Displaying 20 results from an estimated 5000 matches similar to: "output Shapiro-Wild results to a table"
2012 Apr 04
1
Shapiro-Wilk cpoefficients: 2 Qs
Greetings!
I want to have the coefficients that R uses in shapiro.test()
for the Shapiro-Wilk test for a prticular sample size, i.e.
the a[i] in
W = Sum(a[i]*x[i])/(Sum(x[i] - mean(x))^2)
(where the x[i] are sorted). Two questions:
Q1:
Is there a readymade R function from which I can extract these?
Q2:
I was wondering if I might be able to modify the code for the
function shapiro.test() so
2000 Oct 07
2
Possible bug in apply()
In the course of applying Shapiro-Wilk to 100,000 samples of 60 items
from 100,000 different distributions, I encountered a fatal error in
apply(). This can be reconstructed as follows, using the attached data
file distr.dat containing 2 lines of my original 100,000-line file:
> version
_
platform Windows
arch x86
os Win32
system x86, Win32
status
2007 Jun 29
1
Shapiro Test P Value Incorrect? (PR#9768)
Full_Name: Jason Polak
Version: R version 2.5.0 (2007-04-23)
OS: Xubuntu 7.04
Submission from: (NULL) (137.122.144.35)
Dear R group,
I have noticed a strange anomaly with the shapiro.test() function. Unfortunately
I do not know how to calculate the shapiro test P values manually so I don't
know if this is an actual bug.
So, to produce the results, run the following code:
pvalues = 0;
for
2006 Jul 12
2
shapiro.test() output
R Users:
My question is probably more about elementary statistics than the
mechanics of using R, but I've been dabbling in R (version 2.2.0) and
used it recently to test some data .
I have a relatively small set of observations (n = 12) of arsenic
concentrations in background groundwater and wanted to test my
assumption of normality. I used the Shapiro-Wilk test (by calling
shapiro.test()
2005 Nov 09
1
Problems with Shapiro Wilk's test of normality.
Hi,
I am trying to create a table with information from Shapiro Wilk's
test of normality.
However, it fails due to lack of sample size, it says, but the way I
see it, this is not a problem.
(See the table of sample sizes (almost) at the bottom).
Applying a different function using a similar ftable call is not a
problem (See the bottom table).
This is R 2.1.0 on Linux (Gentoo).
/Fredrik
2009 Aug 02
2
Strange column shifting with read.table
Hi,
I am reading in a dataframe from a CSV file. It has 70 columns. I do
not have any kind of unique "row id".
rawdata <- read.table("r_work/train_data.csv", header=T, sep=",",
na.strings=0)
When training an svm, I keep getting an error
So, as an experiment, I wrote the data back out to a new file so that I
could see what the svm function sees.
2001 Jul 02
2
Shapiro-Wilk test
Hi,
does the shapiro wilk test in R-1.3.0 work correctly? Maybe it does, but can
anybody tell me why the following sample doesn't give "W = 1" and
"p-value = 1":
R> x<-1:9/10;x
[1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
R> shapiro.test(qnorm(x))
Shapiro-Wilk normality test
data: qnorm(x)
W = 0.9925, p-value = 0.9986
I can't imagine a sample being
2010 May 26
3
shapiro.test
Hi,
I am not so sure about an error note I got when using shapiro.test.
I imported some data into R by wrinting it into a .txt file via
> tab1<-read.table("etc....txt",header=T)
> attach(tab1)
The following object(s) are masked _by_ .GlobalEnv :
ozon
> ozon$V1 [1] 2.5 3.0 5.6 4.7 6.5 6.7 1.7 5.3 4.6 7.4 5.4 4.1 5.1 5.6 5.4 6.1 7.6[18] 6.2 6.0 5.5 5.8 8.2 3.1 5.8 2.6
Now
2005 Nov 18
1
help on ks.test and shapiro.test
I have three files of data which are available at http://zhangw.com/
R/, varied at the number of data. I tried to use R to analyze using
shapiro.test, ks.test, and t.test. t.test ran as expected, however,
when I run shapiro.test and ks.test commands, error message always
occurred. Error message for
shapiro.test is
"Error in "[.data.frame"(x, complete.cases(x)) :
undefined
2024 Sep 02
2
Adding parameters for Benchmark normal distribution in shapiro.test
Hi,
In ?shapiro.test, there seems to be no option to pass mean and sd
information of the Normal distribution which I want to compare my
sample data to.
For example in the code below, I want to test my sample to N(0, 10).
shapiro.test(rnorm(100, mean = 5, sd = 3))
Is there any way to pass the information of the benchmark normal distribution?
