search for: lawley

...d products for error: (Intercept) (Intercept) 13.05917 Multivariate Tests: Treatment Df test stat approx F num Df den Df Pr(>F) Pillai 1 0.0492517 0.6216377 1 12 0.44574 Wilks 1 0.9507483 0.6216377 1 12 0.44574 Hotelling-Lawley 1 0.0518031 0.6216377 1 12 0.44574 Roy 1 0.0518031 0.6216377 1 12 0.44574 ------------------------------------------ Term: Hormone Response transformation matrix: (Intercept) Diameter.23 1 Diameter.39 1 Sum of squares and product...

Dear all, is there anyone who can help me extracting p-values from an Anova object from the car library? I can't seem to locate the p-values using str(result) or str(summary(result)) in the example below > A <- factor( rep(1:2,each=3) ) > B <- factor( rep(1:3,times=2) ) > idata <- data.frame(A,B) > fit <- lm( cbind(a1_b1,a1_b2,a1_b3,a2_b1,a2_b2,a2_b3) ? sex,

...M=-0.5 N=4.5 Statistic Value F Num DF Den DF Pr > F Wilks' Lambda 0.02042803 32.9813 4 22 0.0001 Pillai's Trace 1.24157026 9.8222 4 24 0.0001 Hotelling-Lawley Trace 35.12690920 87.8173 4 20 0.0001 Roy's Greatest Root 34.75791615 208.5475 2 12 0.0001 ... Manova Test Criteria and Exact F Statistics for the Hypothesis of no Overall TESTEMUNHA vs TURFE Effect...

...Residuals 16????????????????????????????????????? > summary(fit, test="Wilks") ????????? Df?? Wilks approx F num Df den Df Pr(>F) REWGRP???? 1 0.77269? 1.37283????? 3???? 14 0.2918 Residuals 16????????????????????????????????????? > summary(fit, test="Hotelling-Lawley") ????????? Df Hotelling-Lawley approx F num Df den Df Pr(>F) REWGRP???? 1????????? 0.29418? 1.37283????? 3???? 14 0.2918 Residuals 16?????????????????????????????????????????????? > summary(fit, test="Roy") ????????? Df???? Roy approx F num Df den Df Pr(>F) REW...

Dear R-list! My question is related to an Anova including within and between subject factors and unequal group sizes. Here is a minimal example of what I did: library(car) within1 <- c(1,2,3,4,5,6,4,5,3,2); within2 <- c(3,4,3,4,3,4,3,4,5,4) values <- data.frame(w1 = within1, w2 = within2) values <- as.matrix(values) between <- factor(c(rep(1,4), rep(2,6))) betweenanova <-

...cts for error: (Intercept) (Intercept) 34459.4 Multivariate Tests: (Intercept) Df test stat approx F num Df den Df Pr(>F) Pillai 1 0.97137 610.8117 1 18 2.425e-15 Wilks 1 0.02863 610.8117 1 18 2.425e-15 Hotelling-Lawley 1 33.93399 610.8117 1 18 2.425e-15 Roy 1 33.93399 610.8117 1 18 2.425e-15 ------------------------------------------ Term: sex Response transformation matrix: (Intercept) a1_b1 1 a1_b2 1 a1_b3 1 a2_b1 1 a2_b2...

...r(>F) 0.9889929 0.5286555 2.0000000 95.0000000 0.5911205 Code: ------------------------------------------------------------------------------------ mlh.mlm <- function(object, L = null, T = diag(nrow = p), test = c("Pillai", "Wilks", "Hotelling-Lawley", "Roy")) { # test the multivariate linear hypothesis LBT'=0 # B = matrix of regression coefficients # L = matrix, each row is a linear combination of the parameters # T = transformation matrix if(!inherits(object, "mlm")) stop("object...

...blogspot.com/2011/02/r-tutorial-series-one-way-repeated.html Upon running the actual ANOVA I got this error message: "> rmanova=anova(yearmodel, idata=yearframe, idesign=~yearfactor) Warning message: In function (object, ..., test = c("Pillai", "Wilks", "Hotelling-Lawley", : models with response "NULL" removed because response differs from model 1" And the p-values I get in the anova summary are completely off, judging by the t-tests I did on the same data. The dataset is small (n=6) and was arranged so that the data from the two different...

Full_Name: conor dolan Version: 1.4.0 OS: windows98 Submission from: (NULL) (146.50.170.247) On the basis of Lawley and Maxwell's explanation in Factor Analysis as a Statistical Method (London: Butterworth, 1971), the following line in the promax routine (library, mva): attr(z, "covariance") <- crossprod(U) should be attr(z, "covariance") <- solve(crossprod(U)) the attribute as...

