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Measuring averages (Chi squared)

  1. Nov 13, 2005 #1
    Really confused with this entire section of my course involving chi squared and least square fitting.

    I have an assignment question in which 3 measurements of a poisson distributed quantity of counts over a fixed period of time, say N1, N2, N3.The variances associated with these measurements are also N1, N2, N3 respectively. The problem shows two ways of calculating the average. Both involve setting up a Chi Squared function.

    The first gives:

    K^2 = SUM[((t-Ni)^2)/Ni)] from i=1->3

    The 2nd uses the fact that the mean of the poisson dist'n is (N1+N2+N3)/3 which is also equal to the variance. This gives:

    (3/N1+N2+N3) * SUM[(t-Ni)^2]from i=1->3

    The question asks which would be more appropriate to use to find the average given t is our estimate of true poisson mean. (eg which is more biased? in which direction?)

    I know that may have seemed a little confusing, but anything would help, thanks!
  2. jcsd
  3. Nov 17, 2005 #2
    The question asks to investigate the phenomenon of biased and unbiased estimators. Its late and I haven't the inclination to write the equations out, however; a quick read of these concepts should lead to a quick solution. Try it first with a Gaussian distribution with the mean estimators
    1/m sum(qi,i,1,n) and 1/(m+1)sum(qi,i,1,n) to see the idea.
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