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Chapter 12, page 257 shows how to perform an interaction contrast using the data from chapter 10.
use http://www.ats.ucla.edu/stat/stata/examples/da/chap10, clear
This interaction contrast compares a2 with a3 for factor a and compares b1 with b2 for factor b. First, let's create the contrasts for factor a. We will also create a second comparison for factor a that is orthogonal to the first comparison.
char a[user] (0 -1 1\2 -1 -1)
Now let's create the contrast for factor b.
char b[user] (1 -1)
Now, let's perform the interaction comparison by using u.a*u.b after the regress command. The interaction contrast is reflected in the product of the first comparison on a and the first (and only) comparison on b, so it is the term _Ia1Xb1 that is of interest. We show the test command for this simply to show the F value for this test and that it matches the text. You can download the xi3 command by typing findit xi3 (see How can I use the findit command to search for programs and get additional help? for more information about using findit).
xi3: regress errors u.a*u.b
u.a _Ia_1-3 (naturally coded; _Ia_3 omitted)
u.b _Ib_1-2 (naturally coded; _Ib_2 omitted)
Source | SS df MS Number of obs = 24
-------------+------------------------------ F( 5, 18) = 3.05
Model | 280 5 56 Prob > F = 0.0361
Residual | 330 18 18.3333333 R-squared = 0.4590
-------------+------------------------------ Adj R-squared = 0.3087
Total | 610 23 26.5217391 Root MSE = 4.2817
------------------------------------------------------------------------------
errors | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
_Ia_1 | 1 2.140872 0.47 0.646 -3.497805 5.497805
_Ia_2 | -9 3.708099 -2.43 0.026 -16.79043 -1.209573
_Ib_1 | -2 1.748015 -1.14 0.268 -5.672443 1.672443
_Ia1Xb1 | 6 4.281744 1.40 0.178 -2.995611 14.99561
_Ia2Xb1 | -18 7.416198 -2.43 0.026 -33.58085 -2.419145
_cons | 10 .8740074 11.44 0.000 8.163779 11.83622
------------------------------------------------------------------------------
test _Ia1Xb1
( 1) _Ia1Xb1 = 0
F( 1, 18) = 1.96
Prob > F = 0.1781
The effect _Ia2Xb1 reflects the second comparison on a at b1, and this is the comparison shown in the middle of page 259, see below.
test _Ia2Xb1
( 1) _Ia2Xb1 = 0
F( 1, 18) = 5.89
Prob > F = 0.0259
On page 264-265 Keppel shows another interaction contrast. We compute the coefficients for these contrasts below.
use http://www.ats.ucla.edu/stat/stata/examples/da/chap12, clear char a[user] (1 -.5 -.5\0 1 -1) char b[user] (0 1 -1\2 -1 -1)
Now we run the analysis using u.a*u.b and the _Ia1Xb1 term corresponds to the first comparison on a crossed with the first comparison on b. We show the test command for this term and this matches the result shown by Keppel at the top of page 265 (with a bit of rounding error).
xi3: regress memory u.a*u.b
u.a _Ia_1-3 (naturally coded; _Ia_3 omitted)
u.b _Ib_1-3 (naturally coded; _Ib_3 omitted)
Source | SS df MS Number of obs = 45
-------------+------------------------------ F( 8, 36) = 9.26
Model | 129.199997 8 16.1499997 Prob > F = 0.0000
Residual | 62.8200027 36 1.74500008 R-squared = 0.6728
-------------+------------------------------ Adj R-squared = 0.6001
Total | 192.02 44 4.36409091 Root MSE = 1.321
------------------------------------------------------------------------------
memory | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
_Ia_1 | 1.8 .417732 4.31 0.000 .9528003 2.6472
_Ia_2 | .2666666 .4823553 0.55 0.584 -.7115954 1.244929
_Ib_1 | 2.733333 .4823553 5.67 0.000 1.755071 3.711595
_Ib_2 | 2.6 .835464 3.11 0.004 .9056006 4.294399
_Ia1Xb1 | -2.9 1.02323 -2.83 0.007 -4.975207 -.8247928
_Ia1Xb2 | -3.899999 1.772287 -2.20 0.034 -7.494363 -.3056354
_Ia2Xb1 | -.2000004 1.181524 -0.17 0.867 -2.596243 2.196242
_Ia2Xb2 | -1.399999 2.04646 -0.68 0.498 -5.550413 2.750415
_cons | 7.2 .1969207 36.56 0.000 6.800626 7.599374
------------------------------------------------------------------------------
test _Ia1Xb1
( 1) _Ia1Xb1 = 0
F( 1, 36) = 8.03
Prob > F = 0.0075
On page 269/270, Keppel illustrates another interaction contrast. Below we use the chap10 data file.
