[R] exact trend test

David Winsemius dwinsemius at comcast.net
Thu Jan 7 08:31:14 CET 2016

```> On Jan 6, 2016, at 8:16 PM, li li <hannah.hlx at gmail.com> wrote:
>
> Hi all,
>  Is there an R function that does exact randomization trend test?
>  For example, consider the 2 by 5 contingency table below:
>
>            dose0    dose 0.15    dose 0.5    dose 1.5    dose 5       row
> margin
> Yes          4                3                  4               5
>     8                   24
>  No          4                5                   4               3
>       0                  16
> col sum    8                8                   8               8
>   8                   40

Your data presentation has been distorted by your failure to post in plain text. Surely you have been asked in the past to correct this issue?

>
> To do the exact trend test, we need to enumerate all the contingency table
> with the
> row and column margins fixed.

Er, how should that be done? A trend test? What is described above would be a general test of no association rather than a trend test. Please use clear language and be as specific as possible if you choose to respond.

> Find the probability corresponding to
> obtaining
> the corresponding contingency tables based on the multivariate
> hypergeometric distribution. Finally the pvalue is obtained by adding
> relevant probabilities.

If there is a trend under consideration, then I do not understand such a trend would be modeled under a hypergeometric distribution? A hypergeometic distribution would suggest no trend, at least to my current understanding.

>
> Is there an R function that does this? if not, I am wondering whether it is
> possible to
> enumerate all possible contingency tables that has column sun and row sum
> fixed?

Wel, yes, that is possible and routinely done with `fisher.test`, but it is up to you to describe how that activity leads to a trend test.

If you assume Poisson distributed errors a trend test is fairly easy to construct with glm.

--
David.
>
> Thanks very much!!
>
>   Hanna
>
> 	[[alternative HTML version deleted]]
>
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David Winsemius
Alameda, CA, USA

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