search for: slope2

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2005 Jun 08
2
Robustness of Segmented Regression Contributed by Muggeo
...es: Estimated Break-Point(s): Est. St.Err Mean.Vel 1.285 0.05258 1.652 0.01247 Est. St.Err. t value CI(95%).l CI(95%).u slope1 0.4248705 0.3027957 1.403159 -0.1685982 1.018339 slope2 2.3281445 0.3079903 7.559149 1.7244946 2.931794 slope3 9.5425516 0.7554035 12.632390 8.0619879 11.023115 Adjusted R-squared: 0.9924. Result2: Initial break points are 1.5 and 1.7. The estimated break points and slopes: Estimated Break-Point(s):...
2010 Apr 08
2
Overfitting/Calibration plots (Statistics question)
This isn't a question about R, but I'm hoping someone will be willing to help. I've been looking at calibration plots in multiple regression (plotting observed response Y on the vertical axis versus predicted response [Y hat] on the horizontal axis). According to Frank Harrell's "Regression Modeling Strategies" book (pp. 61-63), when making such a plot on new data
2005 Jan 20
1
Windows Front end-crash error
...80,400,80,80,80,80,400,80,80,80,80,400),4,4 ) mu2<-c(0,0,0) LE<-8^2 #Linking Error Sigma2<-diag(LE,3) sample.size<-5000 N<-100 #Number of datasets #Take a single draw from VL distribution vl.error<-mvrnorm(n=N, mu2, Sigma2) intercept1 <- 0 slope1 <- 0 intercept2 <- 0 slope2 <- 0 for(i in 1:N){ temp <- data.frame(ID=seq(1:sample.size),mvrnorm(n=sample.size, mu,Sigma)) temp$X5 <- temp$X1 temp$X6 <- temp$X2 + vl.error[i,1] temp$X7 <- temp$X3 + vl.error[i,2] temp$X8 <- temp$X4 + vl.error[i,3] long<-reshape(temp, idvar="ID", vary...
2006 May 05
1
trouble with step() and stepAIC() selecting the best model
...0.0575099635396228, link = log) Deviance Residuals: Min 1Q Median 3Q Max -7.591e-01 -3.688e-01 -1.828e-01 -8.494e-08 1.520e+00 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 110.729 47.643 2.324 0.02012 * slope2 -34.202 972079.452 -3.52e-05 0.99997 slope3 -1.423 1.371 -1.038 0.29928 log(pH + 1) -51.244 19.544 -2.622 0.00874 ** log(CN + 1) -9.132 6.906 -1.322 0.18602 --- Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 (Dispersion parameter f...
2017 Sep 19
2
symbolic computing example with Ryacas
Hi all, I am trying to implement the following matlab code with Ryacas : syms U x x0 C d1=diff(U/(1+exp(-(x-x0)/C)),x); pretty(d1) d2=diff(U/(1+exp(-(x-x0)/C)),x,2); pretty(d2) solx2 = solve(d2 == 0, x, 'Real', true) pretty(solx2) slope2=subs(d1,solx2) I have tried the following : library(Ryacas) x <- Sym("x");U <- Sym("U");x0 <- Sym("x0");C <- Sym("C") my_func <- function(x,U,x0,C) { return (U/(1+exp(-(x-x0)/C)))} FirstDeriv <- deriv(my_func(x,U,x0,C), x) PrettyFo...
2005 Jun 10
0
Replies of the question about robustness of segmented regression
...es: Estimated Break-Point(s): Est. St.Err Mean.Vel 1.285 0.05258 1.652 0.01247 Est. St.Err. t value CI(95%).l CI(95%).u slope1 0.4248705 0.3027957 1.403159 -0.1685982 1.018339 slope2 2.3281445 0.3079903 7.559149 1.7244946 2.931794 slope3 9.5425516 0.7554035 12.632390 8.0619879 11.023115 Adjusted R-squared: 0.9924. Result2: Initial break points are 1.5 and 1.7. The estimated break points and slopes: Estimated Break-Point(s):...
2017 Sep 19
0
symbolic computing example with Ryacas
...to implement the following matlab code with Ryacas : > > syms U x x0 C > > d1=diff(U/(1+exp(-(x-x0)/C)),x); > > pretty(d1) > > d2=diff(U/(1+exp(-(x-x0)/C)),x,2); > > pretty(d2) > > solx2 = solve(d2 == 0, x, 'Real', true) > > pretty(solx2) > > slope2=subs(d1,solx2) > > > I have tried the following : > > library(Ryacas) > > x <- Sym("x");U <- Sym("U");x0 <- Sym("x0");C <- Sym("C") > > my_func <- function(x,U,x0,C) { > > return (U/(1+exp(-(x-x0)/C)))} > &...
2009 Apr 08
2
Null-Hypothesis
...;- db/sd > 2*pt(-abs(td), df) My value I get by running this test is :[1] 2.305553e-07 Does it mean the two slopes differ significantly, because this value is in the alpha area, so that I have to reject the null- hypothesis and accept the alternative hypothesis? Is the null-hypothesis: slope1=slope2? Thanks for your help, Benedikt --
2017 Sep 19
1
symbolic computing example with Ryacas
...gt;> syms U x x0 C >> >> d1=diff(U/(1+exp(-(x-x0)/C)),x); >> >> pretty(d1) >> >> d2=diff(U/(1+exp(-(x-x0)/C)),x,2); >> >> pretty(d2) >> >> solx2 = solve(d2 == 0, x, 'Real', true) >> >> pretty(solx2) >> >> slope2=subs(d1,solx2) >> >> >> I have tried the following : >> >> library(Ryacas) >> >> x <- Sym("x");U <- Sym("U");x0 <- Sym("x0");C <- Sym("C") >> >> my_func <- function(x,U,x0,C) { >> >...
2011 Apr 22
2
statistic Q
Dear, i am a student and I need help in comparing between different slopes and finding whther there is a significant difference between them? Thanks a lot [[alternative HTML version deleted]]