Section 8c

8.4 Two-Way Classification (Balanced)

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Problem 8c.1
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(a)

Prove that

∑i=1I∑j=1J∑k=1K(Yˉi..−Yˉ…)2=∑i=1IYi.2JK−Y..2IJK \sum _{i = 1}^{I} \sum _{j = 1}^{J} \sum _{k = 1}^{K}\left(\bar{Y}_{i . .}-\bar{Y}_{\ldots }\right)^{2} = \sum _{i = 1}^{I} \frac{Y_{i .}^{2}}{J K}-\frac{Y_{. .}^{2}}{I J K}

Obtain a similar expression for

∑i=1I∑j=1J∑k=1K(Yˉij.−Yˉi..−Yˉ.j.+Yˉ…)2 \sum _{i = 1}^{I} \sum _{j = 1}^{J} \sum _{k = 1}^{K}\left(\bar{Y}_{i j .}-\bar{Y}_{i . .}-\bar{Y}_{. j .}+\bar{Y}_{\ldots }\right)^{2}
(b)

Using Table 8.2, prove the following:

E[(I−1)SA2]=σ2(I−1)+JK∑iαi∗2E[(J−1)SB2]=σ2(J−1)+IK∑jβi∗2E[(I−1)SAB2]=σ2(I−1)(J−1)+K∑i∑j(αβ)ij∗2 \begin{aligned} E\left[(I-1) S_{A}^{2}\right] & = \sigma ^{2}(I-1)+J K \sum _{i} \alpha _{i}^{* 2} \\ E\left[(J-1) S_{B}^{2}\right] & = \sigma ^{2}(J-1)+I K \sum _{j} \beta _{i}^{* 2} \\ E\left[(I-1) S_{A B}^{2}\right] & = \sigma ^{2}(I-1)(J-1)+K \sum _{i} \sum _{j}(\alpha \beta )_{i j}^{* 2} \end{aligned}
Problem 8c.2

Given the population means μij(i=1,2,…,I;j=1,2,…,J)\mu_{i j}\left(i = 1,2, \ldots , I ; j = 1,2, \ldots , J\right), let

Ai=∑jvjμij(∑jvj=1),Bi=∑iuiμij(∑juj=1), \begin{aligned} A_{i} & = \sum _{j} v_{j} \mu _{i j} \quad \left(\sum _{j} v_{j} = 1\right), \\ B_{i} & = \sum _{i} u_{i} \mu _{i j} \quad \left(\sum _{j} u_{j} = 1\right), \end{aligned}

and

μ=∑iuiAi=∑jvjBj=∑i∑juivjμij \mu = \sum _{i} u_{i} A_{i} = \sum _{j} v_{j} B_{j} = \sum _{i} \sum _{j} u_{i} v_{j} \mu _{i j}

Define αi=Ai−μ,βj=Bj−μ\alpha_{i} = A_{i}-\mu , \beta_{j} = B_{j}-\mu, and

(αβ)ij=μij−Ai−Bj+μ (\alpha \beta )_{i j} = \mu _{i j}-A_{i}-B_{j}+\mu
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(a)

Show that ∑iuiαi=∑jvjβj=0,∑iui(αβ)ij=0\sum_{i} u_{i} \alpha_{i} = \sum_{j} v_{j} \beta_{j} = 0, \sum_{i} u_{i}(\alpha \beta )_{i j} = 0 (all jj), and ∑jvj(αβ)ij=0\sum_{j} v_{j}(\alpha \beta )_{i j} = 0 (all ii).

(b)

Conversely, given

μij=μ+αi+βj+(αβ)ij \mu _{i j} = \mu +\alpha _{i}+\beta _{j}+(\alpha \beta )_{i j}

show that the parameters in the equation above are uniquely determined by the constraints in (a).

(c)

Prove that if the interactions {(αβ)ij}\left\{ (\alpha \beta )_{i j}\right\} are all zero for some system of weights {ui}\left\{ u_{i}\right\} and {vi}\left\{ v_{i}\right\}, then they are zero for every system of weights. In that case show that every contrast in the {αi}\left\{ \alpha_{i}\right\}, or {βi}\left\{ \beta_{i}\right\}, has a value that does not depend on the system of weights.

Problem 8c.3

Let Yijk=μ+αi+βj+(αβ)ij+εijkY_{i j k} = \mu +\alpha_{i}+\beta_{j}+(\alpha \beta )_{i j}+\varepsilon_{i j k}, where i=1,2,…,I;j=1,2,…,J;k=1,2,…,Kiji = 1,2, \ldots , I ; j = 1,2, \ldots , J ; k = 1,2, \ldots , K_{i j}; and the εijk\varepsilon_{i j k} are independently distributed as N(0,σ2)N\left(0, \sigma^{2}\right). Given Kij=Ki.K.j/K..K_{i j} = K_{i .} K_{. j} / K_{. .}for all i,ji, j, find a test statistic for testing the hypothesis H:(αβ)ij=0H:(\alpha \beta )_{i j} = 0 (all i,ji, j). Hint: By Exercise 2, the validity of HH does not depend on the weights used in the identifiability constraints ∑iuiαi=∑jvjβj=0\sum_{i} u_{i} \alpha_{i} = \sum_{j} v_{j} \beta_{j} = 0. We can therefore use ui=Ki./K.u_{i} = K_{i .} / K_{.} and vj=K.j/K..v_{j} = K_{. j} / K_{. .}and find the least squares estimates of αi\alpha_{i} and βj\beta_{j} when HH is true.

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