Prediction interval for the AV1R model
In AOV1R: Inference in the Balanced One-Way ANOVA Model with Random Factor

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The balanced case

Consider the balanced one-way random effect ANOVA model: $$ Y_{ij} = \mu + A_i + G_{ij}, \quad i=1, \ldots, I, \quad j=1, \ldots, J, $$ where $A_i \sim \mathcal{N}(0, \sigma^2_b)$ and $G_{ij} \sim \mathcal{N}(0, \sigma^2_w)$.

Denote by $Y^{\text{new}} \sim \mathcal{N}(\mu, \sigma^2_b + \sigma^2_w)$ a future observation.

One has $$ \overline{Y}{\bullet\bullet} \sim \mathcal{N}\left(\mu, \frac{J\sigma^2_b+\sigma^2_w}{IJ}\right) $$ hence $$ \overline{Y}{\bullet\bullet} - Y^{\text{new}} \sim \mathcal{N}\left(0, \left(1+\frac{1}{I}\right)\sigma^2_b + \left(1+\frac{1}{IJ}\right)\sigma^2_w\right). $$

Recall that $I(J-1)\sigma^2_w$ is estimated by $SS_w \sim \sigma^2_w \chi^2_{I(J-1)}$ and $(I-1)(J\sigma^2_b+\sigma^2_w)$ is estimated by $SS_b \sim (J\sigma^2_b+\sigma^2_w)\chi^2_{I-1}$, hence $\sigma^2_w$ is estimated by $\frac{1}{I(J-1)}SS_w$ and $\sigma^2_b$ is estimated by $$ \frac{1}{J}\left(\frac{SS_b}{I-1} - \frac{SS_w}{I(J-1)}\right). $$ Therefore, the variance of $\overline{Y}{\bullet\bullet} - Y^{\text{new}}$ is estimated by $a SS_b + b SS_w$ where $$ a = \frac{1}{J(I-1)}\left(1+\frac{1}{I}\right) $$ and $$ b = \left(1+\frac{1}{IJ}\right)\frac{1}{I(J-1)} - \left(1+\frac{1}{I}\right)\frac{1}{JI(J-1)} = \frac{1}{IJ}. $$ Using the Satterthwaite approximation, we find that $$ \frac{\overline{Y}{\bullet\bullet} - Y^{\text{new}}}{\sqrt{a SS_b + b SS_w}} \approx t_{\hat\nu} $$ with $$ \hat\nu = \frac{{(a SS_b + b SS_w)}^2}{\dfrac{{(a SS_b)}^2}{I-1} + \dfrac{{(b SS_w)}^2}{I(J-1)}}. $$ This yields an approximate prediction interval.

The unbalanced case

Now consider the general case $$ Y_{ij} = \mu + A_i + G_{ij}, \quad i=1, \ldots, I, \quad j=1, \ldots, J_i. $$ We define $N = \sum_{i=1}^I J_i$ and $\widetilde{J}$ as the harmonic mean of the $J_i$'s.

One can check that $SS_w \sim \sigma^2_w \chi^2_{N-I}$ and setting $\overline{\overline{Y}} = \frac{1}{I}\sum_{i=1}^I\overline{Y}_{i\bullet}$, $$ \overline{\overline{Y}} \sim \mathcal{N}\left(\mu, \frac{\widetilde{J}\sigma^2_b + \sigma^2_w}{I\widetilde{J}}\right). $$

Now, set $$ \widetilde{SS}b = \widetilde{J}\sum{i=1}^I(\overline{Y}_{i\bullet} - \overline{\overline{Y}}). $$

It is known that $$ \widetilde{SS}b \approx (\widetilde{J}\sigma^2_b + \sigma^2_w) \chi^2{I-1}, $$ and that $\widetilde{SS}_b$ is independent of $SS_w$. In addition, $SS_w$ is independent of $\overline{\overline{Y}}$, but $\overline{\overline{Y}}$ is not independent of $\widetilde{SS}_b$.

Ignoring this dependence and proceeding as in the balanced case, we find the Satterthwaite degrees of freedom $$ \hat\nu = \frac{{(a \widetilde{SS}_b + b SS_w)}^2}{\dfrac{{(a \widetilde{SS}_b)}^2}{I-1} + \dfrac{{(b SS_w)}^2}{N-I}} $$ with $$ a = \frac{1}{\widetilde{J}(I-1)}\left(1+\frac{1}{I}\right) $$ and $$ b = \left(1+\frac{1}{I\widetilde{J}}\right)\frac{1}{N-I} - \left(1+\frac{1}{I}\right)\frac{1}{\widetilde{J}(N-I)} = \frac{\widetilde{J}-1}{\widetilde{J}(N-I)}. $$