The Four Types of Estimable Functions

Examples

Subsections:

A One-Way Classification Model

For the model

\[ Y = \mu + A_ i + \epsilon \quad \quad i = 1, 2, 3 \]

the general form of estimable functions $\mb{L}\bbeta $ is (from the previous example)

\[ \mb{L} \bbeta = \mathit{L1} \times \mu + \mathit{L2} \times A_1 + \mathit{L3} \times A_2 + (\mathit{L1}-\mathit{L2}-\mathit{L3}) \times A_3 ~ \]

Thus,

\[ \mb{L} = (\mathit{L1}, \mathit{L2}, \mathit{L3}, \mathit{L1}-\mathit{L2}-\mathit{L3}) \]

Tests involving only the parameters $A_1$, $A_2$, and $A_3$ must have an $\mb{L}$ of the form

\[ \mb{L} = (0, \mathit{L2}, \mathit{L3}, -\mathit{L2}-\mathit{L3}) ~ \]

Since this $\mb{L}$ for the A parameters involves only two symbols, hypotheses with at most two degrees of freedom can be constructed. For example, letting $(L2,L3)$ be $(1,0)$ and $(0,1)$, respectively, yields

\[ \mb{L} = \left[ \begin{array}{rrrr} 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{array} \right] \]

The preceding $\mb{L}$ can be used to test the hypothesis that $A_1=A_2=A_3$. For this example, any $\mb{L}$ with two linearly independent rows with column 1 equal to zero produces the same sum of squares. For example, a joint test for linear and quadratic effects of A

\[ \mb{L} = \left[ \begin{array}{rrrr} 0 & 1 & 0 & -1 \\ 0 & 1 & -2 & 1 \end{array} \right] \]

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

\[ \mbox{SS}(H_0\colon ~ \mb{L} \bbeta = 0) = \mbox{SS}(H_0\colon ~ \mb{KL} \bbeta = 0) ~ \]

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 15.1.

Table 15.1: 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 15.2.

Table 15.2: 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 $\times $ 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 $\mb{L}$ matrix with four linearly independent rows according to the preceding rules, then testing $\mb{L}\bbeta =\mb{0}$ is equivalent to testing that $\mr{E}[\mb{Y}]$ is uniformly 0. Symbolically,

\[ \mbox{SS}(H_0\colon ~ \mb{L} \bbeta = 0) = R(\mu , A, B, C) ~ \]

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 unambiguously test such a hypothesis because of confounding: any test for the equality of the parameters for any one of A, B, or C will necessarily involve the parameters for the other two effects. 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 $\mb{L}$ 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

    \[ \mathit{C1} - 2 \times \mathit{C2} +\mathit{C3} = K \quad \quad \mbox{(for any } K\text {)} \]

    or any multiple of the left-hand side equal to K. Furthermore,

    \[ \mbox{SS}(H_0\colon ~ \mathit{C1} - 2 \times \mathit{C2} + \mathit{C3} = 0) = R(C~ |~ \mu , A, B) \]

A Multiple Regression Model

Suppose

\[ \mr{E}[Y] = \beta _0 + \beta _1 x_1 + \beta _2 x_2 + \beta _3 x_3 \]

where the $\mb{X'X}$ matrix has full rank. The general form of estimable functions is as shown in Table 15.3.

Table 15.3: General Form of Estimable Functions for a Multiple Regression Model When $\mb{X'X}$ Matrix Is of Full Rank

Parameter

 

Coefficient

$\beta _0$

 

L1

$\beta _1$

 

L2

$\beta _2$

 

L3

$\beta _3$

 

L4


For example, to test the hypothesis that $\beta _2=0$, let L1 = L2 = L4 = 0 and let L3 = 1. Then SS$(\mb{L}\bbeta =\mb{0}) = R(\beta _2~ |~ \beta _0,\beta _1,\beta _3)$. In this full-rank case, all parameters, as well as any linear combination of parameters, are estimable.

Suppose, however, that $X3=2 x_1 + 3 x_2$. The general form of estimable functions is shown in Table 15.4.

Table 15.4: General Form of Estimable Functions for a Multiple Regression Model When $\mb{X'X}$ Matrix Is Not of Full Rank

Parameter

 

Coefficient

$\beta _0$

 

L1

$\beta _1$

 

L2

$\beta _2$

 

L3

$\beta _3$

 

$2 \times \mathit{L2}+3 \times \mathit{L3}$


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,

\begin{eqnarray*} R(\beta _1~ |~ \beta _0, \beta _2, \beta _3) & = & 0 \\[0.05in] R(\beta _2~ |~ \beta _0, \beta _1, \beta _3) & = & 0 \\[0.05in] R(\beta _3~ |~ \beta _0, \beta _1, \beta _2) & = & 0 ~ \end{eqnarray*}