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This FAQ is for Stata 11. In Stata 12, you can use contrast and pwcompare to compare the levels of categorical variables.

How can I form various tests comparing the different levels of a categorical variable after anova or regress?

Title   Tests comparing levels of a categorical variable after anova or regress
Author Kenneth Higbee, StataCorp
Wesley Eddings, StataCorp
Date October 1999; updated August 2009


Introduction

Often researchers want to test for differences between levels of a factor (categorical variable) or factors after running an anova or regress command. For instance, with one factor the questions might be

There are often other kinds of tests between levels of a factor that are also of interest. For instance, the questions might be of the form:

Many other interesting questions like these can be asked after an estimation command involving a categorical variable (factor).

The test command is one tool to use in answering these questions. There are several variations of the syntax for test depending on if you wish to test coefficients, expressions, terms (after anova), or to test several coefficients at the same time. One form of the test command that we will use is

test [exp = exp] [, accumulate notest ]

Details can be found in the Base Reference Manual ([R] test and in the case of anova the sections in [R] anova postestimation that discuss testing).


One categorical factor

How you parameterize your model makes a difference in what you do to perform your tests. We will examine a couple of ways of parameterizing a simple one-way ANOVA model. We will use the following three approaches:

With each of these approaches, we will show how to use the test command to obtain tests of interest. In particular, we will

  1. test level one against level two;
  2. test the average of levels one and two against level three; and
  3. test that five times level one plus four times level two minus three times level three equals two times level four (a strange linear combination just to show that it can be done).

Here is a 20-observation dataset with two variables—the outcome y and a categorical variable x with four levels.

. list, noobs sepby(x)

x y
1 7
1 5
1 3
1 4
1 3
2 5
2 3
2 5
2 3
2 1
3 6
3 8
3 6
3 4
3 4
4 5
4 8
4 6
4 8
4 5
. table x, c(mean y)
x mean(y)
1 4.4
2 3.4
3 5.6
4 6.4

Regression excluding the intercept

Here is the regression excluding the intercept. We specify ibn.x so there will be no base category: all four levels of x will be included in the model. (See [U] 11.4.3 Factor variables.)

. regress y ibn.x, noconstant

Source SS df MS Number of obs = 20
F( 4, 16) = 48.24
Model 516.2 4 129.05 Prob > F = 0.0000
Residual 42.8 16 2.675 R-squared = 0.9234
Adj R-squared = 0.9043
Total 559 20 27.95 Root MSE = 1.6355
y Coef. Std. Err. t P>|t| [95% Conf. Interval]
x
1 4.4 .7314369 6.02 0.000 2.849423 5.950577
2 3.4 .7314369 4.65 0.000 1.849423 4.950577
3 5.6 .7314369 7.66 0.000 4.049423 7.150577
4 6.4 .7314369 8.75 0.000 4.849423 7.950577

The coefficients agree with the means reported by table. These coefficients are easily interpreted and easily tested. Here are the three tests after this regression:

  1. Test level one against level two
    . test i1.x == i2.x
    
     ( 1)  1bn.x - 2.x = 0
    
           F(  1,    16) =    0.93
                Prob > F =    0.3481
    
  2. Test the average of levels one and two against level three
    . test 0.5*i1.x + 0.5*i2.x == i3.x
    
     ( 1)  .5*1bn.x + .5*2.x - 3.x = 0
    
           F(  1,    16) =    3.60
                Prob > F =    0.0759
    
  3. Test that five times level one plus four times level two minus three times level three equals two times level four
    . test 5*i1.x + 4*i2.x - 3*i3.x == 2*i4.x
    
     ( 1)  5*1bn.x + 4*2.x - 3*3.x - 2*4.x = 0
    
           F(  1,    16) =    1.25
                Prob > F =    0.2808
    

Any of several other strange and/or wonderful tests could be performed. If you wish to test a nonlinear expression, you will want to look at testnl (see [R] testnl). If you want to jointly test two or more of these single degree-of-freedom tests, you can use the accumulate option of test. Another useful command is lincom (see [R] lincom). It can also be used to test many of the same hypotheses as the test command and has the benefit of providing not only the test result but also the estimate of the linear combination and the standard error of the estimate along with confidence intervals. Just as an example, here is the third test done using lincom.

