Emma Smith - 4 years ago 128

R Question

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?

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Answer Source

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)
```

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