csiro.au> writes:
>
> Here is a demo you may like to consider. (I can see what you are trying
> to do with your loops, but I prefer to do it this way.)
This is just for pedagogical purpose,
so I like to keep the simple-minded 'for'
loop.
But what I really wonder is why do I not get the right
I am caught in a mental trap. Why isn't the between groups variance estimated
(0.0038) to be around the value with which I generated the data (0.0002)?
Thanks Toby
set.seed(76589437887)
fph = 0.4
Sigh = sqrt(0.0002)
Sigi = sqrt(0.04)
ci = 1
fpi = matrix(,7200,3)
for (i in 1:90) {
fph =
Hi
I spent hours looking over my formula. Somehow I cant find the reason
why it gives me different answer.
help appreciated.
x =
as.matrix(read.table("http://www.niehs.nih.gov/research/atniehs/core/microarrays/docs/heinloth.txt",1))
x = t(x)#now rows are subjects, cols are genes
x = x[or
uups, ok, I see, thanks
Tim Calkins wrote:
> check the dimensions of your X and P matrices:
>
>> dim(X)
> [1] 21 9
>> dim(P)
> [1] 9 21
>
> what you see is that the final row is *named* 26. It's actually row 21.
>
> cheers,
>
> On 10/29/07, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
>> Hi A
Hi All
Maybe I dont understand it, but I would have expected that the design matrix
has
as many rows as there were observations available to fit the model.
Below a small artificial dataset created, then one model fitted and the design
matrix outputted, having 27 rows. Then I delete 6 obs, and f
only one out of many many
dt = cbind(sim$mean$u2+sim$mean$beta1,sim$sd$u2)
dt = dt[order(dt[,1]),]
bounds =
cbind(c(dt[,1]-1.96*dt[,2],dt[,1]+1.96*dt[,2],dt[,1]),rep(1:length(sim$mean$u2),3))
bounds = bounds[order(bounds[,2]),]
plot(bounds)
T
Matthew Krachey wrote:
> I'm trying to compare the
Hi All
I can specify whatever inits, it has no effect on the estimation. I am
replicating a textbook example. The result is completely trash, having
estimates
of -58.7 (sd=59.3), where it should be closer to an ml estimate of 0.585
(SE=0.063).
The two chains within one run are different, but
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