i'm trying to calculate an eta squared effect size for a repeated measures Manova. However the eta_squared function does not seem to work; I get the following error-message:
Error in [[<-.data.frame(*tmp*, "Mean_Square", value = numeric(0)) :
Replacement has 0 lines, data has 3
Does anyone know an alternative way to calculate eta squared for this kind of model?
Thanks in advance
Janna
That's my model:
modelh3<-manova(cbind(PHQ9_Wert, SF8_Wert) ~ time + Error(ID), data= data.long)
summary(modelh3)```
effectsize::eta_squared(manova1)
Related
I am trying to get a confidence interval for my response in a poisson regression model. Here is my data:
X <- c(1,0,2,0,3,1,0,1,2,0)
Y <- c(16,9,17,12,22,13,8,15,19,11)
What I've done so far:
(i) read my data
(ii) fit a Y by poisson regression with X as a covariate
model <- glm(Y ~ X, family = "poisson", data = mydata)
(iii) use predict()
predict(model,newdata=data.frame(X=4),se.fit=TRUE,interval="confidence",level=0.95, type = "response")
I was expecting to get "fit, lwr, upr" for my response but I got the following instead:
$fit
1
30.21439
$se.fit
1
6.984273
$residual.scale
[1] 1
Could anyone offer some suggestions? I am new to R and struggling with this problem for a long time.
Thank you very much.
First, the function predict() that you are using is the method predict.glm(). If you look at its help file, it does not even have arguments 'interval' or 'level'. It doesn't flag them as erroneous because predict.glm() has the (in)famous ... argument, that absorbs all 'extra' arguments. You can write confidence=34.2 and interval="woohoo" and it still gives the same answer. It only produces the estimate and the standard error.
Second, one COULD then take the fit +/- 2*se to get an approximate 95 percent confidence interval. However, without getting into the weeds of confidence intervals, pivotal statistics, non-normality in the response scale, etc., this doesn't give very satisfying intervals because, for instance, they often include impossible negative values.
So, I think a better approach is to form an interval in the link scale, then transform it (this is still an approximation, but probably better):
X <- c(1,0,2,0,3,1,0,1,2,0)
Y <- c(16,9,17,12,22,13,8,15,19,11)
model <- glm(Y ~ X, family = "poisson")
tmp <- predict(model, newdata=data.frame(X=4),se.fit=TRUE, type = "link")
exp(tmp$fit - 2*tmp$se.fit)
1
19.02976
exp(tmp$fit + 2*tmp$se.fit)
1
47.97273
I have a question regarding how to apply the delta method when I have clustered standard errors. Consider the following dataset and (simple) regression ((Please note that this question is not necessarily about whether it makes sense to cluster around "us" or the correctness / usefulness of this regression).
#Use packages
library(multiwayvcov)
library(sandwich)
library(lmtest)
library(msm)
#load the data
data(mtcars)
# Run the regression
model1<-lm(mpg~cyl+gear+drat, data = mtcars)
#Calculate variance covariance matrix for clustered standard errors
vcov<-cluster.vcov(model1, mtcars$vs)
coeftest(model1, vcov)
# Apply delta method results in error
g<-model1$coefficients[2] / model1$coefficients[1]
deltamethod(g, mean, cov = vcov, ses=TRUE)
# Error I get is this one: "Error in deltamethod(g, mean = g, cov = vcov, ses = TRUE) :
# Covariances should be a 1 by 1 matrix"
Now I want to calculate the standard error for the coefficient (cyl) divided by (intercept) when using my matrix for clustered standard errors around "vs" (i.e. the vcov matrix). Does anyone know how to do this? I looked at this website, but for some reason I got an error when applying this (https://rdrr.io/rforge/msm/man/deltamethod.html). I appreciate any help.
Just editing the deltamethod call to output an answer - I don't know if this answer actually makes sense for what you want to do.
deltamethod(
g = formula('~x2/x1'),
mean = model1$coefficients,
cov = vcov,
ses = TRUE)
I am currently working on a non-linear analysis of various datasets using nls model. On the other hand, I want to calculate the standard error of the regression of the nls model.
The formula of the standard error of regression:
n <- nrow(na.omit((data))
SE = (sqrt(sum(pv-av)^2)/(n-2))
where pv is the predicted value and av is the actual value.
I have a problem on calculating the standard error. Should I calculate the predicted value and actual value first? Are the values based on the dataset? Any help is highly appreciated. Thank You.
R provides this via sigma:
fm <- nls(demand ~ a + b * Time, BOD, start = list(a = 1, b = 1))
sigma(fm)
## [1] 3.085016
This would also work where deviance gives residual sum of squares.
sqrt(deviance(fm) / (nobs(fm) - length(coef(fm))))
## [1] 3.085016
I'm building a penalized multinomial logistic regression, but I'm having trouble coming up with a easy way to get the prediction accuracy. Here's my code:
fit.ridge.cv <- cv.glmnet(train[,-1], train[,1], type.measure="mse", alpha=0,
family="multinomial")
fit.ridge.best <- glmnet(train[,-1], train[,1], family = "multinomial", alpha = 0,
lambda = fit.ridge.cv$lambda.min)
fit.ridge.pred <- predict(fit.ridge.best, test[,-1], type = "response")
The first column of my test data is the response variable and it has 4 categories. And if I look at the result(fit.ridge.pred) it looks like this:
1,2,3,4
0.8743061353, 0.0122328811, 0.004798154, 0.1086628297
From what I understand these are the class probabilities. I want to know if there's a easy way to compute the model accuracy on the test data. Now I'm taking the max for each row and comparing with the original label. Thanks
Something like:
predicted <- colnames(fit.ridge.pred)[apply(fit.ridge.pred,1,which.max)]
table(predicted, test[, 1]
The first line takes the class for which the model outputs the highest probability per row, after which the second line constructs a confusion matrix.
The accuracy is then basically the proportion of observations classified correct (sum of the diagonal / total)
For more details see Glmnet Vignette
fit.ridge.pred <- predict(fit.ridge.best, test[,-1], type = "class") # predict classes, not probability
table(fit.ridge.pred,test[,1]) # confusion matrix
mean(fit.ridge.pred==test[,1]) # accuracy
I have a dataset that contains some missing values (on independent variables). I’m fitting a glm model :
f.model=glm(data = data, formula = y~x1 +x2, "binomial", na.action =na.omit )
After this model I want the ‘null’ model , so I used update:
n.model=update(f.model, . ~ 1)
This seems to work, but the number of observations in both models differ (f.model n=234; n.model n=235). So when I try to estimate a likelihood ratio I get an error: Number of observation not equal!!.
Q: How to update the model so that it accounts for the missing values?
Although it is a bit strange that na.action =na.omit dit not solve the NA problem. I decided to filter out the data.
library(epicalc) # for lrtest
vars=c(“y”, “x1”, “x2”) #variables in the model
n.data=data[,vars] #filter data
f.model=glm(data = data, formula = y~x1 +x2, binomial)
n.model=update(f.model, . ~ 1)
LR= lrtest(n.model,f.model)
If someone has a better solution or an argument way na.action in combination with update results in unequal observations, your answer or solution is more than welcome!