2013 Apr 05
2
How to perform a grouped shapiro wilk test on dataframe
Hello,
I was wandering if it is possible to perform on a dataframe called 'all' a
shapiro wilk normality test for COUNTS by variable Group
ACTIVITY? Could it be done using plyer? I saw an eg that applies to an
array but not to a dataframe:
lapply(split(dataset1$Height,dataset1$Group),shapiro.test)
Any thoughts would be much appreciated.
My dataframe is in shape:
dat ACTIVIT
2008 Nov 21
1
question about shapiro.test()
Hi all!
I tried to perform Shapiro-Wilk test for my sample of 243 values.
> Us
[1] -10.4 -13.1 -12.2 38.1 -18.8 -13.3 -11.7 29.3 49.7 6.8 12.7 16.3
[13] 5.8 -0.7 -29.4 4.1 38.8 -1.4 8.8 15.6 32.9 -5.3 19.1 35.8
[25] 4.0 -1.5 0.6 -4.2 -10.0 -4.0 1.1 48.9 -21.0 -5.3 5.8 -10.8
[37] 21.9 8.2 -3.2 -3.9 -2.3 12.6 -4.7 -8.0 11.8 27.4 -9.5 -20.8
[49]
2009 Feb 06
1
beanplot, Error in shapiro.test(x)
Dear all,
I am trying to create beanplots from a dataset for which boxplot works fine.
(MACOS, R 2.8.1 GUI 1.27 Tiger build 32-bit (5301))
I am getting the following error message:
Error in shapiro.test(x) : sample size must be between 3 and 5000
I am not even sure why the shapiro.test is being used, but is there any
workaround ?
Thanks !
Markus
[[alternative HTML version deleted]]
2000 Sep 25
1
Interpretation of Shapiro-Wilk
Can anybody tell me the exact meaning of the $statistic and $p.value
calculated by shapiro.test? Unfortunately it is not covered in my few
text books, and I cannot find the explanation in the R documentatiom or
on-line.
If I have a test statistic, T, which is Normally distributed with mean=m
and sd=s under the null hypothesis, then I can convert T to a p-value
(one-sided) using:
p <- pnorm(T,
2007 Sep 29
1
Shapiro-Welch W value interpretation
Hello,
I have tested a distribution for normality using the Shapiro-Welch
statistic. The result of this is the following:
Shapiro-Wilk normality test
data: mydata
W = 0.9989, p-value = 0.8791
I know that the p-value > 0.05 (for my purposes) means that the data
IS normally distributed but what I am not sure is with the W value,
what values tell me that the data is normally
2008 Jul 12
5
shapiro wilk normality test
Hi everybody,
somehow i dont get the shapiro wilk test for normality. i just can?t
find what the H0 is .
i tried :
shapiro.test(rnorm(5000))
Shapiro-Wilk normality test
data: rnorm(5000)
W = 0.9997, p-value = 0.6205
If normality is the H0, the test says it?s probably not normal, doesn
?t it ?
5000 is the biggest n allowed by the test...
are there any other test ? ( i know qqnorm
2012 Aug 09
1
POSIXct to ts
Hi,
I have a dataframe (try.1) with date/time and temperature columns, and the date/time is in POSIXct fomat. Sample included below.
I would like to to try decompose () or stl() to look at the trends and seasonality in my data, eventually so that I can look at autocorrelation. The series is 3 years of water temperature with clearly visible seasonal periods.
Right now, if I try decompose,
1997 May 08
1
R-beta: Two simple suggestions.
1) The .RData file (and S .Data area) are cputype dependent. When R is used
from different types of machines (say, HP and Sun) that share a common
filesystem (say afs or dfs) then loading the wrong type of .RData file can
lead to mysterious behaviour. In my environment, undergraduate students are
the most likely to log into different cputypes, and they end up being quite
confused. To
2003 Feb 10
2
shapiro.test
Hi
The shapiro.test function outputs a value of the W statistic, which
should be 1 if the distribution is normal, and a p-value for the test
(as the documentation states).
I'm a bit confused with some results. I'm getting a W=0.9977 and a
p-value=0.1889.
I was expecting that a W of 0.9977 would tell me that the distribution
is normal so p-value should be small ...
What am I missing ?
2011 Sep 30
3
error while using shapiro.test()
hey all, I'm just getting used to R and i'm having issues when it comes to
reading my data in rows rather than columns. any good advice would be much
appreciated !
here is the error:
> data1 <- read.table(file.choose(),header=T)
> x1 <- c(data1[1,1:5])
> shapiro.test(x1)
Error in sort.int(x, na.last = na.last, decreasing = decreasing, ...) :
'x' must be atomic