Re: Chi Square Areas
- To: mathgroup at smc.vnet.net
- Subject: [mg110943] Re: Chi Square Areas
- From: Steve <s123 at epix.net>
- Date: Tue, 13 Jul 2010 05:26:26 -0400 (EDT)
- References: <i199d5$q45$1@smc.vnet.net> <i1c5vc$cgf$1@smc.vnet.net>
On Jul 11, 6:21 am, Norbert Marxer <mar... at mec.li> wrote: > On Jul 10, 10:01 am, Steve <s... at epix.net> wrote: > > > > > > > Hi, > > > Can someone show me how to get Mathematica to provide the areas to the > > right of a given critical value of the Chi Square distribution ? > > > The table entries shown athttp://www2.lv.psu.edu/jxm57/irp/chisquar.html > > are what I need to compute. > > > For example, given 8 degrees of freedom and a probability value of > > 0.05 the result would be 15.51. > > > And given 5 degrees of freedom with probability 0.1 the result is > > 9.24. > > > How can I produce these results in Mathematica ? > > > Thanks so much. > > Hello > > You can use the built-in functions: > > InverseCDF[ChiSquareDistribution[8], 0.95] > InverseCDF[ChiSquareDistribution[5], 0.9] > > which will give you the numbers 15.5073 and 9.23636. > > But note, these numbers represent the critical values! The areas (to > the right) are given by 0.05 (5%) and 0.10 (10%) respectively. > > Best Regards > Norbert Marxer- Hide quoted text - > > - Show quoted text - Thanks to everyone who replied. 15 minutes after posting my question, I figured it out on my own, albeit a not too elegant solution. Norbert provided a particularly elegant solution and also pointed out a small inconsistency in my question. The values I needed returned were the critical values not the areas as I had stated at the beginning of my post. Below is a parameter-consistent comparison of my solution and Norbert's. Thanks again to all, Steve DOF = 8; confidence = .90; p = (1 - confidence)/2 ChiPDF = PDF[ChiSquareDistribution[DOF], x] Plot[ChiPDF, {x, 0, 50}] guessvalue = 2.5; stevesolution = FindRoot[NIntegrate[ChiPDF, {x, CriticalValue, Infinity}] == p, {CriticalValue, guessvalue}]; stevesolution = stevesolution[[1, 2]] norbertsolution = InverseCDF[ChiSquareDistribution[DOF], confidence + p] difference = stevesolution - norbertsolution