Re: st: positive interaction - negative covariance

[Nick Cox <njcoxstata@gmail.com>]

You need to define

local b0 = 42

or whatever.

On Fri, Feb 22, 2013 at 5:47 PM, Nick Cox <njcoxstata@gmail.com> wrote:
> Why not draw a graph of your two curves? There are various ways to do
> it, but here is one
>
> local b1 = .0021067
> local b2 = -.3692713
> local b3 = -.0010758
>
> twoway function `b0' + `b1' * x, ra(0 100) || ///
>        function `b0' + (`b1' + `b3')*x + `b2', ra(0 100)
>
> Naturally, I don't know your -b0- or the range of -x- to use.
>
> You should plot your data too.
>
> As the second two terms seem less important, it's possible that they
> are secondary corrections at best and hardly worth interpreting.
>
> On Fri, Feb 22, 2013 at 5:32 PM, andrea pedrazzani
> <andrea.pedrazzani.piter@gmail.com> wrote:
>
>> I have a simple regression model with an interaction: Y = b0 + (b1)X +
>> (b2)Z + (b3)XZ.
>> Z is a dummy (0 or 1).
>>
>> b1 = .0021067  (SE= .0008513 and p=0.013)
>> b2 = -.3692713  (SE= .2329837 and p=0.113)
>> b3 = -.0010758  (SE= .000926 and p=0.245)
>>
>> Hence, the combined coefficient (i.e., the coefficient on X when Z=1)
>> is positive:
>> b1+b3 =  .0021067 + -.0010758 = .0010309
>>
>> with SE = sqrt( var(b1) + var(b3)*(Z^2) + 2Z*cov(b1,b3)  )
>>             = sqrt( .0000007246 + .0000008574*1 + -.0000007079*2 )
>>             = .00040768
>>
>> To get the p-value for the combinet coefficient, I did
>> .0010309/.00040768 = 2.528699.  The corresponding p = 0.0114.
>>
>> Summing up, X has a positive impact on Y when the condition Z is
>> present (.0010309), and a positive impact also when the condition Z is
>> not present (.0021067).
>> So, what can I say about the interaction? What kind of interaction is
>> it when the impact of X is positive both when the condition is present
>> and when it is absent? Moreover, the coefficients b1 and (b1+b3) are
>> very similar to each other.
>> Also, both b1 and the combined coefficient (b1+b3) are positive, but
>> the covariance between b1 and b3 is negative. It sounds strange to
>> me...
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