CODEWITHSUNDEEP
rmodel-comparison
Emma Smith
Emma Smith

Reputation: 411

Model selection in R, all models giving the same AIC and BIC

So this is the head of my data,

  thickness grains resistivity
1      25.1   14.9      0.0270
2     368.4   58.1      0.0267
3     540.4   77.3      0.0160
4     712.1   95.6      0.0105
5     883.7  113.0      0.0090
6    1055.7  130.0      0.0247

And I want to find AIC and BIC for three different models involving thickness and grains. 

AIC(lm(formula = resistivity ~ (1/thickness), data=z)) #142.194
BIC(lm(formula = resistivity ~ (1/thickness), data=z)) #142.9898

AIC(lm(formula = resistivity ~ (1/grains), data=z)) #142.194
BIC(lm(formula = resistivity ~ (1/grains), data=z)) #142.9898

AIC(lm(formula = resistivity ~ (1/thickness) + (1/grains), data=z)) #142.194
BIC(lm(formula = resistivity ~ (1/thickness) + (1/grains), data=z)) #142.9898

I've commented the output beside each, why are they all the same?

Upvotes: 3

Views: 861

Answers (1)

G5W
G5W

Reputation: 37621

You get the same AIC & BIC because the models are all the same. You are just getting a constant, the mean value of resistivity.

lm(formula = resistivity ~ (1/thickness), data = z)
  Coefficients:
  (Intercept)  
      0.01898

The problem is that if you want a computation like 1/thickness in your formula, you must indicate that in your formula by enclosing the calculation in I(). This is described in help(formula). What you want is 

lm(formula = resistivity ~ I(1/thickness), data=z)
lm(formula = resistivity ~ I(1/grains), data=z)
lm(formula = resistivity ~ I(1/thickness) + I(1/grains), data=z)

Upvotes: 3

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