Section 8a

8.1 Introduction, 8.2 One-Way Classification: 8.2.1 General Theory, 8.2.2 Confidence Intervals, 8.2.3 Underlying Assumptions

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Problem 8a.1

Express H:μ1=⋯=μIH: \mu_{1} = \cdots = \mu_{I} in the form Aμ=0\mathbf{A} \mathbf{\mu } = \mathbf{0}. What is the rank of A\mathbf{A}? As this matrix is not unique, find another matrix, A1\mathbf{A}_{1} say, also satisfying H:A1μ=0H: \mathbf{A}_{1} \mathbf{\mu } = \mathbf{0}. What would be the relationship between A\mathbf{A} and A1\mathbf{A}_{1}?

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Problem 8a.2

Obtain the identity indicated by (8.3), namely,

∑i∑j(Yij−Yˉ..)2=∑i∑i(Yij−Yˉi.)2+∑i∑j(Yˉi.−Yˉ..)2. \sum _{i} \sum _{j}\left(Y_{i j}-\bar{Y} . .\right)^{2} = \sum _{i} \sum _{i}\left(Y_{i j}-\bar{Y}_{i .}\right)^{2}+\sum _{i} \sum _{j}\left(\bar{Y}_{i .}-\bar{Y}_{. .}\right)^{2} .
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Problem 8a.3

Obtain the least squares estimates of the μi\mu_{i} using β^=(XTX)−1XTY\widehat{\mathbf{\beta }} = \left(\mathbf{X}^{T} \mathbf{X}\right)^{-1} \mathbf{X}^{T} \mathbf{Y} directly, where X\mathbf{X} is given by (8.2).

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Problem 8a.4

Prove that

∑i∑j(Yˉi.−Yˉ..)2=∑iYi.2Ji−Y.22n. \sum _{i} \sum _{j}\left(\bar{Y}_{i .}-\bar{Y}_{. .}\right)^{2} = \sum _{i} \frac{Y_{i .}^{2}}{J_{i}}-\frac{Y_{.2}^{2}}{n} .

Obtain a similar expression for ∑i∑j(Yij−Yˉi)2\sum_{i} \sum_{j}\left(Y_{i j}-\bar{Y}_{i}\right)^{2}.

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Problem 8a.5

Find (a) E[∑i∑j(Yˉi.−Yˉ..)2]\mathbb {E}\left[\sum_{i} \sum_{j}\left(\bar{Y}_{i .}-\bar{Y} . .\right)^{2}\right] and (b) E[∑i∑j(Yij−Yˉi.)2]\mathbb {E}\left[\sum_{i} \sum_{j}\left(Y_{i j}-\bar{Y}_{i .}\right)^{2}\right]. Hint: Use Theorem 4.1(ii) for the first expression.

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(a)
(b)