search for: sediff

Hi, based on the sample size I want to calculate whether to correlation coefficients are significantly different or not. I know that as a first step both coefficients have to be converted to z values using fisher's z transformation. I have done this already but I dont know how to further proceed from there. unlike for correlation coefficients I know that the difference for z values is

Dear R users: Suppose I have two different response variables y1, y2 that I regress separately on the different explanatory variables, x1 and x2 respectively. I need to compare points on two regression lines. These are the x and y values for each lines. x1<-c(0.5,1.0,2.5,5.0,10.0) y1<-c(204,407,1195,27404313) x2<-c(2.5,5.0,10.0,25.0) y2<-c(440,713,1520,2634) Suppose we need to

...normal quantile corresponding to the confidence level (e.g., 1.96 for a 95% CI). ciDiff <- function(ci1, ci2, level=0.95){ p1 <- mean(ci1) p2 <- mean(ci2) z <- qnorm((1 - level)/2, lower.tail=FALSE) se1 <- (ci1[2] - ci1[1])/(2*z) se2 <- (ci2[2] - ci2[1])/(2*z) seDiff <- sqrt(se1^2 + se2^2) (p1 - p2) + c(-z, z)*seDiff } > > Example: Prevalence of Diabetes: > ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?2011:?11.0 (95%CI > 10.1-11.9) > ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?2017:?10.1 (95%CI > 9.4-10.9) > ?...

...A simple alternative that doesn't assume equal error variances would be to use something like mod1 <- lm(y1 ~ x1) mod2 <- lm(y2 ~ x2) f1 <- predict(mod1, newdata=data.frame(x1=13), se.fit=TRUE) f2 <- predict(mod2, newdata=data.frame(x2=13), se.fit=TRUE) diff <- f1$fit - f2$fit sediff <- sqrt(f1$se.fit^2 + f2$se.fit^2) diff/sediff The df for the test statistic aren't clear to me and in small samples this could make a difference. I suppose that one could use a Satterthwaite approximation, but a simple alternative would be to take the smaller of the residual df, here 5 - 2...