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Question
A sample of boys and girls were asked to choose their favourite sport, with the following results. Find the value of the χ2 statistic.
| Football | Cricket | Hockey | Basketball | |
| Boys | 86 | 60 | 44 | 10 |
| Girls | 40 | 30 | 25 | 5 |
Solution
Table of observed frequencies.
| Football | Cricket | Hockey | Basketball | Row total (Ri) | |
| Boys | 86 | 60 | 44 | 10 | 200 |
| Girls | 40 | 30 | 25 | 5 | 100 |
| Column total (Cj) | 126 | 90 | 69 | 15 | 300 |
Expected frequencies are given by
Eij = `("R"_"i" xx "C""j")/"N"`
E11 = `(200 xx 126)/300` = 84
E12 = `(200 xx 90)/300` = 60
E13 = `(200 xx 69)/300` = 46
E14 = `(200 xx 15)/300` = 10
E21 = `(100 xx 126)/300` = 42
E22 = `(100 xx 90)/300` = 30
E23 = `(100 xx 69)/300` = 23
E24 = `(100 xx 15)/300` = 5
Table of expected frequencies.
| Football | Cricket | Hockey | Basketball | Total | |
| Boys | 84 | 60 | 46 | 10 | 200 |
| Girls | 42 | 30 | 23 | 5 | 100 |
| Total | 126 | 90 | 69 | 15 | 300 |
Now,
χ2 = `sum[(("o"_"ij" - "E"_"ij")^2)/"E"_"ij"]`
`=((86 - 84)^2)/84 + ((60 - 60)^2)/60 + ((44 - 46)^2)/46 + ((10 - 10)^2)/10 + ((40 - 42)^2)/42 + ((30 - 30)^2)/30 + ((25 - 23)^2)/23 + ((5 - 5)^2)/5`
= `4/84 + 0 + 4/46 + 0 + 4/42 + 0 + 4/23 + 0`
= 0.048 + 0.087 + 0.095 + 0.174
= 0.404
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