The TRANSREG Procedure

ANOVA Codings

This section illustrates several different codings of classification variables and hence several different ways of fitting two-way ANOVA models to some data. Each example fits an ANOVA model, displays the ANOVA table and parameter estimates, and displays the coded design matrix. Note throughout that the ANOVA tables and R squares are identical for all of the models, showing that the codings are equivalent. For each model, the parameter estimates are stated as a function of the cell means. The formulas are appropriate for a design such as this one, which is balanced and orthogonal (every level and every pair of levels occurs equally often). They will not work with unequal frequencies. Since this data set has $3 \times 2 = 6$ cells, the full-rank codings all have six parameters. The following statements create the input data set, and display it in Figure 97.43:

title 'Two-Way ANOVA Models';

data x;
   input a b @@;
   do i = 1 to 2; input y @@; output; end;
   drop i;
   datalines;
1 1   16 14         1 2   15 13
2 1    1  9         2 2   12 20
3 1   14  8         3 2   18 20
;
proc print label;
run;

Figure 97.43: Input Data Set

Two-Way ANOVA Models

Obs a b y
1 1 1 16
2 1 1 14
3 1 2 15
4 1 2 13
5 2 1 1
6 2 1 9
7 2 2 12
8 2 2 20
9 3 1 14
10 3 1 8
11 3 2 18
12 3 2 20


The following statements fit a cell-means model and produce Figure 97.44 and Figure 97.45:

proc transreg data=x ss2 short;
   title2 'Cell-Means Model';
   model identity(y) = class(a * b / zero=none);
   output replace;
run;

proc print label;
run;

Figure 97.44: Cell-Means Model

Two-Way ANOVA Models
Cell-Means Model

The TRANSREG Procedure


Dependent Variable Identity(y)

Class Level Information
Class Levels Values
a 3 1 2 3
b 2 1 2

Number of Observations Read 12
Number of Observations Used 12
Implicit Intercept Model  


The TRANSREG Procedure Hypothesis Tests for Identity(y)

Univariate ANOVA Table Based on the Usual Degrees of Freedom
Source DF Sum of Squares Mean Square F Value Pr > F
Model 5 234.6667 46.93333 3.20 0.0946
Error 6 88.0000 14.66667    
Corrected Total 11 322.6667      

Root MSE 3.82971 R-Square 0.7273
Dependent Mean 13.33333 Adj R-Sq 0.5000
Coeff Var 28.72281    

Univariate Regression Table Based on the Usual Degrees of Freedom
Variable DF Coefficient Type II
Sum of
Squares
Mean Square F Value Pr > F Label
Class.a1b1 1 15.0000000 450.000 450.000 30.68 0.0015 a 1 * b 1
Class.a1b2 1 14.0000000 392.000 392.000 26.73 0.0021 a 1 * b 2
Class.a2b1 1 5.0000000 50.000 50.000 3.41 0.1144 a 2 * b 1
Class.a2b2 1 16.0000000 512.000 512.000 34.91 0.0010 a 2 * b 2
Class.a3b1 1 11.0000000 242.000 242.000 16.50 0.0066 a 3 * b 1
Class.a3b2 1 19.0000000 722.000 722.000 49.23 0.0004 a 3 * b 2


The parameter estimates are

$\displaystyle  \hat{\mu }_{11}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{11} = 15  $
$\displaystyle \hat{\mu }_{12}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{12} = 14  $
$\displaystyle \hat{\mu }_{21}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{21} = 5  $
$\displaystyle \hat{\mu }_{22}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{22} = 16  $
$\displaystyle \hat{\mu }_{31}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{31} = 11  $
$\displaystyle \hat{\mu }_{32}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{32} = 19  $

Figure 97.45: Cell-Means Model, Design Matrix

Two-Way ANOVA Models
Cell-Means Model

Obs _TYPE_ _NAME_ y Intercept a 1 *
b 1
a 1 *
b 2
a 2 *
b 1
a 2 *
b 2
a 3 *
b 1
a 3 *
b 2
a b
1 SCORE ROW1 16 . 1 0 0 0 0 0 1 1
2 SCORE ROW2 14 . 1 0 0 0 0 0 1 1
3 SCORE ROW3 15 . 0 1 0 0 0 0 1 2
4 SCORE ROW4 13 . 0 1 0 0 0 0 1 2
5 SCORE ROW5 1 . 0 0 1 0 0 0 2 1
6 SCORE ROW6 9 . 0 0 1 0 0 0 2 1
7 SCORE ROW7 12 . 0 0 0 1 0 0 2 2
8 SCORE ROW8 20 . 0 0 0 1 0 0 2 2
9 SCORE ROW9 14 . 0 0 0 0 1 0 3 1
10 SCORE ROW10 8 . 0 0 0 0 1 0 3 1
11 SCORE ROW11 18 . 0 0 0 0 0 1 3 2
12 SCORE ROW12 20 . 0 0 0 0 0 1 3 2


The next model is a reference cell model, and the default reference cell is the last cell, which in this case is the (3,2) cell. The following statements fit a reference cell model and produce Figure 97.46 and Figure 97.47:

proc transreg data=x ss2 short;
   title2 'Reference Cell Model, (3,2) Reference Cell';
   model identity(y) = class(a | b);
   output replace;
run;

proc print label;
run;