...rSlope") ## linear.hypothesis does work on this: > print(linear.hypothesis(soils.mod, "ContourSlope"), SSP=FALSE) Multivariate Tests: Df test stat approx F num Df den Df Pr(>F) Pillai 1 0.5844882 3.907418 9 25 0.003312 ** Wilks 1 0.4155118 3.907418 9 25 0.003312 ** Hotelling-Lawley 1 1.4066706 3.907418 9 25 0.003312 ** Roy 1 1.4066706 3.907418 9 25 0.003312 ** --- Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 > I would be grateful if someone could (a) confirm that this is a general problem, not just my system; (b) suggest how to determine what is going w...

...o testing.testing at xxxx.xxx (real domain obscured), I see this in mail.log Dec 23 10:49:24 IBMUBUNTU1 dovecot: auth(default): master in: USER#0111#011testing.testing at xxxx.xxx#011service=deliver Dec 23 10:49:24 IBMUBUNTU1 dovecot: auth(default): ldap(testing.testing): user search: base=dc=lawley, dc=local scope=subtree filter=(&(objectClass=user) (samaccountname=testing.testing)) fields= Dec 23 10:49:24 IBMUBUNTU1 dovecot: auth(default): ldap(testing.testing): result: objectClass(?unknown?)= cn(?unknown?)= sn(?unknown?)= givenName(?unknown?)= distinguishedName(?unknown?)= instanceTy...

...ormula(a2$Edu~a2$Kyph+a2$HL+a2$Vert, method="reverse", overall=T, continuous=5, add=TRUE, test=T) #perform MANOVA a3=manova(cbind(Kyph, Vert, HL)~as.factor(Edu), data=a2) #return results a4=list("Results of Educational Status MANOVA", print(d), summary(a3, test="Hotelling-Lawley"), summary(a3, test="Roy") , summary(a3, test="Pillai"), summary(a3, test="Wilks"), summary.aov(a3) ) print(a4) } This function works as is, but I want to add the mean and standard deviation to my table. When I add the following code to line 36 where I print...

...tion matrix for analysis. The correlation matrix is returned as component \code{correlation} of the result. The fit is done by optimizing the log likelihood assuming multivariate normality over the uniquenesses. (The maximizing loadings for given uniquenesses can be found analytically: Lawley & Maxwell (1971, p. 27).) All the starting values supplied in \code{start} are tried in turn and the best fit obtained is used. If \code{start = NULL} then the first fit is started at the value suggested by \enc{J?reskog}{Joreskog} (1963) and given by Lawley & Maxwell (1971, p....

As mentioned in my previous posting, I've set up a mailserver for a domain of Winboxen, authenticated through Active Directory. After some struggles, I finally got that working. But I also need to receive emails for a few system users on the mailserver (like postmaster, and a couple of accounts set up for maintaining the system). I thought, based on the way I read the documentation,

...new version adds four new criteria ("tau2", "zeta2", "xi2" and "ccr12"), associated with the multivariate linear hypothesis and the well-known test statistics for that hypothesis (Wilks' lambda, Roy's largest root, Pillai's trace and the Hotelling-Lawley statistic). Subselect 0.9-99 also includes two functions that prepare the arguments for the search functions associated with the new criteria in the most frequent contexts of application of the multivariate linear hypothesis - linear regression (function "lmHmat") and linear discriminant...

Sorry if this is a stupid question, but: Is there a routine in R which calculates the Hotellings-Lawley statistic? I searched aov, summary.mlm, CRAN etc. but could not find it. Torsten -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscr...

...new version adds four new criteria ("tau2", "zeta2", "xi2" and "ccr12"), associated with the multivariate linear hypothesis and the well-known test statistics for that hypothesis (Wilks' lambda, Roy's largest root, Pillai's trace and the Hotelling-Lawley statistic). Subselect 0.9-99 also includes two functions that prepare the arguments for the search functions associated with the new criteria in the most frequent contexts of application of the multivariate linear hypothesis - linear regression (function "lmHmat") and linear discriminant...

Hi All, I've installed Xapian and the php module. I've set up a script for use with scriptindex and dbi2omega for getting data from the db into the index easily, the script file is as follows: =============================== id : field=id title : index title: field=title description : index description : truncate=50 field=content ============================= However, when querying

Hello everybody After doing a MANOVA on a bunch of data, I want to be able to make some comment on the amount of variation in the data that is explained by the factor of interest. I want to say this in the following way: XX% of the data is explained by A. I can acheive something like what I want by doing the following: X <- structure(c(9, 6, 9, 3, 2, 7), .Dim = as.integer(c(3,

...0.31281143 3.00 5 48 0.0194 >From R: Df test stat approx F num Df den Df Pr(>F) Pillai 5.0000 0.401037 1.481343 15.0000 144.0000 0.1193910 Wilks 5.0000 0.636571 1.509595 15.0000 127.3871 0.1109964 Hotelling-Lawley 5.0000 0.513082 1.527844 15.0000 134.0000 0.1037152 Roy 5.0000 0.368760 3.540095 5.0000 48.0000 0.0083553 ** Thank you, Ali