search for: lwr

fit lwr upr 1 218.4332 90.51019 346.3561 2 218.3906 90.46133 346.3198 3 218.3906 90.46133 346.3198 4 161.3982 44.85702 277.9394 5 192.4450 68.39903 316.4909 6 179.8056 56.49540 303.1158 7 219.5406 91.52707 347.5542 8 162.6761 46.65760 278.6945 9 193.8506 70.59838 317.1029 10 181.3816 58...

With this data x <- c(0,0,1,1,2,2) y <- c(5,6,4,3,2,6) lwr <- y-1 upr <- y+1 xlab <- c("Low","Low","Med","Med","High","High") mydata <- data.frame(x,xlab,y,lwr,upr) I would like to make a dot plot and use lwr and upr as error bars. Above 0=Low. I would like there to be some space bet...

...dict(mb,data.frame(yrs=yrs),interval='confidence') ##combine the two regressions for a single equation with confidence intervals pr=pra+prb###it couldn't be this simple, could it? #by hand co=coef(ma)+coef(mb) new.sigma=sqrt(summary(ma)$sigma^2+summary(mb)$sigma^2) fit=co[1]+co[2]*yrs lwr= fit - qt(.975,9)*sqrt( new.sigma^2 * ( (1/11) + ( (yrs-mean(yrs))^2)/sum((yrs-mean(yrs))^2) ) )#the df are probably wrong (the 9 in the qt upr= fit + qt(.975,9)*sqrt( new.sigma^2 * ( (1/11) + ( (yrs-mean(yrs))^2)/sum((yrs-mean(yrs))^2) ) )# statement) I can't print the graph here, so here'...

...as follows, where predw is the prediction from the fit that used 0-weights and preds is from using FALSE's in the subset argument. Is this difference proper? predw preds $fit $fit fit lwr upr fit lwr upr 1 1.544302 1.389254 1.699350 1 1.544302 1.194879 1.893724 2 1.935504 1.719482 2.151526 2 1.935504 1.448667 2.422341 $se.fit $se.fit 1 2...

...) ## Calculating t-dist, 2-tailed, 95% prediction intervals for new observations mse <- anova(mod)[2,3] new.x <- obs$X - mean(dat$X) sum.x2 <- sum((dat$X - mean(dat$X))^2) y.hat <- pred$fit var.y.hat <- mse*(1+new.x^2/sum.x2) upr <- y.hat + qt(c(0.975), df = 8) * sqrt(var.y.hat) lwr <- y.hat + qt(c(0.025), df = 8) * sqrt(var.y.hat) hand <- data.frame(cbind(y.hat, lwr, upr)) #The limits are not the same pred hand -- Thank you, Kurt [[alternative HTML version deleted]]

Hi everyone, I am trying to build a table putting standard errors horizontally. I haven't been able to do it. library(memisc) berkeley <- aggregate(Table(Admit,Freq)~.,data=UCBAdmissions) berk0 <- glm(cbind(Admitted,Rejected)~1,data=berkeley,family="binomial") berk1 <- glm(cbind(Admitted,Rejected)~Gender,data=berkeley,family="binomial") berk2 <-

...de estos. Ten en cuenta que el 2do tipo de intervalos de calcula para observaciones _futuras_. En R puedes calcularlos de la siguiente manera: ## IC de confianza## ver ?predict.lm para mas detallesR> data.frame(y, predict(modelo, interval = "confidence")) y fit lwr upr#1 8.35 9.938571 6.580445 13.29670#2 12.42 11.804286 9.134239 14.47433#3 18.00 15.535714 13.664949 17.40648#4 17.58 17.401429 15.396872 19.40599#5 17.97 19.267143 16.798908 21.73538#6 20.76 21.132857 18.014915 24.25080 ## intervalos de prediccion para x = 25 R> predict(modelo, newdat...

...0.4000000 mCREB 7 1.0000000 No Virus 8 1.0000000 No Virus 9 1.0000000 No Virus > TukeyHSD(aov(Values ~ ind, data = nmda456)) Tukey multiple comparisons of means 95% family-wise confidence level Fit: aov(formula = Values ~ ind, data = nmda456) $ind diff lwr upr p adj mCREB-CREB -6.0666126 -6.3289033 -5.8043219 0.0000000 No Virus-CREB -5.3921477 -5.6544383 -5.1298570 0.0000000 No Virus-mCREB 0.6744649 0.4121743 0.9367556 0.0005382 > TukeyHSD(aov(Values ~ ind, data = nmda123)) Tukey multiple comparisons of means 95% famil...

Hi, 1. How do I construct 95% prediction interval for new x values, for example - x = 30000? 2. How do I construct 95% confidence interval? my dataframe is as follows : >dt structure(list(y = c(26100000, 60500000, 16200000, 30700000, 70100000, 57700000, 46700000, 8600000, 10000000, 61800000, 30200000, 52200000, 71900000, 55000000, 12700000 ), x = c(108000, 136000,

...A,cd$Depth, ylim = rev(range(0:100)), xlab="CHAO", ylab="Depth", pch=15, las=2, main="Sep12-RNA", cex.main=1) > lmR <- lm(cd$Depth~cd$CHAOsep12RNA) > abline(lmR) pconfR <- predict(lmR,interval="confidence") matlines(cd$CHAOsep12RNA,pconfR[,c("lwr","upr")], col=1, lty=2) I also tried > newx <- seq(min(cd$CHAOsep12RNA), max(cd$CHAOsep12RNA), length.out=11) > a <- predict(lmR, newdata=data.frame(CHAO=newx), interval=c("confidence")) > plot(cd$CHAOsep12RNA,cd$Depth, ylim = rev(range(0:100)), xlab="...

