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st: Re: Repeated Measures ANOVA: contrasts
From
"Joseph Coveney" <[email protected]>
To
<[email protected]>
Subject
st: Re: Repeated Measures ANOVA: contrasts
Date
Sat, 1 Feb 2014 00:44:53 +0900
Vincent Koppelmans wrote:
I have longitudinal data (7 time points) from 8 subjects belonging to the same
group. I would like to compare time points 1 and 2 versus 3, 4, and 5.
I used:
anova FMT time ID, repeated(time) bse(ID)
contrast time, effect
to compare time points mutually. (ID is subject ID number, FMT is my outcome
measure)
Is it possible to create a contrast (e.g. 1.5 1.5 -1 -1 -1 0 0) to compare time
points 1 and 2 versus 3, 4, and 5?
--------------------------------------------------------------------------------
You can write custom contrasts. Something like that below should work. (Start
at the "Begin Here" section. The stuff before is just set-up.)
Joseph Coveney
. version 13.1
.
. clear *
. set more off
. set seed `=date("2014-01-31", "YMD")'
. quietly set obs 8
.
. generate byte pid = _n
. forvalues i = 1/7 {
2. generate double fmt`i' = rnormal()
3. }
. quietly reshape long fmt, i(pid) j(time)
.
. anova fmt pid time, repeated(time)
Number of obs = 56 R-squared = 0.2208
Root MSE = .931711 Adj R-squared = -0.0204
Source | Partial SS df MS F Prob > F
-----------+----------------------------------------------------
Model | 10.331744 13 .794749535 0.92 0.5453
|
pid | 2.82247193 7 .403210276 0.46 0.8545
time | 7.50927202 6 1.25154534 1.44 0.2218
|
Residual | 36.4595877 42 .868085422
-----------+----------------------------------------------------
Total | 46.7913317 55 .850751485
Between-subjects error term: pid
Levels: 8 (7 df)
Lowest b.s.e. variable: pid
Repeated variable: time
Huynh-Feldt epsilon = 1.0158
*Huynh-Feldt epsilon reset to 1.0000
Greenhouse-Geisser epsilon = 0.5281
Box's conservative epsilon = 0.1667
------------ Prob > F ------------
Source | df F Regular H-F G-G Box
-----------+----------------------------------------------------
time | 6 1.44 0.2218 0.2218 0.2569 0.2689
Residual | 42
----------------------------------------------------------------
.
. *
. * Begin Here
. *
. contrast {time 1.5 1.5 -1 -1 -1 0 0}, lincom
Contrasts of marginal linear predictions
Margins : asbalanced
------------------------------------------------
| df F P>F
-------------+----------------------------------
time | 1 2.43 0.1266
|
Denominator | 42
------------------------------------------------
--------------------------------------------------------------
| Contrast Std. Err. [95% Conf. Interval]
-------------+------------------------------------------------
time |
(1) | -1.406213 .9021253 -3.226775 .4143497
--------------------------------------------------------------
.
. // For the estimate & its CIs, though, you'll probably want this one:
. local mot = -1/3
. contrast {time 0.5 0.5 `mot' `mot' `mot' 0 0}, lincom
Contrasts of marginal linear predictions
Margins : asbalanced
------------------------------------------------
| df F P>F
-------------+----------------------------------
time | 1 2.43 0.1266
|
Denominator | 42
------------------------------------------------
--------------------------------------------------------------
| Contrast Std. Err. [95% Conf. Interval]
-------------+------------------------------------------------
time |
(1) | -.4687376 .3007084 -1.075592 .1381166
--------------------------------------------------------------
.
. exit
end of do-file
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