similar to: Discourage the weights= option of lm with summarized data

Yes. Thank you; I should have quoted it. I suggest to remove this text or to add the word "not" at the beginning. Arie On Sun, Oct 8, 2017 at 4:38 PM, Viechtbauer Wolfgang (SP) <wolfgang.viechtbauer at maastrichtuniversity.nl> wrote: > Ah, I think you are referring to this part from ?lm: > > "(including the case that there are w_i observations equal to y_i and

Indeed: Using 'weights' is not meant to indicate that the same observation is repeated 'n' times. As I showed, this gives erroneous results. Hence I suggested that it is discouraged rather than encouraged in the Details section of lm in the Reference manual. Arie ---Original Message----- On Sat, 7 Oct 2017, wolfgang.viechtbauer at maastrichtuniversity.nl wrote: Using

OK. We have now three suggestions to repair the text: - remove the text - add "not" at the beginning of the text - add at the end of the text a warning; something like: "Note that in this case the standard estimates of the parameters are in general not correct, and hence also the t values and the p value. Also the number of degrees of freedom is not correct. (The parameter

Peter, This is a highly structured text. Just for the discussion, I separate the building blocks, where (D) and (E) and (F) are new: BEGIN OF TEXT -------------------- (A) Non-?NULL? ?weights? can be used to indicate that different observations have different variances (with the values in ?weights? being inversely proportional to the variances); (B) or equivalently, when the elements of

AFAIR, it is a little more subtle than that. If you have replication weights, then the estimates are right, it is "just" that the SE from summary.lm() are wrong. Somehow, the text should reflect this. It is of some importance when you put glm() into the mix, because you can in fact get correct results from things like y <- c(0,1) w <- c(49,51) glm(y~1, weights=w,

My local R-devel version now has (in ?lm) Non-?NULL? ?weights? can be used to indicate that different observations have different variances (with the values in ?weights? being inversely proportional to the variances); or equivalently, when the elements of ?weights? are positive integers w_i, that each response y_i is the mean of w_i unit-weight observations

Since the three posters agree (only) that there is a bug, I propose to file it as a bug, which is the least we can do now. There is more to it: the only other case of a change in the Reference Manual which I know of, is also about the weights option! This is in coxph. The Reference Manual version 3.0.0 (2013) says about coxph: " ... If weights is a vector of integers, then the estimated

Ah, I think you are referring to this part from ?lm: "(including the case that there are w_i observations equal to y_i and the data have been summarized)" I see; indeed, I don't think this is what 'weights' should be used for (the other part before that is correct). Sorry, I misunderstood the point you were trying to make. Best, Wolfgang -----Original Message----- From:

The control flow was: if (!y) { ppix = ... } if (y) { ... } else if (x) { use ppix for something } else { use ppix for something } Merge the if(!y) block with the two else branches. This avoids a false-positive in the clang static analyzer, it can't know that !y and x are mutually exclusive. The result looks something like this: if (y) { ... } else { ppix = ... if (x) {

Hi everyone, ? I am trying to get a point of intersection between a polyline and a straight line ?.. and get the x and y coordinates of this point. For exemplification consider this: ? ? set.seed(123) ? k1 <-rnorm(100, mean=1.77, sd=3.33) ?k1 <- sort(k1) q1 <- rnorm(100, mean=2.37, sd=0.74) q1 <- sort(q1, decreasing = TRUE) plot(k1, q1, xlim <- c((min(k1)-5),

Dear R-List, I've written some code to put measurement values at a position x and y in bins (xb and yb). It works, but I wonder if there isn't a function that would do what I do by hand in "# fill data in bins"? Here is the code: # data x <- c( 1.1, 1.5, 2.3, 2.5, 2.6, 2.9, 3.3, 3.5 ) y <- c( 6.3, 6.2, 5.9, 5.3, 5.4, 4.2, 4.8, 4.6 ) val <- c( 50, 58, 32, 14, 12,

Hi, My english isn't brilliant and my problem is very difficult to describe but I try ;) My first question is: May I write loop "for" like this or similar - for (i in sth : sth[length(sth)], k in sth_else : length(sth_else) ) - I'd like to have two independent conditions in the same loop "for". My secound question depend on program below. I'd like to write every

Hello~   Did any one have used the package 'orcutt' ?   I find that it can not work smoothly in a single variable regression. I use the example following, it function very well.   But when I regress "cons" on "price" (use the "reg1<-lm(cons~price+income+temp)") , then  use "reg11<-cochrane.orcutt(reg1) ". There is an error message “Error in

Dear All, I'm trying to track down a very erratic bug in some fortran; I have an example that quite consistently segfaults on windoz, and more sporadically on mac, all in the course of doing some bootstrap calculations, varying the set.seed call, but I'm now trying to see what is going on on our redhat system: R version 2.15.1 (2012-06-22) -- "Roasted Marshmallows"

Hi Sherri, There are examples of topographic maps which you have been pointed to, however, I suspect that you want to know where you can obtain topographic data from rather than a canned example. There are quite a few intricacies to the process so I will go through them for you. (1) Topography files can be found in the geomapdata library. You will probably want to use the maps package too (if

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Le 31/05/2017 ? 17:30, Serguei Sokol a ?crit : > > More thorough reading revealed that I have overlooked this phrase in the > line's doc: "left and right /thirds/ of the data" (emphasis is mine). Oops. I have read the first ref returned by google and it happened to be tibco's doc, not the R's one. The layout is very similar hence my mistake. The latter does not

I'm new to R, after many years of using Matlab. I've found the R function 'polygon' to be nearly equivalent to the Matlab function 'patch'. For example, the R commands: plot(c(0, 5), c(0, 4), type = 'n', asp = 1, ann = FALSE) x <- c(1, 2, 2, 1.5, 1) z <- c(1, 1, 2, 1.7, 2) polygon(x, z, col = 'green') produce a plot with a small green shape exactly

is there no package/function in R to calculate the conditional power or the bayesian predictive power for trials with binary endpoints? Thanks -- View this message in context: http://www.nabble.com/Conditional-power%2C-predictive-power-tf3603396.html#a10066991 Sent from the R help mailing list archive at Nabble.com.

Seriously, if a method gives a wrong result, it's wrong. line() does NOT implement the algorithm of Tukey, even not after the patch. We're not discussing Excel here, are we? The method of Tukey is rather clear, and it is NOT using the default quantile definition from the quantile function. Actually, it doesn't even use quantiles to define the groups. It just says that the groups