# Examples of Writing CONTRAST and ESTIMATE Statements

**Introduction**

**EXAMPLE 1: A Two-Factor Model with Interaction**

Computing the Cell Means Using the ESTIMATE Statement

Estimating and Testing a Difference of Means

A More Complex Contrast

Comparing One Interaction Mean to the Average of All Interaction Means

**EXAMPLE 2: A Three-Factor Model with Interactions**

**EXAMPLE 3: A Two-Factor Logistic Model with Interaction Using Dummy and Effects Coding**

Dummy Coding

Estimating and Testing Odds Ratios with Dummy Coding

A Nested Model

Effects Coding

Estimating and Testing Odds Ratios with Effects Coding

A More Complex Contrast with Effects Coding

**EXAMPLE 4: Comparing Models**

Comparing Nested Models

Comparing Nonnested Models

**EXAMPLE 5: A Quadratic Logistic Model**

SAS Code from All of These Examples

you might need to print it in landscape mode to avoid truncation of the right edge.

**Printing this document:**Because some of the tables in this document are wide,you might need to print it in landscape mode to avoid truncation of the right edge.

## Introduction

To properly test a hypothesis such as "The effect of treatment A in group 1 is equal to the treatment A effect in group 2," it is necessary to translate it correctly into a mathematical hypothesis using the fitted model. Specifically, you need to construct the linear combination of model parameters that corresponds to the hypothesis. Such linear combinations can be estimated and tested using the CONTRAST and/or ESTIMATE statements available in many modeling procedures. While only certain procedures are illustrated below, this discussion applies to any modeling procedure that allows these statements.Though assisting with the translation of a stated hypothesis into the needed linear combination is beyond the scope of the services that are provided by Technical Support at SAS, we hope that the following discussion and examples will help you. Technical Support can assist you with syntax and other questions that relate to CONTRAST and ESTIMATE statements.

As you'll see in the examples that follow, there are some important steps in properly writing a CONTRAST or ESTIMATE statement:

- Write down the model that you are using the procedure to fit. There are two crucial parts to this:
- Parameterization
- How design variables that are generated by the CLASS statement are coded. The two most commonly used parameterizations that are available in SAS are
*indicator*(or*dummy*) coding and*effects*(or*deviation from mean*) coding. - Parameter Ordering
- The order of the parameters within effects that have multiple parameters (such as CLASS variables and interactions of CLASS variables). The ordering typically depends on the order in which the variables are specified in the CLASS statement and the setting of the ORDER= option in the PROC or CLASS statement. Use the Parameter Estimates or Class Level Information table in the modeling procedure's displayed results to confirm the parameter order.

- Write down the hypothesis to be tested or quantity to be estimated in terms of the model's parameters and simplify. When testing, write the null hypothesis in the form
*contrast = 0*before simplifying the left-hand side. For example, to compare two means, specify the null hypothesis as μ_{1}– μ_{2}= 0 and then write μ_{1}– μ_{2}in terms of the model parameters. - Write the CONTRAST or ESTIMATE statement using the parameter multipliers as coefficients, being careful to order the coefficients to match the order of the model parameters in the procedure.

The CONTRAST and ESTIMATE statements allow for estimation and testing of any linear combination of model parameters. However, a common subclass of interest involves comparison of means and most of the examples below are from this class. While examples in this class provide good examples of the above process for determining coefficients for CONTRAST and ESTIMATE statements, there are other statements available that perform means comparisons more easily. These statements include the LSMEANS, LSMESTIMATE, and SLICE statements that are available beginning with SAS/STAT 9.22 in SAS 9.2 TS2M3 in many procedures. The ODDSRATIO statement in PROC LOGISTIC and the similar HAZARDRATIO statement in PROC PHREG are also available. While the main purpose of this note is to illustrate how to write proper CONTRAST and ESTIMATE statements, these additional statements are also presented when they can provide equivalent analyses.

## Example 1: A Two-Factor Model with Interaction

Consider a model for two factors: A with five levels and B with two levels:*Y*_{ijk}= μ + α_{i}+ β_{j}+ αβ_{ij}+ ε_{ijk}(1)

*i*=1,2,...,5,

*j*=1,2,

*k*=1, 2,...,

*n*

_{ij}. The response,

*Y*, is normally distributed with constant variance. The statements below generate observations from such a model:

data test; seed=6342454; do a=1 to 5; do b=1 to 2; do rep=1 to ceil(ranuni(seed)*5)+5; y=5 + a + b + a*b + rannor(seed); output; end; end; end; run;The following statements fit the main effects and interaction model. The LSMEANS statement computes the cell means for the 10 A*B cells in this example. The E option shows how each cell mean is formed by displaying the coefficient vectors that are used in calculating the LS-means.

proc mixed data=test; class a b; model y=a b a*b / solution; lsmeans a*b / e; run;

Least Squares Means | |||||||

Effect | a | b | Estimate | Standard Error | DF | t Value | Pr > |t| |

a*b | 1 | 1 | 7.5193 | 0.2905 | 74 | 25.88 | <.0001 |

a*b | 1 | 2 | 10.0341 | 0.2598 | 74 | 38.62 | <.0001 |

a*b | 2 | 1 | 10.4189 | 0.2739 | 74 | 38.04 | <.0001 |

a*b | 2 | 2 | 12.5812 | 0.3355 | 74 | 37.50 | <.0001 |

a*b | 3 | 1 | 11.5853 | 0.3355 | 74 | 34.54 | <.0001 |

a*b | 3 | 2 | 15.7347 | 0.2598 | 74 | 60.55 | <.0001 |

a*b | 4 | 1 | 14.5552 | 0.3355 | 74 | 43.39 | <.0001 |

a*b | 4 | 2 | 19.3499 | 0.2598 | 74 | 74.47 | <.0001 |

a*b | 5 | 1 | 16.3459 | 0.2739 | 74 | 59.68 | <.0001 |

a*b | 5 | 2 | 21.6220 | 0.2598 | 74 | 83.21 | <.0001 |

### Computing the Cell Means Using the ESTIMATE Statement

The cell means can also be obtained by using the ESTIMATE statement to compute the appropriate linear combinations of model parameters. Means for the AB_{11}and AB

_{12}cells (highlighted in the above table) are computed below using the ESTIMATE statement. The coefficients that are needed in the ESTIMATE statement are determined by writing what you want to estimate in terms of the fitted model. Using model (1) above, the AB

_{12}cell mean, μ

_{12}, is:

- μ
_{12}= (Σ_{k}Y_{12k})/n_{12}

= (Σ_{k}μ)/*n*_{12}+ (Σ_{k}α_{1})/*n*_{12}+ (Σ_{k}β_{2})/*n*_{12}+ (Σ_{k}αβ_{12})/*n*_{12}+ (Σ_{k}ε_{12k})/*n*_{12}

