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I am working on a data set with two time points and I am interested in the change from baseline. There are no differing treatment groups. First, I realize that this is most simply made using a paired t-test. However, I would like to adjust for baseline response.

My simple question is: Am I right to interpret the intercept term in the following model as the mean baseline adjusted change?

y= score at time2

x= score at time1

$\alpha$ = intercept

$$y - mean(x)= \alpha + \beta (x- mean(x))$$

Nick Sabbe
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E_J
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1 Answers1

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Why are you subtracting the mean of x? All this does is change the meaning of the intercept, which is usually not of interest.

However, here $\alpha$ represents the predicted increase in y above the mean of x when x = the mean of x.

The usual adjustment for baseline adjusts for the individual score at the earlier time; since you seem to have no other variables, this would be

$y - x = \alpha + \beta x$

which would look at whether change is related to initial score.

Peter Flom
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  • Thanks for the reply! I centered on the mean value of X to get that interpretation of the intercept because. I'm interested in the intercept since I do not have a treatment variable and want to examine whether the group has changed when adjusting for baseline response. With the $$y−x=α+β(x-mean(x))$$ setup I get the same intercept. I was however deterred from using that setup by reading this thread: [link](http://stats.stackexchange.com/questions/15713/is-it-valid-to-include-a-baseline-measure-as-control-variable-when-testing-the-e) – E_J May 08 '13 at 11:23