search for: treatmentb

...variable1 variable2; RUN; producing an output with means for each variable by factor groupings as below: *factor1 factor2 obs variable mean* Level A treatmentA 3 variable1 10 variable2 22 treatmentB 3 variable1 12 variable2 30 Level B treatmentA 3 variable1 10 variable2 22 treatmentB 3 variable1 12...

I feel bad even asking, but: Rgames> data(OrchardSprays) Rgames> model<-lm(decrease~.,data=OrchardSprays) Rgames> model Call: lm(formula = decrease ~ ., data = OrchardSprays) Coefficients: (Intercept) rowpos colpos treatmentB treatmentC 22.705 -2.784 -1.234 3.000 20.625 treatmentD treatmentE treatmentF treatmentG treatmentH 30.375 58.500 64.375 63.875 85.625 Rgames> levels(OrchardSprays$treatment) #just double-checking... [1] "A" &q...

Here is a working snippet. library(epitools) mat <- matrix(c(10,15,60,25,98, 12,10,70,28,14, 9,11,68,10,12 ,8,13,20,11,58) ,ncol=2) colnames(mat) <- c("treatmentA","treatmentB") row.names(mat) <- paste("Cond",rep(1:10,1)) dimnames(mat) <- list("Condition" = row.names(mat), "instrument" = colnames(mat)) > mat instrument Condition treatmentA treatmentB Cond 1 10 9 Cond 2 15 11...

Here is a working snippet. library(epitools) mat <- matrix(c(10,15,60,25,98, 12,10,70,28,14, 9,11,68,10,12 ,8,13,20,11,58) ,ncol=2) colnames(mat) <- c("treatmentA","treatmentB") row.names(mat) <- paste("Cond",rep(1:10,1)) dimnames(mat) <- list("Condition" = row.names(mat), "instrument" = colnames(mat)) > mat instrument Condition treatmentA treatmentB Cond 1 10 9 Cond 2 15 11...

Here is a working snippet. library(epitools) mat <- matrix(c(10,15,60,25,98, 12,10,70,28,14, 9,11,68,10,12 ,8,13,20,11,58) ,ncol=2) colnames(mat) <- c("treatmentA","treatmentB") row.names(mat) <- paste("Cond",rep(1:10,1)) dimnames(mat) <- list("Condition" = row.names(mat), "instrument" = colnames(mat)) > mat instrument Condition treatmentA treatmentB Cond 1 10 9 Cond 2 15 11...

Dear R-helpers, I am trying to construct a confidence interval on a prediction of a gam fit. I have the Wood (2006) book, and section 5.2.7 seems relevant but I am not able to apply that to this, different, problem. Any help is appreciated! Basically I have a function Y = f(X) for two different treatments A and B. I am interested in the treatment ratios : Y(treatment = B) / Y(treatment = A) as

Dear R users, I am currently analyzing a dataset using lme(). The model I use has the following structure: model<-lme(response~Covariate+TreatmentA+TreatmentB,random=~1|Block/Plot,method="ML") When I plot the residuals against the fitted values, I see a clear positive trend (meaning that the variance increases with the mean). I tried to solve this issue using weights=varPower(), but it doesn?t change the residual plot at all. How would you...

...IC logLik 365.8327 382.5576 -176.9163 Random effects: Formula: ~1 | block (Intercept) Residual StdDev: 0.4306096 0.9450976 Fixed effects: Cu ~ Treatment Value Std.Error DF t-value p-value (Intercept) 5.587839 0.2632831 104 21.223688 0.0000 *** TreatmentB -0.970384 0.3729675 16 -2.601792 0.0193 *** TreatmentC -1.449250 0.3656351 16 -3.963651 0.0011 *** TreatmentD -1.319564 0.3633837 16 -3.631323 0.0022 *** Correlation: (Intr) TrtmAN TrtmCH TreatmentB -0.706 TreatmentC -0.720 0.508 TreatmentD -0.7...

...ntO -3.0242 0.6561 -4.609 4.05e-06 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Number of iterations in BFGS optimization: 16 Log-likelihood: -363.2 on 10 Df So my question is, where did my "Elev01-Low" and "TreatmentB" go?? Why aren't they appearing in the output table? Any insight would be greatly appreciated! - Jason -- View this message in context: http://r.789695.n4.nabble.com/Can-t-find-all-levels-of-categorical-predictors-in-output-of-zeroinfl-tp4444214p4444214.html Sent from the R help mai...

...: glm(formula = y ~ treatment, family = binomial) Deviance Residuals: Min 1Q Median 3Q Max -2.3134 -0.5712 -0.3288 0.8616 2.4352 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -0.574195 0.036443 -15.756 < 2e-16 *** treatmentB 0.164364 0.051116 3.216 0.00130 ** treatmentC 0.007025 0.054696 0.128 0.89780 treatmentD 0.258135 0.053811 4.797 1.61e-06 *** --- Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 60.467 on...

...8.34 108 3 1 1 6.34 9.34 5.98 As you can see, data for variable3 are missing for the first year, but otherwise there is little data missing. I call upon lmList : lmgroup.data <- lmList (variable1 ~ treatmentB | year/treatmentA, data = data, na.action = na.omit) When I call the object, I see : Call: Model: variable1 ~ treatmentB | year/treatmentA Data: data Coefficients: (Intercept) treatment B year2/treatmentA0 44.08387 81.1...

...17.91 2.0958e-09 41.78 -12.60 Model: Surv(death, censor) ~ treatment * sex + (1 | day/person) + (1 | treatment/box) Fixed coefficients coef exp(coef) se(coef) z p teratmentb -0.0838877 0.9195345 0.3744511 -0.22 0.8200 treatmentb2 -0.4731922 0.6230103 0.3136199 -1.51 0.1300 treatmentn -1.0154149 0.3622521 0.4097754 -2.48 0.0130 sexmale -0.1838885 0.8320286 0.2602169 -0.71 0.4800 treatmentb:sexmale -0.3905856 0.6766605 0.2132936 -1.83 0.0670 treatmentb2:sexmale 0.6742202 1...

Dear list, i'm trying to test if a linear combination of coefficients of glm is equal to 0. For example : class 'cl' has 3 levels (1,2,3) and 'y' is a response variable. We want to test H0: mu1 + mu2 - mu3 =0 where mu1,mu2, and mu3 are the means for each level. for me, the question is how to get the covariance matrix of the estimated parameters from glm. but perhaps there