_{ijk}) are assumed to be zero:

- = μ + α
_{1}+ β_{2}+ αβ_{12}

_{11}cell mean is written this way:

- μ
_{11}= μ + α_{1}+ β_{1}+ αβ_{11}

_{12}mean, you need to add together the estimates of μ, α

_{1}, β

_{2}, and αβ

_{12}. This can be done by multiplying the vector of parameter estimates (the

*solution*vector) by a vector of coefficients such that their product is this sum. The solution vector in PROC MIXED is requested with the SOLUTION option in the MODEL statement and appears as the Estimate column in the Solution for Fixed Effects table:

Solution for Fixed Effects | |||||||

Effect | a | b | Estimate | Standard Error | DF | t Value | Pr > |t| |

Intercept | 21.6220 | 0.2598 | 74 | 83.21 | <.0001 | ||

a | 1 | -11.5879 | 0.3675 | 74 | -31.53 | <.0001 | |

a | 2 | -9.0408 | 0.4243 | 74 | -21.31 | <.0001 | |

a | 3 | -5.8874 | 0.3675 | 74 | -16.02 | <.0001 | |

a | 4 | -2.2722 | 0.3675 | 74 | -6.18 | <.0001 | |

a | 5 | 0 | . | . | . | . | |

b | 1 | -5.2762 | 0.3775 | 74 | -13.97 | <.0001 | |

b | 2 | 0 | . | . | . | . | |

a*b | 1 | 1 | 2.7613 | 0.5426 | 74 | 5.09 | <.0001 |

a*b | 1 | 2 | 0 | . | . | . | . |

a*b | 2 | 1 | 3.1139 | 0.5745 | 74 | 5.42 | <.0001 |

a*b | 2 | 2 | 0 | . | . | . | . |

a*b | 3 | 1 | 1.1268 | 0.5680 | 74 | 1.98 | 0.0510 |

a*b | 3 | 2 | 0 | . | . | . | . |

a*b | 4 | 1 | 0.4815 | 0.5680 | 74 | 0.85 | 0.3994 |

a*b | 4 | 2 | 0 | . | . | . | . |

a*b | 5 | 1 | 0 | . | . | . | . |

a*b | 5 | 2 | 0 | . | . | . | . |

_{1}through α

_{5}. The next two elements are the parameter estimates for the levels of B, β

_{1}and β

_{2}. The last 10 elements are the parameter estimates for the 10 levels of the A*B interaction, αβ

_{11}through αβ

_{52}.

Now choose a coefficient vector, also with 18 elements, that will multiply the solution vector: Choose a coefficient of 1 for the intercept (μ), coefficients of (1 0 0 0 0) for the A term to pick up the α

_{1}estimate, coefficients of (0 1) for the B term to pick up the β

_{2}estimate, and coefficients of (0 1 0 0 0 0 0 0 0 0) for the A*B interaction term to pick up the αβ

_{12}estimate.

The ESTIMATE statement syntax enables you to specify the coefficient vector in sections as just described, with one section for each model effect:

estimate 'AB12' intercept 1 a 1 0 0 0 0 b 0 1 a*b 0 1 0 0 0 0 0 0 0 0;Note that this same coefficient vector is given in the table of LS-means coefficients, which was requested by the E option in the LSMEANS statement. Zeros in this table are shown as blanks for clarity. Notice that Row2 is the coefficient vector for computing the mean of the AB

_{12}cell.

Coefficients for a*b Least Squares Means | ||||||||||||

Effect | a | b | Row1 | Row2 | Row3 | Row4 | Row5 | Row6 | Row7 | Row8 | Row9 | Row10 |

Intercept | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | ||

a | 1 | 1 | 1 | |||||||||

a | 2 | 1 | 1 | |||||||||

a | 3 | 1 | 1 | |||||||||

a | 4 | 1 | 1 | |||||||||

a | 5 | 1 | 1 | |||||||||

b | 1 | 1 | 1 | 1 | 1 | 1 | ||||||

b | 2 | 1 | 1 | 1 | 1 | 1 | ||||||

a*b | 1 | 1 | 1 | |||||||||

a*b | 1 | 2 | 1 | |||||||||

a*b | 2 | 1 | 1 | |||||||||

a*b | 2 | 2 | 1 | |||||||||

a*b | 3 | 1 | 1 | |||||||||

a*b | 3 | 2 | 1 | |||||||||

a*b | 4 | 1 | 1 | |||||||||

a*b | 4 | 2 | 1 | |||||||||

a*b | 5 | 1 | 1 | |||||||||

a*b | 5 | 2 | 1 |

_{12}, because the levels of B change before the levels of A. Had B preceded A in the CLASS statement, the levels of A would have changed before the levels of B, resulting in the second estimate being for αβ

_{21}. In this case, the αβ

_{12}estimate is the sixth estimate in the A*B effect requiring a change in the coefficient vector that you specify in the ESTIMATE statement.

An ESTIMATE statement for the AB

_{11}cell mean can be written as above by rewriting the cell mean in terms of the model yielding the appropriate linear combination of parameter estimates. The result is Row1 in the table of LS-means coefficients. The following statements fit the model and compute the AB

_{11}and AB

_{12}cell means by using the LSMEANS statement and equivalent ESTIMATE statements:

proc mixed data=test; class a b; model y=a b a*b; lsmeans a*b; estimate 'AB11' intercept 1 a 1 0 0 0 0 b 1 0 a*b 1 0 0 0 0 0 0 0 0 0; estimate 'AB12' intercept 1 a 1 0 0 0 0 b 0 1 a*b 0 1 0 0 0 0 0 0 0 0; run;

Estimates | |||||

Label | Estimate | Standard Error | DF | t Value | Pr > |t| |

AB11 | 7.5193 | 0.2905 | 74 | 25.88 | <.0001 |

AB12 | 10.0341 | 0.2598 | 74 | 38.62 | <.0001 |

### Estimating and Testing a Difference of Means

Suppose you want to test that the AB_{11}and AB

_{12}cell means are equal. This is the null hypothesis to test:

- H
_{0}: μ_{11}= μ_{12}

- H
_{0}: μ_{11}– μ_{12}= 0

- μ
_{11}– μ_{12}= (μ + α_{1}+ β_{1}+ αβ_{11}) – (μ + α_{1}+ β_{2}+ αβ_{12}) = β_{1}– β_{2}+ αβ_{11}– αβ_{12}

You can also duplicate the results of the CONTRAST statement with an ESTIMATE statement. Note that the ESTIMATE statement displays the estimated difference in cell means (–2.5148) and a

*t*-test that this difference is equal to zero, while the CONTRAST statement provides only an

*F*-test of the difference. The tests are equivalent.

