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I am having difficulty in how to calculate the Pearson's $\chi^2$ statistic for a contingency table with entries: (3, 0) and (0, 3) (row-wise).

I was studying Agresti's categorical data analysis and found this table. He calculated it as 6.

Through what calculation did he reach this answer?

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Here is the procedure to be followed in order to compute Pearson's Chi-squared statistic.


Assume this is your observed 2x2 contingency table:

$$\begin{array}{c|cc|c} \hline X/Y & \textrm{cat 1} & \textrm{cat 2} & \textrm{tot} \\ \hline \textrm{cat 1} & O_{11} & O_{12} & O_{1.}\\ \textrm{cat 2} & O_{21} & O_{22} & O_{2.}\\ \hline \textrm{tot} & O_{.1} & O_{.2} & O_{..} \end{array}$$

  • there are $O_{11}$ observations that fall within the first category of $X$ $\underline{\textrm{and}}$ within the first category of $Y$;
  • There are $0_{2.}$ observations that fall within the second category of $X$;

  • the total sample size is $O_{..}$.

1st step: Compute the expected contingency table

$$\begin{array}{cc} \hline E_{11} & E_{12} \\ E_{21} & E_{22} \end{array}$$

where

$$E_{ij} = \frac{O_{i.} O_{.j}}{O_{..}} \cdot$$

2nd step Compute Pearson's statistic:

$$\chi^2 = \sum_{ij} \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \cdot$$

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  • $\begingroup$ Note that this can be generalised to higher dimensions... $\endgroup$ – ocram Feb 26 '12 at 11:14
  • $\begingroup$ I'm speculating but if the questioner is entering their data into a stats package rather than calculating the stat from first principles as set out here, it could be that they get an answer different to 6 because the Yates continuity correction might be the default for 2x2 contingency tables. This is the case, for example in R's chisq-test (which will deliver 2.667 if correct=T, the default; or 6 as per the original question if correct=F) $\endgroup$ – Peter Ellis Feb 27 '12 at 4:59
  • $\begingroup$ @Peter Ellis: Yes, good point! $\endgroup$ – ocram Feb 27 '12 at 5:58

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