## 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.