Thread: contingency tables
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Old June 20th, 2009, 11:02 PM
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Our null hypothesis is that preference is independent of gender.

If preference is indeed independent of gender, we would expect, for example, \frac {30}{100}* 60 = 18 men to prefer basektball and \frac {30}{100} *40 = 12 women to prefer basketball.


EDIT: q = \frac {(21-18)^{2}}{18} + \frac{(9-12)^{2}}{12} + \frac{(5-4.8)^{2}}{4.8} + \frac {(3-3.2)^{2}}{3.2} + \frac{(9-6)^{2}}{6} + \frac{(1-4)^{2}}{4} + + \frac{(12-16.2)^{2}}{16.2} + \frac{(15-10.8)^{2}}{10.8} + \frac{(13-15)^{2}} {15} + \frac{(12-10)^{2}}{10}

Calculate q.

Q follows a \chi^{2}-distribution with (2-1)(5-1) = 4 degrees of freedom (one less than the number of rows times one less than the number of columns)

The p-value for this test is P(\chi^{2}_{4} \ge q).

This is the smallest significance level at which the null hypothesis can be rejected. So if this value is less than 0.05, then there is enough evidence to reject the null hypothesis.
Last edited by Random Variable; June 20th, 2009 at 11:33 PM. Reason: fixed a couple of errors
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