. lincom 5*i1.x + 4*i2.x - 3*i3.x - 2*i4.x

 ( 1)  5*1bn.x + 4*2.x - 3*3.x - 2*4.x = 0

y Coef. Std. Err. t P>|t| [95% Conf. Interval]
(1) 6 5.374942 1.12 0.281 -5.394368 17.39437

Regression with the intercept

To fit a model with a constant, we need to remove the noconstant option from our regress command. We also need to specify a base category, because we cannot include all four levels of x and an intercept, too. It would be possible to specify any level of x as the base category, but we will just change ibn.x to i.x and let Stata select the first level of x as the default base.

. regress y i.x

Source SS df MS Number of obs = 20
F( 3, 16) = 3.26
Model 26.15 3 8.71666667 Prob > F = 0.0492
Residual 42.8 16 2.675 R-squared = 0.3793
Adj R-squared = 0.2629
Total 68.95 19 3.62894737 Root MSE = 1.6355
y Coef. Std. Err. t P>|t| [95% Conf. Interval]
x
2 -1 1.034408 -0.97 0.348 -3.192847 1.192847
3 1.2 1.034408 1.16 0.263 -.9928471 3.392847
4 2 1.034408 1.93 0.071 -.1928471 4.192847
_cons 4.4 .7314369 6.02 0.000 2.849423 5.950577

If you compare the coefficients from this regression with the output from the regression without a constant (and to the table showing the mean for each category), you see that the mean of the base category (one) corresponds to the coefficient for the constant, and if you add the coefficient for the constant to the other category coefficients, you get the mean for those categories. The coefficients for the included levels are relative to the base level.

This result provides a clue on how to obtain our three example tests of interest when the constant is in the model. It is a matter of doing some simple algebra to take a test from the first regression model and produce an equivalent test in this regression model. For reference, here is the correspondence between the two regression models:

No constant With constant
i1.x _cons
i2.x _cons + i2.x
i3.x _cons + i3.x
i4.x _cons + i4.x

After our first regression (without a constant), the first test was test i1.x == i2.x. After this latest regression (with a constant), the same test is test _cons = _cons + i2.x. This test can be simplified to test i2.x. The equivalent second and third tests can be similarly determined. Here are the three tests after regress with the constant included:

  1. Test level one against level two
    . test i2.x
    
     ( 1)  2.x = 0
    
           F(  1,    16) =    0.93
                Prob > F =    0.3481
    
  2. Test the average of levels one and two against level three
    . test 0.5*i2.x == i3.x
    
     ( 1)  .5*2.x - 3.x = 0
    
           F(  1,    16) =    3.60
                Prob > F =    0.0759
    
  3. Test that five times level one plus four times level two minus three times level three equals two times level four
    . test 4*_cons + 4*i2.x - 3*i3.x == 2*i4.x
    
     ( 1)  4*2.x - 3*3.x - 2*4.x + 4*_cons = 0
    
           F(  1,    16) =    1.25
                Prob > F =    0.280811
    

When you have the constant in the model (and in more complicated designs), it is important to understand how to interpret the coefficients from the regression model so that you can form the correct tests. The formation of the test above is not very intuitive but becomes clearer after doing some algebra starting with your understanding of the meaning of each coefficient and how they relate to the quantities you wish to test.

ANOVA

The anova command is a natural choice for analyzing this same dataset.

. anova y x

                           Number of obs =      20     R-squared     =  0.3793
                           Root MSE      = 1.63554     Adj R-squared =  0.2629

Source Partial SS df MS F Prob > F
Model 26.15 3 8.71666667 3.26 0.0492
x 26.15 3 8.71666667 3.26 0.0492
Residual 42.8 16 2.675
Total 68.95 19 3.62894737

We will examine obtaining individual degrees-of-freedom tests shortly. First, we will take a look at the underlying regression for this anova.

. regress

Source SS df MS Number of obs = 20
F( 3, 16) = 3.26
Model 26.15 3 8.71666667 Prob > F = 0.0492
Residual 42.8 16 2.675 R-squared = 0.3793
Adj R-squared = 0.2629
Total 68.95 19 3.62894737 Root MSE = 1.6355
y Coef. Std. Err. t P>|t| [95% Conf. Interval]
x
2 -1 1.034408 -0.97 0.348 -3.192847 1.192847
3 1.2 1.034408 1.16 0.263 -.9928471 3.392847
4 2 1.034408 1.93 0.071 -.1928471 4.192847
_cons 4.4 .7314369 6.02 0.000 2.849423 5.950577

anova used the first level of x as the base category, so the output matches the output of our command regress y i.x. All the same test commands will work after anova just as they worked after regress. The lincom command can also be used after anova.