For simple pairwise contrasts like this involving a single effect, there are several other ways to obtain the test. You can use the DIFF option in the LSMEANS statement. You can specify a contrast of the LS-means themselves, rather than the model parameters, by using the LSMESTIMATE statement. Finally, you can use the SLICE statement. Note that the CONTRAST and ESTIMATE statements are the most flexible allowing for any linear combination of model parameters. Only these two statements may be flexible enough to estimate or test sufficiently complex linear combinations of model parameters.

The following statements show all five ways of computing and testing this contrast. The DIFF option in the LSMEANS statement provides all pairwise comparisons of the ten LS-means. The contrast of the ten LS-means specified in the LSMESTIMATE statement estimates and tests the difference between the AB11 and AB12 LS-means. The DIFF and SLICEBY(A='1') options in the SLICE statement estimate the differences in LS-means at A=1. Finally, the CONTRAST and ESTIMATE statements use the contrast determined above to compute the AB11 - AB12 difference. All produce equivalent results.

proc mixed data=test; class a b; model y= a b a*b; lsmeans a*b / diff; lsmestimate a*b 'AB11 - AB12' 1 -1 0 0 0 0 0 0 0 0; slice a*b / sliceby(a='1') diff; contrast 'AB11 - AB12' b 1 -1 a*b 1 -1 0 0 0 0 0 0 0 0; estimate 'AB11 - AB12' b 1 -1 a*b 1 -1 0 0 0 0 0 0 0 0; run;These results are from the SLICE statement:

Simple Differences of a*b Least Squares Means | |||||||
---|---|---|---|---|---|---|---|

Slice | b | _b | Estimate | Standard Error | DF | t Value | Pr > |t| |

a 1 | 1 | 2 | -2.5148 | 0.3898 | 74 | -6.45 | <.0001 |

Least Squares Means Estimate | ||||||
---|---|---|---|---|---|---|

Effect | Label | Estimate | Standard Error | DF | t Value | Pr > |t| |

a*b | AB11 - AB12 | -2.5148 | 0.3898 | 74 | -6.45 | <.0001 |

Estimates | |||||

Label | Estimate | Standard Error | DF | t Value | Pr > |t| |

AB11 - AB12 | -2.5148 | 0.3898 | 74 | -6.45 | <.0001 |

Contrasts | ||||

Label | Num DF | Den DF | F Value | Pr > F |

AB11 - AB12 | 1 | 74 | 41.63 | <.0001 |

Differences of Least Squares Means | |||||||||

Effect | a | b | _a | _b | Estimate | Standard Error | DF | t Value | Pr > |t| |

a*b | 1 | 1 | 1 | 2 | -2.5148 | 0.3898 | 74 | -6.45 | <.0001 |

### A More Complex Contrast

Suppose you want to test the average of AB_{11}and AB

_{12}versus the average of AB

_{21}and AB

_{22}.

- H
_{0}: ½(μ_{11}+ μ_{12}) = ½(μ_{21}+ μ_{22})

_{11}and AB

_{12}are again determined by writing them in terms of the model. The individual AB

_{11}and AB

_{12}cell means are:

- μ
_{11}= μ + α_{1}+ β_{1}+ αβ_{11}

- μ
_{12}= μ + α_{1}+ β_{2}+ αβ_{12}

- ½(μ
_{11}+μ_{12}) = ½[ (μ + α_{1}+ β_{1}+ αβ_{11}) + (μ + α_{1}+ β_{2}+ αβ_{12}) ]

= μ + α_{1}+ ½β_{1}+ ½β_{2}+ ½αβ_{11}+ ½αβ_{12}

_{21}and AB

_{22}cells are determined in the same fashion. Then, as before, subtracting the two coefficient vectors yields the coefficient vector for testing the difference of these two averages. The final coefficients appear in ESTIMATE and CONTRAST statements below. Note that within a set of coefficients for an effect you can leave off any trailing zeros.

The LSMESTIMATE statement can also be used. Since the contrast involves only the ten LS-means, it is much more straight-forward to specify. Again, trailing zero coefficients can be omitted. The SLICE and LSMEANS statements cannot be used for this more complex contrast.

proc mixed data=test; class a b; model y= a b a*b; estimate 'avg AB11,AB12' intercept 1 a 1 b .5 .5 a*b .5 .5; estimate 'avg AB21,AB22' intercept 1 a 0 1 b .5 .5 a*b 0 0 .5 .5; contrast 'avg AB11,AB12 - avg AB21+AB22' a 1 -1 a*b .5 .5 -.5 -.5; lsmestimate a*b 'avg AB11,AB12 - avg AB21+AB22' 1 1 -1 -1 / divisor=2; run;

Estimates | |||||

Label | Estimate | Standard Error | DF | t Value | Pr > |t| |

avg AB11,AB12 | 8.7767 | 0.1949 | 74 | 45.04 | <.0001 |

avg AB21,AB22 | 11.5001 | 0.2165 | 74 | 53.11 | <.0001 |

Contrasts | ||||

Label | Num DF | Den DF | F Value | Pr > F |

avg AB11,AB12 - avg AB21+AB22 | 1 | 74 | 87.39 | <.0001 |

*t*statistic value is the square root of the F statistic from the CONTRAST statement producing an equivalent test.

Least Squares Means Estimate | ||||||
---|---|---|---|---|---|---|

Effect | Label | Estimate | Standard Error | DF | t Value | Pr > |t| |

a*b | avg AB11,AB12 - avg AB21+AB22 | -2.7233 | 0.2913 | 74 | -9.35 | <.0001 |

### Comparing One Interaction Mean to the Average of All Interaction Means

Suppose A has two levels and B has three levels and you want to test if the AB_{12}cell mean is different from the average of all six cell means.

- H
_{0}: μ_{12}– 1/6 Σ_{ij}μ_{ij}= 0

_{12}cell mean in terms of the model:

- μ
_{12}= μ + α_{1}+ β_{2}+ αβ_{12}

- Σ
_{ij}μ_{ij}= 1/6 Σ_{i}Σ_{j}(μ + α_{i}+ β_{j}+ αβ_{ij}) = μ + 3/6 Σ_{i}α_{i}+ 2/6 Σ_{j}β_{j}+ 1/6 Σ_{i}Σ_{j}αβ_{ij}

estimate 'avg ABij' intercept 1 a .5 .5 b .333 .333 .333 a*b .167 .167 .167 .167 .167 .167;then the procedure provides no results, either displaying Non-est in the table of results or issuing this message in the log:

NOTE: avg ABij is not estimable.The estimate is declared nonestimable simply because the coefficients 1/3 and 1/6 are not represented precisely enough. To avoid this problem, use the DIVISOR= option. The value that you specify in the option divides all the coefficients that are provided in the ESTIMATE statement.