Figure 97.46: Reference Cell Model, (3,2) Reference Cell

Two-Way ANOVA Models
Reference Cell Model, (3,2) Reference Cell

The TRANSREG Procedure


Dependent Variable Identity(y)

Class Level Information
Class Levels Values
a 3 1 2 3
b 2 1 2

Number of Observations Read 12
Number of Observations Used 12


The TRANSREG Procedure Hypothesis Tests for Identity(y)

Univariate ANOVA Table Based on the Usual Degrees of Freedom
Source DF Sum of Squares Mean Square F Value Pr > F
Model 5 234.6667 46.93333 3.20 0.0946
Error 6 88.0000 14.66667    
Corrected Total 11 322.6667      

Root MSE 3.82971 R-Square 0.7273
Dependent Mean 13.33333 Adj R-Sq 0.5000
Coeff Var 28.72281    

Univariate Regression Table Based on the Usual Degrees of Freedom
Variable DF Coefficient Type II
Sum of
Squares
Mean Square F Value Pr > F Label
Intercept 1 19.0000000 722.000 722.000 49.23 0.0004 Intercept
Class.a1 1 -5.0000000 25.000 25.000 1.70 0.2395 a 1
Class.a2 1 -3.0000000 9.000 9.000 0.61 0.4632 a 2
Class.b1 1 -8.0000000 64.000 64.000 4.36 0.0817 b 1
Class.a1b1 1 9.0000000 40.500 40.500 2.76 0.1476 a 1 * b 1
Class.a2b1 1 -3.0000000 4.500 4.500 0.31 0.5997 a 2 * b 1


The parameter estimates are

$\displaystyle  \hat{\mu }_{32}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{32} = 19  $
$\displaystyle \hat{\alpha }_{1}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{12} - \overline{y}_{32} = 14 - 19 = -5  $
$\displaystyle \hat{\alpha }_{2}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{22} - \overline{y}_{32} = 16 - 19 = -3  $
$\displaystyle \hat{\beta }_{1}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{31} - \overline{y}_{32} = 11 - 19 = -8  $
$\displaystyle \hat{\gamma }_{11}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{11} - (\hat{\mu }_{32} + \hat{\alpha }_{1} + \hat{\beta }_{1}) = 15 - (19 + -5 + -8) = 9  $
$\displaystyle \hat{\gamma }_{21}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{21} - (\hat{\mu }_{32} + \hat{\alpha }_{2} + \hat{\beta }_{1}) = 5 - (19 + -3 + -8) = -3  $

Figure 97.47: Reference Cell Model, (3,2) Reference Cell, Design Matrix

Two-Way ANOVA Models
Reference Cell Model, (3,2) Reference Cell

Obs _TYPE_ _NAME_ y Intercept a 1 a 2 b 1 a 1 *
b 1
a 2 *
b 1
a b
1 SCORE ROW1 16 1 1 0 1 1 0 1 1
2 SCORE ROW2 14 1 1 0 1 1 0 1 1
3 SCORE ROW3 15 1 1 0 0 0 0 1 2
4 SCORE ROW4 13 1 1 0 0 0 0 1 2
5 SCORE ROW5 1 1 0 1 1 0 1 2 1
6 SCORE ROW6 9 1 0 1 1 0 1 2 1
7 SCORE ROW7 12 1 0 1 0 0 0 2 2
8 SCORE ROW8 20 1 0 1 0 0 0 2 2
9 SCORE ROW9 14 1 0 0 1 0 0 3 1
10 SCORE ROW10 8 1 0 0 1 0 0 3 1
11 SCORE ROW11 18 1 0 0 0 0 0 3 2
12 SCORE ROW12 20 1 0 0 0 0 0 3 2


The next model is a deviations-from-means model. This coding is also called effects coding. The default reference cell is the last cell (3,2). The following statements produce Figure 97.48 and Figure 97.49:

proc transreg data=x ss2 short;
   title2 'Deviations from Means, (3,2) Reference Cell';
   model identity(y) = class(a | b / deviations);
   output replace;
run;

proc print label;
run;

Figure 97.48: Deviations-from-Means Model, (3,2) Reference Cell

Two-Way ANOVA Models
Deviations from Means, (3,2) Reference Cell

The TRANSREG Procedure


Dependent Variable Identity(y)

Class Level Information
Class Levels Values
a 3 1 2 3
b 2 1 2

Number of Observations Read 12
Number of Observations Used 12


The TRANSREG Procedure Hypothesis Tests for Identity(y)

Univariate ANOVA Table Based on the Usual Degrees of Freedom
Source DF Sum of Squares Mean Square F Value Pr > F
Model 5 234.6667 46.93333 3.20 0.0946
Error 6 88.0000 14.66667    
Corrected Total 11 322.6667      

Root MSE 3.82971 R-Square 0.7273
Dependent Mean 13.33333 Adj R-Sq 0.5000
Coeff Var 28.72281    

Univariate Regression Table Based on the Usual Degrees of Freedom
Variable DF Coefficient Type II
Sum of
Squares
Mean Square F Value Pr > F Label
Intercept 1 13.3333333 2133.33 2133.33 145.45 <.0001 Intercept
Class.a1 1 1.1666667 8.17 8.17 0.56 0.4837 a 1
Class.a2 1 -2.8333333 48.17 48.17 3.28 0.1199 a 2
Class.b1 1 -3.0000000 108.00 108.00 7.36 0.0349 b 1
Class.a1b1 1 3.5000000 73.50 73.50 5.01 0.0665 a 1 * b 1
Class.a2b1 1 -2.5000000 37.50 37.50 2.56 0.1609 a 2 * b 1


