoscedasticity), if a sum of squares (SS) has \(v\) degrees of freedom, then the quantity \( \frac{\text{SS}}{\sigma_e^2} \) follows a Chi-squared distribution with \(v\) degrees of freedom, i.e., \( \chi^2_v \). A Chi-squared distribution is a special case of the Gamma distribution. Specifically, \( \chi^2_v \equiv \text{Gamma}(\text{shape}=v/2, \text{scale}=2) \). The rate would be 1/scale = 1/2. We need to find the degrees of freedom for the interaction sum of squares, \(S_{AB}^2\).

Step 3: Detailed Explanation:
The model is \(x_{ijk} = \mu + \alpha_i + \beta_j + \gamma_{ij} + e_{ijk}\). - Factor A has \(p\) levels. - Factor B has \(q\) levels. - There are \(m\) replications per cell. The degrees of freedom for the main and interaction effects are: - df(A) = \(p-1\) - df(B) = \(q-1\) - df(Interaction AB) = \((p-1)(q-1)\) - df(Error) = \(pq(m-1)\) - df(Total) = \(pqm-1\) The sum of squares for the interaction, \(S_{AB}^2\), has \(v = (p-1)(q-1)\) degrees of freedom. Therefore, the random variable \( y = \frac{S_{AB}^2}{\sigma_e^2} \) follows a Chi-squared distribution with \((p-1)(q-1)\) degrees of freedom. \[ y \sim \chi^2_{(p-1)(q-1)} \] Now, we express this in terms of a Gamma distribution. The relationship is \( \chi^2_v \equiv \text{Gamma}(\text{shape}=k, \text{scale}=\theta) \) where \(k = v/2\) and \(\theta=2\). So, the shape parameter is \( k = \frac{(p-1)(q-1)}{2} \). The scale parameter is \( \theta = 2 \). The PDF provided in the options uses a form related to the Gamma distribution. The PDF for \( Z \sim \text{Gamma}(k, \theta) \) is \( f(z) = \frac{1}{\Gamma(k)\theta^k} z^{k-1} e^{-z/\theta} \). The form in the options is \( f(y) = \frac{e^{-y/2} y^{k-1}}{2^k \Gamma(k)} \). This corresponds to a Gamma distribution with shape \(k\) and scale \(\theta=2\). For our variable y, the shape parameter is \( k = \frac{(p-1)(q-1)}{2} \). So the distribution is Gamma with parameters \( (\frac{(p-1)(q-1)}{2}, 2) \). This matches option (C).
Step 4: Final Answer:
The p.d.f. of y is that of a Gamma distribution with parameters shape = \(\frac{(p-1)(q-1)}{2}\) and scale = 2.
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