Displaying 20 results from an estimated 20000 matches similar to: "basic question about t-test with adjusted p value"
2010 Sep 01
3
how to represent error bar in a plot legend?
I have a simple barplot of 4 mean values, each mean value has an
associated 95% confidence interval drawn on the plot as an error bar.
I want to make a legend on the plot that uses the error bar symbol, and
explains "95% C.I."
How do I show the error bar symbol in the legend? I could not find any
"pch" values that are appropriate
Thanks
[[alternative HTML version
2010 Jun 14
2
how to make a barplot similar to Excel’s “clustered column chart”.
I have a matrix with 12 rows (one for each month), 2 columns (baseflow,
runoff). I would like to make a barplot similar to Excel’s “clustered
column chart”.
Here is my matrix ‘x’
8.258754 13.300710
10.180953 10.760465
11.012184 13.954887
10.910870 13.839839
9.023519 11.511129
7.189241 12.519830
5.925576 17.101491
5.211613 13.585175
2009 May 29
1
Problem loading rJava
I am running R on Windows XP
I first tried to install rJava and got this message
utils:::menuInstallLocal()
package 'rJava' successfully unpacked and MD5 sums checked
updating HTML package descriptions
then i tried this
require(rJava)
Loading required package: rJava
Error in if (!nchar(javahome)) stop("JAVA_HOME is not set and could not be
determined from the registry") :
2010 Apr 29
1
any feedback on XL Solutions Courses for R?
About a week ago, XL Solutions Corporation posted to this list with a
solicitation for course offerings in R. I looked at their site, and many
courses do look interesting. Can anyone provide feedback on the quality
of their courses?
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2009 May 28
1
plotting time series with data gap using type line- but do not want to connect gap with line
I have a time series of about 1500 measurements. There are sporadic data
gaps. I would like to plot the time series with type "line". I only want
a line to connect the periods with data; I don't want a line to connect
the points across a data gap. Is there a function or recommended method
to deal with this?
For example
there are 2000 days
days 1:1000 each have one
2010 Apr 19
1
How to make a boxplot with exclusion of certain groups
This seems like a simple thing, but I have been stuck for some time. My
data has 2 columns. Column 1 is the value, and column 2 is the Site where
data was collected. Column 2 contains 7 different Sites (i.e. groups). I
am only interested in showing 3 groups on a single boxplot.
I have tried various methods of subsetting the data, in order to only have
the 3 groups in my subset. However
2013 Jul 20
1
BH correction with p.adjust
Dear List,
I have been trying to use p.adjust() to do BH multiple test correction and have gotten some unexpected results. I thought that the equation for this was:
pBH = p*n/i
where p is the original p value, n is the number of tests and i is the rank of the p value. However when I try and recreate the corrected p from my most significant value it does not match up to the one computed by the
2009 Jun 17
1
problem with axis alignment when plotting 2 time series on same graph
I am trying to plot 2 time series on the same graph.
For example, X1 is the vector of dates and times, its class is POSIXt. Y1
is an environmental parameter, e.g. salinity. X2 is a second vector of
dates and times, also of class POSIXt. X2 has a different length than
X1, but they have the exact same range. Y2 is another environmental
parameter, e.g. stream flow rate.
My approach is
2010 Dec 16
2
moving average with gaps in time series
I have a time series with interval of 2.5 minutes, or 24 observations per
hour. I am trying to find a 1 hr moving average, looking backward, so
that moving average at n = mean(n-23 : n)
The time series has about 1.5 million rows, with occasional gaps due to
poor data quality. I only want to take a 1 hour moving average for those
periods that are complete, i.e. have 24 observations in the
2010 Aug 08
1
p.adjust( , fdr)
Hello,
I am not sure about the p.adjust( , fdr). How do these adjusted p-values
get?
I have read papers of BH method. For independent case, we compare the
ordered p-values with the alfa*i/m, where m is the number of tests. But I
have checked that result based on the adjusted p-values is different with
that by using the independent case method.
Then how do the result of p.adjust( , fdr) come?
And
2005 Jul 14
2
Partek has Dunn-Sidak Multiple Test Correction. Is this the same/similar to any of R's p.adjust.methods?