The parameter estimates are

$\displaystyle  \hat{\mu }  $
$\displaystyle  =  $
$\displaystyle  \overline{y} = 13.3333  $
$\displaystyle \hat{\alpha }_{1}  $
$\displaystyle  =  $
$\displaystyle  (\overline{y}_{11} + \overline{y}_{12}) / 2 - \overline{y} = (15 + 14) / 2 - 13.3333 = 1.1667  $
$\displaystyle \hat{\alpha }_{2}  $
$\displaystyle  =  $
$\displaystyle  (\overline{y}_{21} + \overline{y}_{22}) / 2 - \overline{y} = (5 + 16) / 2 - 13.3333 = -2.8333  $
$\displaystyle \hat{\beta }_{1}  $
$\displaystyle  =  $
$\displaystyle  (\overline{y}_{11} + \overline{y}_{21} + \overline{y}_{31}) / 3 - \overline{y} = (15 + 5 + 11) / 3 - 13.3333 = -3  $
$\displaystyle \hat{\gamma }_{11}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{11} - (\overline{y} + \hat{\alpha }_{1} + \hat{\beta }_{1}) = 15 - (13.3333 + 1.1667 + -3) = 3.5  $
$\displaystyle \hat{\gamma }_{21}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{21} - (\overline{y} + \hat{\alpha }_{2} + \hat{\beta }_{1}) = 5 - (13.3333 + -2.8333 + -3) = -2.5  $

Figure 97.49: Deviations-from-Means Model, (3,2) Reference Cell, Design Matrix

Two-Way ANOVA Models
Deviations from Means, (3,2) Reference Cell

Obs _TYPE_ _NAME_ y Intercept a 1 a 2 b 1 a 1 * b 1 a 2 * b 1 a b
1 SCORE ROW1 16 1 1 0 1 1 0 1 1
2 SCORE ROW2 14 1 1 0 1 1 0 1 1
3 SCORE ROW3 15 1 1 0 -1 -1 0 1 2
4 SCORE ROW4 13 1 1 0 -1 -1 0 1 2
5 SCORE ROW5 1 1 0 1 1 0 1 2 1
6 SCORE ROW6 9 1 0 1 1 0 1 2 1
7 SCORE ROW7 12 1 0 1 -1 0 -1 2 2
8 SCORE ROW8 20 1 0 1 -1 0 -1 2 2
9 SCORE ROW9 14 1 -1 -1 1 -1 -1 3 1
10 SCORE ROW10 8 1 -1 -1 1 -1 -1 3 1
11 SCORE ROW11 18 1 -1 -1 -1 1 1 3 2
12 SCORE ROW12 20 1 -1 -1 -1 1 1 3 2


The next model is a less-than-full-rank model. The parameter estimates are constrained to sum to zero within each effect. The following statements produce Figure 97.50 and Figure 97.51:

proc transreg data=x ss2 short;
   title2 'Less-Than-Full-Rank Model';
   model identity(y) = class(a | b / zero=sum);
   output replace;
run;

proc print label;
run;

Figure 97.50: Less-Than-Full-Rank Model

Two-Way ANOVA Models
Less-Than-Full-Rank Model

The TRANSREG Procedure


Dependent Variable Identity(y)

Class Level Information
Class Levels Values
a 3 1 2 3
b 2 1 2

Number of Observations Read 12
Number of Observations Used 12


The TRANSREG Procedure Hypothesis Tests for Identity(y)

Univariate ANOVA Table Based on the Usual Degrees of Freedom
Source DF Sum of Squares Mean Square F Value Pr > F
Model 5 234.6667 46.93333 3.20 0.0946
Error 6 88.0000 14.66667    
Corrected Total 11 322.6667      

Root MSE 3.82971 R-Square 0.7273
Dependent Mean 13.33333 Adj R-Sq 0.5000
Coeff Var 28.72281    

Univariate Regression Table Based on the Usual Degrees of Freedom
Variable DF Coefficient Type II
Sum of
Squares
Mean Square F Value Pr > F Label
Intercept 1 13.3333333 2133.33 2133.33 145.45 <.0001 Intercept
Class.a1 1 1.1666667 8.17 8.17 0.56 0.4837 a 1
Class.a2 1 -2.8333333 48.17 48.17 3.28 0.1199 a 2
Class.a3 1 1.6666667 16.67 16.67 1.14 0.3274 a 3
Class.b1 1 -3.0000000 108.00 108.00 7.36 0.0349 b 1
Class.b2 1 3.0000000 108.00 108.00 7.36 0.0349 b 2
Class.a1b1 1 3.5000000 73.50 73.50 5.01 0.0665 a 1 * b 1
Class.a1b2 1 -3.5000000 73.50 73.50 5.01 0.0665 a 1 * b 2
Class.a2b1 1 -2.5000000 37.50 37.50 2.56 0.1609 a 2 * b 1
Class.a2b2 1 2.5000000 37.50 37.50 2.56 0.1609 a 2 * b 2
Class.a3b1 1 -1.0000000 6.00 6.00 0.41 0.5461 a 3 * b 1
Class.a3b2 1 1.0000000 6.00 6.00 0.41 0.5461 a 3 * b 2

The sum of the regression table DF's, minus one for the intercept, will be greater than the model df when there are ZERO=SUM constraints.



