similar to: Forcing a negative slope in linear regression?

Dear All: Could you please tell me how I can draw figure formatted like figure 4.18 of MASS (4th) with the attached data set? Thanks Zhongming Yang --------------------------------- -------------- next part -------------- An embedded and charset-unspecified text was scrubbed... Name: sample.txt Url: https://stat.ethz.ch/pipermail/r-help/attachments/20050316/abfdb85e/sample.txt

Hello: I am looking for a recommendation for a PCI-e RAID card for my server. The server has a PCI-e x16 low profile slot so the card has to be at most 6.6 inches long x 2.536 inches high. I would like to use RAID 5 with 3 drives so I have to have those capabilities. It has to be CentOS 5.4 compatible (Of course!). I took a look at the offerings from 3Ware, but their cards are too long. If

Hello, We use drc to fit dose-response curves, recently we discovered that there are quite different standard error values returned for the same dataset depending on the drc-version / R-version that was used (not clear which factor is important) On R 2.9.0 using drc_1.6-3 we get an IC50 of 1.27447 and a standard error on the IC50 of 0.43540 Whereas on R 2.7.0 using drc_1.4-2 the IC50 is

Hi all, I'm quite new to R and have a very basic question regarding how one gets the standard error of the mean for factor levels under aov. I was able to get the factor level means using: summary(print(model.tables(rawfixtimedata.aov,"means"),digits=3)), where rawfixtimedata.aov is my aov model. It doesn't appear that there is an equivalent function to get the standard

I have a data set for different time intervals. The data has three comment lines before data for each time interval. For each time interval there are 500 data points. I want to change the dataset such that I have the following format: t1 t2 t3 ................ 0.00208 0.00417 0.00625 ................. a1 a2 a3 ...................

Hi R-users, I would like to calculate weighted mean of several variables by two factors where the weight vector is the same for all variables. Below, there is a simple example where I have only two variables: "v1","v2" both weighted by "wt" and my factors are "gender" and "year". set.seed(1) df <- data.frame(gender = rep(c("M",

Sorry. Let me try again then. I am trying to find "significant" predictors" from a list of about 44 independent variables. So I started with all 44 variables and ran drop1(sep22lr, test="Chisq")... and then dropped the highest p value from the run. Then I reran the drop1. Model: MIN_Mstocked ~ ORG_CODE + BECLBL08 + PEM_SScat + SOIL_MST_1 + SOIL_NUTR + cE + cN +

This is on Solaris 10 x86, do not see this behavior on Solaris 10 sparc. Seen on multiple machines. Sshd debug: debug1: server_input_channel_open: ctype session rchan 256 win 16384 max 16384 debug1: input_session_request debug1: channel 0: new [server-session] debug2: session_new: allocate (allocated 0 max 10) debug3: session_unused: session id 0 unused debug1: session_new: session 0 debug1:

Hello, I tried to do a 'sem' analysis for data of how blueberry consumption by birds is influenced by a pollution gradient, using distance and vegetation structural and composition variables, but I got the following error message: Error in sem.default(ram = ram, S = S, N = N, param.names = pars, var.names = vars, : S must be a square triangular or symmetric matrix This may be very

Hi all: Can someone advice me on how to hold the residuals Mean sq value on a string so it can be used in other calculations. I was trying something like this: Msquare<-dfr$Mean sq but fails..Thanks dfr <- read.table(textConnection("percentQ Efficiency 1.565 0.0125 1.94 0.0213 0.876 0.003736 1.027 0.006 1.536 0.0148 1.536 0.0162 2.607 0.02 1.456 0.0157 2.16 0.0103

AF_E PS_E 90.838 2.206 83.139 1.751 134.272 3.710 84.043 2.076 105.184 2.788 157.249 3.783 50.280 1.027 96.973 2.355 123.582 3.398 60.417 1.236 123.501 3.315 90.128 1.566 193.783 5.167 116.036 2.994 100.289 2.216 56.943 1.106 102.272 2.692 145.579 3.810 53.105 1.202 127.212 3.061 102.838 2.383 126.352 2.723 13.661 0.190 164.352 4.870 159.945 4.160 54.382 0.884 128.253 3.598 181.208 4.767 145.118

I am trying to use t.test on the following data: date type INTERVAL nCASES MTF SDF MTO SDO nFST MF nOBS MO MB BIASCV BIASEV ME MAE RMSE CRCF 2001-06-15 avn GE1.00 4385 0.246 0.300 1.502 0.556 1367 1.373 4385 1.502 1.471 0.285 0.164 -1.256 1.266 1.399 0.056 2001-06-15 avn

> I will do a run with --merge-strings. This should probably the the > default to match other ELF linkers. Trying --merge-strings with today's trunk I got * comment got 77 797 bytes smaller. * rodata got 9 394 257 bytes smaller. Comparing with gold, comment now has the same size and rodata is 55 021 bytes bigger. Amusingly, merging strings seems to make lld a bit faster. With

Dear R-users, I'm studing a DB, structured like this (just a little part of my dataset): _____________________________________________________________________________________________________________ Site Latitude Longitude Year Tot-Prod Total_Density dmp Dendoudi-1 15.441964 -13.540179 2005 3271.16 1007 16993.25 Dendoudi-2 15.397321 -13.611607

Hello, my test.R file contains two huge arrays (>3000 entries), from which R needs to calculate the Pearson Correlation, if I look at the file the numbers look correct. if I run R R < test.R --no-save I see things like this: 0.723, 0.838, 1.002, 0.364, 0.357, 0.227, 0.982+ , 0.963, 0.535, 1.214, 1.270, 0.832, 1.033, 0.632, 2.482, 1.239, 0.743, 1.077, 0.962, 1.052, 1.075, 1.427, 1.395,

Hi Everyone, I just don't know how to extract the information I want from the summary of a linear regression model fitting. For example, I fit the following simple linear regression model: results = lm(y_var ~ x_var) summary(results) gives me: Call: lm(formula = y_var ~ x_var) Residuals: Min 1Q Median 3Q Max -5.9859 -1.5849 0.4574 2.0163 4.6015 Coefficients:

Hii, I will create boxplots from matrices. I have the following data sets: 5.0 1.78 2.99 2.019 0 10.0 1.79 3.00 1.744 0 15.0 1.78 2.98 1.936 0 20.0 1.78 2.99 1.975 0 25.0 1.73 2.91 3.591 0 30.0 1.79 3.00 1.966 0 35.0 1.79 3.00 2.451 0 40.0 1.79 3.00

Hi I am trying to use active record to copy some data. One of the entry I am copying is type interval (Postgres); however, it lose some precision after I copied that column. Here is what I did: @newData.elapsedtime = @oldData.elapsedtime @newData.save Result: both displayed as 00:00:02.453 in the table, however, if I use the following command to convert its value into float ( SELECT

I'm trying to solve a homework problem using R. The problem gives a list of cricket chirps per second and corresponding temperature, and asks to give the equation for the linear model and then predict the temperature to produce 18 chirps per second. So far, I have: > # Homework 11.2.1 and 11.3.3 > chirps <- scan() 1: 20 2: 16 3: 19.8 4: 18.4 5: 17.1 6: 15.5 7: 14.7 8: 17.1 9: 15.4

Dear R users, I'm trying to fit a set an ODE to an experimental time series. In the attachment you find the R code I wrote using modFit and modCost of FME package and the file of the time series. When I run summary(Fit) I obtain this error message, and the values of the parameters are equal to the initial guesses I gave to them. The problem is not due to the fact that I have only one