[R] survival object

2011-04-09 Thread Eugenio Larios
Hi All,

I am trying to do a survivorship analysis with library(survival)from a data
set that looks like this:
I followed a bunch of naturally germinated seedlings of an annual plant from
germination to death (none made it to reproduce, and died in a period of ~60
days after germination.)
I also know the size of the seed of every individual censused. So I am
trying to analyze seedling survival as a function of seed size.

I performed 5 censuses in unequal intervals of time starting 15 days after
germination until everyone died. Does that make my data right censored?
So I have the following variables: seed size (as a continuous variable and
as a categorized variable in big and small with 0=small and 1=big), the 5
census events (with 0=dead 1=survivors)

First, I want to make a survival object with Surv() but apparently this
function only takes two intervals of time (time, and time2). Is there a way
to include my five census events in it?
With the survival object I can go on and fit a model with
survfit(survivalobject~seedsize,data=mydata), right?

thanks
-- 
Eugenio Larios
PhD Student
University of Arizona.
Ecology  Evolutionary Biology.
elari...@email.arizona.edu

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[R] SMATR common slopes test

2010-11-06 Thread Eugenio Larios
Hi All,

I am confused with SMATR's test for common slope. My null hypothesis here is
that all slopes are parallel (common slopes?), right?
So if I get a p value  0.05 means that we can have confidence to reject it?
That slopes are different?
Or the other way around? it means that we have statistical confidence that
the slopes are parallel?
thanks
-- 
Eugenio Larios
PhD Student
University of Arizona.
Ecology  Evolutionary Biology.
(520) 481-2263
elari...@email.arizona.edu

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Re: [R] SMATR common slopes test

2010-11-06 Thread Eugenio Larios
great thanks a lot!

On Sat, Nov 6, 2010 at 3:54 PM, Kevin Middleton k...@csusb.edu wrote:


 Eugenio -

  I am confused with SMATR's test for common slope. My null hypothesis here
 is
  that all slopes are parallel (common slopes?), right?
  So if I get a p value  0.05 means that we can have confidence to reject
 it?
  That slopes are different?
  Or the other way around? it means that we have statistical confidence
 that
  the slopes are parallel?

 Try this:

 set.seed(5)
 n - 20
 x - rnorm(n)

 y1 - 2 * x + rnorm(n)
 y2 - 2 * x + rnorm(n)
 y3 - 4 * x + rnorm(n)

 # Slopes approximately equal
 slope.com(x = c(x, x), y = c(y1, y2), groups = rep(c(1,2), each = n))

 #$p
 #[1] 0.4498037

 # Slopes of 2 and 4
 slope.com(x = c(x, x), y = c(y1, y3), groups = rep(c(1,2), each = n))

 #$p
 #[1] 0.0003850332


 Cheers,
 Kevin

 -
 Kevin M. Middleton
 Department of Biology
 California State University San Bernardino




-- 
Eugenio Larios
PhD Student
University of Arizona.
Ecology  Evolutionary Biology.
(520) 481-2263
elari...@email.arizona.edu

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Re: [R] help with an unbalanced split plot

2010-10-15 Thread Eugenio Larios
Hi Dennis,

The first thing I did with my data was to explore it with 6 graphs
(wet-high, med, and solo-; dry-high, med, and solo-) and gave me very
interesting patterns: seed size in wet treatments is either negatively
correlated (high and medium densities) or flat (solo). But dry treatments
are all positively correlated! There is a very interesting switch there.

I also figured out why I can't do three way interactions. I explored the
structure of my data with str(mydata) and it shows that water treatment has
three levels when it should have just two. Then I went back to the excel
sheet, tried to sort the data by water treatment and I discover a single
data point from the wet treatment sticking out by itself. That is why R
reads three levels and since it is only one point, there cannot be any stats
of course.

thanks
E

On Thu, Oct 14, 2010 at 9:27 PM, Dennis Murphy djmu...@gmail.com wrote:

 Hi:

 On Thu, Oct 14, 2010 at 7:50 PM, Eugenio Larios 
 elari...@email.arizona.edu wrote:

 Hi Dennis,

 thank you very much for your help, I really appreciate it.

 I forgot to say about the imbalance, yes. I only explained the original
 set up, sorry. Let me explain.

 It is because in the process of the experiment which lasted 3 months I
 lost individuals within the plots and I actually ended up losing 2 whole
 plots (one dry and one wet) and some other individuals in other plots.


 That still leaves you balanced at the plot level :)  Fortunately, you have
 enough replication. If you have missing subplots within the remaining plots,
 that would be another source of imbalance at the subplot level, but you
 should have enough subplots to be able to estimate all of the interactions
 unless an entire treatment in one set of plots was missing.