The Partek package (www.partek.com) allows only two selections for Multiple
Test Correction: Bonferroni and Dunn-Sidak. Can anyone suggest why Partek
implemented Dunn-Sidak and not the other methods that R has? Is there any
particular advantage to the Dunn-Sidak method?
R knows about these methods (in R 2.1.1):
> p.adjust.methods
[1] "holm" "hochberg" "hommel"
2005 Jan 16
1
p.adjust(<NA>s), was "Re: [BioC] limma and p-values"
I append below a suggested update for p.adjust().
1. A new method "yh" for control of FDR is included which is valid for any
dependency structure. Reference is Benjamini, Y., and Yekutieli, D. (2001).
The control of the false discovery rate in multiple testing under
dependency. Annals of Statistics 29, 1165-1188.
2. I've re-named the "fdr" method to "bh" but
2010 Feb 07
1
p.adjust.Rd sugggestion
L.S.
In the current version of ?p.adjust.Rd, one needs
to scroll down to the examples section to find
confirmation of one's guess that "fdr" is an
alias of "BH".
Please find a patch in attachment which mentions
this explicitly.
Best,
Tobias
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2008 Mar 09
2
p-adjust using Benjamn and Hochberg
Hello,
I am trying to use the p.adjust function for multiple testing.
here is what i have
9997 201674_s_at 0.327547396
9998 221013_s_at 0.834211067
9999 221685_s_at 0.185099475
I import them from excel have have the gene symbol as well as the pvalue
here is the issue
> pa<-p.adjust(pt,method="BH")
Error in p[nna] : object is not
2007 Feb 28
2
topTable function from LIMMA
Dear R-Help,
I am using the function "topTable" from the LIMMA package. To estimate
adjusted P-values there are several options (adjust="fdr" , adjust="BH")
as shown below:
topTable(fit, number = 10, adjust = "BH", fit$Name)
I guess any of these options (fdr, BH, etc.) is using a default of
FDR=0.05 which is quite conservative (i.e., very
2018 Mar 15
1
Adjusting OHCL data via quantmod
Hello,
I'm trying to do two things:
-1. Ensure that I understand how quantmod adjust's OHLC data
-2. Determine how I ought to adjust my data.
My overarching-goal is to adjust my OHLC data appropriately to minimize the
difference between my backtest returns, and the returns I would get if I
was trading for real (which I'll be doing shortly).
Background:
-1. I'm using Alpha
2009 Mar 18
0
p.adjust(p, n) for n>length(p)
Hi all,
I am having a problem with the function "p.adjust" in stats. I have looked at the manuals and searched the R site, but didn't get anything that seems directly relevant. Can anybody throw any light on it or confirm my suspicion that this might be a bug?
I am trying to use the p.adjust() function to do Benjamini/Hochberg FDR control on a vector of p-values that are the
2004 Dec 20
1
[BioC] limma, FDR, and p.adjust
You asked the same question on the Bioconductor mailing list back in August. At that time, you
suggested yourself a solution for how the adjusted p-values should be interpreted. I answered
your query and told you that your interpretation was correct. So I'm not sure what more can be
said, except that you should read the article Wright (1992), which is cited in the help entry for
p.adjust(),
2005 May 15
3
adjusted p-values with TukeyHSD?
hi list,
i have to ask you again, having tried and searched for several days...
i want to do a TukeyHSD after an Anova, and want to get the adjusted
p-values after the Tukey Correction.
i found the p.adjust function, but it can only correct for "holm",
"hochberg", bonferroni", but not "Tukey".
Is it not possbile to get adjusted p-values after
2008 Jun 09
1
Student Distribution and Funtion qt
Hello,
I am trying to calculate and plot mean and confidence intervall for a set of data. This is the code that I am currently using:
means <- sapply(data, mean, na.rm=TRUE)
n <- sapply(data,length)
stdev <- sqrt(sapply(data, var, na.rm=TRUE))
ciw <- qt(0.98, n) * stdev / sqrt(n)
par(mgp=c(2,0.6,0), las=2, fin=c(7,3), mai=c(1,0.5,0.2,0.2), cex=0.8)
plotCI(x=means, uiw=ciw,