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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
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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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Copyright 1995-2010 David W. A. Bourne (david@boomer.org)