Examples

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The Four Types of Estimable Functions

Examples

A One-Way Classification Model

For the model

$Y = \mu + A_i + \epsilon i = 1, 2, 3$

the general form of estimable functions Lb is (from the previous example)

$L {\beta}= L1 x \mu + L2 x A_1 + L3 x A_2 + (L1-L2-L3) x A_3$

Thus,

L = (L1, L2, L3, L1-L2-L3)

Tests involving only the parameters A₁, A₂, and A₃ must have an L of the form

L = (0, L2, L3, -L2-L3)

Since the preceding L involves only two symbols, hypotheses with at most two degrees-of-freedom can be constructed. For example, let L2=1 and L3=0; then let L2=0 and L3=1:

$L = [ 0 & 1 & 0 & -1 \ 0 & 0 & 1 & -1 ]$

The preceding L can be used to test the hypothesis that A₁=A₂=A₃. For this example, any L with two linearly independent rows with column 1 equal to zero produces the same Sum of Squares. For example, a pooled linear quadratic

$L = [ 0 & 1 & 0 & -1 \ 0 & 1 & -2 & 1 ]$

gives the same SS. In fact, for any L of full row rank and any nonsingular matrix K of conformable dimensions,

${SS}(H_0\colon L {\beta}= 0) = {SS}(H_0\colon {KL} {\beta}= 0)$

A Three-Factor Main Effects Model

Consider a three-factor main effects model involving the CLASS variables A, B, and C, as shown in Table 12.1.

Table 12.1: Three-Factor Main Effects Model

Obs	A	B	C
1	1	2	1
2	1	1	2
3	2	1	3
4	2	2	2
5	2	2	2

The general form of an estimable function is shown in Table 12.2.

Table 12.2: General Form of an Estimable Function for Three-Factor Main Effects Model

Parameter		Coefficient
$\mu$ (Intercept)		L1
A1		L2
A2		L1-L2
B1		L4
B2		L1-L4
C1		L6
C2		L1+L2-L4-2 ×L6
C3		-L2+L4+L6

Since only four symbols (L1, L2, L4, and L6) are involved, any testable hypothesis will have at most four degrees of freedom. If you form an L matrix with four linearly independent rows according to the preceding rules, then

${SS}(H_0\colon L {\beta}= 0) = R(\mu, A, B, C)$

In a main effects model, the usual hypothesis of interest for a main effect is the equality of all the parameters. In this example, it is not possible to test such a hypothesis because of confounding. One way to proceed is to construct a maximum rank hypothesis (MRH) involving only the parameters of the main effect in question. This can be done using the general form of estimable functions. Note the following:

To get an MRH involving only the parameters of A, the coefficients of L associated with $\mu$ , B1, B2, C1, C2, and C3 must be equated to zero. Starting at the top of the general form, let L1=0, then L4=0, then L6=0. If C2 and C3 are not to be involved, then L2 must also be zero. Thus, A1-A2 is not estimable; that is, the MRH involving only the A parameters has zero rank and $R(A|\mu,B,C)=0$ .
To obtain the MRH involving only the B parameters, let L1=L2=L6=0. But then to remove C2 and C3 from the comparison, L4 must also be set to 0. Thus, B1-B2 is not estimable and $R(B|\mu,A,C)=0$ .
To obtain the MRH involving only the C parameters, let L1=L2=L4=0. Thus, the MRH involving only C parameters is
C1 - 2 ×C2 + C3 = K (for any K)
or any multiple of the left-hand side equal to K. Furthermore,
${SS}(H_0\colon C1 = 2 x C2 - C3 = 0) = R(C|\mu, A, B)$

A Multiple Regression Model

Suppose

$E(Y) = \beta_0 + \beta_1 x X1 + \beta_2 x X2 + \beta_3 x X3$

If the X'X matrix is of full rank, the general form of estimable functions is as shown in Table 12.3.

Table 12.3: General Form of Estimable Functions for a Multiple Regression Model When X'X Matrix Is of Full Rank

Parameter		Coefficient
$\beta_0$		L1
$\beta_1$		L2
$\beta_2$		L3
$\beta_3$		L4

To test, for example, the hypothesis that $\beta_2=0$ , let L1=L2=L4=0 and let L3=1. Then SS $(L {\beta}=0)=R(\beta_2|\beta_0,\beta_1,\beta_3)$ . In the full-rank case, all parameters, as well as any linear combination of parameters, are estimable.

Suppose, however, that X3 = 2 ×X1+3 ×X2. The general form of estimable functions is shown in Table 12.4.

Table 12.4: General Form of Estimable Functions for a Multiple Regression Model When X'X Matrix Is Not of Full Rank

Parameter		Coefficient
$\beta_0$		L1
$\beta_1$		L2
$\beta_2$		L3
$\beta_3$		2 ×L2+3 ×L3

For this example, it is possible to test $H_0\colon \beta_0 = 0$ . However, $\beta_1$ , $\beta_2$ , and $\beta_3$ are not jointly estimable; that is,

$R(\beta_1|\beta_0, \beta_2, \beta_3) & = & 0 \ R(\beta_2|\beta_0, \beta_1, \beta_3) & = & 0 \ R(\beta_3|\beta_0, \beta_1, \beta_2) & = & 0$

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