Finally, writing the hypothesis μ

_{12}– 1/6 Σ

_{ij}μ

_{ij}in terms of the model results in these contrast coefficients: 0 for μ, 1/2 and –1/2 for A, –1/3, 2/3, and –1/3 for B, and –1/6, 5/6, –1/6, –1/6, –1/6, and –1/6 for AB. The statements below fit the model, estimate each part of the hypothesis, and estimate and test the hypothesis. The DIVISOR= option is used to ensure precision and avoid nonestimability.

The LSMESTIMATE statement again makes this easier. The necessary contrast coefficients are stated in the null hypothesis above: (0 1 0 0 0 0) - (1/6 1/6 1/6 1/6 1/6 1/6) , which simplifies to the contrast shown in the LSMESTIMATE statement below.

proc mixed data=test; class a b; model y=a b a*b; estimate 'AB12' intercept 1 a 1 0 b 0 1 0 a*b 0 1 0 0 0 0; estimate 'avg ABij' intercept 6 a 3 3 b 2 2 2 a*b 1 1 1 1 1 1 / divisor=6; estimate 'AB12 vs avg ABij' a 3 -3 b -2 4 -2 a*b -1 5 -1 -1 -1 -1 / divisor=6; lsmestimate a*b 'AB12 vs avg ABij' -1 5 -1 -1 -1 -1 / divisor=6; run;

Parameter | Estimate | Standard Error | t Value | Pr > |t| |
---|---|---|---|---|

AB12 | 10.0341436 | 0.25521923 | 39.32 | <.0001 |

avg ABij | 11.3122544 | 0.11799152 | 95.87 | <.0001 |

AB12 vs avg ABij | -1.2781108 | 0.23947144 | -5.34 | <.0001 |

## Example 2: A Three-Factor Model with Interactions

Now consider a model in three factors, with five, two, and three levels, respectively. Here is the model that includes main effects and all interactions:*Y*_{ijkl}= μ + α_{i}+ β_{j}+ γ_{k}+ αβ_{ij}+ αγ_{ik}+ βγ_{jk}+ αβγ_{ijk}+ ε_{ijkl}(2)

*i*=1,2,...,5,

*j*=1,2,

*k*=1,2,3, and

*l*=1,2,...,

*N*.

_{ijk}These statements generate data from the above model:

data test; seed=8422636; do a=1 to 5; do b=1 to 2; do c=1 to 3; do rep=1 to ceil(ranuni(seed)*3)+3; y=5 + a + b + c + a*b + a*c + b*c + a*b*c + rannor(seed); output; end; end; end; end; run;The following statements fit model (2) and display the solution vector and cell means. Note that there are 5 × 2 × 3 = 30 cell means.

proc mixed data=test; class a b c; model y=a|b|c / solution; lsmeans a*b*c; run;Suppose it is of interest to test the null hypothesis that cell means ABC

_{121}and ABC

_{212}are equal — that is, H

_{0}: μ

_{121}- μ

_{212}= 0. Note that these are the fourth and eighth cell means in the Least Squares Means table. Writing the means and their difference in terms of model (2):

- μ
_{121}= μ + α_{1}+ β_{2}+ γ_{1}+ αβ_{12}+ αγ_{11}+ βγ_{21}+ αβγ_{121}

- μ
_{212}= μ + α_{2}+ β_{1}+ γ_{2}+ αβ_{21}+ αγ_{22}+ βγ_{12}+ αβγ_{212}

- μ
_{121}– μ_{212}= α_{1}– α_{2}– β_{1}+ β_{2}+ γ_{1}– γ_{2}+ αβ_{12}– αβ_{21}+ αγ_{11}– αγ_{22}+ βγ_{21}– βγ_{12}+ αβγ_{121}– αβγ_{212}

proc mixed data=test; class a b c; model y=a|b|c; estimate 'ABC121' intercept 1 a 1 b 0 1 c 1 a*b 0 1 a*c 1 b*c 0 0 0 1 a*b*c 0 0 0 1; estimate 'ABC212' intercept 1 a 0 1 b 1 c 0 1 a*b 0 0 1 a*c 0 0 0 0 1 b*c 0 1 a*b*c 0 0 0 0 0 0 0 1; contrast 'ABC121 - ABC212' a 1 -1 b -1 1 c 1 -1 a*b 0 1 -1 a*c 1 0 0 0 -1 b*c 0 -1 0 1 a*b*c 0 0 0 1 0 0 0 -1; estimate 'ABC121 - ABC212' a 1 -1 b -1 1 c 1 -1 a*b 0 1 -1 a*c 1 0 0 0 -1 b*c 0 -1 0 1 a*b*c 0 0 0 1 0 0 0 -1; lsmestimate a*b*c 'ABC121 - ABC212' 0 0 0 1 0 0 0 -1; run;

Estimates | |||||

Label | Estimate | Standard Error | DF | t Value | Pr > |t| |

ABC121 | 17.0454 | 0.4118 | 125 | 41.40 | <.0001 |

ABC212 | 21.8270 | 0.4118 | 125 | 53.01 | <.0001 |

ABC121 - ABC212 | -4.7816 | 0.5823 | 125 | -8.21 | <.0001 |

Contrasts | ||||

Label | Num DF | Den DF | F Value | Pr > F |

ABC121 - ABC212 | 1 | 125 | 67.42 | <.0001 |

Least Squares Means | ||||||||

Effect | a | b | c | Estimate | Standard Error | DF | t Value | Pr > |t| |

a*b*c | 1 | 2 | 1 | 17.0454 | 0.4118 | 125 | 41.40 | <.0001 |

a*b*c | 2 | 1 | 2 | 21.8270 | 0.4118 | 125 | 53.01 | <.0001 |

Differences of Least Squares Means | |||||||||||

Effect | a | b | c | _a | _b | _c | Estimate | Standard Error | DF | t Value | Pr > |t| |

a*b*c | 1 | 2 | 1 | 2 | 1 | 2 | -4.7816 | 0.5823 | 125 | -8.21 | <.0001 |

## Example 3: A Two-Factor Logistic Model with Interaction Using Dummy and Effects Coding

Logistic models are in the class of generalized linear models. You can fit many kinds of logistic models in many procedures including LOGISTIC, GENMOD, GLIMMIX, PROBIT, CATMOD, and others. For these models, the response is no longer modeled directly. Instead, you model a function of the response distribution's mean. In logistic models, the response distribution is binomial and the log odds (or*logit*of the binomial mean,

