search for: h0

Hi There, Using t.test to test hypothesis about which one is greater, A or B? where A={10,20,30},B={25,30,15}. My question is which of the following conclusions is right? #################hypothesis testing 1 h0: A greater than or equal to B h1: A less than B below is splus code A=c(10,20,30) B=c(25,30,15) t.test(c(10,20,30),c(25,30,15),alternative="less") output: p-value=0.3359 because p-value is not less than alpha (0.05), we cannot reject h0. so A greater than or equal to B. ############...

...now want to see what sort of results I get when applying LLVM's optimisations and some more intelligence to the code generation. (Plus, Sparse is buggy and really awkward to work with.) For giggles, here's some example Javascript produced by Clue. function _dtime(fp, stack) { var sp; var H0; var H1; var H2; var H3; var H4; var state = 0; for (;;) { switch (state) { case 0: sp = 2; sp = fp + sp; H1 = null; H0 = 0; H2 = fp; H3 = _gettimeofday; H4 = H3(sp, stack, H2, stack, H0, H1); H0 = fp; H1 = stack[H0 + 0]; H0 = fp; H2 = stack[H0 + 1]; H0 = 1000000.000000; H3 = H2 / H0; H0 = H1 + H3;...

David Given <dg at cowlark.com> writes: > The obvious place to start on this is the C backend, except in these 2.8 > days the C backend is so hedged about with caveats I'm rather wary of > basing anything on it. I also recall seeing comments here that it's due > for a rewrite from scratch, and that various people were looking into > it. Can anyone go into more detail

...1,ncol=n) tvec<-vector("numeric",N) h<-vector("numeric",m) t<-0 xmat[1,]<-x for (i in 1:N) { h[1]<-cvec[1]*x[1] h[2]<-cvec[2]*x[1]*x[2] h[3]<-cvec[3]*x[2] h0=sum(h) tp<-rexp(1,h0) t<-t+tp u<-runif(1,0,1) if ( u < h[1]/h0 ) { x[1] <- x[1]+1 } else if ( u < (h[1]+h[2])/h0 ) { x[1] <- x[1]-1...

Hi everybody, i'm dealing with some percentage tables, of which i should test rowwise if the entries are sgnificantly equal or not. Namely, on row 1, test H0: element 1= element2, H0: element 1= element3...H0: element 2= element3...H0: element n-1= element n. The same on the other rows. Anybody knows how this can be done in quick way? I don't have large matrices, but it seems quite boring... Thank you very much in advance for your answering, Em...

Hello. How can I compute the Bootstrap p-Value for a one- and two sided test, when I have a bootstrap sample of a statistic of 1000 for example? My hypothesis are for example: 1. Two-Sided: H0: mean=0 vs. H1: mean!=0 2. One Sided: H0: mean>=0 vs. H1: mean<0 I hope you can help me Thanks in advance Regards, Andreas

Hi everybody, somehow i dont get the shapiro wilk test for normality. i just can?t find what the H0 is . i tried : shapiro.test(rnorm(5000)) Shapiro-Wilk normality test data: rnorm(5000) W = 0.9997, p-value = 0.6205 If normality is the H0, the test says it?s probably not normal, doesn ?t it ? 5000 is the biggest n allowed by the test... are there any other test ? ( i know qqnorm alre...

Hi, is there a test for the H0 skewness=0 (or with skewness as test statistic and normality as H0) implemented in R? Thank you, Christian *********************************************************************** Christian Hennig Fachbereich Mathematik-SPST/ZMS, Universitaet Hamburg hennig at math.uni-hamburg.de, http://www.math...

...kernel f1=apply(x1,2,mean)/h return(f1) } ################################################################################################################################# Simulation for different bandiwidths and different kernels n=1010 # n=1010 ker=1 # ker=1=>Epan; ker=0=> Gaussian h0=c(0.00452,0.001984) # set initial bandwidths z=seq(-0.05,0.05,by=10) # grid points nz=length(z) # number of grid points x=rnorm(1010, mean=0, sd=0.0077) # simulate x-N(0,0.0077^2) if(ker==1){h_o=2.34*n^{-0.2}} # bandwidth for Epanechnikov kernel if(ker==0){...

Hi I was wondering if there is a permutation test available in R for linear models with continuous dependent covariates. I want to do a test like the one shown here. bmi<-rnorm(100,25) x<-c(rep(0,75),rep(1,25)) y<-rnorm(100)+bmi^(1/2)+rnorm(100,2)*x+bmi*x H0<-lm(y~1+x+bmi) H1<-lm(y~1+x+bmi+x*bmi) anova(H0,H1) summary(lm(y~1+x+bmi)) But I want to use permutation testing to avoid an inflated p-value due to a y that is not totally normal distributed and I do not want to log transform y. Thanks Anders

...nute post hoc analyses, meaning that my sample size (N=3404) has been long fixed, and I'm interested in assessing the ES and Power after the fact.. As far as I can deduce from the implementation of the ES.w2 formula or Cohen's (1992) own article, it seems to me that the probabilities p(H0) and p(H1) would simply be the expected and observed absolute frequencies divided by the sample size N, in that the 'true' probablities are the observed proportions and the null probabilities the expected ones. If this is correct, then the effect size and the power statistics can natura...

