How to carry out multiple post-hoc chi-square tests on a 2 X 3 table? My data set is comprised of either total mortality or survival of an organism at three site types, inshore, midchannel and offshore. The numbers in the table below represent the number of sites. 
              100% Mortality            100% Survival
Inshore             30                       31 
Midchannel          10                       20 
Offshore             1                       10

I would like to know if the # of sites where 100% mortality occurred is significant based on site type.
If I run a 2 x 3 chi-square, I get a significant result. Is there a post-hoc pairwise comparison that I can run or should I actually be using a logistical ANOVA or regression with binomial distribution? 
Thanks!
 A: A contingency table should contain all the mutually exclusive categories on both axes. Inshore/Midchannel/Offshore look fine, however unless "less than 100% mortality" means "100% survival" in this biological setting you may need to construct tables that account for all the cases observed or explain why you restrict your analysis to the extreme ends of the sample.
As 100% survival means 0% mortality, you could have a table with columns 100%=mortality / 100%>mortality>0% / mortality=0%. In this case you wouldn't any more compare percentages, but compare ordinal mortality measures across three site type categories. (What about using the original percentage values instead of categories?) A version of Kruskal-Wallis test may be appropriate here that takes ties appropriately into consideration (maybe a permutation test).
There are established post hoc tests for the Kruskal-Wallis test: 1, 2, 3. (A resampling approach may help tackling with ties.)
Logistic regression and binomial regression may be even better as they not only give you p values, but also useful estimates and confidence intervals of the effect sizes. However to set up those models more details would be needed concerning the 100%>mortality>0% sites.
A: Here is the code to do the chi square tests as well as generate a variety of test statistics. However, statistical tests of association of the table margins are useless here; the answer is obvious. No one does a statistical test to see if summer is hotter than winter. 
Chompy<-matrix(c(30,10,1,31,20,10), 3, 2)
Chompy
chisq.test(Chompy)
chisq.test(Chompy, simulate.p.value = TRUE, B = 10000)
chompy2<-data.frame(matrix(c(30,10,1,31,20,10,1,2,1,2,1,2,1,2,3,1,2,3), 6,3))
chompy2
chompy2$X2<-factor(chompy2$X2) 
chompy2$X3<-factor(chompy2$X3)
summary(fit1<-glm(X1~X2+X3, data=chompy2, family=poisson))
summary(fit2<-glm(X1~X2*X3, data=chompy2, family=poisson)) #oversaturated
summary(fit3<-glm(X1~1, data=chompy2, family=poisson)) #null
anova(fit3,fit1)
library(lmtest)
waldtest(fit1)
waldtest(fit2) #oversaturated
kruskal.test(X1~X2+X3, data=chompy2)
kruskal.test(X1~X2*X3, data=chompy2)

A: I believe you could use the "simultaneous confidence intervals" for doing multiple comparisons. The reference is Agresti et al. 2008 Simultaneous confidence intervals for comparing binomial parameters. Biometrics 64 1270-1275. 
You could find the corresponding R code in http://www.stat.ufl.edu/~aa/cda/software.html
A: You can do proportion testing for 100%Mortality vs totals:
                Inshore  Midchannel  Offshore  Inshore%  Midchannel%  Offshore%   Pvalue
 FullMortality       30          10         1    49.180       33.333      9.091    0.029 
 FullSurvival        31          20        10 
 Total               61          30        11 

P=0.029 indicates that there is a significant difference between different sites for 100%Mortality.
