Displaying 20 results from an estimated 3000 matches similar to: "convert p-values into z-scores"
2008 Jul 23
2
Warning message in if else statement
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
2002 Nov 03
1
Ansari-Bradley test (PR#2252)
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
>
2019 Jun 21
4
Calculation of e^{z^2/2} for a normal deviate z
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
2019 Jun 21
4
Calculation of e^{z^2/2} for a normal deviate z
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()
2019 Jun 23
2
Calculation of e^{z^2/2} for a normal deviate z
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
2004 Mar 06
2
normal scores test
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,
2019 Jun 24
2
Calculation of e^{z^2/2} for a normal deviate z
>>>>> 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
2006 Mar 24
4
How to capture t-score and p-values from t.test
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
2006 Nov 25
3
OT: P(Z <= -1.46).
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
2003 Jul 07
1
P-value for F from summary.lm (was RE: (no subject))
[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],
2000 Jan 12
1
Usage of p/d/qnorm
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)
2003 Jul 16
1
The two chisq.test p values differ when the contingency table is transposed! (PR#3486)
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
2002 Sep 12
4
normal score transform
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
2006 May 24
0
the computation of exact p-value for the nonparametric cor-test with ties
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",
2005 Jul 11
4
exact values for p-values
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
2010 Feb 12
2
Function Fstats and p value
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,
2013 Mar 03
2
Kolmogorov-Smirnov: calculate p value given as input the test statistic
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
2012 Mar 16
2
variable values in plotmath expressions
## 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 *
2011 Aug 29
3
how to start R script editor by default
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]]
2004 Aug 06
3
Bug in qnorm or pnorm?
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))