The test() option of test after anova provides a convenient shorthand for specifying these kinds of tests using a matrix to denote the equation. test, showorder lists the order of the terms, and hence, the order of columns for the matrix.

. test, showorder

 Order of columns in the design matrix
      1: (x==1)
      2: (x==2)
      3: (x==3)
      4: (x==4)
      5: _cons

Here are the same three tests using the test() option.

  1. Test level one against level two
    . mat c1 = (0,1,0,0,0)
    
    . test, test(c1)
    
     ( 1)  2.x = 0
    
           F(  1,    16) =    0.93
                Prob > F =    0.3481
    
  2. Test the average of levels one and two against level three
    . mat c2 = (0,0.5,-1,0,0)
    
    . test, test(c2)
    
     ( 1)  .5*2.x - 3.x = 0
    
           F(  1,    16) =    3.60
                Prob > F =    0.0759
    
  3. Test that five times level one plus four times level two minus three times level three equals two times level four
    . mat c3 = (0,4,-3,-2,4)
    
    . test, test(c3)
    
     ( 1)  4*2.x - 3*3.x - 2*4.x + 4*_cons = 0
    
           F(  1,    16) =    1.25
                Prob > F =    0.280811
    

Two categorical factors

When there are two (or more) categorical factors in our model, we again may want to test various single degrees-of-freedom hypotheses that compare various levels of the two (or more) factors in the model. As with the one-way ANOVA model, how you parameterize your two-way (or higher) model affects how you go about performing individual tests.

To demonstrate how to obtain single degrees-of-freedom tests after a two-way ANOVA, we will use the following 24-observation dataset where the variables a and b are categorical variables with 4 and 3 levels, respectively, and there is a response variable, y.

. list, noobs sepby(a)

a b y
1 1 26
1 1 30
1 2 54
1 2 50
1 3 34
1 3 46
2 1 16
2 1 20
2 2 36
2 2 24
2 3 50
2 3 34
3 1 48
3 1 28
3 2 28
3 2 28
3 3 50
3 3 46
4 1 50
4 1 46
4 2 48
4 2 44
4 3 48
4 3 28

The following table shows the mean of y for each cell of a by b as well as the means for each level of a and b (the column and row titled “Total”):

. table a b, c(mean y) row col

b
a 1 2 3 Total
1 28 52 40 40
2 18 30 42 30
3 38 28 48 38
4 48 46 38 44
Total 33 39 42 38

This dataset is balanced (two observations per cell). If you are dealing with unbalanced data (including the case where you have missing cells) you will want to also read the technical notes in the section titled Two-way ANOVA in the [R] anova manual entry.

The standard way of performing an ANOVA on this dataset is with

. anova y a b a#b

                           Number of obs =      24     R-squared     =  0.7606
                           Root MSE      = 7.74597     Adj R-squared =  0.5412

Source Partial SS df MS F Prob > F
Model 2288 11 208 3.47 0.0214
a 624 3 208 3.47 0.0509
b 336 2 168 2.80 0.1005
a#b 1328 6 221.333333 3.69 0.0259
Residual 720 12 60
Total 3008 23 130.782609

This is the overparameterized two-way ANOVA model (which we will discuss in more detail later). When it comes to single degrees-of-freedom tests, some people prefer to use a different parameterization—the cell means model.

Cell means ANOVA model

In the cell means ANOVA model, we first create one categorical variable that corresponds to the cells in the two-way table (or higher-order table if more than two categorical variables are involved). The egen group() function is useful for creating the single categorical variable.

. egen c = group(a b)

. table a b, c(mean c)

b
a 1 2 3
1 1 2 3
2 4 5 6
3 7 8 9
4 10 11 12

The table above reminds us how the c variable relates to the original a and b variables. For instance, when c is 8, it means that a is 3 and b is 2. (If a and b had been reversed in the egen group() option, then the table above would show a different relationship.)

The cell means ANOVA model is then obtained by using the noconstant option of anova and the newly created c variable in place of a and b.

(You could also fit the model by typing regress y ibn.c, noconstant.)

