Pearson's chisquare test versus chisquared distributionby nomadreid Tags: chisquare, chisquared, distribution, pearson, test, versus 

#1
Feb1813, 06:37 AM

P: 509

I know that there are many websites that explain Pearson's chisquare 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 s^{2} the chisquare statistic = k*the ratio of the sample variance to the population variance = k*(s^{2}/σ^{2}), 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 chisquared distribution is then the graph for all samples with the chisquared statistic on the xaxis and the probability density on the y. 3) The chisquared test uses the statistic Ʃ (O_{i}E_{i})^{2}/E_{i} for i values, with O_{i} being the observed frequency of the i'th value, E_{i} 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 chisquared statistic approaches the chisquare 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
Feb1813, 08:28 AM

Sci Advisor
P: 3,380

In case of the chisquare 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/Converg...ndom_variables 



#3
Feb1813, 09:46 AM

P: 509

Dr. Du: Thank you, that adequately answers my first question.
Now, if I am lucky, someone will answer my second question. 



#4
Feb1913, 07:40 AM

Sci Advisor
P: 3,380

Pearson's chisquare test versus chisquared distribution
Calculate ##p(t)=\int\ldots\int dy_1\ldots dy_n p(y_1)\ldots p(y_n) \delta(tf(\vec{y}))## and use a saddle point approximation for large n.




#5
Feb1913, 07:58 AM

P: 509

Dr. Du: thanks very much. Makes sense. Enlightening.



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