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



Luis A. Afonso
Posts:
4,758
From:
LIsbon (Portugal)
Registered:
2/16/05


The Wilson  Hilferty (1931) upsidedown
Posted:
Jun 15, 2013 5:40 PM


The Wilson  Hilferty (1931) upsidedown
The direct way provides a normal standard r.v. z from a chisquare y with r degrees of freedom: z = [(y/r)^(1/3)  (1 2/(9*r)]/sqrt(2/(9*r)) Inversely, z*sqrt(2/(9*r)) + (12/(9*r)) = (y/r)^(1/3) _______ Corollary: With s=2/(9*r) we get y= r*[z*sqrt(s) + (1s)]^3 ___(1) y being a chisquare 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. semiamplitude ssdX ,ssdY, the sum of squared deviations, +/ t(nX+ nY2) * k * sqrt(ssdX+ ssdY) Here k= sqrt[1/(nX+ nY 2)*(1/nX+ 1/nY)] Because the quantity ssdX+ ssdY have a Chisquare distribution, nX+ nY 2 df, the result, although approximate, do follow demands only one step.
Luis A. Afonso



