1
$\begingroup$

This is a question to reflect some very basic understanding of the logic of the chi-square for a planned introductory essay.

[update] I've tried to improve the question with clearer examples and better focusing of the problems upon which I stumbled. It might still be a bit weak but I can't do better at the moment

I'm used to the chi-square as a measure for the deviance from an expected discrete distribution, where the expected frequencies are computed via the marginal frequencies of a crosstab over the empirical data of two categorical items. Let's discuss the 2x2-table here which has one degree of freedom. Assume the marginal row frequencies being [53,47] and the marginal column frequencies being [40,60] giving the expected frequencies as $$ {\bf A}: \text{ expectation for cell-frequencies by equal ratios} \\ \small \begin{array} {r|rr|r} &0&1&all\\ \hline 0&21.2&18.8&40\\ 1&31.8&28.2&60\\ \hline all&53&47&100 \end{array} $$

First I began to think what it means, that the empirical contingency table is bounded by the minimal entry of the marginal frequencies: in this case the cell [0,0] can vary between 0 and 40 only, so we have at most 41 possible outcomes depending on the possible frequencies in that cell. If we assume a normal random process, which generates that frequencies, the (expected) frequency in this cell should be centered around the mean of them 20:

$$ {\bf B}: \text{ means of possible ranges for frequencies in cells}\\ \small \begin{array} {r|rr|r} &0&1&all\\ \hline 0&20&20&40\\ 1&33&27&60\\ \hline all&53&47&100 \end{array} $$

This expectation in cell[0,0] is not equal to the expected frequency of 21.2 . First question: can the latter concept be brought/translated into the first (and standard) one? How could the difference (and relation or possible non-relation) between that two concepts be best explained?

To answer this myself I went back one further step and asked, what does it mean at all to base the chi-square computation on the sample's marginal distribution if this distribution is itself subject of random...

So I generated a "population" data set with N=10000 cases distributed exactly like our table A and took 1000 random samples each with n=100. Over the 1000 samples I've got different marginal frequencies and consequently, each sample has different parameters for its expected frequencies and so for its possible chi-square.
The table of the deviances of the empirical marginal frequencies from the population marginal frequencies was

$$ {\bf C}: \text{ deviations of empirical marginal frequencies from population} \\ \small \begin{array} {r|rrrr} \text{dev } & \text{dev from}& \text{dev from}& \text{dev from}& \text{dev from}\\ \text{value} & 53 & 47 & 40 & 60 \\ \hline -18&0&0&0&1\\ -15&1&1&1&1\\ -14&1&1&0&0\\ -13&2&4&1&1\\ -12&2&6&5&4\\ -11&10&9&5&5\\ -10&8&12&7&14\\ -9&13&15&17&17\\ -8&27&20&21&23\\ -7&34&38&26&31\\ -6&45&35&31&37\\ -5&53&43&62&50\\ -4&46&46&50&49\\ -3&81&67&67&63\\ -2&55&83&84&67\\ -1&85&71&97&80\\ 0&86&86&83&83\\ 1&71&85&80&97\\ 2&83&55&67&84\\ 3&67&81&63&67\\ 4&46&46&49&50\\ 5&43&53&50&62\\ 6&35&45&37&31\\ 7&38&34&31&26\\ 8&20&27&23&21\\ 9&15&13&17&17\\10&12&8&14&7\\11&9&10&5&5\\12&6&2&4&5\\ 13&4&2&1&1\\14&1&1&0&0\\15&1&1&1&1\\18&0&0&1&0 \end{array} $$ The table of occurrences of chi-square for those deviations of the empirical marginal frequencies from the population's is $$ {\bf D}: \text{ deviations of the empirical marginal frequencies} \\ \text{ from that of the population }\\ \text{(in terms of}\ \chi^2\ \text{values)} \\ \begin{array} {r|rr} & & \text{backwards} \\ \chi^2 \text{-value} & \text{freq} & \text{cum freq} \\ \hline 0.0&12.8&100.0\\ 0.5&22.0&87.2\\ 1.0&12.7&65.2\\ 1.5&11.2&52.5\\ 2.0&10.5&41.3\\ 2.5&7.8&30.8\\ 3.0&3.3&23.0\\ 3.5&4.8&19.7\\ 4.0&3.2&14.9\\ 4.5&2.5&11.7\\ 5.0&2.3&9.2\\ 5.5&1.0&6.9\\ 6.0&1.7&5.9\\ 6.5&1.2&4.2\\ 7.0&0.9&3.0\\ 7.5&0.6&2.1\\ 8.0&0.5&1.5\\ 9.0&0.1&1.0\\ 9.5&0.5&0.9\\ 10.0&0.1&0.4\\ 10.5&0.2&0.3\\ 17.0&0.1&0.1 \end{array} $$

I'm stumbling at the simple fact that we compute the chi-square as deviation from an expected frequency - but where the expected frequency is based on an empirical marginal frequency which is itself subject of a random process - and for instance has some confidence-interval when I infer from the sample onto the population.
So - in reverse - having an empirical marginal distribution in our single empirical sample the conclusion to the population's marginal distribution is a matter of confidence intervals. (Is here also a maximum-likelihood aspect lurking around anywhere ?)
This reminds me of the praxis, that we use the sample's variation as estimate of the population's variation-parameter, and do tests based on this assumption.

Q: In the justification/formula for the chi-square-distribution as basis for the significance test - can we find some point, where the randomness of the marginal frequencies is reflected in the formulae?

$\endgroup$
  • $\begingroup$ I haven't read your whole question but you lost me in your third paragraph with " in this case the cell [1,1] can vary between 0 and 40 only" - I can't think of what you're thinking here but it must be incorrect and perhaps might help explain the rest? $\endgroup$ – Peter Ellis Aug 3 '13 at 6:43
  • $\begingroup$ @Peter: Hmm, now this makes me asking what your misunderstanding is... With the given marginal frequencies the cell in row 1/column 1 can vary only between the value 0 and at most 40 - because if it had a bigger number it would be bigger than the whole rowsum which is 40. $\endgroup$ – Gottfried Helms Aug 3 '13 at 7:15
  • $\begingroup$ ok, I was confused by the fact that the columns and rows had labels of 0 and 1 - so what you call [1, 1] I thought was [0, 0]. $\endgroup$ – Peter Ellis Aug 3 '13 at 7:36
  • $\begingroup$ @Peter - ok, sorry for that unnessecary confusion... $\endgroup$ – Gottfried Helms Aug 3 '13 at 8:34
  • 1
    $\begingroup$ I think this needs both (a) one paragraph summarizing your question (b) details of what you did (your discussion in various places seems to hint that cell frequencies are normally distributed or at least symmetrically distributed, neither of which is even possible, let alone assumed by the theory here. $\endgroup$ – Nick Cox Aug 3 '13 at 8:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.