search for: beta_0

Hi, I'm trying to get maximum likelihood estimates of \alpha, \beta_0 and \beta_1, this can be achieved by solving the following three equations: n / \alpha + \sum\limits_{i=1}^{n} ln(\psihat(i)) - \sum\limits_{i=1}^{n} ( ln(x_i + \psihat(i)) ) = 0 \alpha \sum\limits_{i=1}^{n} 1/(psihat(i)) - (\alpha+1) \sum\limits_{i=1}^{n} ( 1 / (x_i + \psihat(i)) ) = 0 \alph...

...as I cannot turn up a solution anywhere else. I will frame it in the context of an R related issue. To illustrate the problem, consider student achievement test score data with multiple observations available for each student. One way of modeling these data might be Y_{ti} = (\mu + \mu_{i} ) + (\beta_0 + \beta_{i} )*(time) + \epsilon_{ti} ; t indexes time and i indexes student The nlme code is tt<-lme(reponse~time, data, random=~time|ID) With this, I can extract the growth rate for each individual in the data set. Conceptually this is the sum of the main effect for time plus the empirical...

In the logistic regression model, there is no residual log (pi/(1-pi)) = beta_0 + beta_1*X_1 + ..... But glm model will return residuals What is that? How to understand this? Can we put some residual in the logistic regression model by replacing pi with pi' (the estimated pi)? log (pi'/(1-pi')) = beta_0 + beta_1*X_1 + .....+ ei Thanks! [[alternative HTML v...

Hola, Tengo una base de datos con estructura jerárquica en la que quiero clasificar observaciones en distintas categorías. En el caso más simple, tengo una variable con dos categorías (variable Y1) y dentro de cada una de ellas hay otras dos categorías (variable Y2). Además tengo una variable explicativa cuantitativa discreta X. El banco de datos sería de este tipo: X Y1 Y2 5 0 1 9 0 0 2

Hi Listers, seems I have forgotten some basics re offset in glm: data: counts (y) from locations off different size (area), explanatory variable: x Model: y ~ x+offset(area) Predictions (pred) using Poisson errors plot(x,y) and points(x,pred) gives neat "line" of estimated values. However, for ease of understanding graphs are better using plot(x,y/area). Question: How to display

...ails on this: > str(model.mat) num [1:18, 1:4] -0.170 -0.304 -2.617 2.025 -1.610 ... - attr(*, "dimnames")=List of 2 ..$ : chr [1:18] "Amount: CP for VF" "Amount: CP for F" "Amount: CP for M" "Amount: RD for VF" ... ..$ : chr [1:4] "beta_0" "P-value" "beta_1" "P-value" It contains the following: > model.mat[1:2,] beta_0 P-value beta_1 P-value Amount: CP for VF -0.1702877 0.7716153 0.5148616 2.020602e-03 Amount: CP for F -0.3042522 0.1966133 0.8795982 6.058299e-...

Hi all Assume I have a data set xx; Group: 1=group1 ?, 2=group2 IQ: ?1= High, 0 =low fit <- glm(IQ ~group, data = xx, family = binomial()) summary(fit) Results ?????? ????????????Estimate Std. Error z value Pr(>|z|) (Intercept) -2.55456??? 0.210 -12.273? < 5e-16 *** group????????? 0.36180 ?????0.076?? 3.952 ????5.24e-05 *** the odd ratio = exp(0.36180 )= 1.435912 My question

...wonder how to relate the estimated parameters to the Weibull regression model. Here is the model specification from Harrel, Regression Modeling Strategies, p. 422: S(t|X)=exp[-\alpha*t^\gamma exp(X\beta)] This is the model without intercept, and it is indicated that \alpha can be replaced by exp(\beta_0) in the model with intercept. Now I am puzzled by the fact that \alpha (or \alpha exp(X\beta)) is usually referred to as "scale parameter" in the context of the Weibull distribution. If this would be the case, I could get the Weibull scale from the \beta/\beta_0 estimators and I would ne...

