similar to: R for simple stats

Hello everyone, Thanks to all who offered suggestions after my initial query to the list. I've been busy trying to absorb the docs as fast as possible, but this little thing has me stumped for the moment. I'm creating a random sample of 100 item and saving it to a vector named 'iq'. I'd like to create a second vector (iq_prime, say) by adding 20 to each element of the

Hello everyone, Thanks to all who offered suggestions after my initial query to the list. I've been busy trying to absorb the docs as fast as possible, but this little thing has me stumped for the moment. I'm creating a random sample of 100 item and saving it to a vector named 'iq'. I'd like to create a second vector (iq_prime, say) by adding 20 to each element of the

Hi everyone, I've got a dataframe with columns of different types. A certain number of columns in the dataframe hold the results of a series of Likert-type items. I've got a function that will print a simple table of frequencies and I want to apply that function to those columns of the dataframe only. What's the best approach? -Tim -- Tim Wilson | Visit Sibley online: |

Hi everyone, I wonder if there's a patient soul out there who has a minute to look at the following. I've got a set of summary statistics I need to perform many times. Naturally, I've looked at writing a function to automate the process as much as possible. (These are the data I mentioned recently in my question about weighted means.) I'm having trouble figuring out the proper

Hi everyone, I've got a dataframe called 'faculty'. I want to do a weighted mean on the column called 'Q8' weighted by the contents of column 'CETP'. In addition, I need to operate on the result of splitting 'faculty' according to the contents of a column 'FACULTY'. For example > lapply(split(faculty$Q8, faculty$FACULTY), mean) $"1" [1]

Hi everyone, I want to use the Fisher r-to-z transformation as part of a hypothesis test of r. I can't find an R function that can do that. Am I missing it? -Tim -- Tim Wilson | Visit Sibley online: | Check out: Henry Sibley HS | http://www.isd197.org | http://www.zope.com W. St. Paul, MN | | http://slashdot.org wilson at visi.com | <dtml-var

Hi everyone, I've got a problem with an analysis of variance where it appears that my independent variable is being treated as an integer when it should be nominal. The data are being loaded from an SPSS file and the independent variable 'YearColl' corresponds (surprisingly enough) to a student's year in college. :-) The integers 1-4 are used in the dataset for this purpose. My

Hi everyone, I've downloaded the lmtest package, but I'm having trouble building it. Here's the output: copland:/home/wilson/tmp# R CMD INSTALL -l /usr/lib/R/library lmtest_0.9-0.tar.gz * Installing *source* package 'lmtest' ... ** libs g77 -fPIC -g -O2 -c pan.f -o pan.o gcc -shared -o lmtest.so pan.o -L/usr/lib/gcc-lib/i386-linux/2.95.4 -lreadline -ldl -lncurses

Hi everyone, I'm having trouble figuring out how to split a dataframe more than once. Let's say I have a dataframe d with a certain column called splitcol composed of four possible ordinal values. The same dataframe has two other columns, col1 and col2, that have one of two possible values each. I'd like to split d$splitcol based on col1 and col2 so I can report frequencies of the

Software information: R Version 1.5.1 (2002-06-17) ESS 5.1.20 Emacs 21.1.1 Mozilla 1.0 Linux Mandrake 8.2 Colleagues I am having a little trouble installing packages. Having downloaded the gzipped tarball of a package, put it in a /tmp directory, and using this command: [root at localhost /]# R CMD INSTALL -l 'usr/lib/R/library/' '/home/smc/tmp/sgeostat_1.0-18.tar.gz' I

try ?power.prop.test > -----Messaggio originale----- > Da: Tim Wilson [mailto:wilson at visi.com] > Inviato: sabato 6 luglio 2002 6.05 > A: R-help > Oggetto: [R] one-sample binomial test > > > Hi everyone, > > Here's how I solved a problem for my stats class. I'm pretty sure I > understand what's going on, but I wonder if there's a more >

Hi everyone, Here's a question about extracting data from a dataframe: Let's say I've got a dataframe with two vectors, TEST and GENDER. GENDER contains a 1 for males and 2 for females. I want to separate the results of TEST by GENDER so I can compare their means. What's the most efficient way to do this with R? -Tim -- Tim Wilson | Visit Sibley online: | Check out:

Dear R People I have a function that I wrote. It's not part of a library, it's just a nice function. However, I would like to have a "Help" page for it. How do I go about doing that please? Thanks so much in advance! R version 1.5.1 for Windows. Sincerely, Erin -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

Hi everyone, Here's how I solved a problem for my stats class. I'm pretty sure I understand what's going on, but I wonder if there's a more direct way to solve it. Problem summary: A recent poll indicated that Candidate A is leading B with 55% of the vote. How many voters need to be surveyed to ensure a margin of error of +/- 2.5% with 99% confidence. Here's what I did:

Dear list users, I have trouble with covariance analysis. I measured nitrate concentrations in the soil (NO3) and the percentage of legumes (LEG, continuous), affected by 2 different CO2 concentrations (CO2, discrete). I suspect that CO2 has an effect on LEG and NO3, but also that LEG has an effect on NO3, so this is the formula I wrote to test this: NO3 ~ CO2 + LEG + CO2:LEG Will LEG be

I wonder whether the functions for skewness and kurtosis in the e1071 package are based on correct formulas. The functions in the package e1071 are: # -------------------------------------------- skewness <- function (x, na.rm = FALSE) { if (na.rm) x <- x[!is.na(x)] sum((x - mean(x))^3)/(length(x) * sd(x)^3) } # -------------------------------------------- and #

Hello everyone, Does anyone know what would cause the skewness() function (from e1071), as well as skew() from psych, to return a value of NaN? I have a vector of positively-skewed data (https://docs.google.com/file/d/0B6-m45Jvl3ZmYzlHRVRHRURzbVk/edit?usp=sharing) which these functions return a value for like normal: > skewness( data ) # returns 1.400405 but when I instead give those

Greetings Everybody: I generated a 1.2MB dta file based on the general social survey with Stata8 for linux. The file can be re-opened with Stata, but when I bring it into R, it says all the values are missing for most of the variables. This dataset is called "morgen.dta" and I dropped a copy online in case you are interested http://www.ku.edu/~pauljohn/R/morgen.dta looks like this

Dear R users, I hope to explain the concepts of skewness and kurtosis by generating series of distributions with same skewness and different kurtosis or with same kurtosis and different skewness, but it seems that i cannot find the right functions. I have searched the mailing list, but no answers were found. Is it possible to do that in R? Which function could be used? Thanks a lot. --

Hi.. i have an expression of the form: model1<-nls(y~beta1*(x1+(k1*x2)+(k1*k1*x3)+(k2*x4)+(k2*k1*x5)+(k2*k2*x6)+(k3*x7)+(k3*k4*x8)+(k3*k2*x9)+(k3*k3*x10)+ (k4*x11)+(k4*k1*x12)+(k4*k2*x13)+(k4*k3*x14)+(k4*k4*x15)+(k5*x16)+(k5*k1*x17)+(k5*k2*x18)+(k5*k3*x19)+