The parameter estimates are

$\displaystyle  \hat{\mu }  $
$\displaystyle  =  $
$\displaystyle  \overline{y} = 13.3333  $
$\displaystyle \hat{\alpha }_{1}  $
$\displaystyle  =  $
$\displaystyle  (\overline{y}_{11} + \overline{y}_{12}) / 2 - \overline{y} = (15 + 14) / 2 - 13.3333 = 1.1667  $
$\displaystyle \hat{\alpha }_{2}  $
$\displaystyle  =  $
$\displaystyle  (\overline{y}_{21} + \overline{y}_{22}) / 2 - \overline{y} = (5 + 16) / 2 - 13.3333 = -2.8333  $
$\displaystyle \hat{\alpha }_{3}  $
$\displaystyle  =  $
$\displaystyle  (\overline{y}_{31} + \overline{y}_{32}) / 2 - \overline{y} = (11 + 19) / 2 - 13.3333 = 1.6667  $
$\displaystyle \hat{\beta }_{1}  $
$\displaystyle  =  $
$\displaystyle  (\overline{y}_{11} + \overline{y}_{21} + \overline{y}_{31}) / 3 - \overline{y} = (15 + 5 + 11) / 3 - 13.3333 = -3  $
$\displaystyle \hat{\beta }_{2}  $
$\displaystyle  =  $
$\displaystyle  (\overline{y}_{12} + \overline{y}_{22} + \overline{y}_{32}) / 3 - \overline{y} = (14 + 16 + 19) / 3 - 13.3333 = 3  $
$\displaystyle \hat{\gamma }_{11}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{11} - (\overline{y} + \hat{\alpha }_{1} + \hat{\beta }_{1}) = 15 - (13.3333 + 1.1667 + -3) = 3.5  $
$\displaystyle \hat{\gamma }_{12}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{12} - (\overline{y} + \hat{\alpha }_{1} + \hat{\beta }_{2}) = 14 - (13.3333 + 1.1667 + 3) = -3.5  $
$\displaystyle \hat{\gamma }_{21}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{21} - (\overline{y} + \hat{\alpha }_{2} + \hat{\beta }_{1}) = 5 - (13.3333 + -2.8333 + -3) = -2.5  $
$\displaystyle \hat{\gamma }_{22}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{22} - (\overline{y} + \hat{\alpha }_{2} + \hat{\beta }_{2}) = 16 - (13.3333 + -2.8333 + 3) = 2.5  $
$\displaystyle \hat{\gamma }_{31}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{31} - (\overline{y} + \hat{\alpha }_{3} + \hat{\beta }_{1}) = 11 - (13.3333 + 1.6667 + -3) = -1  $
$\displaystyle \hat{\gamma }_{32}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{32} - (\overline{y} + \hat{\alpha }_{3} + \hat{\beta }_{2}) = 19 - (13.3333 + 1.6667 + 3) = 1  $

The constraints are

\[  \alpha _1 + \alpha _2 + \alpha _3 \equiv \beta _1 + \beta _2 \equiv 0  \]

\[  \gamma _{11} + \gamma _{12} \equiv \gamma _{21} + \gamma _{22} \equiv \gamma _{31} + \gamma _{32} \equiv \gamma _{11} + \gamma _{21} + \gamma _{31} \equiv \gamma _{12} + \gamma _{22} + \gamma _{32} \equiv 0  \]

Only four of the five interaction constraints are needed. The fifth constraint is implied by the other four. (Given a $2 \times 3$ table with four marginal sum-to-zero constraints, you can freely fill in only two cells. The values in the other four cells are determined from the first two cells and the constraints.) A full-rank model has six estimable parameters. This less-than-full-rank model has one parameter for the intercept, two for the first main effect (plus one more as determined by the first constraint), one for the second main effect (plus one more as determined by the second constraint), and two for the interactions (plus four more as determined by the next four constraints). Six of the twelve parameters are determined given the other six and the constraints. Notice that $\hat{\mu }, \hat{\alpha }_{1}, \hat{\alpha }_{2}, \hat{\beta }_{1}, \hat{\gamma }_{11},$ and $\hat{\gamma }_{21}$ match the corresponding estimates from the effects coding.

Figure 97.51: Less-Than-Full-Rank Model, Design Matrix

Two-Way ANOVA Models
Less-Than-Full-Rank Model

Obs _TYPE_ _NAME_ y Intercept a 1 a 2 a 3 b 1 b 2 a 1 *
b 1
a 1 *
b 2
a 2 *
b 1
a 2 *
b 2
a 3 *
b 1
a 3 *
b 2
a b
1 SCORE ROW1 16 1 1 0 0 1 0 1 0 0 0 0 0 1 1
2 SCORE ROW2 14 1 1 0 0 1 0 1 0 0 0 0 0 1 1
3 SCORE ROW3 15 1 1 0 0 0 1 0 1 0 0 0 0 1 2
4 SCORE ROW4 13 1 1 0 0 0 1 0 1 0 0 0 0 1 2
5 SCORE ROW5 1 1 0 1 0 1 0 0 0 1 0 0 0 2 1
6 SCORE ROW6 9 1 0 1 0 1 0 0 0 1 0 0 0 2 1
7 SCORE ROW7 12 1 0 1 0 0 1 0 0 0 1 0 0 2 2
8 SCORE ROW8 20 1 0 1 0 0 1 0 0 0 1 0 0 2 2
9 SCORE ROW9 14 1 0 0 1 1 0 0 0 0 0 1 0 3 1
10 SCORE ROW10 8 1 0 0 1 1 0 0 0 0 0 1 0 3 1
11 SCORE ROW11 18 1 0 0 1 0 1 0 0 0 0 0 1 3 2
12 SCORE ROW12 20 1 0 0 1 0 1 0 0 0 0 0 1 3 2


