Li, Jari, and Mike,

If you *really* don't like nonparametics, try a permutation t-test.  Here's the code for Mike's data, as well as the output I got after running it with 9999 permutations:

> #input the data into x, use g to determine the groups that data go into
> x<-c(13,0,10,2,0,0,1,0,0,1,5,0,0,1,0,0,0,0,0,0,0,1)
> g<-factor(c(1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2))
>
> #set up a matix a with nothing in it
> a<-c()
>
> #split x into two groups, defined by g
> x.g <- split(x,g)
> x.g
$`1`
 [1] 13  0 10  2  0  0  1  0  0  1  5

$`2`
 [1] 0 0 1 0 0 0 0 0 0 0 1

>
> # do a t-test, report the results
> z <- t.test(x.g$"1",x.g$"2", var.equal = TRUE)
> z

Two Sample t-test

data:  x.g$"1" and x.g$"2"
t = 1.9807, df = 20, p-value = 0.06154
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.1449011  5.5994465
sample estimates:
mean of x mean of y
2.9090909 0.1818182

>
> #permute the data, split again and report
> xperm <- sample(x,NROW(x))
> xperm
 [1]  0  1  1  0 13  1  0  0 10  0  0  0  1  0  0  0  0  5  0  0  0  2
> xperm.g <- split(xperm,g)
> xperm.g
$`1`
 [1]  0  1  1  0 13  1  0  0 10  0  0

$`2`
 [1] 0 1 0 0 0 0 5 0 0 0 2

>
> #permute the data 9999 times, and every time,
> # compute a t-value, store its absolute value in matrix a
> for(i in 1:9999) {
+   xperm <- sample(x,NROW(x))
+   xperm.g <- split(xperm,g)
+   z <- t.test(xperm.g$"1",xperm.g$"2", var.equal = TRUE)
+   a[i] <- abs(z$statistic)
+ }
>
> #recompute the absolute value of the t-value,
> # store it in the 10000 position of the array, report the matrix a
> x.g <- split(x,g)
> z <- t.test(x.g$"1",x.g$"2")
> a[10000] <- abs(z$statistic)
> #a
> #show rank of your actual t-value against
> # all the other t-values from the permutations
> rank(a)[10000]
[1] 9517.5
>
> #calculate p-value as proportion of t-values from permuations that
> # are greater than your actual t-value;
> # what is the chance that your t could be found just by chance?, or
> # "How likely is it that if the null hypothesis were true, I would
> # observe a value this extreme just due to chance?"
> # It is worth knowing that Fisher used randomization tests to test the value of
> # the t-test, F-tests, etc.
> p <- (10000-rank(a)[10000])/10000
> p
[1] 0.04825

So note that the P you get from this is less than 0.05!

Like any permutation test, the P is only approximate.

Cheers, Rick

On Mar 24, 2014, at 8:45 AM, Li Wen <Li.Wen@environment.nsw.gov.au<mailto:Li.Wen@environment.nsw.gov.au>> wrote:

HI, Jari

The default in the Welch t test (an adaptive student's t test) doesn't assume equal variance; but student's t-test do assume normal distribution.

Cheers
Li

-----Original Message-----
From: r-sig-ecology-bounces@r-project.org<mailto:r-sig-ecology-bounces@r-project.org> [mailto:r-sig-ecology-bounces@r-project.org] On Behalf Of Jari Oksanen
Sent: Monday, 24 March 2014 10:43 PM
To: Richard Boyce; r-sig-ecology@r-project.org<mailto:r-sig-ecology@r-project.org>
Subject: Re: [R-sig-eco] report out by t.test

Except that t-test does not assume that *observations* are normally distributed, nor that variances are equal.

Avoid non-parametric tests: they assume too much of data properties.

For var.equal assumption in t.test, see ?t.test.

Cheers, Jari Oksanen
________________________________________
From: r-sig-ecology-bounces@r-project.org<mailto:r-sig-ecology-bounces@r-project.org> [r-sig-ecology-bounces@r-project.org<mailto:r-sig-ecology-bounces@r-project.org>] on behalf of Richard Boyce [boycer@nku.edu<mailto:boycer@nku.edu>]
Sent: 24 March 2014 13:23
To: r-sig-ecology@r-project.org<mailto:r-sig-ecology@r-project.org>
Subject: Re: [R-sig-eco] report out by t.test

Mike,

There is no way that your data meet the assumptions of a t-test (normal distributions, equal variance). A nonparametric Mann-Whitney (aka Wilcoxon) test is much better suited to your data.

Here's what I got when I ran it:

Q<-c(13,0,10,2,0,0,1,0,0,1,5)
WD<-c(0,0,1,0,0,0,0,0,0,0,1)
wilcox.test(Q,WD)

Wilcoxon rank sum test with continuity correction

data:  Q and WD
W = 86.5, p-value = 0.05119
alternative hypothesis: true location shift is not equal to 0

Warning message:
In wilcox.test.default(Q, WD) : cannot compute exact p-value with ties

This has a p-value quite close to 0.05, giving some evidence that there's a difference between your groups. Note that this you have different null and alternative hypothesis: groups are the same vs. groups are different.

Rick Boyce

On Mar 24, 2014, at 7:00 AM, r-sig-ecology-request@r-project.org<mailto:r-sig-ecology-request@r-project.org><mailto:r-sig-ecology-request@r-project.org> wrote:

Message: 1
Date: Sun, 23 Mar 2014 14:21:41 -0700
From: Michael Marsh <swamp@blarg.net<mailto:swamp@blarg.net><mailto:swamp@blarg.net>>
To: r-sig-ecology@r-project.org<mailto:r-sig-ecology@r-project.org><mailto:r-sig-ecology@r-project.org>
Subject: [R-sig-eco] report out by t.test
Message-ID: <532F5065.7030102@blarg.net<mailto:532F5065.7030102@blarg.net><mailto:532F5065.7030102@blarg.net>>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed

I test differences between frequency of hits of exotic annual forbs in plots on  two sites, Q and WD.

Q<-c(13,0,10,2,0,0,1,0,0,1,5)
WD<-c(0,0,1,0,0,0,0,0,0,0,1)
t.test(Q,WD)

       Welch Two Sample t-test

data:  Q and WD
t = 1.9807, df = 10.158, p-value = 0.07533 alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.3342006  5.7887460
sample estimates:
mean of x mean of y
2.9090909 0.1818182

The p-value is greater than 0.05, thus does not reach the 95% confidence level, yet the difference in means is reported as not equal to 0.
Am I encountering a one-sided versus two sided comparison that I don't understand, or is ther another explanation?

Mike Marsh




================================
Richard L. Boyce, Ph.D.
Director, Environmental Science Program
Professor
Department of Biological Sciences, SC 150 Northern Kentucky University Nunn Drive Highland Heights, KY  41099  USA

859-572-1407 (tel.)
859-572-5639 (fax)
boycer@nku.edu<mailto:boycer@nku.edu><mailto:boycer@nku.edu>
http://www.nku.edu/~boycer/
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"One of the advantages of being disorderly is that one is constantly making exciting discoveries." - A.A. Milne






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Richard L. Boyce, Ph.D.
Director, Environmental Science Program
Professor
Department of Biological Sciences, SC 150
Northern Kentucky University
Nunn Drive
Highland Heights, KY  41099  USA

859-572-1407 (tel.)
859-572-5639 (fax)
boycer@nku.edu<mailto:boycer@nku.edu>
http://www.nku.edu/~boycer/
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"One of the advantages of being disorderly is that one is constantly making exciting discoveries." - A.A. Milne






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