anova

Using Kolmogorov Smirnov Test in R

流过昼夜 提交于 2019-12-01 09:21:16
I designed 3000 experiments, so that in one experiment there are 4 groups (treatment), in each group there are 50 individuals (subjects). For each experiment I do a standard one way ANOVA and proof if their p.values has a uni probability function under the null-hypothesis, but ks.test rejects this assumption and I cant see why? subject<-50 treatment<-4 experiment<-list() R<-3000 seed<-split(1:(R*subject),1:R) for(i in 1:R){ e<-c() for(j in 1:subject){ set.seed(seed[[i]][j]) e<-c(e,rmvnorm(mean=rep(0,treatment),sigma=diag(3,4),n=1,method="chol")) } experiment<-c(experiment,list(matrix(e,subject

Why do R and statsmodels give slightly different ANOVA results?

限于喜欢 提交于 2019-12-01 05:45:59
Using a small R sample dataset and the ANOVA example from statsmodels , the degrees of freedom for one of the variables are reported differently, & the F-values results are also slightly different. Perhaps they have slightly different default approaches? Can I set up statsmodels to use R's defaults? import pandas as pd import statsmodels.api as sm from statsmodels.formula.api import ols ##R code on R sample dataset #> anova(with(ChickWeight, lm(weight ~ Time + Diet))) #Analysis of Variance Table # #Response: weight # Df Sum Sq Mean Sq F value Pr(>F) #Time 1 2042344 2042344 1576.460 < 2.2e-16 *

ANOVA: Degrees of freedom almost all equal 1

∥☆過路亽.° 提交于 2019-12-01 04:37:34
I have a data set that begins like this: > d.weight R N P C D.weight 1 1 0 0 GO 45.3 2 2 0 0 GO 34.0 3 3 0 0 GO 19.1 4 4 0 0 GO 26.6 5 5 0 0 GO 23.5 6 1 45 0 GO 22.1 7 2 45 0 GO 15.5 8 3 45 0 GO 23.4 9 4 45 0 GO 15.8 10 5 45 0 GO 42.9 ... and so on. R is replicate and there are 5 of them (1-5). N is nitrogen level, and there are 5 as well (0, 45, 90, 180, 360). P is phosphorus level, and there are 5 as well (0, 35, 70, 140, 280). C is plant combination, and there are 4 (GO, GB, LO, LB). D.weight is dry weight in grams. However, when I do an ANOVA I get the wrong degrees of freedom. I usually

Why do R and statsmodels give slightly different ANOVA results?

ⅰ亾dé卋堺 提交于 2019-12-01 03:13:45
问题 Using a small R sample dataset and the ANOVA example from statsmodels, the degrees of freedom for one of the variables are reported differently, & the F-values results are also slightly different. Perhaps they have slightly different default approaches? Can I set up statsmodels to use R's defaults? import pandas as pd import statsmodels.api as sm from statsmodels.formula.api import ols ##R code on R sample dataset #> anova(with(ChickWeight, lm(weight ~ Time + Diet))) #Analysis of Variance

Posthoc labels on anova boxplot in R

一个人想着一个人 提交于 2019-11-30 14:23:41
问题 If I have some data and do an ANOVA and post-hoc tests, how do I make a boxplot that adds the post-hoc classification automatically, rather than having to edit the figure outside of R? For example, here are some data to get started: install.packages("reshape", dependencies=T) library(reshape) x <- rnorm(30) y <- rnorm(30)+1 z <- rnorm(30)+0.5 data.1 <- data.frame(x, y, z) data.2 <- melt(data.1) Here's code for running a simple one way ANOVA and all of the unplanned comparision post-hoc tests:

Posthoc labels on anova boxplot in R

假如想象 提交于 2019-11-30 09:58:11
If I have some data and do an ANOVA and post-hoc tests, how do I make a boxplot that adds the post-hoc classification automatically, rather than having to edit the figure outside of R? For example, here are some data to get started: install.packages("reshape", dependencies=T) library(reshape) x <- rnorm(30) y <- rnorm(30)+1 z <- rnorm(30)+0.5 data.1 <- data.frame(x, y, z) data.2 <- melt(data.1) Here's code for running a simple one way ANOVA and all of the unplanned comparision post-hoc tests: linear.model <- lm(value~variable, data=data.2) anova(linear.model) # Analysis of Variance Table #

Running scipy's oneway anova in a script

蓝咒 提交于 2019-11-30 09:32:59
I have a problem. I want to run the scipy.stats f_oneway() ANOVA in a script that loads a data-archive containing groups with numpy arrays like so: archive{'group1': array([ 1, 2, 3, ..., ]), 'group2': array([ 9, 8, 7, ..., ]), ...} Now my problem is that the number of groups is not fixed for different data-archives. In other words, I don't know beforehand, how many groups there are in an archive (and also not necessarily what their names are). The scipy implementation of a oneway ANOVA only accepts comma delimited arrays as input like so: a = array([ 1, 2, 3, ..., ]) b = array([ 9, 8, 7, ...,

How to extract coefficients' standard error from an “aov” model

≯℡__Kan透↙ 提交于 2019-11-30 08:50:31
问题 I did a aov model and I just want to extract the standard errors of each coefficient. model <- aov(Molecule ~ Comorbidity + Age + BMI + Sex, data = mydata) I can see the estimated marginal mean with: allEffects(model) But I can't extract standard error of each coefficient. I saw on the internet the function se.coef() but it doesn't work. summary(model)$coefficients[, "Std. Error"] does not work either. I've read effects and other packages, but I don't find what I want. Any idea? 回答1: Use lm

Fitting linear model / ANOVA by group [duplicate]

安稳与你 提交于 2019-11-30 06:04:14
问题 This question already has answers here : Linear Regression and group by in R (10 answers) Closed 3 years ago . I'm trying to run anova() in R and running into some difficulty. This is what I've done up to now to help shed some light on my question. Here is the str() of my data to this point. str(mhw) 'data.frame': 500 obs. of 5 variables: $ r : int 1 2 3 4 5 6 7 8 9 10 ... $ c : int 1 1 1 1 1 1 1 1 1 1 ... $ grain: num 3.63 4.07 4.51 3.9 3.63 3.16 3.18 3.42 3.97 3.4 ... $ straw: num 6.37 6.24

Running scipy's oneway anova in a script

三世轮回 提交于 2019-11-29 14:26:12
问题 I have a problem. I want to run the scipy.stats f_oneway() ANOVA in a script that loads a data-archive containing groups with numpy arrays like so: archive{'group1': array([ 1, 2, 3, ..., ]), 'group2': array([ 9, 8, 7, ..., ]), ...} Now my problem is that the number of groups is not fixed for different data-archives. In other words, I don't know beforehand, how many groups there are in an archive (and also not necessarily what their names are). The scipy implementation of a oneway ANOVA only