similar to: Maen squared deviation

R List, Please excuse this ultra-newbie post. I looked at this page but it's a bit beyond me. https://www2.kenyon.edu/Depts/Math/hartlaub/Math305%20Fall2011/R.htm I'm interested in R construct(s) to be entered at the command line that would output slope, y-intercept, and r-squared values read from a csv or other filename entered at the command line, and the same for standard deviation

Hi all, I've got clusters and would like to match individual records to each cluster based on a sum of squares deviation. For each cluster and individual, I've got 50 variables to use (measured in the same way). Matrix 1 is individuals and is 25000x50. Matrix 2 is the cluster centroids and is 100x50. The same variables are found in each matrix in the same order. I'd like to

? Thu, 22 Aug 2024 13:07:37 -0600 Keith Christian <keith1christian at gmail.com> ?????: > I'm interested in R construct(s) to be entered at the command > line that would output slope, y-intercept, and r-squared values read > from a csv or other filename entered at the command line, and the same > for standard deviation calculations, namely the standard deviation, >

Dear R colleagues, for a statistics tutorial I want to develop a nice 3d-graphic of the well known target comparison/analogy of accuracy and precision (see i.e. http://en.wikipedia.org/wiki/Accuracy_and_precision for a simple hand made 2d graphic). The code for a really beautiful graphic is already provided as demo(bivar) in the rgl package (for a picture see i.e

Greetings Listers! the R-squared value reported by summary of lm is calculated as 1 - RSS/RSS_m where RSS_m is the residual sum of squares of a minimal model. In most cases, the minimal model is simply y = mean(y), but when a constant is left out of the model, the minimal model is y = 0. However, if you manually add a constant, R still considers y = 0 the minimal model. This also causes

Greetings R Users, I have a system of equations for which I would like to output all the R-squares. Assume there are four equations in my system, the only way I found to output all the R-squares is by calling them out one by one as this: summary(fitSUR$eq[[1]])$r.squared summary(fitSUR$eq[[2]])$r.squared summary(fitSUR$eq[[3]])$r.squared summary(fitSUR$eq[[4]])$r.squared But isn't there a

Hi, Attention chi-squared distribution, unlike F distribution, has only df1 as parameter, not df1 and df2. So correct into: outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1)) ^^^^^^^^^^^^^^^^^^^^ Regards, Vito you wrote: Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F

Hi, pchisq -> distribution function dchisq -> density function pval is the area under the curve, to calculte it you use distribution function which is the integral of density function. See: http://www.itl.nist.gov/div898/handbook/eda/section3/eda362.htm http://mathworld.wolfram.com/DistributionFunction.html f(x) density function F(x) distribution function =Pr(X<x)= integral(f(x))

Hi, R seems to use the "1/n-1"-factor calculating the standard-deviation sd(). If i wat to get the "non-sample" standard-deviation i use sqrt(sd(x)^2*((n-1)/n)) Is there a parameter to get the sd()-function using the "1/n" factor directly? Or is there any other function to do so? Thank you in advance :-) Best, Eli

Hello everyone, I know that to calculate the linear regression between x and y I have to use the function "lm". But how to do if x and y have a standard deviation like in the following example? How to compute the chi square test in this case? Thank you, Fulvio. example: x +- dx 5 2 13 4 17 4 23 6 y +- dy 6.3 0.6 9.2 0.9 14 1 21 4

Hi, Standard correlations (Pearson's, Spearman's, Kendall's Tau) do not accurately reflect how closely the model (GAM) fits the data. I was told that the accuracy of the correlation can be improved using a root mean square deviation (RMSD) calculation on binned data. For example, let 'o' be the real, observed data and 'm' be the model data. I believe I can calculate

Dear UseRs, I'm running a chi-squared test where the expected matrix is the same as the observed, after rounding. R reports a X-squared of zero with a p value of one. I can justify this because any other result will deviate at least as much from the expected because what we observe is the expected, after rounding. But the formula for X-squared, sum (O-E)^2/E gives a positive value. What

Dear researchers I wish to plot a box plot without the mean line (the black line) and plot only the mean (red square). Futhermore, is it possible to add standard error and/or stadard deviation? This is an example mytest <- c(2.1,2.6,2.7,3.2,4.1,4.3,5.2,5.1,4.8,1.8,1.4,2.5,2.7,3.1,2.6,2.8) boxplot(mytest,lty = "solid") means <- mean(mytest,na.rm=TRUE) points(means, pch = 22, col

Dear R-experts! I am currently running a meta-analysis with the help of the great metafor package. However I have some difficulties setting up my raw data to enter it to the meta-analysis models. I have a group of subjects that have been measured in two continuous variables (A & B). I have the information about the mean of the two variables, the group size (n) and the standard deviations of

Hello, I'm trying to calculate a chi-squared test to see if my data are different from the theoretical distribution or not: chisq.test(rbind(c(79, 52, 69, 71, 82, 87, 95, 74, 55, 78, 49, 60),c(80,80,80, 80, 80, 80, 80, 80, 80, 80, 80, 80))) Pearson's Chi-squared test data: rbind(c(79, 52, 69, 71, 82, 87, 95, 74, 55, 78, 49, 60), c(80, 80, 80, 80, 80, 80, 80, 80, 80, 80, 80,

Hi. I'm trying to create a 3d plot for a teaching example of finding a least-squares estimate of the parameters to fit a line to some data. I was hoping to get a nice plot with a clear, single minima where the derivative of the surface is zero. No matter how much I tinker, I can't seem to get a simple straightforward plot. Am I doing something wrong? Thanks in anticipation, Ross-c x

Dear R users, This is more a statistical question rather than an R question. I'd appreciate it if you can give me some suggestions. I have a sample of a time series (sample size 500, fat tail in density). I am trying to calculate the Standard error of standard deviation of a sub-block-sample (sample size 250). I take 100 this kind of sub-block-sample, randomly. For these 100 subsamples, I

Hi, I have problem getting the standard deviation from the manova output. I have used the manova function: myfit <- manova(cbind(y1, y2) ~ x1 + x2 + x3, data=mydata) . I tried to get the predicted values and their standard deviation by using: predict(myfit, type="response", se.fit=TRUE) But the problem is that I don't get the standard deviation values, I only

I have a graph of residuals and I am attempting to get a list of the indexes of each time the residual is greater than 2 standard deviations or less than -2 standard deviations, but only the first point of the section. And then I'd also need the first point where the point returns to the range between +/- 2 standard deviations. So basically if my standard deviation=1 and my residuals=c(1, 2.1,

Hello, I am attempting to get the standard deviation in multiple distance bins in my spatial data. It appears as though the 'variog' command in the geoR package will do the trick, as one of the outputs from 'variog' is 'variog$sd', which, according to the manual, is the "standard deviation of the values in each bin". However, when I run this command, the