use http://www.ats.ucla.edu/stat/stata/examples/da/chap10, clear
The comparisons on a are orthogonal polynomial comparisons and can be achieved with the o. suffix (i.e., o.a) and the comparison on be is a simple comparison comparing b1 with b2 so we can use s.b, as shown below. The effect _Ia1Xb2 is the linear effect of a crossed with b and corresponds to that shown by Keppel at the top of page 270, see the test command below.
xi3: regress errors o.a*g.b
o.a _Ia_1-3 (_Ia_3 for a==3 omitted)
g.b _Ib_1-2 (naturally coded; _Ib_1 omitted)
Source | SS df MS Number of obs = 24
-------------+------------------------------ F( 5, 18) = 3.05
Model | 280 5 56 Prob > F = 0.0361
Residual | 330 18 18.3333333 R-squared = 0.4590
-------------+------------------------------ Adj R-squared = 0.3087
Total | 610 23 26.5217391 Root MSE = 4.2817
------------------------------------------------------------------------------
errors | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
_Ia_1 | 2.041241 .8740074 2.34 0.031 .2050201 3.877463
_Ia_2 | -.7071068 .8740074 -0.81 0.429 -2.543328 1.129115
_Ib_2 | 2 1.748015 1.14 0.268 -1.672443 5.672443
_Ia1Xb2 | -4.898979 1.748015 -2.80 0.012 -8.571422 -1.226537
_Ia2Xb2 | -1.52e-16 1.748015 -0.00 1.000 -3.672443 3.672443
_cons | 10 .8740074 11.44 0.000 8.163779 11.83622
------------------------------------------------------------------------------
test _Ia1Xb2
( 1) _Ia1Xb2 = 0
F( 1, 18) = 7.85
Prob > F = 0.0118
On pages 270 to 274 Keppel shows how to perform a partial interaction that compares a1 versus a2 and a3 by b. We start by creating the contrast corresponding to this comparison, and creating a second contrast which is orthogonal to acomp1.
use http://www.ats.ucla.edu/stat/stata/examples/da/chap12, clear char a[user] (-2 1 1\0 -1 1)
Now, we use the user-defined contrasts that we made with the char command above. When we combine the all of the tests of the first comparison on a crossed with b, we get the Acomp by B effect described in Keppel. So we test _Ia1Xb2 _Ia1Xb3 and we get the result shown in the top portion of page 274.
xi3: regress memory u.a*g.b
u.a _Ia_1-3 (naturally coded; _Ia_3 omitted)
g.b _Ib_1-3 (naturally coded; _Ib_1 omitted)
Source | SS df MS Number of obs = 45
-------------+------------------------------ F( 8, 36) = 9.26
Model | 129.199997 8 16.1499997 Prob > F = 0.0000
Residual | 62.8200027 36 1.74500008 R-squared = 0.6728
-------------+------------------------------ Adj R-squared = 0.6001
Total | 192.02 44 4.36409091 Root MSE = 1.321
------------------------------------------------------------------------------
memory | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
_Ia_1 | -3.6 .835464 -4.31 0.000 -5.294399 -1.905601
_Ia_2 | -.2666666 .4823553 -0.55 0.584 -1.244929 .7115954
_Ib_2 | .0666667 .4823553 0.14 0.891 -.9115953 1.044929
_Ib_3 | -2.666667 .4823553 -5.53 0.000 -3.644929 -1.688405
_Ia1Xb2 | -.9999996 2.04646 -0.49 0.628 -5.150414 3.150414
_Ia1Xb3 | -6.799999 2.04646 -3.32 0.002 -10.95041 -2.649585
_Ia2Xb2 | -.5999994 1.181524 -0.51 0.615 -2.996242 1.796243
_Ia2Xb3 | -.7999998 1.181524 -0.68 0.503 -3.196242 1.596243
_cons | 7.2 .1969207 36.56 0.000 6.800626 7.599374
------------------------------------------------------------------------------
test _Ia1Xb2 _Ia1Xb3
( 1) _Ia1Xb2 = 0
( 2) _Ia1Xb3 = 0
F( 2, 36) = 6.44
Prob > F = 0.0041
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