tag:blogger.com,1999:blog-6977755959349847093.post3151817947601572717..comments2024-03-01T21:53:15.921-08:00Comments on The Pith of Performance: Confidence Bands for Universal Scalability ModelsNeil Guntherhttp://www.blogger.com/profile/11441377418482735926noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6977755959349847093.post-89553793899676596992014-04-16T09:53:20.851-07:002014-04-16T09:53:20.851-07:00I believe the short answer goes something like thi...I believe the short answer goes something like this. The DOF is associated with the Student's t dsn qt() b/c the sample size is n<30 measurements. That can be regarded as a single parameter dsn b/c the mean is assumed to be zero. More generally, the Chi-Sq dsn, for example, estimates the population mean and var. In the large-N limit this becomes a Gaussian, which has 2 pop parameters. The figure of merit for the USL nonlinear model (rational function of a single variable N and 2 coefficients α and β) is the Adjusted R^2 value.Neil Guntherhttps://www.blogger.com/profile/11441377418482735926noreply@blogger.comtag:blogger.com,1999:blog-6977755959349847093.post-88690766857618201562014-04-15T04:54:12.648-07:002014-04-15T04:54:12.648-07:00It doesn't make a huge difference, but shouldn...It doesn't make a huge difference, but shouldn't <i>degfree</i> be <i>samples - 2</i> since we are estimating two different parameters in the nonlinear regression?<br /><br />Regards,<br />StefanStefan Mödinghttps://www.blogger.com/profile/08301431613941005857noreply@blogger.comtag:blogger.com,1999:blog-6977755959349847093.post-14538617215875587482014-02-25T08:35:10.520-08:002014-02-25T08:35:10.520-08:00Thank you for the post, extremely helpful.
I kept...Thank you for the post, extremely helpful.<br /><br />I kept getting the following error in the nls line:<br />"Error in nls(Cp ~ p/(1 + A * (p - 1) + B * p * (p - 1)), data = bmark, : <br /> number of iterations exceeded maximum of 50"<br /><br />I changed the command to:<br />usl <- nls( Cp ~ p/(1 + A * (p-1) + B * p * (p-1)), data=bmark, start=c(A=0.1, B=0.01), control = list(maxiter = 500))<br /><br />Hope this helps anyone that encounters the same problem. <br /><br />Thanks again!<br /><br />Best,<br />HarrisHarrishttps://www.blogger.com/profile/18068332297683203546noreply@blogger.comtag:blogger.com,1999:blog-6977755959349847093.post-80885731775422926172010-09-07T12:49:15.017-07:002010-09-07T12:49:15.017-07:00So why not just use a polygon to plot the confiden...So why not just use a polygon to plot the confidence region? Doesn't it seem a bit easier that way?Anonymoushttps://www.blogger.com/profile/00721276325087523351noreply@blogger.com