Question

A test for independence is applied to a contingency table with 3 rows and 5 columns for two nominal variables. The degrees of freedom for this test will be
Degrees of freedom = (r - 1)(c - 1) ...where r = number of rows and c = number of columns.
15
It's 8 not 15. 15 is way off.

Related Questions

A test for independence is applied to a contingency table with 3 rows and 5 columns for two nominal... A chi-square test for independence with 8 degrees of freedom results in a test statistic of 18.21. U... A 3 x 5 table is considered to test the association of two variables having 3 and 5 categories respe...