The next model is a reference cell model, but this time the reference cell is the first cell (1,1). The following statements produce Figure 97.52 and Figure 97.53:

proc transreg data=x ss2 short;
   title2 'Reference Cell Model, (1,1) Reference Cell';
   model identity(y) = class(a | b / zero=first);
   output replace;
run;

proc print label;
run;

Figure 97.52: Reference Cell Model, (1,1) Reference Cell

Two-Way ANOVA Models
Reference Cell Model, (1,1) Reference Cell

The TRANSREG Procedure


Dependent Variable Identity(y)

Class Level Information
Class Levels Values
a 3 1 2 3
b 2 1 2

Number of Observations Read 12
Number of Observations Used 12


The TRANSREG Procedure Hypothesis Tests for Identity(y)

Univariate ANOVA Table Based on the Usual Degrees of Freedom
Source DF Sum of Squares Mean Square F Value Pr > F
Model 5 234.6667 46.93333 3.20 0.0946
Error 6 88.0000 14.66667    
Corrected Total 11 322.6667      

Root MSE 3.82971 R-Square 0.7273
Dependent Mean 13.33333 Adj R-Sq 0.5000
Coeff Var 28.72281    

Univariate Regression Table Based on the Usual Degrees of Freedom
Variable DF Coefficient Type II
Sum of
Squares
Mean Square F Value Pr > F Label
Intercept 1 15.000000 450.000 450.000 30.68 0.0015 Intercept
Class.a2 1 -10.000000 100.000 100.000 6.82 0.0401 a 2
Class.a3 1 -4.000000 16.000 16.000 1.09 0.3365 a 3
Class.b2 1 -1.000000 1.000 1.000 0.07 0.8027 b 2
Class.a2b2 1 12.000000 72.000 72.000 4.91 0.0686 a 2 * b 2
Class.a3b2 1 9.000000 40.500 40.500 2.76 0.1476 a 3 * b 2


The parameter estimates are

$\displaystyle  \hat{\mu }_{11}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{11} = 15  $
$\displaystyle \hat{\alpha }_{2}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{21} - \overline{y}_{11} = 5 - 15 = -10  $
$\displaystyle \hat{\alpha }_{3}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{31} - \overline{y}_{11} = 11 - 15 = -4  $
$\displaystyle \hat{\beta }_{2}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{12} - \overline{y}_{11} = 14 - 15 = -1  $
$\displaystyle \hat{\gamma }_{22}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{22} - (\hat{\mu }_{11} + \hat{\alpha }_{2} + \hat{\beta }_{2}) = 16 - (15 + -10 + -1) = 12  $
$\displaystyle \hat{\gamma }_{32}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{32} - (\hat{\mu }_{11} + \hat{\alpha }_{3} + \hat{\beta }_{2}) = 19 - (15 + -4 + -1) = 9  $

Figure 97.53: Reference Cell Model, (1,1) Reference Cell, Design Matrix

Two-Way ANOVA Models
Reference Cell Model, (1,1) Reference Cell

Obs _TYPE_ _NAME_ y Intercept a 2 a 3 b 2 a 2 *
b 2
a 3 *
b 2
a b
1 SCORE ROW1 16 1 0 0 0 0 0 1 1
2 SCORE ROW2 14 1 0 0 0 0 0 1 1
3 SCORE ROW3 15 1 0 0 1 0 0 1 2
4 SCORE ROW4 13 1 0 0 1 0 0 1 2
5 SCORE ROW5 1 1 1 0 0 0 0 2 1
6 SCORE ROW6 9 1 1 0 0 0 0 2 1
7 SCORE ROW7 12 1 1 0 1 1 0 2 2
8 SCORE ROW8 20 1 1 0 1 1 0 2 2
9 SCORE ROW9 14 1 0 1 0 0 0 3 1
10 SCORE ROW10 8 1 0 1 0 0 0 3 1
11 SCORE ROW11 18 1 0 1 1 0 1 3 2
12 SCORE ROW12 20 1 0 1 1 0 1 3 2


The next model is a deviations-from-means model, but this time the reference cell is the first cell (1,1). This coding is also called effects coding. The following statements produce Figure 97.54 and Figure 97.55:

proc transreg data=x ss2 short;
   title2 'Deviations from Means, (1,1) Reference Cell';
   model identity(y) = class(a | b / deviations zero=first);
   output replace;
run;

proc print label;
run;

Figure 97.54: Deviations-from-Means Model, (1,1) Reference Cell

Two-Way ANOVA Models
Deviations from Means, (1,1) Reference Cell

The TRANSREG Procedure


Dependent Variable Identity(y)

Class Level Information
Class Levels Values
a 3 1 2 3
b 2 1 2

Number of Observations Read 12
Number of Observations Used 12


The TRANSREG Procedure Hypothesis Tests for Identity(y)