*p*) is the response function that you model:

- logit(
*p*_{i}) ≡ log(Odds_{i}) ≡ log[*p*_{i}/ (1–*p*_{i})]

Consider the following medical example in which patients with one of two diagnoses (

*complicated*or

*uncomplicated*) are treated with one of three treatments (

*A*,

*B*, or

*C*) and the result (

*cured*or

*not*cured) is observed.

data uti; input diagnosis : $13. treatment $ response $ count @@; datalines; complicated A cured 78 complicated A not 28 complicated B cured 101 complicated B not 11 complicated C cured 68 complicated C not 46 uncomplicated A cured 40 uncomplicated A not 5 uncomplicated B cured 54 uncomplicated B not 5 uncomplicated C cured 34 uncomplicated C not 6 ;

### Dummy Coding

*Indicator*or

*dummy*coding of a predictor replaces the actual variable in the design matrix (or model matrix) with a set of variables that use values of 0 or 1 to indicate the level of the original variable. One variable is created for each level of the original variable. A main effect parameter is interpreted as the difference in the level's effect compared to the reference level. This is the default coding scheme for CLASS variables in most procedures including GLM, MIXED, GLIMMIX, and GENMOD. Some procedures allow multiple types of coding. In PROC LOGISTIC, use the PARAM=GLM option in the CLASS statement to request dummy coding of CLASS variables. A full-rank version of indicator coding (called

*reference*coding) that omits the indicator variable for the reference level (by default, the last level) is also available in PROC LOGISTIC, PROC GENMOD, PROC CATMOD, and some other procedures via the PARAM=REF option.

Using dummy coding, the right-hand side of the logistic model looks like it does when modeling a normally distributed response as in Example 1:

- log(Odds
_{ij}) = μ + α_{i}+ β_{j}+ αβ_{ij}(3a)

*i*=1,2,...,5,

*j*=1,2,

*k*=1, 2,...,

*N*. But an equivalent representation of the model is:

_{ij}- log(Odds
_{ij}) = μ + Σ_{i}α_{i}A_{i}+ Σ_{j}β_{j}B_{j}+ Σ_{ij}γ_{ij}A_{i}B_{j}(3b)

_{i}and B

_{j}are sets of design variables that are defined as follows using dummy coding:

- A
_{i}= 1 if A =*i*; otherwise A_{i}= 0 ,

where*i*= 1, 2, 3, 4, or 5 (although the levels might be coded differently). - B
_{j}= 1 if B =*j*; otherwise B_{j}= 0 ,

where*j*= 1 or 2 (although the levels might be coded differently).

- log(Odds
_{dt}) = μ +*d*_{1}O +*d*_{2}U +*t*_{1}A +*t*_{2}B +*t*_{3}C +*g*_{1}OA +*g*_{2}OB +*g*_{3}OC +*g*_{4}UA +*g*_{5}UB +*g*_{6}UC (3c)

- O is the dummy variable for the complicated diagnosis
- U is the dummy variable for the uncomplicated diagnosis
- A, B, and C are the dummy variables for the three treatments
- OA through UC are the products of the diagnosis and treatment dummy variables, jointly representing the diagnosis by treatment interaction

Note that the difference in log odds is equivalent to the log of the odds ratio:

- log(Odds
_{i}) – log(Odds_{j}) = log( Odds_{i}/ Odds_{j}) = log(OR_{ij})

For the medical example, suppose we are interested in the odds ratio for treatment A versus treatment C in the complicated diagnosis. We write the null hypothesis this way:

- H
_{0}: log(Odds_{OA}) = log(Odds_{OC})

- H
_{0}: log(Odds_{OA}) - log(Odds_{OC}) = 0

Table of treatment by response | |||

treatment | response | Total | |

cured | not | ||

A | 78 | 28 | 106 |

C | 68 | 46 | 114 |

Total | 146 | 74 | 220 |

78/28 78·46 ----- = ----- = 1.8845 68/46 68·28This means that, when the diagnosis is complicated, the odds of being cured by treatment A are 1.8845 times the odds of being cured by treatment C. The following statements display the table above and compute the odds ratio:

proc freq data=uti; where diagnosis = "complicated" and treatment in ("A","C"); table treatment * response / relrisk norow nocol nopercent; weight count; run;

Estimates of the Relative Risk (Row1/Row2) | |||

Type of Study | Value | 95% Confidence Limits | |

Case-Control (Odds Ratio) | 1.8845 | 1.0643 | 3.3367 |

Cohort (Col1 Risk) | 1.2336 | 1.0210 | 1.4906 |

Cohort (Col2 Risk) | 0.6546 | 0.4440 | 0.9652 |

The log odds for treatment A in the complicated diagnosis are:

- log(Odds
_{OA}) = μ +*d*_{1}+*t*_{1}+*g*_{1}

- log(Odds
_{OC}) = μ +*d*_{1}+*t*_{3}+*g*_{3}

- log(Odds
_{OA}) - log(Odds_{OC}) =*t*_{1}-*t*_{3}+*g*_{1}-*g*_{3}

proc logistic data=uti; freq count; class diagnosis treatment / param=glm; model response(event='cured') = diagnosis treatment diagnosis*treatment; contrast 'trt A vs C in comp' treatment 1 0 -1 diagnosis*treatment 1 0 -1 0 0 0 / estimate=both; output out=out xbeta=xbeta; run;

Contrast Rows Estimation and Testing Results | |||||||||

Contrast | Type | Row | Estimate | StandardError | Alpha | Lower Limit | Upper Limit | WaldChi-Square | Pr > ChiSq |

trt A vs C in comp | PARM | 1 | 0.6336 | 0.2915 | 0.05 | 0.0623 | 1.2050 | 4.7246 | 0.0297 |

trt A vs C in comp | EXP | 1 | 1.8845 | 0.5493 | 0.05 | 1.0643 | 3.3367 | 4.7246 | 0.0297 |

**x′β**, for each observation. This is the log odds. The following statements print the log odds for treatments A and C in the complicated diagnosis.

