[cc'ing back to r-sig-mixed: it's generally a good idea to keep the list "in the loop"]
On 11-03-16 12:12 PM, Brian Edward wrote:
Thank you, I am sorry the tables got messed up. It was due to the Html->Text conversion. The hit-score is an integer between 0 and 10, so its not a 0/1 binary outcome but a linear score each time (I guess the binomial wont work here)
Binomial will exactly work here.
I know there might not be any big differences in between individuals (for this dataset), but each individual in my sample is scored 10 times with one eye and 10 times with two eyes used, so I dont think I can do the paired t-test. I also noted that the distribution of scores in not normally distributed (in the population) but I ignore the fact because I have no idea how to do that without a much more complicated model (bootstrapping etc.)
Unless there is something non-exchangeable about the individual tests (e.g. you think the probability of a hit might depend on the sequence of the test), then I don't see why you can't do the paired t-test: that is, calculate the differences per individual in the number of two-eye and one-eye successes and do a t test against a population mean of zero. The t-test is reasonably robust to non-normality. You could also do a paired Wilcoxon test (?wilcox.test).
The NumEyesUsed p-value for my lme is 0.3217 and for the equivalent aov it is 0.1772. So the actual question is if my conversion was totally wrong: i.e anova(lme(Hits~NumEyesUsed, random=~1|PersonID/NumEyesUsed,data=y)) <totally different>
Hits ~ b_i + eps_person + eps_(person:numeyes) + eps_resid where the first term is the fixed effect of number of eyes, the second is the among-subject effect, the third is the among-tests-within-subject effect, and the the fourth is residual variation (I'm still not sure if you're using total number of hits per person/eye combination as your data (0-10), or whether you're using success on a particular trial (0/1). If the latter, you're violating normality pretty badly. If the former, then the third and fourth terms above are completely confounded [as I suggested based on your zero residual error].)
summary(aov(Hits~NumEyesUsed+Error(PersonID/NumEyesUsed),data=y)) in the lme model you suggest: random=~NumEyesUsed|PersonID
This is actually a slightly more complicated model, because it allows for different variances in the success with one vs two eyes.
the p-Value I obtain is: 0.0788 How would you convert your random lme term back into the normal aov formula? The idea to express/convert the lme -> aov formula in this fashion I found on: http://blog.gribblelab.org/2009/03/09/repeated-measures-anova-using-r/ Thanks again, Brian
Date: Wed, 16 Mar 2011 11:44:00 -0400 From: bbolker at gmail.com To: b.edward at live.com CC: r-sig-mixed-models at r-project.org Subject: Re: [R-sig-ME] aov() -> lme() conversion difficulty On 11-03-16 10:41 AM, Brian Edward wrote:
according to a couple of PDFs and webpages in came up with this
aov() <- lme() conversion:
I think you want random = ~NumEyesUsed|PersonID
numDF denDF F-value p-value (Intercept) 1 36 56.06667 <.0001 NumEyesUsed 1 1 3.26667 0.3217
Your table got mangled. Was it supposed to look like this? If so, then it's clear from the denominator DF that something is wrong ...
summary(aov(Hits~NumEyesUsed+Error(PersonID/NumEyesUsed),data=y))
Error: PersonID
Df Sum Sq Mean Sq F value Pr(>F)Residuals 1 1.7514e-33 1.7514e-33
Also note that your residual errors are essentially zero, which means you don't have any variation left after you have included experimental blocks -- some are aliased with the residual error.
Error: PersonID:NumEyesUsed Df Sum Sq Mean Sq F value Pr(>F) NumEyesUsed 1 4.9 4.9 12.25 0.1772 Residuals 1 0.4 0.4 Error: Within Df Sum Sq Mean Sq F value Pr(>F) Residuals 36 56.6 1.5722 In this study each person is allowed try ten times with one eye and
then ten times with two eyes to score hits.
The question is if there is a difference in hits between using one
or two eyes.
I get different p-values in my aov() <- lme() conversion, which one
answers the question more closely?
What are your actual response variables? Number of successes out of 10 tries? In that case I might suggest lmer(cbind(Hits,10-Hits)~NumEyesUsed+(NumEyesUsed|PersonID), family=binomial, data= ...) although that will make it hard for you get p-values. For that matter, since you have only two treatment levels, what's wrong with a paired t-test ... ??? Ben Bolker