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Re: st: interpretation for negative and positive slope combination of interaction term


From   Nahla Betelmal <nahlaib@gmail.com>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: interpretation for negative and positive slope combination of interaction term
Date   Fri, 10 May 2013 12:29:13 +0100

Thanks again, I will give it a go. However, back to the original
model, instead of describing the effect of MV on earnings management
by overconfident mangers in ratio terms, I think it is sensible to say
excessively overconfident bidders have a greater incentive to manage
earnings via accruals, approximately 5.96 cents more than rational
bidders for each dollar increase, for MV.

What bothers me is that MV has negative effect on earnings management
for rational managers (negative slope -0.056) and positive intercept,
while the effect is positive but really small (positive slope 0.003)
and it does not make sense to divide the two slopes in this case to
get the effect ratio. Well, I wanted to present the result in the
common way of ratios but I have to accept the mechanical aspect of
mathematics.

Many thanks for your reply and for the log advice

Nahla

On 10 May 2013 12:09, David Hoaglin <dchoaglin@gmail.com> wrote:
> Dear Nahla,
>
> If the dependent variable is y, my basic suggestion is to use log(y)
> instead as the dependent variable, because you are interested in
> ratios.  In a simple example, if the model is
> log(y) = b0 + b1x
> and x is an indicator variable (i.e., 0 for one group and 1 for the
> other), then the difference between the mean of log(y) in the two
> groups is b1.  In the original scale of the data, that difference
> translates into a ratio, the "antilog" of b1.  I usually use logs base
> 10 when data are transformed, so the ratio would be 10^b1.  For logs
> base e, it would be exp(b1).
>
> Whether the continuous explanatory variables should be transformed (to
> the log scale or some other scale) is a question of choosing a scale
> in which their contributions to log(y) are linear.
>
> David Hoaglin
>
> On Fri, May 10, 2013 at 5:20 AM, Nahla Betelmal <nahlaib@gmail.com> wrote:
>> Hi David, thank you for your reply, but can you kindly explain more or
>> give me a link or reference how having a log function can solve the
>> issue and how to do it plz. Do you mean taking the log of the
>> dependent variable only or both the dependent and all continuous
>> independent variables? And how can this help plz
>>
>> Many thanks in advance
>> Nahla
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