proc print data=out noobs; where diagnosis="complicated" and response="cured" and treatment in ("A","C"); var diagnosis treatment xbeta; run;

diagnosis | treatment | xbeta |

complicated | A | 1.02450 |

complicated | C | 0.39087 |

PROC GENMOD can also be used to estimate this odds ratio. Specify the DIST=BINOMIAL option to specify a logistic model

proc genmod data=uti; freq count; class diagnosis treatment; model response = diagnosis treatment diagnosis*treatment / dist=binomial; estimate 'trt A vs C in comp' treatment 1 0 -1 diagnosis*treatment 1 0 -1 0 0 0 / exp; run;As shown in Example 1, tests of simple effects within an interaction can be done using any of several statements other than the CONTRAST and ESTIMATE statements. In PROC LOGISTIC, odds ratio estimates for variables involved in interactions can be most easily obtained using the ODDSRATIO statement. The following ODDSRATIO statement provides the same estimate of the treatment A vs. treatment C odds ratio in the complicated diagnosis as above (along with odds ratio estimates for the other treatment pairs in that diagnosis).

oddsratio treatment / at(diagnosis='complicated');As in Example 1, you can also use the LSMEANS, LSMESTIMATE, and SLICE statements in PROC LOGISTIC, PROC GENMOD, and PROC GLIMMIX when dummy coding (PARAM=GLM) is used. The ILINK option in the LSMEANS statement provides estimates of the probabilities of cure for each combination of treatment and diagnosis. The DIFF option estimates and tests each pairwise difference of log odds. The EXP option exponentiates each difference providing odds ratio estimates for each pair. The LSMESTIMATE statement allows you to request specific comparisons. Since treatment A and treatment C are the first and third in the LSMEANS list, the contrast in the LSMESTIMATE statement estimates and tests their difference. The EXP option provides the odds ratio estimate by exponentiating the difference. Similarly, the SLICEBY, DIFF, and EXP options in the SLICE statement estimate and test differences and odds ratios in the complicated diagnosis.

lsmeans diagnosis*treatment / ilink exp diff; lsmestimate diagnosis*treatment 'A vs C complicated' 1 0 -1 / exp; slice diagnosis*treatment / sliceby(diagnosis='complicated') diff exp;An example of using the LSMEANS and LSMESTIMATE statements to estimate odds ratios in a repeated measures (GEE) model in PROC GENMOD is available.

Rather than the usual main effects and interaction model (3c), the same tasks can be accomplished using an equivalent nested model:

- log(Odds
_{dt}) = μ +*d*_{d}+*t*(*d*)_{dt}

- log(Odds
_{dt}) = μ +*d*_{1}O +*d*_{2}U +*g*_{1}OA +*g*_{2}OB +*g*_{3}OC +*g*_{4}UA +*g*_{5}UB +*g*_{6}UC (3d)

You write the contrast of log odds in terms of the nested model (3d):

- log(Odds
_{OA}) = μ +*d*_{1}+*g*_{1}

- log(Odds
_{OC}) = μ +*d*_{1}+*g*_{3}

- log(Odds
_{OA}) – log(Odds_{OC}) =*g*_{1}–*g*_{3}

proc logistic data=uti; freq count; class diagnosis treatment / param=glm; model response(event='cured') = diagnosis treatment(diagnosis) / expb; contrast 'trt A vs C in comp' treatment(diagnosis) 1 0 -1 0 0 0 / estimate=both; run;The contrast table that shows the log odds ratio and odds ratio estimates is exactly as before. Notice that the parameter estimate for treatment A within complicated diagnosis is the same as the estimated contrast and the exponentiated parameter estimate is the same as the exponentiated contrast.

Analysis of Maximum Likelihood Estimates | ||||||||

Parameter | | | DF | Estimate | StandardError | WaldChi-Square | Pr > ChiSq | Exp(Est) |

Intercept | | | 1 | 1.7346 | 0.4428 | 15.3451 | <.0001 | 5.667 |

diagnosis | complicated | | 1 | -1.3437 | 0.4822 | 7.7653 | 0.0053 | 0.261 |

diagnosis | uncomplicated | | 0 | 0 | . | . | . | . |

treatment(diagnosis) | A | complicated | 1 | 0.6336 | 0.2915 | 4.7246 | 0.0297 | 1.884 |

treatment(diagnosis) | B | complicated | 1 | 1.8262 | 0.3705 | 24.3005 | <.0001 | 6.210 |

treatment(diagnosis) | C | complicated | 0 | 0 | . | . | . | . |

treatment(diagnosis) | A | uncomplicated | 1 | 0.3448 | 0.6489 | 0.2824 | 0.5952 | 1.412 |

treatment(diagnosis) | B | uncomplicated | 1 | 0.6445 | 0.6438 | 1.0020 | 0.3168 | 1.905 |

treatment(diagnosis) | C | uncomplicated | 0 | 0 | . | . | . | . |

proc genmod data=uti; freq count; class diagnosis treatment; model response = diagnosis treatment(diagnosis) / dist=binomial; estimate 'trt A vs C in comp' treatment(diagnosis) 1 0 -1 0 0 0 / exp; run;

### Effects Coding

*Effects*or

*Deviation from mean*coding of a predictor replaces the actual variable in the design matrix (or model matrix) with a set of variables that use values of –1, 0, or 1 to indicate the level of the original variable. The number of variables that are created is one fewer than the number of levels of the original variable, yielding one fewer parameters than levels, but equal to the number of degrees of freedom. For this reason, it is known as a

*full-rank*parameterization. Reference parameterization (using the PARAM=REF option) is also a full-rank parameterization. A main effect parameter is interpreted as the deviation of the level's effect from the average effect of all the levels. This coding scheme is used by default by PROC CATMOD and PROC LOGISTIC and can be specified in these and some other procedures such as PROC GENMOD with the PARAM=EFFECT option in the CLASS statement.

Using effects coding, the model still looks like model 3b, but the design variables for diagnosis and treatment are defined differently as you can see in the following table. Notice that if you add up the rows for diagnosis (or treatments), the sum is zero. With effects coding, the parameters are constrained to sum to zero. Therefore, the estimate of the last level of an effect, A, is α

_{a}= –(α

_{1}+ α

_{2}+ ... + α

_{a–1}).

Class Level Information | |||

Class | Value | Design Variables | |

1 | 2 | ||

diagnosis | complicated | 1 | |

| uncomplicated | -1 | |

treatment | A | 1 | 0 |

| B | 0 | 1 |

| C | -1 | -1 |

Although the coding scheme is different, you still follow the same steps to determine the contrast coefficients. First, write the model, being sure to verify its parameters and their order from the procedure's displayed results:

- log(Odds
_{dt}) = μ +*d*O +*t*_{1}A +*t*_{2}B +*g*_{1}OA +*g*_{2}OB (3e)

- log(Odds
_{OA}) = μ +*d*+*t*_{1}+*g*_{1}

- log(Odds
_{OC}) = μ +*d*–*t*_{1}–*t*_{2}–*g*_{1}–*g*_{2}

- log(Odds
_{OA}) – log(Odds_{OC}) = 2*t*_{1}+*t*_{2}+ 2*g*_{1}+*g*_{2}

proc logistic data=uti; freq count; class diagnosis treatment; model response(event='cured') = diagnosis treatment diagnosis*treatment; contrast 'trt A vs C in comp' treatment 2 1 diagnosis*treatment 2 1 / estimate=both; run;The same log odds ratio and odds ratio estimates are obtained as from the dummy-coded model. The change in coding scheme does not affect how you specify the ODDSRATIO statement. The ODDSRATIO statement used above with dummy coding provides the same results with effects coding.

