similar to: Minitab Parametric Distribution Analysis in R

I'm having trouble using the read.mtp function in the foreign package to import datasets from minitab (.mtw) format. Specifically, each file I try to import fails to load any data beyond a header row stating the version of Minitab that saved the dataset. I get this error: incomplete final line found by readtableHeader on 'income.mtw' The dataset appears to be complete (ie, using

Hi, I ran a small data set from a factorial experiment through R, Minitab and JMP... the result from R is significantly different from what Minitab or JMP give... The data set is at the following link: http://www.personal.psu.edu/nug107/Uploads/2x3_16repsANOVA.txt The first 5 columns are the factors and the next three are responses. In particular, for the response beta11MSE, two of the

Hi all. I got a question about eigenvector. I've tried input a symmetric matrix to both R (using eigen function) and minitab, but the result is really different. Can anyone tell me what's wrong with that? Thanks. [[alternative HTML version deleted]]

hi folks, Bit of a newbie, but I've spent a fair bit of time looking for an answer on this, with no joy. Can anyone help me? Dataset: A single column of values in a csv file (eg. 52, 53, 54, 85, etc) Goal: In Minitab, you have what they call a dot plot. It's a histogram, where a single dot represents a set of identical values (eg. 57, 57, 57 would be one dot). Multiple dots are

Dear R People: I ran an ARIMA(1,0,1) on a particular series in R and got a negative MA(1) estimate. Then I ran an ARIMA(1,0,1) on the same series in Minitab and got a positive MA(1) estimate. The values are about -0.69 and 0.70. Does R show the opposite value, please? Thanks, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston -

Hello, I'm quite new to R and want to make a Weibull-regression with the survival package. I know how to build my "Surv"-object and how to make a standard-weibull regression with "survreg". However, I want to fit a translated or 3-parametric weibull dist to account for a failure-free time. I think I would need a new object in survreg.distributions, but I don't know how

Greetings -- I would like to estimate a proportional hazard model with a weibull or lognormal baseline. I have looked at both the coxph() and survreg() functions and neither appear (to me ) to do it. Am I missing something in the docs or is there another terrific package out there that will do this. Many Thanks. Carl Mason

Hola, LLevo buscando desde hace tiempo como hacer el Goodness of fit test en R. Es decir, me explico, intento hacer una cosa parecida que se hace en Minitab, por ejemplo, yo tengo un conjunto de datos, y lo que quiero es sabes que tipo de distibución es, en minitab se hace un histograma para ver si se ajusta bien o no a la campana de Gauss, luego vemos si aproximar la distribución de la muestra

Greetings, For the following quarterly data I did a classical decomposition by hand in a spreadsheet and got reasonably similar results using Minitab 15. x 1 36 2 44 3 45 4 106 5 38 6 46 7 47 8 112 9 42 10 49 11 48 12 118 13 42 14 50 15

hi, I am trying to get minitab demo work with wine but can't:-( here's what happened so far: installation worked, the program seems to be installed properly, on windows partition (wine c:\Program Files directory) here's what happens when I run it: jojda:~>wine /mnt/win/Program\ Files/MTBDEMO/Mtb13GE.exe FIXME:pthread_rwlock_rdlock FIXME:pthread_rwlock_unlock

Hello, Given a probability function: p(x) = 12 / (25(x+1)) , x=0, 1, 2, 3 we generate the following values: C1 C2 0 0.48 1 0.24 2 0.16 3 0.12 Now, I'm supposed to create 50 random values using this table. In MiniTab, I simply selected Calc -> Random Data -> Discrete, and selected the columns, and it created 50 random values in a new column.[1] How do I do the

 Please, Help me, How do I generate data from the weibull distribution if the data contain both failure and interval censored, For example, I want to generate n=100, shape=2 and scale =4 with 30% interval censored.  What about right censoring Thank you  [[alternative HTML version deleted]]

Hi all, I'm trying to generate a Weibull distribution including four covariates in the model. Here is the code I used: T = rweibull(200, shape=1.3, scale=0.004*exp(-(-2.5*b1+2.5*b2+0.9*x1-1.3*x2)/1.3)) C = rweibull(n, shape=1.5, scale=0.008) #censoring time time = pmin(T,C) #observed time is min of censored and true event = time==T # set to 1 if event is observed

Dear All, I would like to fit some parametric survival models using left truncated, right censored data in R. However I am having problems finding a function to fit parametric survival models which can handle left truncated data. I have tested both the survreg function in package survival: fit1 <- survreg(Surv(start, stop, status) ~ X + Y + Z, data=data1) and the psm function in package

Hi there, can somebody give me a guide as to how to generate data from weibull distribution with censoring for example, the code below generates only failure data, what do i add to get the censored data, either right or interval censoring q<-rweibull(100,2,10). Thank you Grace Kam student, University of Ghana [[alternative HTML version deleted]]

The function wtd.var(x,w) in Hmisc calculates the weighted variance of x where w are the weights. It appears to me that wtd.var(x,w) = var(x) if all of the weights are equal, but this does not appear to be the case. Can someone point out to me where I am going wrong here? Thanks. Tom La Bone [[alternative HTML version deleted]]

Hello, I have interval censored data, censored between (0, 100). I used the tobit function in the AER package which in turn backs on survreg. Actually I'm struggling with the distribution. Data is asymmetrically distributed, so first choice would be a Weibull distribution. Unfortunately the Weibull doesn't allow for zero values in time data, as it requires x > 0. So I tried the

The R code to reproduce the solutions to the example problems in Design of Experiments with MINITAB by Paul Mathews (ASQ Quality Press, 2005) is now posted at http://www.mmbstatistical.com/DOEwithMINITAB.html. Enjoy! Paul Mathews -- Paul Mathews Mathews Malnar and Bailey, Inc. 217 Third Street, Fairport Harbor, OH 44077 Phone: 440-350-0911 Fax: 440-350-7210 E-mail: Paul: pmathews at

Hi I am performing an analysis of variance with two factors, each with two levels. I have differing numbers of observations in each of the four combinations, but all four combinations *are* present (2 of the factor combinations have 3 observations, 1 has 4 and 1 has 5) I have used both anova(aov(...)) and anova(lm(...)) in R and it gave the same result - as expected. I then plugged this into

Dear All, Would like to ask the inconsistency in the autocorrelation from R with SPSS/Minitab. I have tried a dataset x with 20 data (1-20) and ask R to give the autocorrelation of different lags using the command < acf(x, lag.max=100, type = "correlation"), However while SPSS and Minitab give the same answers (0.85 for lag1), R gives 0.3688 which is much smaller. Obviously, the