Univariate ANOVA Table Based on the Usual Degrees of Freedom
Source DF Sum of Squares Mean Square F Value Pr > F
Model 5 234.6667 46.93333 3.20 0.0946
Error 6 88.0000 14.66667    
Corrected Total 11 322.6667      

Root MSE 3.82971 R-Square 0.7273
Dependent Mean 13.33333 Adj R-Sq 0.5000
Coeff Var 28.72281    

Univariate Regression Table Based on the Usual Degrees of Freedom
Variable DF Coefficient Type II
Sum of
Squares
Mean Square F Value Pr > F Label
Intercept 1 13.3333333 2133.33 2133.33 145.45 <.0001 Intercept
Class.a2 1 -2.8333333 48.17 48.17 3.28 0.1199 a 2
Class.a3 1 1.6666667 16.67 16.67 1.14 0.3274 a 3
Class.b2 1 3.0000000 108.00 108.00 7.36 0.0349 b 2
Class.a2b2 1 2.5000000 37.50 37.50 2.56 0.1609 a 2 * b 2
Class.a3b2 1 1.0000000 6.00 6.00 0.41 0.5461 a 3 * b 2


The parameter estimates are

$\displaystyle  \hat{\mu }  $
$\displaystyle  =  $
$\displaystyle  \overline{y} = 13.3333  $
$\displaystyle \hat{\alpha }_{2}  $
$\displaystyle  =  $
$\displaystyle  (\overline{y}_{21} + \overline{y}_{22}) / 2 - \overline{y} = (5 + 16) / 2 - 13.3333 = -2.8333  $
$\displaystyle \hat{\alpha }_{3}  $
$\displaystyle  =  $
$\displaystyle  (\overline{y}_{31} + \overline{y}_{32}) / 2 - \overline{y} = (11 + 19) / 2 - 13.3333 = 1.6667  $
$\displaystyle \hat{\beta }_{2}  $
$\displaystyle  =  $
$\displaystyle  (\overline{y}_{12} + \overline{y}_{22} + \overline{y}_{32}) / 3 - \overline{y} = (14 + 16 + 19) / 3 - 13.3333 = 3  $
$\displaystyle \hat{\gamma }_{22}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{22} - (\overline{y} + \hat{\alpha }_{2} + \hat{\beta }_{2}) = 16 - (13.3333 + -2.8333 + 3) = 2.5  $
$\displaystyle \hat{\gamma }_{32}  $
$\displaystyle  =  $
$\displaystyle  \overline{y}_{32} - (\overline{y} + \hat{\alpha }_{3} + \hat{\beta }_{2}) = 19 - (13.3333 + 1.6667 + 3) = 1  $

Notice that all of the parameter estimates match the corresponding estimates from the less-than-full-rank coding.

Figure 97.55: Deviations-from-Means Model, (1,1) Reference Cell, Design Matrix

Two-Way ANOVA Models
Deviations from Means, (1,1) Reference Cell

Obs _TYPE_ _NAME_ y Intercept a 2 a 3 b 2 a 2 * b 2 a 3 * b 2 a b
1 SCORE ROW1 16 1 -1 -1 -1 1 1 1 1
2 SCORE ROW2 14 1 -1 -1 -1 1 1 1 1
3 SCORE ROW3 15 1 -1 -1 1 -1 -1 1 2
4 SCORE ROW4 13 1 -1 -1 1 -1 -1 1 2
5 SCORE ROW5 1 1 1 0 -1 -1 0 2 1
6 SCORE ROW6 9 1 1 0 -1 -1 0 2 1
7 SCORE ROW7 12 1 1 0 1 1 0 2 2
8 SCORE ROW8 20 1 1 0 1 1 0 2 2
9 SCORE ROW9 14 1 0 1 -1 0 -1 3 1
10 SCORE ROW10 8 1 0 1 -1 0 -1 3 1
11 SCORE ROW11 18 1 0 1 1 0 1 3 2
12 SCORE ROW12 20 1 0 1 1 0 1 3 2


The following statements fit a model with an orthogonal-contrast coding and produce Figure 97.56 and Figure 97.57:

proc transreg data=x ss2 short;
   title2 'Orthogonal Contrast Coding';
   model identity(y) = class(a | b / orthogonal);
   output replace;
run;

proc print label;
run;

Figure 97.56: Orthogonal-Contrast Coding

Two-Way ANOVA Models
Orthogonal Contrast Coding

The TRANSREG Procedure


Dependent Variable Identity(y)

Class Level Information
Class Levels Values
a 3 1 2 3
b 2 1 2

Number of Observations Read 12
Number of Observations Used 12


The TRANSREG Procedure Hypothesis Tests for Identity(y)

Univariate ANOVA Table Based on the Usual Degrees of Freedom
Source DF Sum of Squares Mean Square F Value Pr > F
Model 5 234.6667 46.93333 3.20 0.0946
Error 6 88.0000 14.66667    
Corrected Total 11 322.6667      

Root MSE 3.82971 R-Square 0.7273
Dependent Mean 13.33333 Adj R-Sq 0.5000
Coeff Var 28.72281    

