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Topic: The Wilson - Hilferty (1931) upside-down
Replies: 1   Last Post: Jun 16, 2013 6:17 AM

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Luis A. Afonso

Posts: 4,758
From: LIsbon (Portugal)
Registered: 2/16/05
The Wilson - Hilferty (1931) upside-down
Posted: Jun 15, 2013 5:40 PM
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The Wilson - Hilferty (1931) upside-down

The direct way provides a normal standard r.v. z from a chi-square y with r degrees of freedom:
z = [(y/r)^(1/3) - (1- 2/(9*r)]/sqrt(2/(9*r))
z*sqrt(2/(9*r)) + (1-2/(9*r)) = (y/r)^(1/3)
With s=2/(9*r) we get y= r*[z*sqrt(s) + (1-s)]^3 ___(1)
y being a chi-square r.v. obtained from a z~N(0,1).

Numerical Example
With z= 1, r= 5
s=0.0444. . . y= 7.9338 (from Kalkulator)
Then p(chi,5df < y) = 0.8401
In fact p(z< 1) = 0.8413 is the exact value.

A direct application, which avoids the simulation of the samples, is that we can obtain the results from difference of normal sample means, same variance, by the C.I. semi-amplitude ssdX ,ssdY, the sum of squared deviations,
+/- t(nX+ nY-2) * k * sqrt(ssdX+ ssdY)
Here k= sqrt[1/(nX+ nY- 2)*(1/nX+ 1/nY)]
Because the quantity ssdX+ ssdY have a Chi-square distribution, nX+ nY- 2 df, the result, although approximate, do follow demands only one step.

Luis A. Afonso

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