Let's break down and solve the problem step by step: --- ### **Problem Restatement** - **Design:** Balanced three-stage nested design. - **Factors:** - \(A\): Fixed effect - \(B\) (nested in \(A\)): Random effect - \(C\) (nested in \(B\)): Random effect - **Objective:** 1. Derive the **expected mean squares (EMS)** for each source. 2. Obtain **formulas for estimating variance components**. 3. Assume the **restricted (reduced) form of the mixed model**. --- ## **Step 1: Model Specification** Let: - \( a \): Number of levels of \(A\) - \( b \): Number of levels of \(B\) within each \(A\) - \( c \): Number of levels of \(C\) within each \(B\) - \( n \): Number of observations per \(C\) The model for observation \( Y_{ijkl} \) is: \[ Y_{ijkl} = \mu + \alpha_i + \beta_{j(i)} + \gamma_{k(ij)} + \epsilon_{l(kij)} \] where: - \( \mu \): Overall mean - \( \alpha_i \): Effect of the i-th level of \(A\) (fixed) - \( \beta_{j(i)} \): Effect of j-th level of \(B\) within i-th \(A\) (random, variance \(\sigma^2_B\)) - \( \gamma_{k(ij)} \): Effect of k-th level of \(C\) within j-th \(B\) and i-th \(A\) (random, variance \(\sigma^2_C\)) - \( \epsilon_{l(kij)} \): Random error (variance \(\sigma^2\)) Assumptions: - All random effects and errors are independent and have mean 0. --- ## **Step 2: ANOVA Table Sources** | Source | df | Mean Square | |-----------------|-----------------------|------------------| | \(A\) | \(a-1\) | \(MS_A\) | | \(B(A)\) | \(a(b-1)\) | \(MS_{B(A)}\) | | \(C(B(A))\) | \(ab(c-1)\) | \(MS_{C(B(A))}\) | | Error | \(abc(n-1)\) | \(MS_E\) | --- ## **Step 3: Expected Mean Squares (EMS)** ### **Error:** \[ E[MS_E] = \sigma^2 \] ### **C(B(A)):** \[ E[MS_{C(B(A))}] = \sigma^2 + n\sigma^2_C \] ### **B(A):** \[ E[MS_{B(A)}] = \sigma^2 + n\sigma^2_C + nc\sigma^2_B \] ### **A:** \[ E[MS_A] = \sigma^2 + n\sigma^2_C + nc\sigma^2_B + nbc \left[ \frac{1}{a-1} \sum_{i=1}^a \alpha_i^2 \right] \] Since \(A\) is fixed, the expectation involves the fixed effects. If you want the deviation from the mean, \( \sum_{i=1}^a \alpha_i^2 / (a-1) \). --- ## **Step 4: Estimation of Variance Components** We use the method of moments (equate observed MS to their expectations and solve): ### **Solve recursively:** #### **1. Error variance:** \[ \hat{\sigma}^2 = MS_E \] #### **2. \(C(B(A))\) variance:** \[ \hat{\sigma}_C^2 = \frac{MS_{C(B(A))} - MS_E}{n} \] #### **3. \(B(A)\) variance:** \[ \hat{\sigma}_B^2 = \frac{MS_{B(A)} - MS_{C(B(A))}}{nc} \] #### **(No estimate for \(\sigma_A^2\) since \(A\) is fixed)** --- ## **Step 5: Final Formulas** ### **Expected Mean Squares:** \[ \boxed{ \begin{align*} E[MS_A] &= \sigma^2 + n\sigma^2_C + nc\sigma^2_B + nbc \left( \frac{1}{a-1} \sum_{i=1}^a \alpha_i^2 \right) \\ E[MS_{B(A)}] &= \sigma^2 + n\sigma^2_C + nc\sigma^2_B \\ E[MS_{C(B(A))}] &= \sigma^2 + n\sigma^2_C \\ E[MS_E] &= \sigma^2 \\ \end{align*} } \] ### **Variance Component Estimators:** \[ \boxed{ \begin{align*} \hat{\sigma}^2 &= MS_E \\ \hat{\sigma}_C^2 &= \frac{MS_{C(B(A))} - MS_E}{n} \\ \hat{\sigma}_B^2 &= \frac{MS_{B(A)} - MS_{C(B(A))}}{nc} \\ \end{align*} } \] --- ### **Summary Table** | Source | EMS | Variance Component Estimate | |-----------------|----------------------------------------------------------------|-------------------------------------------| | \(A\) | \(\sigma^2 + n\sigma^2_C + nc\sigma^2_B + nbc\frac{1}{a-1}\sum_{i=1}^a \alpha_i^2\) | -- (fixed) | | \(B(A)\) | \(\sigma^2 + n\sigma^2_C + nc\sigma^2_B\) | \(\frac{MS_{B(A)} - MS_{C(B(A))}}{nc}\) | | \(C(B(A))\) | \(\sigma^2 + n\sigma^2_C\) | \(\frac{MS_{C(B(A))} - MS_E}{n}\) | | Error | \(\sigma^2\) | \(MS_E\) | --- **This completes the derivation and estimation as requested. If you need a worked example with numbers or clarification, just ask!**