## Example implementing duplicate (regenotyped) data in a test 
## of genetic association 

## Please use the following reference when using this software:

## Reference: NL Tintle, D Gordon, F McMahon, SJ Finch
## Using duplicate genotyped data in genetic analyses: 
## testing association and estimating error rates, SAGMB, 2007

## Statistical Applications in Genetics and Molecular Biology. 
## Accepted December 4, 2006.

## Script last updated 6-3-08
 
## Script Author: Dirk VanBruggen, Hope College

## Contact: tintle@hope.edu

## Example: Consider a situation where 100 individuals are genotyped once
## and the other 100 individuals are genotyped twice. 
## Assume we have a case-control study and so there are 100 cases 
## and 100 controls.  Thus, there are 50 cases that are genotyped once, 
## 50 cases that are genotyped twice, 50 controls that
## are genotyped once and 50 controls that are genotyped twice.

## Assume that of the singly genotyped cases: 35 are classified AA, 
## 10 are classified AB, and 5 are classified BB.  While, of the 
## duplicate genotpyed cases, 27 are genotyped to AA both times,
## 5 are genotyped AA once and AB once, 7 are genotyped to AB twice, 
## 0 are genotyped to AA once and BB once, 2 are genotyped to AB once 
## and BB once, and 9 are genotpyed to BB both times.

## Assume that of the singly classified controls: 20 are classified AA,
## 20 are classified AB, and 10 are classified BB.  While, of the 
## duplicate genotpyed controls, 17 are genotyped to AA both times,
## 3 are genotyped AA once and AB once, 19 are genotyped to AB twice, 
## 1 is genotyped to AA once and BB once, 6 are genotyped to AB once 
## and BB once, and 4 are genotpyed to BB both times.

## Then, we have a 2x3 table of singly genotyped data as follows:

## 				 |   AA	  |   AB   |   BB   |   Total   |
## 	-----------------------------------------------------------------
## 	Cases  		|   35   |   10   |   5    |   50   |   100     |
## 	-----------------------------------------------------------------
## 	Controls	|   20   |   20	  |   10   |   50   |   100     |
## 	-----------------------------------------------------------------
## 	Total		|   55	 |   30   |   15   |   100  |   200     |


## And, we also have a 2x6 table of duplicate genotyped data as follows:
## 
##	   |  AA both  |  AA once  |  AB both  |  AA once   |  AB once  |  BB both  | Total|
##	   |  times    |and AB once|   times   |and BB once |and BB once|  times    |      |
##   ---------------------------------------------------------------------------------------
## Cases   |    27     |    5	   |     7     |    0       |     2     |    9      |  50  |
## 	------------------------------------------------------------------------------------
##Controls |    17     |    3      |     19    |    1       |     6     |    4      |  50  |
## 	------------------------------------------------------------------------------------
## Total   |    44     |    8      |     26    |    1       |     8     |    13     | 100  |
 
## To run the MANOVA program, create a vector of values containing the 
## cell counts in the following order: 
## na1,na2,na3,nu1,nu2,nu3,na11,na12,na22,na13,na23,na33,nu11,nu12,nu22,
## nu13,nu23,nu33.  Thus, 

data=c(35,10,5,20,20,10,27,5,7,0,2,9,17,3,19,1,6,4)


## Now you can run the Manova test on the data but entering this command:

->permtest(data,0)

## Zero is used to indicate zero permutations.  
## The permutation test functionality will be added in
## the near future.
## THE CODE ABOVE YIELDS THE FOLLOWING OUTPUT
##
##       typicalchisq  typicalpval manovats manovalpval
##  [1,] 14.91082 0.0005783059 8.414398 0.000311461
##
## Note: typicalchisq and typical pval is the value of the chisquared
## test statistic and p-value ignoring inconsistently identified
## individuals.  manovats and manovapval are the values of the MANOVA
## F test statistic and p-value which includes inconsistently identified 
## individuals.