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Pearson's chi-square test versus chi-squared distribution

  1. Feb 18, 2013 #1
    I know that there are many web-sites that explain Pearson's chi-square test, but they all leave the same questions unanswered. First, to make sure I have the definitions right:
    1) for a fixed population with standard deviation σ,a fixed number of degrees of freedom df=k, and a fixed sample with variance s2
    the chi-square statistic = k*the ratio of the sample variance to the population variance = k*(s22), also expressed as the sum of the squares of (the difference between an observation to the expected value, as expressed in terms in units of population standard deviation).
    2) For this population and this df, the chi-squared distribution is then the graph for all samples with the chi-squared statistic on the x-axis and the probability density on the y.
    3) The chi-squared test uses the statistic
    Ʃ (Oi-Ei)2/Ei for i values, with Oi being the observed frequency of the i'th value, Ei being its expected frequency.

    OK, so far so good. But now what I do not get is the next comment: that as i goes to infinity, the chi-squared statistic approaches the chi-square distribution. First and foremost, how does a statistic, which is a single number, approach a distribution? Does it mean the cumulative distribution? Second (but not as important), is there a relatively short proof of this fact? Or at least a way to see the connection between the formulas? Thanks in advance.
     
  2. jcsd
  3. Feb 18, 2013 #2

    DrDu

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    A statistic T isn't a single number, it is a function of a (usually vectorial) random variable T=f(Y). That means that T itself is a random variable. Its realizations ##t_j## in repetitions of an experiment follow a distribution function.
    In case of the chi-square statistic, the realization y of Y is the vector ##(O_i)^T##.
    If you repeat the experiment, you will get different realizations ##y_j## and different realizations of the statistic ##t_j##.
    When the dimension of the vector, i.e. the maximal i, goes to infinity, the distribution of ##T## converges in distribution to the chisquare distribution.
    See
    http://en.wikipedia.org/wiki/Convergence_of_random_variables
     
  4. Feb 18, 2013 #3
    Dr. Du: Thank you, that adequately answers my first question.
    Now, if I am lucky, someone will answer my second question.
     
  5. Feb 19, 2013 #4

    DrDu

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    Calculate ##p(t)=\int\ldots\int dy_1\ldots dy_n p(y_1)\ldots p(y_n) \delta(t-f(\vec{y}))## and use a saddle point approximation for large n.
     
  6. Feb 19, 2013 #5
    Dr. Du: thanks very much. Makes sense. Enlightening.
     
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