# Re: [R] How are interaction terms computed in lm's result / problems with interaction terms in lm?

```> On Sep 18, 2016, at 12:39 PM, mviljamaa <mvilja...@kapsi.fi> wrote:
>
>> On Sep 18, 2016, at 11:01 AM, mviljamaa <mvilja...@kapsi.fi> wrote:
>> Also if you, rather than doing what's done below, do:
>> fit3 <- lm(kidmomhsage\$kid_score ~ kidmomhsage\$mom_age + kidmomhsage\$mom_hs
>> + kidmomhsage\$mom_age * kidmomhsage\$mom_hs)
>> Then this gives the result:
>> Call:
>> lm(formula = kidmomhsage\$kid_score ~ kidmomhsage\$mom_age +
>> kidmomhsage\$mom_hs +
>>   kidmomhsage\$mom_age * kidmomhsage\$mom_hs)
>> Coefficients:
>>                          (Intercept)
>>                              110.542
>>                  kidmomhsage\$mom_age
>>                               -1.522
>>                   kidmomhsage\$mom_hs
>>                              -41.287
>> kidmomhsage\$mom_age:kidmomhsage\$mom_hs
>>                                2.391
>> Where the interaction term now seems properly interpretable. So perhaps this
>> is the way to use interaction terms with lm.
>
> But why does
>
> fit3 <- lm(kidmomhsage\$kid_score ~ kidmomhsage\$mom_age * kidmomhsage\$mom_hs)
>
> also give exactly the same result:
>
> Call:
> lm(formula = kidmomhsage\$kid_score ~ kidmomhsage\$mom_age * kidmomhsage\$mom_hs)
>
> Coefficients:
>                           (Intercept)
>                               110.542
>                   kidmomhsage\$mom_age
>                                -1.522
>                    kidmomhsage\$mom_hs
>                               -41.287
> kidmomhsage\$mom_age:kidmomhsage\$mom_hs
>                                 2.391
>
> It's as if lm is interpreting there having to also be "independent" mom_age
> and mom_hs variables, if there's just the interaction term. Why does it work
> this way?```
```
kidmomhsage\$mom_age * kidmomhsage\$mom_hs

... is expanded by the formula-engine so that it is exactly:

kidmomhsage\$mom_age + kidmomhsage\$mom_hs +
kidmomhsage\$mom_age:kidmomhsage\$mom_hs

(That's essentially the definiton of the `*`-operator in the formula-world.)

David Winsemius
Alameda, CA, USA

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