Dear R users, I am using the function "constrOptim" to minimize the -1*log-likelihood where \beta_i>=0 i=1,...,p and \beta_0 is unconstrained. I construct u_i as 0 0 0 ... 0 0 1 0 ... 0 0 0 1 ... 0 . . . ... 0 . . . ... 0 . . . ......

Dear All, I have a binomial response with one continuous predictor (d) and one factor (g) (8 levels dummy-coded). glm(resp~d*g, data, family=binomial) Y=b0+b1*X1+b2*X2 ... b7*X7 how can I get the inflection point per group, e.g., P(d)=.5 I would be grateful for any help. Thanks in advance, Ren?

Hi R users! I am trying to learn some multilevel analysis, but unfortunately i am now very confused. The reason: http://www.ats.ucla.edu/stat/hlm/seminars/hlm_mlm/mlm_hlm_seminar.htm http://www.ats.ucla.edu/stat/sas/seminars/sas_mlm/mlm_sas_seminar.htm and MlmSoftRev. pdf from mlmRev package. >From what i see, the first two links seem to declare the level one variable as a random part (i

Hi Dr. Snow, I am a graduate student working on analyzing data for my thesis and came across your post on an R forum: The default link function for the glm poisson family is a log link, which means that it is fitting the model: log(mu) ~ b0 + b1 * x But the data that you generate is based on a linear link. Therefore your glm analysis does not match with how the data was generated

Hello, I currently get anova results out of the aov function (see below) I use the model.tables and I believe it gives me back the model parameters of the fit (betas), however I don't see the intercept (beta_0) and don't understand what the "rep" output means and there is no description in the documentation. Another question: is there a function that outputs the results in a more meaningful way e.g. show the percentage of variation of each factor towards the response I believe the formula...

...- I regressed y to A, and calculated the residuals (e_y) (reg1) - I regressed x to A, and calculated the residuals (e_x) (reg2) - I regressed e_y to e_x (reg5) It looks like this: y = a_0 + a_1 A (reg1) x = b_0 + b_1 A (reg2) e_y = y - (a_0 + a_1 A) (3) e_x = x - (b_0 + b_1 A) (4) e_y = beta_0 + beta_1 e_x (reg5) Then, to predict a y_0 from a new x_0 and A_0, we would: Calculate e_x0 with the equation (4). Calculate e_y0 with the equation (reg5) and then: y_0 = e_y0 + (a_0 + a_1 A_0) Now, I would like to know how different in Area (A) must be two new observations with the same x_0 value...

...505e-01 -4.461005e-02 -1.728046e-09 -5.154932e-04 Now, by changing my data set(final), I repeat the process 100 times and obtain 100 such vectors which I store as 100 rows in a 100X5 matrix: matrix[i,]=coef(reg)[best,] (i varying from 1 to 100) Now my final estimates for the beta's are: Beta_0=median(matrix[,1]) Beta_1=median(matrix[,2]) Beta_2=median(matrix[,3]) Beta_3=median(matrix[,4]) Beta_4=median(matrix[,5]) I want to find the p-values of each of the estimated beta's. I am confused how to extract these p values in R (may be we need to go back to the reg= lm.ridge model corresp...

I have the following script for generating a dataset. It works like a champ except for a couple of things. 1. I need the variables "itbs" and "map" to be negatively correlated with the binomial variable "lunch" (around -0.21 and -0.24, respectively). The binomial variable "lunch" needs to remain unchanged. 2. While my generated variables do come out

Hello, Is this a bug or a feature? I am using R 2.7.1 on Apple OS X. > y <- matrix(1:3,nrow=3) # y is a single-column matrix > df <-data.frame(x=1:3,y=y) > sapply(df,data.class) x y "numeric" "numeric" > df$yy <- y > sapply(df,data.class) x y yy "numeric" "numeric" "matrix"

Dear list: I am attempting to use what I thought would be a pretty straightforward practical application of Cox regression. I figure users of the survival package must have come across this problem before, so I would like to ask you how you dealt with it. I have set up an illustrative example and included it at the end of this post. I took a sample of 100 data points from each of two populations