...res[test,distn,sample.size,mu1]<-result }}}} output(res) } hypoth.test<-function(sample.size,distn,test,mu1){ #Purpose: #Samples n values from the chosen distribution #Inputs: #n - sample size of data to generate #distn - the distribution to generate data from #test - which test to use #H0 - the null hypothesis #mu.true - the true value of the mean under the chosen distribution #sigma.true - the true value of the variance under the chosen distribution #alpha - trimming level #K - number of simulations to perform #Outputs: #reject.null - the number of times we reject the null hypothes...

...VAR(cbind(x,y1),ic="SC") ); results=lapply(models,function(x) causality(x,cause="y1")); print(results)} Count<-Granger(y1,y2) which produces the following output (I have printed only part of it (for Granger causality of bs on ml)): $ml $ml$Granger Granger causality H0: y1 do not Granger-cause x data: VAR object x F-Test = 0.2772, df1 = 1, df2 = 122, p-value = 0.5995 $ml$Instant H0: No instantaneous causality between: y1 and x data: VAR object x Chi-squared = 19.7429, df = 1, p-value = 8.859e-06 My questions: 1)How can I edit the function above s...

> -----Original Message----- > From: r-help-bounces at stat.math.ethz.ch > [mailto:r-help-bounces at stat.math.ethz.ch]On Behalf Of Thomas Petzoldt > Sent: 06 September 2005 06:34 > Cc: petzoldt at rcs.urz.tu-dresden.de; R-Help > Subject: Re: [R] model selection vs. H0 based testing > > > Hello, > > I wish to thank Douglas Bates very much for clarification and > pointing me to the MCMC simulation method to get p values even for cases where > Wald tests are inappropriate. > > One question however remains when publishing statistical...

...n=na.exclude, data=LeaderPG.data, control=coxph.control(eps=1e-09,iter.max=100,outer.max=100)) library(aod) # To test if Military Leaders are equally sensitive to the outcome of WAR as Civilian leaders we need a JOINT test. wald.test(b=coef(modelPG2), Sigma = vcov(modelPG2), Terms= c(19:21), H0=c(-2.9101, 2.4028, -1.6504)) #wald.test(b=coef(modelPG2), Sigma = vcov(modelPG2), Terms= c(19:21), H0=c(0, 2.4028, 0)) wald.test(b=coef(modelPG2), Sigma = vcov(modelPG2), Terms= c(25:27), H0=c(-8.2330,2.3041,-0.2626)) Any help would be very much appreciated. Hein Goemans.

...amplesizes)))dimnames(res)<-list(distributions,tests,samplesizes)for(distribution in distributions){for(test in tests){for(samplesize in samplesizes){for(i in 1:k){results<-size()res[distribution,test,samplesize]<-results}output<-size.power.testreturn(output)}}}}size<-function(k=1000,H0=0,mu.true=0,sigma.true=1,alpha=0.05,x1=-sqrt(3),y1=sqrt(3)){distributions<-c("Normal","Uniform")tests<-c("t","Wilcoxon")samplesizes<-c(10,30,40)reject1<-numeric(k)for(distribution in distributions){for(test in tests){for(samplesize in samplesizes)...

...ta (MASS, data(hills)): one dependent variable: time two independent var (metric): dist, climb if I am interested, after (!) fitting a lm: my. lm <- lm(time ~ dist + climb, data = hills) in the equivalence (or non-equivalence) of the two predictors "dist" and "climb": H0: dist = climb Is there any function in R, which lets me calculate this, in just giving the lm-object "my.lm" and e.g. a vector such as c(1, -1), operationalizing the hypothesis H0: t(c(1, -1)) %*% c(dist, climb) = 0 ? many thanks Cheers! Christoph

...B0 + B1*X1 + B2*X2 + B3*X3 and Model 2: Y = B0 + B2*X2 + B3*X3 I.E. Model1 = lm(Y~X1+X2+X3) Model2 = lm(Y~X2+X3) The Ajusted R-Square for Model1 is 0.9 and the Ajusted R-Square for Model2 is 0.99, among many other significant improvements. And I want to do the anova test to choose the best one: H0: B1 = 0 H1: B1 != 0 Test = Anova(Model2,Model1) How do I know what model wins? (I'm using a confidence level of 0.1)... My guess is that: If p-value of summary(Test) is greater than 0.1 then I don't reject H0 so Model2 is better and otherwise I reject H0 so Model1 is better? My teacher...

Hi, I had a question about specifying the Null hypothesis in a significance test. Advance apologies if this has already been asked previously or is a naive question. I have two samples A and B, and I want to test whether A and B come from the same distribution. The default Null hypothesis would be H0: A=B But since I am trying to prove that A and B indeed come from the same distribution, I think this is not the right choice for the null hypothesis (it should be one that is set up to be rejected) How do I specify a null hypothesis H0: A not equal to B for say a KS test. An example to do this in...

I am interested in using LLVM to translate C and C++ into high-level language code. (As an update to an earlier project of mine, Clue, which used the Sparse compiler library to do this: it targets Lua, Javascript, Perl 5, C, Java and Common Lisp, with a disturbing amount of success. See http://cluecc.sourceforge.net for details.) The obvious place to start on this is the C backend, except in