Univariate Regression Table Based on the Usual Degrees of Freedom
Variable DF Coefficient Type II
Sum of
Squares
Mean Square F Value Pr > F Label
Intercept 1 13.3333333 2133.33 2133.33 145.45 <.0001 Intercept
Class.a1 1 -0.2500000 0.50 0.50 0.03 0.8596 a 1
Class.a2 1 -1.4166667 48.17 48.17 3.28 0.1199 a 2
Class.b1 1 -3.0000000 108.00 108.00 7.36 0.0349 b 1
Class.a1b1 1 2.2500000 40.50 40.50 2.76 0.1476 a 1 * b 1
Class.a2b1 1 -1.2500000 37.50 37.50 2.56 0.1609 a 2 * b 1


The parameter estimates are

$\displaystyle  \hat{\mu }  $
$\displaystyle  =  $
$\displaystyle  \overline{y} = 13.3333  $
$\displaystyle \hat{\alpha }_{1}  $
$\displaystyle  =  $
$\displaystyle  ((\overline{y}_{11} + \overline{y}_{12}) - (\overline{y}_{31} + \overline{y}_{32})) / 4 = ((15 + 14) - (11 + 19)) / 4 = -0.25  $
$\displaystyle \hat{\alpha }_{2}  $
$\displaystyle  =  $
$\displaystyle  ((\overline{y}_{21} + \overline{y}_{22}) - (\overline{y}_{11} + \overline{y}_{12} + \overline{y}_{31} + \overline{y}_{32}) / 2) / 6  $
$\displaystyle  $
$\displaystyle  =  $
$\displaystyle  ((5 + 16) - (15 + 14 + 11 + 19) / 2) / 6 = -1.417  $
$\displaystyle \hat{\beta }_{1}  $
$\displaystyle  =  $
$\displaystyle  ((\overline{y}_{11} + \overline{y}_{21} + \overline{y}_{31}) - (\overline{y}_{12} + \overline{y}_{22} + \overline{y}_{32})) / 6  $
$\displaystyle  $
$\displaystyle  =  $
$\displaystyle  ((15 + 5 + 11) - (14 + 16 + 19)) / 6 = -3  $
$\displaystyle \hat{\gamma }_{11}  $
$\displaystyle  =  $
$\displaystyle  (\overline{y}_{11} - \overline{y}_{12} - \overline{y}_{31} + \overline{y}_{32}) / 4 = (15 - 14 - 11 + 19) / 4 = 2.25  $
$\displaystyle \hat{\gamma }_{21}  $
$\displaystyle  =  $
$\displaystyle  ((-\overline{y}_{11} + \overline{y}_{12} - \overline{y}_{31} + \overline{y}_{32}) / 2 + (\overline{y}_{21} - \overline{y}_{22})) / 6  $
$\displaystyle  $
$\displaystyle  =  $
$\displaystyle  ((-15 + 14 - 11 + 19) / 2 + (5 - 16)) / 6 = -1.25  $

Figure 97.57: Orthogonal-Contrast Coding, Design Matrix

Two-Way ANOVA Models
Orthogonal Contrast Coding

Obs _TYPE_ _NAME_ y Intercept a 1 a 2 b 1 a 1 * b 1 a 2 * b 1 a b
1 SCORE ROW1 16 1 1 -1 1 1 -1 1 1
2 SCORE ROW2 14 1 1 -1 1 1 -1 1 1
3 SCORE ROW3 15 1 1 -1 -1 -1 1 1 2
4 SCORE ROW4 13 1 1 -1 -1 -1 1 1 2
5 SCORE ROW5 1 1 0 2 1 0 2 2 1
6 SCORE ROW6 9 1 0 2 1 0 2 2 1
7 SCORE ROW7 12 1 0 2 -1 0 -2 2 2
8 SCORE ROW8 20 1 0 2 -1 0 -2 2 2
9 SCORE ROW9 14 1 -1 -1 1 -1 -1 3 1
10 SCORE ROW10 8 1 -1 -1 1 -1 -1 3 1
11 SCORE ROW11 18 1 -1 -1 -1 1 1 3 2
12 SCORE ROW12 20 1 -1 -1 -1 1 1 3 2


The following statements fit a model with a standardized-orthogonal coding and produce Figure 97.58 and Figure 97.59:

proc transreg data=x ss2 short;
   title2 'Standardized-Orthogonal Coding';
   model identity(y) = class(a | b / standorth);
   output replace;
run;

proc print label;
run;

Figure 97.58: Standardized-Orthogonal Coding

Two-Way ANOVA Models
Standardized-Orthogonal Coding

The TRANSREG Procedure


Dependent Variable Identity(y)

Class Level Information
Class Levels Values
a 3 1 2 3
b 2 1 2

Number of Observations Read 12
Number of Observations Used 12


The TRANSREG Procedure Hypothesis Tests for Identity(y)

Univariate ANOVA Table Based on the Usual Degrees of Freedom
Source DF Sum of Squares Mean Square F Value Pr > F
Model 5 234.6667 46.93333 3.20 0.0946
Error 6 88.0000 14.66667    
Corrected Total 11 322.6667      

Root MSE 3.82971 R-Square 0.7273
Dependent Mean 13.33333 Adj R-Sq 0.5000
Coeff Var 28.72281    