. anova y ibn.c, noconstant

                           Number of obs =      24     R-squared     =  0.9809
                           Root MSE      = 7.74597     Adj R-squared =  0.9618

Source Partial SS df MS F Prob > F
Model 36944 12 3078.66667 51.31 0.0000
c 36944 12 3078.66667 51.31 0.0000
Residual 720 12 60
Total 37664 24 1569.33333
. regress
Source SS df MS Number of obs = 24
F( 12, 12) = 51.31
Model 36944 12 3078.66667 Prob > F = 0.0000
Residual 720 12 60 R-squared = 0.9809
Adj R-squared = 0.9618
Total 37664 24 1569.33333 Root MSE = 7.746
y Coef. Std. Err. t P>|t| [95% Conf. Interval]
c
1 28 5.477226 5.11 0.000 16.06615 39.93385
2 52 5.477226 9.49 0.000 40.06615 63.93385
3 40 5.477226 7.30 0.000 28.06615 51.93385
4 18 5.477226 3.29 0.007 6.066151 29.93385
5 30 5.477226 5.48 0.000 18.06615 41.93385
6 42 5.477226 7.67 0.000 30.06615 53.93385
7 38 5.477226 6.94 0.000 26.06615 49.93385
8 28 5.477226 5.11 0.000 16.06615 39.93385
9 48 5.477226 8.76 0.000 36.06615 59.93385
10 48 5.477226 8.76 0.000 36.06615 59.93385
11 46 5.477226 8.40 0.000 34.06615 57.93385
12 38 5.477226 6.94 0.000 26.06615 49.93385

Compare the 12 coefficients for c in the table above with the table of means presented earlier. The coefficients from the cell means ANOVA model are the cell means from the two-way table. This correspondence makes creating meaningful test statements easy.

You can recreate an F test from the overparameterized ANOVA model by using appropriate combinations of single degree-of-freedom tests after the cell means ANOVA model. For example, the test for the term a with 3 degrees of freedom can be obtained by accumulating 3 single degree-of-freedom tests. Below I combine the tests of level 1 versus 2, level 1 versus 3, and level 1 versus 4 of the a variable. (Remember that c 1, 2, and 3 correspond to level 1 of a; c 4, 5, and 6 corresponds to level 2 of a; and so on.)

. mat amat = (1,1,1,-1,-1,-1,0,0,0,0,0,0 \ 
>             1,1,1,0,0,0,-1,-1,-1,0,0,0 \
>             1,1,1,0,0,0,0,0,0,-1,-1,-1)

. test, test(amat)

 ( 1)  1bn.c + 2.c + 3.c - 4.c - 5.c - 6.c = 0
 ( 2)  1bn.c + 2.c + 3.c - 7.c - 8.c - 9.c = 0
 ( 3)  1bn.c + 2.c + 3.c - 10.c - 11.c - 12.c = 0

       F(  3,    12) =    3.47
            Prob > F =    0.0509

This F of 3.47 agrees with the F test for the term a in the overparameterized ANOVA presented earlier. Of course, it is easier to obtain the test of a term like a by running the overparameterized model. I just wanted to show that you could also obtain the result with a little work starting from the cell means model.

For the sake of completeness, here is the test command to produce the F test for the b term.


. mat bmat = (1,-1,0,1,-1,0,1,-1,0,1,-1,0 \ 
>             1,0,-1,1,0,-1,1,0,-1,1,0,-1)

. test, test(bmat)

 ( 1)  1bn.c - 2.c + 4.c - 5.c + 7.c - 8.c + 10.c - 11.c = 0
 ( 2)  1bn.c - 3.c + 4.c - 6.c + 7.c - 9.c + 10.c - 12.c = 0

       F(  2,    12) =    2.80
            Prob > F =    0.1005

Here is the test command to produce the F test for the a by b interaction term.