These are the equivalent PROC GENMOD statements:

proc genmod data=uti; freq count; class diagnosis treatment / param=effect; model response = diagnosis treatment diagnosis*treatment / dist=binomial; estimate 'trt A vs C in comp' treatment 2 1 diagnosis*treatment 2 1 / exp; run;Suppose you want to test whether the effect of treatment A in the complicated diagnosis is different from the average effect of the treatments in the complicated diagnosis. The null hypothesis, in terms of model 3e, is:

- H
_{0}: log(Odds_{OA}) = Σ_{j}log(Odds_{Oj})/3

- H
_{0}: log(Odds_{OA}) – Σ_{j}log(Odds_{Oj})/3 = 0

_{OA}) = μ +

*d*+

*t*

_{1}+

*g*

_{1}. You use model 3e to expand the average treatment effect:

- Σ
_{j}log(Odds_{Oj})/3 = [(μ +*d*+*t*_{1}+*g*_{1}) + (μ +*d*+*t*_{2}+*g*_{2}) + (μ +*d*–*t*_{1}–*t*_{2}–*g*_{1}–*g*_{2})]/3 = (3μ + 3*d*)/3 = μ +*d*

_{OA}):

- (μ +
*d*+*t*_{1}+*g*_{1}) – (μ +*d*) =*t*_{1}+*g*_{1}

- H
_{0}:*t*_{1}+*g*_{1}= 0

contrast 'trt A vs avg trt in comp' treatment 1 0 diagnosis*treatment 1 0 / estimate=both;In PROC GENMOD, use this equivalent ESTIMATE statement:

estimate 'trt A vs avg trt in comp' treatment 1 0 diagnosis*treatment 1 0 / exp;

Contrast Rows Estimation and Testing Results | |||||||||

Contrast | Type | Row | Estimate | StandardError | Alpha | Lower Limit | Upper Limit | WaldChi-Square | Pr > ChiSq |

trt 1 vs avg trt in comp | PARM | 1 | -0.1863 | 0.1919 | 0.05 | -0.5624 | 0.1898 | 0.9428 | 0.3316 |

trt 1 vs avg trt in comp | EXP | 1 | 0.8300 | 0.1593 | 0.05 | 0.5698 | 1.2090 | 0.9428 | 0.3316 |

Because PROC CATMOD also uses effects coding, you can use the following CONTRAST statement in that procedure to get the same results as above. The WEIGHT statement in PROC CATMOD enables you to input data summarized in cell count form.

proc catmod data=uti; weight count; model response = diagnosis treatment diagnosis*treatment; contrast 'trt 1 vs avg trt in comp' treatment 1 0 diagnosis*treatment 1 0 / estimate=both; run;PROC CATMOD has a feature that makes testing this kind of hypothesis even easier. You can specify

*nested-by-value*effects in the MODEL statement to test the effect of one variable within a particular level of another variable. This is an extension of the nested effects that you can specify in other procedures such as GLM and LOGISTIC. For more information, see the "Generation of the Design Matrix" section in the CATMOD documentation.

In the medical example, you can use nested-by-value effects to decompose treatment*diagnosis interaction as follows:

proc catmod data=uti; weight count; model response = diagnosis treatment(diagnosis='complicated') treatment(diagnosis='uncomplicated'); run;The model effects, treatment(diagnosis='complicated') and treatment(diagnosis='uncomplicated'), are nested-by-value effects that test the effects of treatments within each of the diagnoses. In the following output, the first parameter of the treatment(diagnosis='complicated') effect tests the effect of treatment A versus the average treatment effect in the complicated diagnosis. This is exactly the contrast that was constructed earlier.

Analysis of Maximum Likelihood Estimates | |||||

Parameter | | Estimate | StandardError | Chi-Square | Pr > ChiSq |

Intercept | | 1.6377 | 0.1514 | 116.98 | <.0001 |

diagnosis | complicated | -0.4268 | 0.1514 | 7.95 | 0.0048 |

treat(diagn=complicated) | A | -0.1864 | 0.1919 | 0.94 | 0.3315 |

| B | 1.0064 | 0.2329 | 18.67 | <.0001 |

trea(diag=uncomplicated) | A | 0.0149 | 0.3822 | 0.00 | 0.9689 |

| B | 0.3150 | 0.3793 | 0.69 | 0.4063 |

## Example 4: Comparing Models

The CONTRAST statement can also be used to compare competing*nested*models. Models are nested if one model results from restrictions on the parameters of the other model. The most commonly used test for comparing nested models is the likelihood ratio test, but other tests (such as Wald and score tests) can also be used. The next section illustrates using the CONTRAST statement to compare nested models. Other methods must be used to compare nonnested models and this is discussed in the section that follows.

### Comparing Nested Models

In an example from Ries and Smith (1963), the choice of detergent brand (Brand= M or X) is related to three other categorical variables: the softness of the laundry water (Softness= soft, medium, or hard); the temperature of the water (Temperature= high or low); and whether the subject was a previous user of Brand M (Previous= yes or no). Two logistic models are fit in this example: The first model is*saturated*, meaning that it contains all possible main effects and interactions using all available degrees of freedom. The second model is a reduced model that contains only the main effects.

