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I have linear model

lmB <- lm(vsnIntensity ~  Donor + Condition  ,data = check)

Donor and Condition are both factors and vsnIntensity is continous The model in mathematical notation, (I think) is $$ y_{ij} = \alpha + \beta_i + \gamma_j + \epsilon_{ij} $$

Where $\alpha$ is the intercept, which is the mean of the reference donor and condition. $\beta_i$ are the coefficients (effects) for all conditions except the reference condition. $\gamma_j$ are the coefficients (effects) for all the donors except the reference donor. Is the mathematical notation correct?

When I fit the model I am getting:

  > lmB <- lm(vsnIntensity ~  Donor + Condition  ,data = check)
  > data.frame(coefficients(lmB))
                 coefficients.lmB.
  (Intercept)        18.15866653
  Donor185            0.06377651
  Donor234            0.30834387
  Donor235            0.36166529
  Donor236            0.09642398
  ConditionCMP       -0.01566147
  ConditionGMP        0.20452979
  ConditionMEP        0.06511231

However, computing the mean for the reference donor and condition using aggregate I have different values:

   Group.1 Group.2        x
   1      HSC     132 18.06667
   2      CMP     132 18.26274
   3      GMP     132 18.31288
   4      MEP     132 18.24636
   5      HSC     185 18.14692
   6      CMP     185 18.20435

Why Do I have a different value for the Intercept than the group mean for the reference factors and why when I am using only a single factor e.g.

lmB <- lm(vsnIntensity ~  Condition  ,data = check)

the intercept is the group mean for the reference condition?

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1 Answer 1

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I believe aggregate() is giving you the actual means by condition. That is, it should equal what you get from doing something like: with(check, tapply(donor, condition, mean)).

However, the intercept is the model-implied or predicted value for when the subjects are in the reference conditions for both donor and condition.

The model-implied solution does not always equal the actual observed values, especially if donor and condition might be correlated.

I believe this is correct, but I would need to see the data and/or what you typed in for the aggregate() command to know for sure.

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