R help - Jul 2008 - Interpretation of EXACT Statistical Test in finding the probability as Std. Deviations (SumP)

Okay, so I'm fairly retarded, and asked a question about finding the T-Value
in the Fisher Exact method.  I suppose what I'm truly after can best be
explained by the Biddle Consulting site that has a program setup to deal
with this kind of thing.  Unfortunately, it is not currently functioning,
and has not been for some time.  As a result, I'm trying to figure out how
to do this on my own.

What I'm after is the "Probability as Std. Deviations (SumP)"
which can be
seen here; 
http://www.biddle.com/adverseimpacttoolkit/SelectionRateComparison.aspx

I appreciate the time everyone took in getting me straightened out on my
previous question.  Hopefully I've provided a somewhat more meaningful
scenario here.
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On 10-Jul-08 20:43:12, DaveFrisch wrote:
> 
> Okay, so I'm fairly retarded, and asked a question about finding
> the T-Value in the Fisher Exact method.  I suppose what I'm truly
> after can best be explained by the Biddle Consulting site that has
> a program setup to deal with this kind of thing.  Unfortunately,
> it is not currently functioning, and has not been for some time.
> As a result, I'm trying to figure out how to do this on my own.
It is working now.
> What I'm after is the "Probability as Std. Deviations (SumP)"
> which can be seen here;
> http://www.biddle.com/adverseimpacttoolkit/SelectionRateComparison.aspx
> 
> I appreciate the time everyone took in getting me straightened out on
> my previous question.  Hopefully I've provided a somewhat more
> meaningful scenario here.
>From the explanation in Note [4] of that web-page, it seems that
what you are after corresponds to:

  fisher.test(matrix(c(11,5,7,12),ncol=2))
  #      Fisher's Exact Test for Count Data
  # data:  matrix(c(11, 5, 7, 12), ncol = 2) 
  # p-value = 0.09222
  # alternative hypothesis: true odds ratio is not equal to 1 
  # 95 percent confidence interval:
  # 0.7619524 19.6952184 
  # sample estimates:
  # odds ratio 
  # 3.621276 

  fisher.test(matrix(c(11,5,7,12),ncol=2))$p
  # [1] 0.09221518

  qnorm(1-fisher.test(matrix(c(11,5,7,12),ncol=2))$p/2)
  # [1] 1.683827

Hoping this helps!
Ted.

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E-Mail: (Ted Harding) <Ted.Harding at manchester.ac.uk>
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Date: 10-Jul-08                                       Time: 22:52:31
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