. forvalues i = 1/3 {
  2.    forvalues j = 1/2 {
  3.       mat abmat = nullmat(abmat) \ vecdiag(amat[`i',1...]'*bmat[`j',1...])
  4.    }
  5. }

. mat list abmat

abmat[6,12]
     c1   c2   c3   c4   c5   c6   c7   c8   c9  c10  c11  c12
r1    1   -1    0   -1    1    0    0    0    0    0    0    0
r1    1    0   -1   -1    0    1    0    0    0    0    0    0
r1    1   -1    0    0    0    0   -1    1    0    0    0    0
r1    1    0   -1    0    0    0   -1    0    1    0    0    0
r1    1   -1    0    0    0    0    0    0    0   -1    1    0
r1    1    0   -1    0    0    0    0    0    0   -1    0    1

. test, test(abmat)

 ( 1)  1bn.c - 2.c - 4.c + 5.c = 0
 ( 2)  1bn.c - 3.c - 4.c + 6.c = 0
 ( 3)  1bn.c - 2.c - 7.c + 8.c = 0
 ( 4)  1bn.c - 3.c - 7.c + 9.c = 0
 ( 5)  1bn.c - 2.c - 10.c + 11.c = 0
 ( 6)  1bn.c - 3.c - 10.c + 12.c = 0

       F(  6,    12) =    3.69
            Prob > F =    0.0259

There are actually many different ways I could have obtained the overall F tests for the a, b, and a by b terms.

Now that we have demonstrated that you can reproduce the results from the overparameterized model with an appropriate series of test statements after a cell means model, let us now look at a few different single degrees-of-freedom tests. (We will later see how to obtain these same single degrees-of-freedom tests after the overparameterized ANOVA.) You will want to look back at the table showing how c relates to a and b to see how these tests were constructed.

  1. Test level two against level four of factor a
    . test i4.c + i5.c + i6.c = i10.c + i11.c + i12.c
    
     ( 1)  4.c + 5.c + 6.c - 10.c - 11.c - 12.c = 0
    
           F(  1,    12) =    9.80
                Prob > F =    0.0087
    
  2. Test the average of levels one and two against level three of factor b
    . test (i1.c + i4.c + i7.c + i10.c + i2.c + i5.c + i8.c + i11.c)/2 
    >    = i3.c + i6.c + i9.c + i12.c
    
     ( 1)  .5*1bn.c + .5*2.c - 3.c + .5*4.c + .5*5.c - 6.c + .5*7.c + .5*8.c - 9.c +
           .5*10.c + .5*11.c - 12.c = 0
    
           F(  1,    12) =    3.20
                Prob > F =    0.0989
    
  3. Test the average of levels one and two of a, when b is also at level one or two, against the average of levels three and four of a, when b is at level three
    . test (i1.c + i4.c + i2.c + i5.c)/2 = i9.c + i12.c
    
     ( 1)  .5*1bn.c + .5*2.c + .5*4.c + .5*5.c - 9.c - 12.c = 0
    
           F(  1,    12) =    5.38
                Prob > F =    0.0388
    
  4. Test that three times a at one and b at one, minus four times a at three and b at two, plus six times a at four and b at three, equals a at two and b at two, minus two times a at two and b at three
    . test 3*i1.c - 4*i8.c + 6*i12.c = i5.c - 2*i6.c
    
     ( 1)  3*1bn.c - 5.c + 2*6.c - 4*8.c + 6*12.c = 0
    
           F(  1,    12) =   32.58
                Prob > F =    0.0001
    

Those same four tests using the test() option are

  1. . mat m1 = (0,0,0,1,1,1,0,0,0,-1,-1,-1)
    
    . test, test(m1)
    
     ( 1)  4.c + 5.c + 6.c - 10.c - 11.c - 12.c = 0
    
           F(  1,    12) =    9.80
                Prob > F =    0.0087
    
  2. . mat m2 = (.5,.5,-1,.5,.5,-1,.5,.5,-1,.5,.5,-1)
    
    . test, test(m2)
    
     ( 1)  .5*1bn.c + .5*2.c - 3.c + .5*4.c + .5*5.c - 6.c + .5*7.c + .5*8.c - 9.c +
           .5*10.c + .5*11.c - 12.c = 0
    
           F(  1,    12) =    3.20
                Prob > F =    0.0989
    
  3. . mat m3 = (.5,.5,0,.5,.5,0,0,0,-1,0,0,-1)
    
    . test, test(m3)
    
     ( 1)  .5*1bn.c + .5*2.c + .5*4.c + .5*5.c - 9.c - 12.c = 0
    
           F(  1,    12) =    5.38
                Prob > F =    0.0388
    
  4. . mat m4 = (3,0,0,0,-1,2,0,-4,0,0,0,6)
    
    . test, test(m4)
    
     ( 1)  3*1bn.c - 5.c + 2*6.c - 4*8.c + 6*12.c = 0
    
           F(  1,    12) =   32.58
                Prob > F =    0.0001
    

Constructing various single degrees-of-freedom tests after a cell means ANOVA model is relatively easy. You pick the appropriate linear combination of the coefficients based on how the single categorical variable (c in our example) relates to the original categorical variables (a and b in our example) and based on the hypothesis of interest.