 It's worth graphing your data to anticipate which effects/interactions
 should be significant; graphs involving the spatial configuration of the
 plots and subplots would also be worthwhile.


 My study system has this special feature that allows me to track parental
 seed sizes in plants germinated in the field, a persistent ring that stays
 attached to the root even when the plant has germinated, so some of the
 plants I lost did not have this ring anymore. It happens sometimes but most
 of the time they have it. Also, some plants disappeared probably due to
 predation, etc That made my experiment imbalanced.


 That's common. No big deal.


 Do you think that will change the analysis? Also, do you think I should
 use least squares ANOVA  (perhaps type III due to the imbalance?) instead of
 LMM? What about the random effects that my blocking has created?


 Actually, with unbalanced data it's to your advantage to use lme() over
 ANOVA. Just don't place too much importance on the p-values of tests; even
 the degrees of freedom are debatable. With unbalanced data, it's hard to
 predict what the sampling distribution of a given statistic will actually
 be, so the p-values aren't as trustworthy.

 You mentioned that you couldn't fit a three-way interaction; given your
 data configuration, that shouldn't happen.

 (1) Get two-way tables of water * density, one for the counts and one for
 the averages, something like

 with(mydata, table(water, density))
 aggregate(log(fitness) ~ water + density, data = mydata, FUN = mean, na.rm
 = TRUE)

 In the first table, unless you have very low frequencies in some category,
 your data 'density' should be enough to estimate all the main effects and
 interactions of interest. The second table is to check that you don't have
 NaNs or missing cells, etc.


 I am new to R-help website so I wrote you this message to your email but I
 would like to post it on the R website, do you know how?


 Wag answer: I hope so, since I managed to view and respond to your message
 :)

 More seriously, in gmail, the window that opens to produce replies has an
 option 'Reply to all'. I don't know if your e-mail client at UofA has that
 feature, but if not, you could always cc R-help and put the e-mail address
 in by hand if necessary. Most mailers are smart enough to auto-complete an
 address as you type in the name, so you could see if that applies on your
 system.

 I keep a separate account for R-help because of the traffic volume - if you
 intend to subscribe to the list, you might want to do the same. It's not
 unusual for 75-100 e-mails a weekday to enter your inbox...


 Thanks again!

 Eugenio


 On Thu, Oct 14, 2010 at 5:34 PM, Dennis Murphy djmu...@gmail.com wrote:

 Hi:

 On Thu, Oct 14, 2010 at 3:58 PM, Eugenio Larios 
 elari...@email.arizona.edu wrote:

 Hi Everyone,

 I am trying to analyze a split plot experiment in the field that was
 arranged like this:
 I am trying to measure the fitness consequences of seed size.

 Factors (X):
 *Seed size*: a continuous variable, normally distributed.
 *Water*: Categorical Levels- wet and dry.
 *Density*: Categorical Levels- high, medium and solo
 *Plot*: Counts from 1 to 20
 The *response variable *(Y

[R] help with an unbalanced split plot

2010-10-14 Thread Eugenio Larios
Hi Everyone,

I am trying to analyze a split plot experiment in the field that was
arranged like this:
I am trying to measure the fitness consequences of seed size.

Factors (X):
*Seed size*: a continuous variable, normally distributed.
*Water*: Categorical Levels- wet and dry.
*Density*: Categorical Levels- high, medium and solo
*Plot*: Counts from 1 to 20
The *response variable *(Y) was the number of seeds produced at the end of
the season.

The experiment started 15 days after plants germinated in the field.
20 plots were chosen where there was high enough density so I could
manipulate it. In an area where artificial irrigation was possible for the
wet treatment, dry treatment was natural precip.
Water was blocked so 10 plots were wet and the other 10 were dry. Randomly
assigned.
Within those 20 plots 6 focal plants were chosen and randomly assigned the
three densities. (split plot design)
I did not control for seed size since it is continuous and normally
distributed, hoping that with 120 plants total (6 in each 20 blocks) I could
get all kind of sizes for every treatment. It worked ok.

I have been trying to analyze this with lme (library NLME). I am not quiet
sure which are my random variables. models I have used are:

m-lme(log(fitness)~seedsize*density,random=~1|plot,data=dataset)
m-lme(log(fitness)~seedsize+density+water,random=~1|plot,data=dataset)

I have also tried to include plot and water as random effects:

m-lme(log(fitness)~seedsize+density+water,random=~1|plot/water,data=dataset)

I am actually not sure if I am using the right random variables here. Also
for some reason, it won't let me include seedsize*density*water triple
interaction
help!
thanks

-- 
Eugenio Larios
PhD Student
University of Arizona.
Ecology  Evolutionary Biology.
(520) 481-2263
elari...@email.arizona.edu

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