OK, I should have coffee before sitting at the computer... sorry about
that.  Please ignore the original email as I'm sure you noticed an obvious
lapse in judgement.  I will try this again in a minute.

On Thu, May 10, 2012 at 11:06 AM, Charlie Brown <charliethebrown77@gmail.com
> wrote:

> Hello,
> I have a repeated measures situation and would like some help to determine
> if this is 1) appropriate, and 2) how to properly interpret the results.
> Unfortunately, these data are very limited (but I have been asked to
> analyze them anyway).  Here is an example of the data:
>
> Basically, I have repeated measures of "Y" recorded in consecutive years
> (94:98) taken at approximately the same time (July) at the same 6 sites; Y
> is a percentage that I would like to regress along the continuous variable
> called "env.gradient".
>
> # Data
> #######################################################################
> dat <- structure(list(site = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L,
> 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L,
> 5L, 5L, 6L, 6L, 6L, 6L, 6L), .Label = c("C1", "C2", "C3", "Q1",
> "Q2", "Q3"), class = "factor"), year = c(94L, 95L, 96L, 97L,
> 98L, 94L, 95L, 96L, 97L, 98L, 94L, 95L, 96L, 97L, 98L, 94L, 95L,
> 96L, 97L, 98L, 94L, 95L, 96L, 97L, 98L, 94L, 95L, 96L, 97L, 98L
> ), env.gradient = c(18, 18, 18, 18, 18, 3.5, 3.5, 3.5, 3.5, 3.5,
> 14, 14, 14, 14, 14, 13, 13, 13, 13, 13, 8, 8, 8, 8, 8, 6.5, 6.5,
> 6.5, 6.5, 6.5), Y = c(0.171158638926407, 0.169713288641782,
> 0.152153933837569,
> 0.197575217418616, 0.203910975101867, 0.0750279002284801,
> 0.170743241890423,
> 0.124641793138257, 0.114178607721316, 0.094511173181886,
> 0.246397132453605,
> 0.204852712848377, 0.127234745915353, 0.250078422865994,
> 0.167826032256711,
> 0.179431760128959, 0.197893902312921, 0.129699846863689,
> 0.149706090438842,
> 0.138852795878536, 0.125088313507783, 0.178352158019673, 0.116281351771267,
>
> 0.147501950163454, 0.118865506981108, 0.0924414290963082,
> 0.144984499149044,
> 0.132532609540349, 0.137715018373314, 0.134704824360581)), .Names =
> c("site",
> "year", "env.gradient", "Y"), class = "data.frame", row.names = c(NA,
> -30L))
>
> # Analysis
> #########################################################################
> # I am trying to follow Venables and Ripley (2002), Chapter 10.
> # I am likely compleely misguided on how to proceed, so please correct me
>
> dat1.lm <- with(dat, lm(Y~as.factor(year)/env.gradient - 1))
> summary(dat1.lm)
>
> # if I understand this right, and looking at a graph of the data, this
> seems correct, the 1st 6 rows (e.g. "as.factor(year)94  0.0485") are the
> intercepts, and the rest are the slope estimates (which makes sense, I've
> never run a regression like this). It looks like there is a relatively
> large variation in both slopes and intercepts with these data.  I will
> assume that I cannot adopt a parallel-line or even same slope model.
>
> # What does "-1" mean in that formula?
>
> # now to the fun stuff
> library(nlme)
>
> dat1.lme <- lme(Y ~ as.factor(year) + env.gradient , random=~1|site,
> data=dat)
> summary(dat1.lme)
>
> # What does "~1" mean in that formula.... random is the 1st variable in
> the formula given the site ??
>
> # interpretation
> #############################################################################
> From the looks of this summary, all null hypotheses regarding slope are
> rejected at the 5% significance level except for 1994, which looking at the
> graph, I can see that it likely isn't sig. different from zero.  Further,
> all of the slopes are significant except for 1995 and 1996, as these are
> relatively flat.  The overall model is significant, but I don't really
> understand what that means regarding 95 and 96 since they were not...
>
> Can you please tell me what I have done wrong, need to do additionally,
> and how to properly interpret this.
>
> I also have checked none of the assumptions of a linear model at this
> time, are these the same as with a simple linear model (which should I be
> most concerned about).
>
>
> THANK YOU for any advice.
> Chuck
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