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MathGroup Archive 2006

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Re: The D'Agostino Pearson k^2 test implemented in mathematica / variance of difference sign test

  • To: mathgroup at smc.vnet.net
  • Subject: [mg65000] Re: [mg64990] The D'Agostino Pearson k^2 test implemented in mathematica / variance of difference sign test
  • From: leigh pascoe <leigh at cephb.fr>
  • Date: Sat, 11 Mar 2006 05:15:36 -0500 (EST)
  • References: <200603101015.FAA22041@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

john.hawkin at gmail.com wrote:
> Hello,
>
> I have two questions.
>
> 1.  Are there any resources of .nb files available on the internet
> where I might find an implementation of the D'Agostino Pearson k^2 test
> for normal variates?
>
> 2.  In the mathematica time series package (an add-on), the
> "difference-sign" test of residuals is mentioned (url:
> http://documents.wolfram.com/applications/timeseries/UsersGuidetoTimeSeries/1.6.2.html).
>  It says that the variance of this test is (n+1) / 2.  However, it
> would seem to me that a simple calculation gives a variance of (n-1)/4.
>  It goes as follows:
>
> If the series is differenced once, then the number of positive and
> negative values in the difference should be approximately equal.  If Xi
> denotes the sign of each value in the differenced series, then
> Mean(Xi) = 0.5(1) + 0.5(0) = 0.5
> Var(Xi) = Expectation( (Xi - Mean(Xi))^2 )
> = Expectation( Xi^2 -Xi + 0.25 )
> = 0.5 - 0.5 + 0.25
> = 0.25
>
> And assuming independence of each sign from the others, the total
> variance should be the sum of the individual variances, up to n-1 for n
> data points (since there are only n-1 changes in sign), thus
>
> Variance = (n-1) / 4
>
> There is an equivalent problem in Lemon's "Stochastic Physics" about
> coin flips, for which the answer is listed, without proof, as (n-1)/8.
> Because of these three conficting results I am wondering if I have made
> an error in my calculation, and if anyone can find one please let me
> know.
>
> Thank you very much,
>
> -John Hawkin
>
>
>
>   
Dear John,

I agree with your calculation. We assume an indicator variable that is 1 
if the succeeding value is greater or zero if it is less than the 
preceding one. Under the null hypothesis this is just a binomial 
variable with p=.5 and n-1 trials. The mean and variance of the number 
of successes must be proportional to n-1 and are in fact the values you 
calculated.

I don't have Lemon's book, so I can't comment on the equivalence of the 
problems. Perhaps the factor of 2 is due to the way the problem is 
stated w.r.t heads and tails.

LP


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