similar to: convert p-values into z-scores

Hi, I am using an if else statement inside a function ?. If I use that function I have no problems ?. If I use the function with the if else statement inside a second function I get the following waring: Warning message: In if (pval == 0) p_value <- "< 2.2e-16" else p_value <- pval : the condition has length> 1 and only the first element will be used Using the second

Full_Name: Wei Xu Version: 1.5.1 OS: WindowsME Submission from: (NULL) (63.215.238.92) The P-value for a two.sided test is not consistent with the confidence interval. For example, P-value=0.1372, but the 95% CI doesn't include the H0 value(1). > x [1] 0.80 0.83 1.89 1.04 1.45 1.38 1.91 1.64 0.73 1.46 > y [1] 1.15 0.88 0.90 0.74 1.21 >

Hello, Well, try it: p <- .Machine$double.eps^seq(0.5, 1, by = 0.05) z <- qnorm(p/2) pnorm(z) # [1] 7.450581e-09 1.228888e-09 2.026908e-10 3.343152e-11 5.514145e-12 # [6] 9.094947e-13 1.500107e-13 2.474254e-14 4.080996e-15 6.731134e-16 #[11] 1.110223e-16 p/2 # [1] 7.450581e-09 1.228888e-09 2.026908e-10 3.343152e-11 5.514145e-12 # [6] 9.094947e-13 1.500107e-13 2.474254e-14 4.080996e-15

You may want to look into using the log option to qnorm e.g., in round figures: > log(1e-300) [1] -690.7755 > qnorm(-691, log=TRUE) [1] -37.05315 > exp(37^2/2) [1] 1.881797e+297 > exp(-37^2/2) [1] 5.314068e-298 Notice that floating point representation cuts out at 1e+/-308 or so. If you want to go outside that range, you may need explicit manipulation of the log values. qnorm()

I agree with many the sentiments about the wisdom of computing very small p-values (although the example below may win some kind of a prize: I've seen people talking about p-values of the order of 10^(-2000), but never 10^(-(10^8)) !). That said, there are a several tricks for getting more reasonable sums of very small probabilities. The first is to scale the p-values by dividing the

Hello, I need help in performing a Van_der_Waerden normal scores test in R. I have two arrays of scores(final on therapy scores from drug and placebo) and want to use the normal scores procdeure to test for significance. (observations are unequal in number - due to dropouts). Could you please help me out with the coding or let me know if there is a package that can be used (for example,

>>>>> William Dunlap via R-devel >>>>> on Sun, 23 Jun 2019 10:34:47 -0700 writes: >>>>> William Dunlap via R-devel >>>>> on Sun, 23 Jun 2019 10:34:47 -0700 writes: > include/Rmath.h declares a set of 'logspace' functions for use at the C > level. I don't think there are core R functions that call

When I do t.test on two distributions (see example below), it outputs numerous data about the t.test. What I'd like to do is individually capture some of this data and assign it to other variables. However, I am unable to find anything in the help section. In the example below, the t value is -4.0441 and the p-value is 0.006771 How can I assign these values to two variables, let's

In checking over the solutions to some homework that I had assigned I observed the fact that in R (version 2.4.0) pnorm(-1.46) gives 0.07214504. The tables in the text book that I am using for the course give the probability as 0.0722. Fascinated, I scanned through 5 or 6 other text books (amongst the dozens of freebies from publishers that lurk on my shelf) and found that some agree with R

[Please use the subject line!] In the help page for summary.lm, the "Value" section says that the returned object has a component called "fstatistic", which has the F-statistic and the associated numerator and denominator degrees of freedom. You can get the p-value by something like: fstat <- summary(speciallinearmodel)$fstatistic pval <- pf(fstat[1], fstat[2],

Hello, could You please help: I am looking for a way to formulate test accuracy measures such as test sensitivity, specificity, predictive values, and correct classification rate using p/d/qnorm. The tests' primary values follow a bimodal distribution, which is modelled by a mixture of two normal distributions: p * dnorm ((x - u1) / s1) / s1 + (1 - p) * dnorm ((x - u2) / s2) / s2)

Full_Name: Tao Shi Version: 1.7.0 OS: Windows XP Professional Submission from: (NULL) (149.142.163.65) > x [,1] [,2] [1,] 149 151 [2,] 1 8 > c2x<-chisq.test(x, simulate.p.value=T, B=100000)$p.value > for(i in (1:20)){c2x<-c(c2x,chisq.test(x, simulate.p.value=T,B=100000)$p.value)} > c2tx<-chisq.test(t(x), simulate.p.value=T, B=100000)$p.value > for(i in

Dear list, could someone point me to the normal score transform and it's use in R? I would like to transform my data with normal score transform, do some geostatsitical predictions and would like to transform the estimated values back including the estimation variance. Any suggestions? Ulrich -- __________________________________________________ Ulrich Leopold MSc. Department of

Hello, I wuold like to propose my modifications of the original cor.test to you : I tried to calcolate the correct p-value for Spearman and Kendall's test with ties. Let me know what you think. Thanks you for your time. Antonietta di Salvatore test <- function(x, ...) UseMethod("test") test.default <- function(x, y, alternative = c("two.sided",

Hi there, If I do an lm, I get p-vlues as p-value: < 2.2e-16 Suppose am interested in exact value such as p-value = 1.6e-16 (note = and not <) How do I go about it? stephen

Hello, I used the function Fstats (in the package strucchange) and would like to transform the F probability given by Fstats in P value. This transformation can be made while making a plot, but I need to have the numerical P value which are ploted... and I can't find out how to do. Here a is an exemple, to plot the P value. let's take data as a array fs <-fstats(data ~ 1, from = 4,

Dear all, I calculate the test statistic for the KS test outside R, and wish to use R only to calculate the corresponding p-value. Is there a way for doing this? (as far as I see, ks.test() requires raw data as input). Alternatively, is there a way to provide the ks.test() the two CDFs (two samples test) rather than the (x, y) data vectors? Thanks in advance, Rani

## I would like help in using variable values in plotmath expressions ## in lattice x <- 1:10 y <- 1:10 pval <- .95 plot(y ~ x, ## works as I want in base graphics main=substitute(list(alpha * " = " * group("",list(p),"")), list(p=pval))) plot(y ~ x, ## doesn't work as intended: "pval" is displayed main=substitute(list(alpha *

Hi All, 1) Is it possible to set the options such that R opens a new script editor every time I start the R and 2) specify the size of windows. Thanks for the suggestion and Best regards, Krishna [[alternative HTML version deleted]]

I found the following strange behavior using qnorm() and pnorm(): > x<-8.21;x-qnorm(pnorm(x)) [1] 0.0004638484 > x<-8.22;x-qnorm(pnorm(x)) [1] 0.01046385 > x<-8.23;x-qnorm(pnorm(x)) [1] 0.02046385 > x<-8.24;x-qnorm(pnorm(x)) [1] 0.03046385 > x<-8.25;x-qnorm(pnorm(x)) [1] 0.04046385 > x<-8.26;x-qnorm(pnorm(x)) [1] 0.05046385 > x<-8.27;x-qnorm(pnorm(x))