Overparameterized ANOVA model

Most people are used to the results presented by the overparameterized ANOVA model. As we saw when we discussed the cell means ANOVA model, the F tests for terms in the ANOVA model are obtained directly from the overparameterized model ANOVA table. Compare this with computing an F test for a term after the cell means ANOVA model. However, when it comes to obtaining single degrees-of-freedom tests, most people find the cell means model approach to be the easiest.

Here again is the overparameterized ANOVA model for our example data. Also, I use the regress command to replay the ANOVA as a regression table.

. anova y a b a#b

                           Number of obs =      24     R-squared     =  0.7606
                           Root MSE      = 7.74597     Adj R-squared =  0.5412

Source Partial SS df MS F Prob > F
Model 2288 11 208 3.47 0.0214
a 624 3 208 3.47 0.0509
b 336 2 168 2.80 0.1005
a#b 1328 6 221.333333 3.69 0.0259
Residual 720 12 60
Total 3008 23 130.782609
. regress, noheader
y Coef. Std. Err. t P>|t| [95% Conf. Interval]
a
2 -10 7.745967 -1.29 0.221 -26.87701 6.877012
3 10 7.745967 1.29 0.221 -6.877012 26.87701
4 20 7.745967 2.58 0.024 3.122988 36.87701
b
2 24 7.745967 3.10 0.009 7.122988 40.87701
3 12 7.745967 1.55 0.147 -4.877012 28.87701
a#b
2 2 -12 10.95445 -1.10 0.295 -35.8677 11.8677
2 3 12 10.95445 1.10 0.295 -11.8677 35.8677
3 2 -34 10.95445 -3.10 0.009 -57.8677 -10.1323
3 3 -2 10.95445 -0.18 0.858 -25.8677 21.8677
4 2 -26 10.95445 -2.37 0.035 -49.8677 -2.132301
4 3 -22 10.95445 -2.01 0.068 -45.8677 1.867699
_cons 28 5.477226 5.11 0.000 16.06615 39.93385

Now the important question is how the coefficients in this model relate to the cell means. To refresh your memory, here is the table of cell means (and marginal means).

. table a b, c(mean y) row col

b
a 1 2 3 Total
1 28 52 40 40
2 18 30 42 30
3 38 28 48 38
4 48 46 38 44
Total 33 39 42 38

The cell mean for level i of a and level j of b is equal to the coefficient for the constant plus the coefficient for a at level i, plus the coefficient for b at level j, plus the coefficient for a and b at i and j. When a coefficient is omitted from the regression table, the corresponding coefficient is zero. The table below shows the relationship.

a b Cell mean Cell mean (simplified)
a = 1 b = 1 _cons+i1.a+i1.b+i1.a#i1.b _cons
a = 1 b = 2 _cons+i1.a+i2.b+i1.a#i2.b _cons+i2.b
a = 1 b = 3 _cons+i1.a+i3.b+i1.a#i3.b _cons+i3.b
a = 2 b = 1 _cons+i2.a+i1.b+i2.a#i1.b _cons+i2.a
a = 2 b = 2 _cons+i2.a+i2.b+i2.a#i2.b _cons+i2.a+i2.b+i2.a#i2.b
a = 2 b = 3 _cons+i2.a+i3.b+i2.a#i3.b _cons+i2.a+i3.b+i2.a#i3.b
a = 3 b = 1 _cons+i3.a+i1.b+i3.a#i1.b _cons+i3.a
a = 3 b = 2 _cons+i3.a+i2.b+i3.a#i2.b _cons+i3.a+i2.b+i3.a#i2.b
a = 3 b = 3 _cons+i3.a+i3.b+i3.a#i3.b _cons+i3.a+i3.b+i3.a#i3.b
a = 4 b = 1 _cons+i4.a+i1.b+i4.a#i1.b _cons+i4.a
a = 4 b = 2 _cons+i4.a+i2.b+i4.a#i2.b _cons+i4.a+i2.b+i4.a#i2.b
a = 4 b = 3 _cons+i4.a+i3.b+i4.a#i3.b _cons+i4.a+i3.b+i4.a#i3.b

The simplifications shown at the far right of the table are due to the coefficients that are omitted from the overparameterized model being zero. The marginal means can easily be built up by averaging appropriate cell means together.