...t;) I am trying to plot the results in lattice for publication purposes so I need to figure this out. I have been struggling but I think I have reached a dead end.? Here is what I have been able to code: M<-predict(model$gam,type="response",se.fit=T) upr<- M$fit + (1.96 * M$se.fit)lwr<- M$fit - (1.96 * M$se.fit) library(lattice)xyplot(fitted(model$gam) ~ Q95 |super.end.group, data = spring, gm=model,? ? ? ?prepanel=function (x,y,...)list(ylim=c(min(upr),max(lwr))),? ? ? ?panel = function(x,y, gm, ...){ ? ??? ? ? ? ?panel.xyplot(x,y, type="smooth")? ? ? ? ?panel.line...

.../TukeyHSD.R 2005-07-15 02:43:25.055207448 -0300 @@ -66,10 +66,12 @@ center <- center[keep] width <- qtukey(conf.level, length(means), x$df.residual) * sqrt((MSE/2) * outer(1/n, 1/n, "+"))[keep] - dnames <- list(NULL, c("diff", "lwr", "upr")) + est <- center/(sqrt((MSE/2) * outer(1/n, 1/n, "+"))[keep]) + pvals <- ptukey(abs(est),length(means),x$df.residual,lower.tail=FALSE) + dnames <- list(NULL, c("diff", "lwr", "upr","p adj"))...

...f. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 > TukeyHSD(fm1,conf.level=0.90) Tukey multiple comparisons of means 90% family-wise confidence level Fit: aov(formula = breaks ~ wool * tension, data = warpbreaks) $wool diff lwr upr p adj B-A -5.777778 -10.77183 -0.7837284 0.058213 $tension diff lwr upr p adj M-L -10.000000 -17.66710 -2.332900 0.0228554 H-L -14.722222 -22.38932 -7.055122 0.0005595 H-M -4.722222 -12.38932 2.944878 0.4049442 $`wool:tension` diff l...

Como usas la función image puedes consultar la ayuda ?image o help(image) y encontrarás el siguiente ejemplo donde se usa un diferente color Palette (mencionada por pepeceb en su respuesta). x <- 10*(1:nrow(volcano)) y <- 10*(1:ncol(volcano)) image(x, y, volcano, col = terrain.colors(100), axes = FALSE) # O puedes usar directamente el número para indicar el color image(x, y, volcano, col =

Hello, I have a data frame with the following columns: "date","name","value" the name is the same for each date I would like to get TukeyHSD p-value for the differences of "value" between "name"s in each "date" separately I tried different ANOVA (aov()) but can only get either tukey by "name" or by "data" but not

Hello I'm having trouble figuring out how to use the output of "segmented()" with a new set of predictor values. Using the example of the help file: ??set.seed(12) xx<-1:100 zz<-runif(100) yy<-2+1.5*pmax(xx-35,0)-1.5*pmax(xx-70,0)+15*pmax(zz-.5,0)+rnorm(100,0,2) dati<-data.frame(x=xx,y=yy,z=zz) out.lm<-lm(y~x,data=dati) o<-## S3

Buenas, una pregunta. Si yo estoy calculando un modelo lineal, el caso más simple, 1 variable respuesta y una variable explicativa y creo un modelo, me da un R2 del 80% y quiero ver como es esa relacion entre las variables, para calcular el error de predicción del modelo, basta con ver el intervalo de confianza del modelo e irme a los extremos? Por si no me he expresado bien, un ejemplo tonto:

...image(cps_bkde$x1, cps_bkde$x2, cps_bkde$fhat, col = rev(gray.colors(10, gamma = 1)), ############################AQUI PARA CAMBIAR Y PONER TONOS ROJIZOS MAS FUERTES xlab = "experience", ylab = "log(wage)") box() lines(fit ~ experience, data = cps2) lines(lwr ~ experience, data = cps2, lty = 2) lines(upr ~ experience, data = cps2, lty = 2) image(cps_bkde$x1, cps_bkde$x2, cps_bkde$fhat, col = rev(terrain.colors(10, alpha = 1)), xlab = "experience", ylab = "log(wage)") box() lines(fit ~ experience, dat...

...;out is %d\n", out); p = &b; __asm volatile ( "lw %0, %1\n\t" : "=r"(out) : "R"(*p) ); printf("out is %d\n", out); p = &c; __asm volatile ( "lwl %0, 1 + %1\n\t" "lwr %0, 2 + %1\n\t" : "=r"(out) : "R"(*p) ); printf("out is %x\n", out); return 0; } LLVM-MIPS-BE diff: diff --git a/lib/Target/Mips/MipsISelLowering.cpp b/lib/Target/Mips/MipsISelLowering.cpp index 36e1a15..4a5d045 100644 --- a/lib/T...

Dear community, I have developed a backend of new 32-bit RISC ISA, which does not have unaligned memory access instructions (e.g., LWL, LWR, SWL, and SWR in MIPS). Since char and short variables are not 32-bit alignment, these variables cannot be correctly accessed. Therefore, alignment member functions, especially getCharAlign() and getShortAlign() of TargetInfo class in clang/include/clang/Basic/TargetInfo.h, should be virtual, in or...