SAS Code Fragments
Analyzing changes in trend over time
Create fake data set with given intercepts and slopes.
Imagine a treatment and control group measured 5 times, then an intervention
with 5 more observations. The intercepts and slopes are as shown in the
data step.
NOTE: The analyses here simply focus on the
meaning of the parameter estimates, and do NOT take into account correlations
over time, covariance structures etc...
data hosp;
do trt = 0 to 1;
do int = 0 to 1;
do year = 1 to 5;
if int = 0 and trt = 0 then y = 0+0*year ;
if int = 1 and trt = 0 then y = 3+1*year ;
if int = 0 and trt = 1 then y = 2+2*year ;
if int = 1 and trt = 1 then y = 14+4*year ;
year2 = year + 5*int;
output;
end;
end;
end;
run;
show plot of data
SYMBOL1 V=circle C=blue I=join;
SYMBOL2 V=circle C=red I=join;
proc gplot data=hosps;
plot y*year2=trt;
run;
We use simple coding int as ints, trt as trts
data hosps;
set hosp;
ints = int - .5;
trts = trt - .5;
run;
Show data
proc print data=hosps;
run;
Obs trt int year y year2 ints trts
1 0 0 1 0 1 -0.5 -0.5
2 0 0 2 0 2 -0.5 -0.5
3 0 0 3 0 3 -0.5 -0.5
4 0 0 4 0 4 -0.5 -0.5
5 0 0 5 0 5 -0.5 -0.5
6 0 1 1 4 6 0.5 -0.5
7 0 1 2 5 7 0.5 -0.5
8 0 1 3 6 8 0.5 -0.5
9 0 1 4 7 9 0.5 -0.5
10 0 1 5 8 10 0.5 -0.5
11 1 0 1 4 1 -0.5 0.5
12 1 0 2 6 2 -0.5 0.5
13 1 0 3 8 3 -0.5 0.5
14 1 0 4 10 4 -0.5 0.5
15 1 0 5 12 5 -0.5 0.5
16 1 1 1 18 6 0.5 0.5
17 1 1 2 22 7 0.5 0.5
18 1 1 3 26 8 0.5 0.5
19 1 1 4 30 9 0.5 0.5
20 1 1 5 34 10 0.5 0.5
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Show slopes by int
b(int=0) is 1, b(int=1) is 2.5
proc sort data=hosps;
by int;
run;
proc glm data=hosps;
by int;
model y = year / solution;
run;
<output abbreviated>
int=0
Standard
Parameter Estimate Error t Value Pr > |t|
Intercept 1.000000000 3.51781182 0.28 0.7834
year 1.000000000 1.06066017 0.94 0.3734
int=1
Standard
Parameter Estimate Error t Value Pr > |t|
Intercept 8.500000000 8.47606925 1.00 0.3453
year 2.500000000 2.55563104 0.98 0.3566
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show slopes by trt
b(trt=0) is .5, b(trt=1) is 3
proc sort data=hosps;
by trt;
run;
proc glm data=hosps;
by trt;
model y = year / solution;
run;
trt=0
Standard
Parameter Estimate Error t Value Pr > |t|
Intercept 1.500000000 2.55563104 0.59 0.5734
year 0.500000000 0.77055175 0.65 0.5346
trt=1
Standard
Parameter Estimate Error t Value Pr > |t|
Intercept 8.000000000 7.55397246 1.06 0.3205
year 3.000000000 2.27760839 1.32 0.2243
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run mixed analysis via simple coding
ints*trts*year is (4-1) - (2-0) = 1
ints*year is (2.5-1) = 1.5
trts*year is (3-.5) = 2.5
proc mixed data=hosps;
model y = ints|trts|year / solution;
run;
<output abbreviated>
Standard
Effect Estimate Error DF t Value Pr > |t|
Intercept 4.7500 0 12 Infty <.0001
ints 7.5000 0 12 Infty <.0001
trts 6.5000 0 12 Infty <.0001
ints*trts 9.0000 0 12 Infty <.0001
year 1.7500 0 12 Infty <.0001
ints*year 1.5000 0 12 Infty <.0001
trts*year 2.5000 0 12 Infty <.0001
ints*trts*year 1.0000 0 12 Infty <.0001
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run mixed analysis via default (dummy) coding
after sas revises dummy codes (0=1, 1=0) slopes are
i=0 i=1
t=0 4 2
t=1 1 0
year = slope when i=0, t=0, = 4
y*i = change in slope for i, when t=0, 2-4 = -2
y*t = change in slope for t, when i=0, 1-4 = -3
i*t*y = (0-1) - (2-4) = -1 --2 = -1+2 = 1
proc mixed data=hosps;
class int trt ;
model y = int|trt|year / solution;
run;
<output abbreviated>
Solution for Fixed Effects
Standard
Effect int trt Estimate Error DF t Value Pr > |t|
Intercept 14.0000 0 12 Infty <.0001
int 0 -12.0000 0 12 -Infty <.0001
int 1 0 . . . .
trt 0 -11.0000 0 12 -Infty <.0001
trt 1 0 . . . .
int*trt 0 0 9.0000 0 12 Infty <.0001
int*trt 0 1 0 . . . .
int*trt 1 0 0 . . . .
int*trt 1 1 0 . . . .
year 4.0000 0 12 Infty <.0001
year*int 0 -2.0000 0 12 -Infty <.0001
year*int 1 0 . . . .
year*trt 0 -3.0000 0 12 -Infty <.0001
year*trt 1 0 . . . .
year*int*trt 0 0 1.0000 0 12 Infty <.0001
year*int*trt 0 1 0 . . . .
year*int*trt 1 0 0 . . . .
year*int*trt 1 1 0 . . . .
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