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How to work out chi square independence in the following table?

Below is the observed and expected data concerning 7 themes displayed in a newspaper over a period of 3 months. I understand how to work out chi square but not with this number of columns and rows. Thank you!

        May     June    July    Total           May     June    July    Total 
Gov     16      34      28      78      Gove    18.525  29.25   30.225  78
Mort    1       0       1       2       Mort    0.475   0.75    0.775   2
Damage  0       2       1       3       Damage  0.7125  1.125   1.1625  3
Recon   0       3       7       10      Recon   2.375   3.75    3.875   10
Relief  18      16      18      52      Relief  12.35   19.5    20.15   52
Health  0       0       2       2       Health  0.475   0.75    0.775   2
Survi   2       3       4       9       Survi   2.1375  3.375   3.4875  9
Other   1       2       1       4       Other   0.95    1.5     1.55    4
Total   38      60      62      160     Total   38      60      62      160

$$ \small \begin{array} {r|rrr|r||r|rrr|r} obsrv. & May & June & July & Total & expct. & May & June & July & Total \\ \hline Gov & 16 & 34 & 28 & 78 & Gove & 18.525 & 29.25 & 30.225 & 78 \\ Mort & 1 & 0 & 1 & 2 & Mort & 0.475 & 0.75 & 0.775 & 2 \\ Damage & 0 & 2 & 1 & 3 & Damage & 0.7125 & 1.125 & 1.1625 & 3 \\ Recon & 0 & 3 & 7 & 10 & Recon & 2.375 & 3.75 & 3.875 & 10 \\ Relief & 18 & 16 & 18 & 52 & Relief & 12.35 & 19.5 & 20.15 & 52 \\ Health & 0 & 0 & 2 & 2 & Health & 0.475 & 0.75 & 0.775 & 2 \\ Survi & 2 & 3 & 4 & 9 & Survi & 2.1375 & 3.375 & 3.4875 & 9 \\ Other & 1 & 2 & 1 & 4 & Other & 0.95 & 1.5 & 1.55 & 4 \\ \hline Total & 38 & 60 & 62 & 160 & Total & 38 & 60 & 62 & 160 \\ \end{array} $$

Gottfried Helms
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owain
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    Hi I've formatted your table as text and as LaTex. Hope this did not frustrate you... feel free to redo/delete if/what is convenient – Gottfried Helms Apr 29 '13 at 23:04

1 Answers1

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The test statistic would be $$ \sum \frac{(\text{observed}-\text{expected})^2}{\text{expected}}. $$ It should have a chi-square distribution with $(8-1)(3-1)=14$ degrees of freedom if the hypothesis of independence is true. It it has an improbably large value, you reject the null hypothesis.

Michael Hardy
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