We can obtain the same four single degrees-of-freedom tests as were obtained with the cell means ANOVA model by examining the relationship (shown in the table above) between the coefficients of the overparameterized model and the quantities of real interest—the cell means. We could simply plug in all the coefficients for each cell involved in the test and let Stata’s test command do the algebra, or we can do the simplifying ourselves. For the same four tests that were performed for the cell means model, I will show the results when you plug everything into test (based on the simplification in the far right column of the table above) and let Stata’s test command do the algebra. (It also would work if I plugged the unsimplified cell mean expressions into test.)

  1. Test level two against level four of factor a

    . test (_cons+i2.a) 
    >      + (_cons+i2.a+i2.b+i2.a#i2.b)
    >      + (_cons+i2.a+i3.b+i2.a#i3.b)
    >    = (_cons+i4.a)                 
    >      + (_cons+i4.a+i2.b+i4.a#i2.b)
    >      + (_cons+i4.a+i3.b+i4.a#i3.b)
    
     ( 1)  3*2.a - 3*4.a + 2.a#2.b + 2.a#3.b - 4.a#2.b - 4.a#3.b = 0
    
           F(  1,    12) =    9.80
                Prob > F =    0.0087
    
  2. Test the average of levels one and two against level three of factor b

    . test ((_cons) 
    >       + (_cons+i2.a)                
    >       + (_cons+i3.a)               
    >       + (_cons+i4.a)              
    >       + (_cons+i2.b)                 
    >       + (_cons+i2.a+i2.b+i2.a#i2.b)  
    >       + (_cons+i3.a+i2.b+i3.a#i2.b) 
    >       + (_cons+i4.a+i2.b+i4.a#i2.b))
    >         /2                          
    >     = (_cons+i3.b)                 
    >       + (_cons+i2.a+i3.b+i2.a#i3.b)
    >       + (_cons+i3.a+i3.b+i3.a#i3.b)
    >       + (_cons+i4.a+i3.b+i4.a#i3.b)
    
     ( 1)  2*2.b - 4*3.b + .5*2.a#2.b - 2.a#3.b + .5*3.a#2.b - 3.a#3.b + .5*4.a#2.b
           - 4.a#3.b = 0
    
           F(  1,    12) =    3.20
                Prob > F =    0.0989
    
  3. Test the average of levels one and two of a, when b is also at level one or two, against the average of levels three and four of a, when b is at level three

    . test ((_cons)                          
    >       + (_cons+i2.a)                  
    >       + (_cons+i2.b)                 
    >       + (_cons+i2.a+i2.b+i2.a#i2.b))/2
    >     = (_cons+i3.a+i3.b+i3.a#i3.b)   
    >       + (_cons+i4.a+i3.b+i4.a#i3.b) 
    
     ( 1)  2.a - 3.a - 4.a + 2.b - 2*3.b + .5*2.a#2.b - 3.a#3.b - 4.a#3.b = 0
    
           F(  1,    12) =    5.38
                Prob > F =    0.0388
    
  4. Test that three times a at one and b at one, minus four times a at three and b at two, plus six times a at four and b at three, equals a at two and b at two, minus two times a at two and b at three

    . test 3*(_cons)                      
    >       - 4*(_cons+i3.a+i2.b+i3.a#i2.b)
    >       + 6*(_cons+i4.a+i3.b+i4.a#i3.b)
    >     = (_cons+i2.a+i2.b+i2.a#i2.b)     
    >       - 2*(_cons+i2.a+i3.b+i2.a#i3.b)
    
     ( 1)  2.a - 4*3.a + 6*4.a - 5*2.b + 8*3.b - 2.a#2.b + 2*2.a#3.b - 4*3.a#2.b +
           6*4.a#3.b + 6*_cons = 0
    
           F(  1,    12) =   32.58
                Prob > F =    0.0001
    

You can compare these results with those obtained after the cell means ANOVA model to see that they are the same.

The test() option of test may also be used. First we check the order of the columns.