The following statements create the data set and fit the saturated logistic model.

data detergent; input Softness $ Brand $ Previous $ Temperature $ Count @@; datalines; soft X yes high 19 soft X yes low 57 soft X no high 29 soft X no low 63 soft M yes high 29 soft M yes low 49 soft M no high 27 soft M no low 53 med X yes high 23 med X yes low 47 med X no high 33 med X no low 66 med M yes high 47 med M yes low 55 med M no high 23 med M no low 50 hard X yes high 24 hard X yes low 37 hard X no high 42 hard X no low 68 hard M yes high 43 hard M yes low 52 hard M no high 30 hard M no low 42 ; ods select modelfit type3; ods output modelfit=full; proc genmod data=detergent; class Softness Previous Temperature; freq Count; model Brand = Softness|Previous|Temperature / dist=binomial type3; run;The partial results shown below suggest that interactions are not needed in the model:

Criteria For Assessing Goodness Of Fit | |||
---|---|---|---|

Criterion | DF | Value | Value/DF |

Deviance | 996 | 1364.4956 | 1.3700 |

Scaled Deviance | 996 | 1364.4956 | 1.3700 |

Pearson Chi-Square | 996 | 1008.0000 | 1.0120 |

Scaled Pearson X2 | 996 | 1008.0000 | 1.0120 |

Log Likelihood | -682.2478 |

LR Statistics For Type 3 Analysis | |||
---|---|---|---|

Source | DF | Chi-Square | Pr > ChiSq |

Softness | 2 | 0.10 | 0.9522 |

Previous | 1 | 22.13 | <.0001 |

Softness*Previous | 2 | 3.79 | 0.1506 |

Temperature | 1 | 3.64 | 0.0564 |

Softness*Temperature | 2 | 0.20 | 0.9066 |

Previous*Temperature | 1 | 2.26 | 0.1327 |

Softne*Previo*Temper | 2 | 0.74 | 0.6917 |

ods select modelfit type3; ods output modelfit=reduced; proc genmod data=detergent; class Softness Previous Temperature; freq Count; model Brand = Softness Previous Temperature / dist=binomial type3; run;This partial output summarizes the main-effects model:

Criteria For Assessing Goodness Of Fit | |||
---|---|---|---|

Criterion | DF | Value | Value/DF |

Deviance | 1003 | 1372.7236 | 1.3686 |

Scaled Deviance | 1003 | 1372.7236 | 1.3686 |

Pearson Chi-Square | 1003 | 1007.9360 | 1.0049 |

Scaled Pearson X2 | 1003 | 1007.9360 | 1.0049 |

Log Likelihood | -686.3618 |

LR Statistics For Type 3 Analysis | |||
---|---|---|---|

Source | DF | Chi-Square | Pr > ChiSq |

Softness | 2 | 0.22 | 0.8976 |

Previous | 1 | 19.89 | <.0001 |

Temperature | 1 | 3.74 | 0.0532 |

Any estimable linear combination of model parameters can be tested using the procedure's CONTRAST statement. The difficulty is constructing combinations that are estimable and that jointly test the set of interactions. This can be particularly difficult with dummy (PARAM=GLM) coding. The problem is greatly simplified using effects coding, which is available in some procedures via the PARAM=EFFECT option in the CLASS statement. The CONTRAST statement tests the hypothesis

**Lβ**=

**0**, where

**L**is the hypothesis matrix and

**β**is the vector of model parameters. With effects coding, each row of

**L**can be written to select just one interaction parameter when multiplied by

**β**. In the CONTRAST statement, the rows of

**L**are separated by commas. The CONTRAST statement below defines seven rows in

**L**for the seven interaction parameters resulting in a 7 DF test that all interaction parameters are zero.

ods select contrasts; proc genmod data=detergent; class Softness Previous Temperature / param=effect; freq Count; model Brand = Softness|Previous|Temperature / dist=binomial; contrast 'lrt' softness*previous 1 0, softness*previous 0 1, softness*temperature 1 0, softness*temperature 0 1, previous*temperature 1, softness*previous*temperature 1 0, softness*previous*temperature 0 1; run;By default, PROC GENMOD computes a likelihood ratio test for the specified contrast. As expected, the results show that there is no significant interaction (

*p*=0.3129) or that the reduced model fits as well as the saturated model.

Contrast Results | ||||
---|---|---|---|---|

Contrast | DF | Chi-Square | Pr > ChiSq | Type |

lrt | 7 | 8.23 | 0.3129 | LR |

ods select contrasttest; proc logistic data=detergent; class Softness Previous Temperature / param=effect; freq Count; model Brand = Softness|Previous|Temperature; contrast 'lrt' softness*previous 1 0, softness*previous 0 1, softness*temperature 1 0, softness*temperature 0 1, previous*temperature 1, softness*previous*temperature 1 0, softness*previous*temperature 0 1; run;

Contrast Test Results | |||
---|---|---|---|

Contrast | DF | Wald Chi-Square | Pr > ChiSq |

lrt | 7 | 8.1794 | 0.3170 |

**Limitations on constructing valid LR tests**

The likelihood ratio test can be used to compare any two nested models that are fit by maximum likelihood. It is not necessary that the larger model be saturated. So, this test can be used with models that are fit by many procedures such as GENMOD, LOGISTIC, MIXED, GLIMMIX, PHREG, PROBIT, and others, but there are cases with some of these procedures in which a LR test cannot be constructed:

- With any procedure, models that are not nested cannot be compared using the LR test.
- Models fit with the GENMOD or GEE procedure using the REPEATED statement are estimated using the generalized estimating equations (GEE) method and not by maximum likelihood so a LR test cannot be constructed. However, the CONTRAST statement can be used in PROC GENMOD as shown above to produce a score test of the hypothesis.
- With mixed models fit in PROC MIXED, if the models are nested in the covariance parameters and have identical fixed effects, then a LR test can be constructed using results from REML estimation (the default) or from ML estimation. Basing the test on the REML results is generally preferred. However, if the nested models do not have identical fixed effects, then results from ML estimation must be used to construct a LR test. See this note on comparing covariance structures in PROC MIXED.
- In most cases, models fit in PROC GLIMMIX using the RANDOM statement do not use a true log likelihood. When the procedure reports a log pseudo-likelihood you cannot construct a LR test to compare models. In some cases, the Laplace or quadrature estimation methods (METHOD=LAPLACE or METHOD=QUAD, first available in SAS 9.2) can be used which compute and report an approximate log likelihood making construction of a LR test possible. See this note on comparing covariance structures in PROC GLIMMIX.

### Comparing Nonnested Models

Nonnested models can still be compared using information criteria such as AIC, AICC, and BIC (also called SC). These statistics are provided in most procedures using maximum likelihood estimation. Models with smaller values of these criteria are considered better models. However, no statistical tests comparing criterion values is possible.Tests to compare nonnested models are available, but not by using CONTRAST statements as discussed above. See this sample program for discussion and examples of using the Vuong and Clarke tests to compare nonnested models. For a more detailed definition of nested and nonnested models, see the Clarke (2001) reference cited in the sample program.

## Example 5: A Quadratic Logistic Model

This example shows the use of the CONTRAST and ODDSRATIO statements to compare the response at two levels of a continuous predictor when the model contains a higher-order effect. Specifically, PROC LOGISTIC is used to fit a logistic model containing effects X and X^{2}. The correct coefficients are determined for the CONTRAST statement to estimate two odds ratios: one for an increase of one unit in X, and the second for a two unit increase. It is shown how this can be done more easily using the ODDSRATIO and UNITS statements in PROC LOGISTIC.

## No comments:

## Post a Comment