similar to: Calculating p-value for 1-tailed test in a linear model

Hi, I'm trying to fit the following model to data using 'nls': y = alpha_1 * beta_1 * exp(-beta_1 * x) + alpha_2 * beta_2 * exp(-beta_2 * x) and the call I've been using is: nls(y ~ alpha_1 * beta_1 * exp(-beta_1 * x) + alpha_2 * beta_2 * exp(-beta_2 * x), start=list(alpha_1=4, alpha_2=2, beta_1=3.5, beta_2=2.5), trace=TRUE, control=nls.control(maxiter =

How do we get 2-tailed p-values for the rlm summary? I'm using the following: > fit <- rlm(oatRT ~ oatoacData$erp, psi=psi.bisquare, maxit=100, na.action='na.omit') > fitsum <- summary(fit, cor=F) > print(fitsum) Call: rlm(formula = oatRT ~ oatoacData$erp, psi = psi.bisquare, maxit = 100, na.action = "na.omit") Residuals: Min 1Q Median

I would like to be able to place a normal distribution surrounding the predicted values at various places on a plot. Below is some toy code that creates a scatterplot and plots a regression line through the data. library(MASS) mu <- c(0,1) Sigma <- matrix(c(1,.8,.8,1), ncol=2) set.seed(123) x <- mvrnorm(50,mu,Sigma) plot(x) abline(lm(x[,2] ~ x[,1])) Say I want to add a normal

Hello, I would like to create a matrix with one of the columns named $\delta$. I have also created columns $\beta_1$ , $\beta_2$, etc. However, it seems like \d is an escape sequence which gets automatically removed. (Using these names such that they work right in xtable -> latex) colnames(simpleReg.mat) <- c("$\beta_1$","$SE(\beta_1)$", "$\beta_2$",

Hi *, is there a way to obtain some kind of p-value for a model fitted with RMA using the lmodel2 package? I know that p-values are discussed and criticized a lot and as you can image from my question I'm not very much of a statistican (only writing my bachelor thesis). As fare as I understood the confidence interval statistic correctly, a coefficient is regarded as statistically significant

I believe the binom.test procedure is producing one tailed p values rather than the two tailed value implied by the alternative hypothesis language. A textbook and SAS both show 2*9.94e-07 = 1.988e-06 as the two tailed value. As does the R summation syntax from R below. It looks to me like the alternative hypothesis language should be revised to something like " ... greater than or equal

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 \alpha \sum\limits_{i=1}^{n} (

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 version deleted]]

Hi all I know about producing a minimal example to show my problem. But I'm having trouble producing a minimal example that displays this behaviour, so please bear with me to begin with. Observe: I create an array called model.mat. Some details 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

I am fitting a logistic binomial model of the form glm(y ~ a*x,family=binomial) where a is a factor (with 5 levels) and x is a continuous predictor. To assess how much ``impact'' x has, I want to compare the fitted success probability when x = its maximum value with the fitted probability when x = its mean value. (The mean and the max are to be taken by level of the factor

Dear R users, I am analyzing several response variables (all scaled to [0;1]) using a multivariate linear model. After fitting the model, I set up a hypothesis matrix to test specific contrasts for these response variables; for example: "a always increases significantly more than b when regressed against x". What I am stuck with now is how to calculate the correct F-values (and

I want to estimate the following fixed effect model: y_i,t = alpha_i + beta_1 x1_t + beta_2 x2_i,tx2_i,t = gamma_i + gamma_1 x1_t + gamma_2 Z1_i + gamma_3 Z2_i I can use ivreg from AER to do the iv regression. fm <- ivreg(y_i,t ~ x1_t + x2_i,t | x1_t + Z1_i + Z2_i, data = DataSet) But, I'm not sure how can I add the fixed effects. Thanks! [[alternative HTML

Dear R users, If anyone can tell me how to extract the p values from the output of gee? Many thanks in advance. Yu-Kang _________________________________________________________________ §K¶O¸ÕÅ¥ MSN ­^»y¾Ç²ß¡G©M¯u¤H¦Ñ®v½u¤W¾Ç­^¤å http://www.msn.com.tw/english/

Dear stats experts: Me and my little brain must be missing something regarding bootstrapping. I understand how to get a 95%CI and how to hypothesis test using bootstrapping (e.g., reject or not the null). However, I'd also like to get a p-value from it, and to me this seems simple, but it seems no-one does what I would like to do to get a p-value, which suggests I'm not understanding

On 16/02/18 15:28, Bert Gunter wrote: > This is really a statistical issue. What do you think the Intercept term > represents? See ?contrasts. > > Cheers, > Bert > > > > Bert Gunter > > "The trouble with having an open mind is that people keep coming along and > sticking things into it." > -- Opus (aka Berkeley Breathed in his "Bloom

Hi guys Is there a way in R to simulate/generate random numbers from a negative skewed and fat tailed distribution ? I would like to simulate a set of (discrete) data. Regards, Carlos Carlos http://www.nabble.com/file/p25783889/graph.png graph.png -- View this message in context: http://www.nabble.com/Simulate-negative-skewed%2C-fat-tailed-distribution-tp25783889p25783889.html Sent from the

To give a short answer to the original question: > On 16 Feb 2018, at 05:02 , Rolf Turner <r.turner at auckland.ac.nz> wrote: > > In order to ascribe unique values to the parameters, one must apply a "constraint". With the "treatment contrasts" the constraint is that > beta_1 = 0. ...and consequently, being a constant, has an s.e. of 0. -- Peter

producing a multi-figure plot, i am try to add beta_1, beta_2,.. beta_9 to ylab using expression or substitution, but cannot work out like for (i in 1:9){ plot(seq(1/m, 1-1/m, 1/m), beta.q[,i], type="l", col=1, ylim=range(beta.q), xlab="quantile", ylab=expresion(beta[i])) } any suggestions will be greatly appreciated. DL [[alternative HTML version deleted]]

When I perform a two-tailed sign test with the following simple syntax, binom.test(59,100) R returns a P-value (0.088) but nothing else. As I want the result for a one-tailed test I take P/2 = 0.044). However, the journal to which I've submitted my results requests the test statistics, not just the P-values. How can I make R return the test statistics? Best regards, Johan Stenberg, Umea

I am trying to find a definition for the two-tailed exact binomial test but have been unsuccessful. Can you help? --------------------------------- [[alternative HTML version deleted]]