Univariate Regression Table Based on the Usual Degrees of Freedom
Variable DF Coefficient Type II
Sum of
Squares
Mean Square F Value Pr > F Label
Intercept 1 13.3333333 2133.33 2133.33 145.45 <.0001 Intercept
Class.a1 1 -0.2041241 0.50 0.50 0.03 0.8596 a 1
Class.a2 1 -2.0034692 48.17 48.17 3.28 0.1199 a 2
Class.b1 1 -3.0000000 108.00 108.00 7.36 0.0349 b 1
Class.a1b1 1 1.8371173 40.50 40.50 2.76 0.1476 a 1 * b 1
Class.a2b1 1 -1.7677670 37.50 37.50 2.56 0.1609 a 2 * b 1


The parameter estimates are

$\displaystyle  \hat{\mu }  $
$\displaystyle  =  $
$\displaystyle  \overline{y} = 13.3333  $
$\displaystyle \hat{\alpha }_{1}  $
$\displaystyle  =  $
$\displaystyle  (((\overline{y}_{11} + \overline{y}_{12}) - (\overline{y}_{31} + \overline{y}_{32})) / 4) \times \sqrt {2 / 3}  $
$\displaystyle  $
$\displaystyle  =  $
$\displaystyle  (((15 + 14) - (11 + 19)) / 4) \times \sqrt {2 / 3} = -0.2041  $
$\displaystyle \hat{\alpha }_{2}  $
$\displaystyle  =  $
$\displaystyle  (((\overline{y}_{21} + \overline{y}_{22}) - (\overline{y}_{11} + \overline{y}_{12} + \overline{y}_{31} + \overline{y}_{32}) / 2) / 6) \times \sqrt {6 / 3}  $
$\displaystyle  $
$\displaystyle  =  $
$\displaystyle  (((5 + 16) - (15 + 14 + 11 + 19) / 2) / 6) \times \sqrt {6 / 3} = -2.0035  $
$\displaystyle \hat{\beta }_{1}  $
$\displaystyle  =  $
$\displaystyle  (((\overline{y}_{11} + \overline{y}_{21} + \overline{y}_{31}) - (\overline{y}_{12} + \overline{y}_{22} + \overline{y}_{32})) / 6) \times \sqrt {2 / 2}  $
$\displaystyle  $
$\displaystyle  =  $
$\displaystyle  (((15 + 5 + 11) - (14 + 16 + 19)) / 6) \times \sqrt {2 / 2} = -3  $
$\displaystyle \hat{\gamma }_{11}  $
$\displaystyle  =  $
$\displaystyle  ((\overline{y}_{11} - \overline{y}_{12} - \overline{y}_{31} + \overline{y}_{32}) / 4) \times \sqrt {2 / 3} \times \sqrt {2 / 2}  $
$\displaystyle  $
$\displaystyle  =  $
$\displaystyle  ((15 - 14 - 11 + 19) / 4) \times \sqrt {2 / 3} \times \sqrt {2 / 2} = 1.8371  $
$\displaystyle \hat{\gamma }_{21}  $
$\displaystyle  =  $
$\displaystyle  (((-\overline{y}_{11} + \overline{y}_{12} - \overline{y}_{31} + \overline{y}_{32}) / 2 + (\overline{y}_{21} - \overline{y}_{22})) / 6) \times \sqrt {6 / 3} \times \sqrt {2 / 2}  $
$\displaystyle  $
$\displaystyle  =  $
$\displaystyle  (((-15 + 14 - 11 + 19) / 2 + (5 - 16)) / 6) \times \sqrt {6 / 3} \times \sqrt {2 / 2} = -1.7678  $

The numerators in the square roots are sums of squares of the coded values for the unstandardized-orthogonal codings, and the denominators are the numbers of levels. These terms convert the estimates from the orthogonal contrast coding to the standardized-orthogonal coding. The term $\sqrt {2 / 2}$, which is 1 and could be dropped, is included in the preceding formulas to show the general pattern. Notice the regression tables for the orthogonal-contrast coding and the standardized-orthogonal coding. Some of the coefficients are different, but the rest of the table is the same since the coded variables for the two models differ only by a constant.

Figure 97.59: Standardized-Orthogonal Coding, Design Matrix

Two-Way ANOVA Models
Standardized-Orthogonal Coding

Obs _TYPE_ _NAME_ y Intercept a 1 a 2 b 1 a 1 * b 1 a 2 * b 1 a b
1 SCORE ROW1 16 1 1.22474 -0.70711 1 1.22474 -0.70711 1 1
2 SCORE ROW2 14 1 1.22474 -0.70711 1 1.22474 -0.70711 1 1
3 SCORE ROW3 15 1 1.22474 -0.70711 -1 -1.22474 0.70711 1 2
4 SCORE ROW4 13 1 1.22474 -0.70711 -1 -1.22474 0.70711 1 2
5 SCORE ROW5 1 1 0.00000 1.41421 1 0.00000 1.41421 2 1
6 SCORE ROW6 9 1 0.00000 1.41421 1 0.00000 1.41421 2 1
7 SCORE ROW7 12 1 0.00000 1.41421 -1 0.00000 -1.41421 2 2
8 SCORE ROW8 20 1 0.00000 1.41421 -1 0.00000 -1.41421 2 2
9 SCORE ROW9 14 1 -1.22474 -0.70711 1 -1.22474 -0.70711 3 1
10 SCORE ROW10 8 1 -1.22474 -0.70711 1 -1.22474 -0.70711 3 1
11 SCORE ROW11 18 1 -1.22474 -0.70711 -1 1.22474 0.70711 3 2
12 SCORE ROW12 20 1 -1.22474 -0.70711 -1 1.22474 0.70711 3 2