- On 2 Sep 2002 at 09:11:33, "Bonate, Peter" (pbonate.-a-.ilexonc.com) sent the message

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If I may add my two cents into this topic on R2 and SE of the slope.

They go hand in hand. R2 is equal to 1 - SSE/SStotal. The standard

error of the slope is equal to sqrt(MSE/Sxx) where Sxx=sum((x -

mean(x))^2)). After a little algebra, the standard error of the

slope can be written as sqrt((-(R2-1)/(n-2)*SStotal)/Sxx). Hence,

for high R2, SE(slope) will be small. For small R2, SE(slope) will

be large. So implicit in the calculation of the standard error is

the use of R2.

pete bonate - On 4 Sep 2002 at 12:37:35, "J.H.Proost" (J.H.Proost.-a-.farm.rug.nl) sent the message

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Dear Martin Schumacher and Pete Bonate,

Thank you for your contributions, which provide a more complete

picture of the topic.

Martin Schumacher explained the rather large confidence interval of

the estimated variance. I completely agree. However, in general, we

are not interested in the confidence interval of the variance, but in

the confidence interval of the estimated parameter. The fact that

these two are related does not matter. By using the t-distribution in

the calculation of the confidence interval of the estimated

parameter, we take into account the uncertainty in the estimated

variance. That is exactly what the t-distribution is for. Or am I

wrong here?

Pete Bonate explained the relationship between R2 and the standard

error of the slope. I completely agree (please note that MSE =

SSE/(n-2) ). It is clear that SE(slope) is dependent of R2. However,

it is also dependent on SStotal, Sxx and n. So, knowledge on R2 does

not tell the whole story about SE(slope). I still do not see a key

role for R2, in contrast to SE(slope). To say it more pathetically:

"I can live without R2, but not without SE(slope)".

Best regards,

Hans Proost

Johannes H. Proost

Dept. of Pharmacokinetics and Drug Delivery

University Centre for Pharmacy

Antonius Deusinglaan 1

9713 AV Groningen, The Netherlands

Email: j.h.proost.-at-.farm.rug.nl

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