In his message Toby Lipman suggested the best way to be realistic would be
to calculate a confidence interval for the NNT.
One option is to calculate it as the inverse of the absolute risk reduction
CI with a SE as the square root of : (p1(1-p1)/n1) + (p2(1-p2)/n2);
in wich p1 and p2 are the risks in study and n1 and n2 sample sizes.
I'm not sure this way is the best one.
In your opinion: ich is the best procedure to perform such calculation?
Sincerely,
Marķa
Marķa Viniegra
Instituto Alexander Fleming
Buenos Aires, Argentina
----- Mensaje original -----
De: Toby Lipman 7, Collingwood Terrace, Jesmond, Newcastle upon Tyne. Tel
0191-2811060 (home), 0191-2869178 (surgery) <[log in to unmask]>
Para: Julie Brown <[log in to unmask]>
CC: <[log in to unmask]>
Enviado: Martes 30 de Noviembre de 1999 06:09 AM
Asunto: Re: a funny little question
> In message <[log in to unmask]
> on.edu>, Julie Brown <[log in to unmask]> writes
> >Maybe this should be obvious to me, but it isn't, and I just know someone
> >out there can give me a quick answer.
> >
> >If I do a calculation, and end up with a number needed to treat of
1730.1,
> >do I round this down to 1730, as I would normally do, or do I round up to
> >1731, since I can't treat part of a person?
> >
> I don't think you can be as precise as that with a point estimate - if
> you really want to be realistic you should calculate the confidence
> interval of the NNT
>
> Toby
> --
> Toby Lipman
> General practitioner, Newcastle upon Tyne
> Northern and Yorkshire research training fellow
>
> Tel 0191-2811060 (home), 0191-2437000 (surgery)
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