Solution:
We have to find large of a X2 value is needed to get a p-value of 0.05 for testing independence in a 4 x 6 contingency table.
Thus df = ( m - 1) X ( n - 1) = ( 4-1) X ( 6-1) = ( 3 ) X( 5) =15
Thus look in Chi-square table for df = 15 and level of significance = 0.05 and find Chi-square value.
Chi-square value for df = 15 and level of significance 0.05 is 24.996
That is:
(Rounded to one decimal place)
Thus correct option is: a) 25.0
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