While RR is between 0 and 1, RRR can be negative, indicating adverse effect
of treatment. Of course, if there is a null effect, RRR may fluctuate to
either side of 0 without there actually being a real adverse effect of
treatment.
David L. Doggett, Ph.D.
Senior Medical Research Analyst
Technology Assessment Group
ECRI, a non-profit health services research organization
5200 Butler Pike
Plymouth Meeting, PA 19462-1298, USA
Phone: +1 (610) 825-6000 ext.5509
Fax: +1(610) 834-1275
E-mail: [log in to unmask]
> -----Original Message-----
> From: Jeanne Lenzer [SMTP:[log in to unmask]]
> Sent: Tuesday, August 22, 2000 3:47 AM
> To: EBM (E-mail)
> Subject: When treatment harms
>
> Just an uneducated question, re: a listmember's query:
> (snip) - "I am getting some negative RRR [with CAT snipper]. What would be
>
> their interpretation?. I know that RRR = 0 is null effect and RRR= 1 is
> cure."
>
> Here's my question: If - as must be presumed by the range for RRR of 0-1 -
>
> the term RRR assesses only efficacy of intervention and not risks of
> intervention - is there a number that incorporates risk of treatment? (In
>
> other words a more global assessment of intervention that would allow a
> negative outcome)?
>
> Without such a global value, how do we measure the many thrombolytic
> trials
> in stroke, for example, that show more deaths with treatment intervention
> than without? I am increasingly concerned by the ways sponsored
> researchers do "spin control" and it seems to me that RRR is one such term
>
> that can be so abused if the reader is not alert. For example, an
> intervention that reduces a primary endpoint (fatal MI in the pt with
> diabetes) but increases total death rate (from pancreatitis/hepatitis
> whatever - [shades of Rezulin]) the outcome may be spun as positive
> through
> a set of maneuvers from a.) referring to a positive RRR (correctly) and
> then b.) combining endpoints (death and disability) to a more neutral - or
>
> even positive effect (as has been done in the thrombolytic trials).
>
> Have I misunderstood something here?
>
> jeanne
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