. test, showorder
      
 Order of columns in the design matrix
      1: (a==1)
      2: (a==2)
      3: (a==3)
      4: (a==4)
      5: (b==1)
      6: (b==2)
      7: (b==3)
      8: (a==1)*(b==1)
      9: (a==1)*(b==2)
     10: (a==1)*(b==3)
     11: (a==2)*(b==1)
     12: (a==2)*(b==2)
     13: (a==2)*(b==3)
     14: (a==3)*(b==1)
     15: (a==3)*(b==2)
     16: (a==3)*(b==3)
     17: (a==4)*(b==1)
     18: (a==4)*(b==2)
     19: (a==4)*(b==3)
     20: _cons

Now here are the same four tests.

  1. . mat x1 = (0,3,0,-3,0,0,0,0,0,0,1,1,1,0,0,0,-1,-1,-1,0)
    
    . test, test(x1)
    
     ( 1)  3*2.a - 3*4.a + 2o.a#1b.b + 2.a#2.b + 2.a#3.b - 4o.a#1b.b - 4.a#2.b -
           4.a#3.b = 0
    
           F(  1,    12) =    9.80
                Prob > F =    0.0087
    
  2. . mat x2 = (0,0,0,0,2,2,-4,.5,.5,-1,.5,.5,-1,.5,.5,-1,.5,.5,-1,0)
    
    . test, test(x2)
    
     ( 1)  2*1b.b + 2*2.b - 4*3.b + .5*1b.a#1b.b + .5*1b.a#2o.b - 1b.a#3o.b +
           .5*2o.a#1b.b + .5*2.a#2.b - 2.a#3.b + .5*3o.a#1b.b + .5*3.a#2.b - 3.a#3.b
           + .5*4o.a#1b.b + .5*4.a#2.b - 4.a#3.b = 0
    
           F(  1,    12) =    3.20
                Prob > F =    0.0989
    
  3. . mat x3 = (1,1,-1,-1,1,1,-2,.5,.5,0,.5,.5,0,0,0,-1,0,0,-1,0)
    
    . test, test(x3)
    
     ( 1)  1b.a + 2.a - 3.a - 4.a + 1b.b + 2.b - 2*3.b + .5*1b.a#1b.b + .5*1b.a#2o.b
           + .5*2o.a#1b.b + .5*2.a#2.b - 3.a#3.b - 4.a#3.b = 0
    
           F(  1,    12) =    5.38
                Prob > F =    0.0388
    
  4. . mat x4 = (3,1,-4,6,3,-5,8,3,0,0,0,-1,2,0,-4,0,0,0,6,6)
    
    . test, test(x4)
    
     ( 1)  3*1b.a + 2.a - 4*3.a + 6*4.a + 3*1b.b - 5*2.b + 8*3.b + 3*1b.a#1b.b -
           2.a#2.b + 2*2.a#3.b - 4*3.a#2.b + 6*4.a#3.b + 6*_cons = 0
    
           F(  1,    12) =   32.58
                Prob > F =    0.0001
    

How did we come up with matrices x1x4? Let me illustrate with x4.

column definition 3*a[1]b[1] -4*a[3]b[2] 6*a[4]b[3] -1*a[2]b[2] 2*a[2]b[3] Sum
1 a==1 3 0 0 0 0 3
2 a==2 0 0 0 -1 2 1
3 a==3 0 -4 0 0 0 -4
4 a==4 0 0 6 0 0 6
5 b==1 3 0 0 0 0 3
6 b==2 0 -4 0 -1 0 -5
7 b==3 0 0 6 0 2 8
8 a==1,b==1 3 0 0 0 0 3
9 a==1,b==2 0 0 0 0 0 0
10 a==1,b==3 0 0 0 0 0 0
11 a==2,b==1 0 0 0 0 0 0
12 a==2,b==2 0 0 0 -1 0 -1
13 a==2,b==3 0 0 0 0 2 2
14 a==3,b==1 0 0 0 0 0 0
15 a==3,b==2 0 -4 0 0 0 -4
16 a==3,b==3 0 0 0 0 0 0
17 a==4,b==1 0 0 0 0 0 0
18 a==4,b==2 0 0 0 0 0 0
19 a==4,b==3 0 0 6 0 0 6
20 _cons 3 -4 6 -1 2 6

The column titled “Sum” is the sum over the previous five columns and provides the row elements of the matrix x4.

Somtimes it is easiest to specify the equation, and other times it is easier to specify the corresponding matrix in the test() option. Either method works.

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