Command tutorial

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mcri...@gmail.com

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Feb 8, 2018, 9:16:06 AM2/8/18
to dnenrich-users
I am trying to access to the command tutoria (https://psychgen.u.hpc.mssm.edu/dnenrich/command_tutorial.shtml) in your website and it is completely impossible. I would appreciate if anyone could sent me the dnenrich command tutorial
Thank you very much in advance
Best regards
Cristina

https://psychgen.u.hpc.mssm.edu/dnenrich/

Menachem Fromer

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Feb 8, 2018, 11:27:17 PM2/8/18
to mcri...@gmail.com, dnenrich-users
OK, this should all be working now - I fixed some transient errors.

Enjoy!
Menachem

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mcri...@gmail.com

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Feb 22, 2018, 4:05:06 AM2/22/18
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We've been trying to run Dnenrich in different data sets. We would like to answer you some questions regarding the p-value results. In some cases, the p-value is around 9,999e-005 which doesn't make any sense. Therefore, we have many analysis which have resulted in the same p-value. We have increased the number of permutation to 1000000 but the results have only changed to  9,999e-007. We would appreciate any help regarding this issue. We are not completely sure about this.
Thank you very much
Best regards
Cristina 

Menachem Fromer

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Feb 22, 2018, 9:50:08 AM2/22/18
to mcri...@gmail.com, dnenrich-users
Since dnenrich uses permutation, the p-value (under the dnenrich null model) of the particular event you are testing is less than 10^-4 and even 10^-6. You can either report that p < 10^-6, or keep increasing the number of permutations until it gives you a better estimate that is not the lowest possible for that number of permutations. Of course, the latter may require you to wait for a while. You could however manually parallelize it by running 100 runs of 10^6 permutations each (10^8 permutations in total) and then:
1. multiply each run’s p-value by (10^6 + 1)
2. Subtract 1 from each result in 1
3. Add up all the results from step 2
4. Add 1 to the sum of events from step 3
5. Divide the result from 4 by (10^8 + 1)

Note that the whole thing here with 1 (adding or subtracting) is because dnenrich always counts the original observation of the event in question once in the numerator and once in the denominator.
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