2 added 76 characters in body edited Oct 28 '16 at 21:11 Ami Tavory 4,01577 silver badges1818 bronze badges Even before rounding, someparts of your question describes things which seem to me problematic. IMHO, it pays to consider them, as they relate to the cause of rounding. Scaling Say my model just divides everything by 3 The rationale behind this test involves the multinomial distribution, and contains combinatorical terms of the form $${n \choose n_1 \cdot n_k}$$ These terms are not invariant to scaling. I.e., you cannot replace this with $${\alpha n \choose \alpha^k n_1 \cdot n_k} = {\alpha n \choose \alpha n_1 \cdot \alpha n_k}$$ and expect to get the same results. Test Assumptions It's possible your division of 3 is caused by this being an average of three observations. In this case, though, there's a problem with assuming that the test is relevant here: A common rule is 5 or more in all cells of a 2-by-2 table Post the division by 3, this does not hold, and the numbers are not in the range where it can be assumed that this test is applicable. Even before rounding, some of your question describes things which seem to me problematic. Scaling Say my model just divides everything by 3 The rationale behind this test involves the multinomial distribution, and contains combinatorical terms of the form $${n \choose n_1 \cdot n_k}$$ These terms are not invariant to scaling. I.e., you cannot replace this with $${\alpha n \choose \alpha^k n_1 \cdot n_k} = {\alpha n \choose \alpha n_1 \cdot \alpha n_k}$$ and expect to get the same results. Test Assumptions It's possible your division of 3 is caused by this being an average of three observations. In this case, though, there's a problem with assuming that the test is relevant here: A common rule is 5 or more in all cells of a 2-by-2 table Even before rounding, parts of your question describes things which seem to me problematic. IMHO, it pays to consider them, as they relate to the cause of rounding. Scaling Say my model just divides everything by 3 The rationale behind this test involves the multinomial distribution, and contains combinatorical terms of the form $${n \choose n_1 \cdot n_k}$$ These terms are not invariant to scaling. I.e., you cannot replace this with $${\alpha n \choose \alpha^k n_1 \cdot n_k} = {\alpha n \choose \alpha n_1 \cdot \alpha n_k}$$ and expect to get the same results. Test Assumptions It's possible your division of 3 is caused by this being an average of three observations. In this case, though, there's a problem with assuming that the test is relevant here: A common rule is 5 or more in all cells of a 2-by-2 table Post the division by 3, this does not hold, and the numbers are not in the range where it can be assumed that this test is applicable. 1 answered Oct 28 '16 at 20:18 Ami Tavory 4,01577 silver badges1818 bronze badges Even before rounding, some of your question describes things which seem to me problematic. Scaling Say my model just divides everything by 3 The rationale behind this test involves the multinomial distribution, and contains combinatorical terms of the form $${n \choose n_1 \cdot n_k}$$ These terms are not invariant to scaling. I.e., you cannot replace this with $${\alpha n \choose \alpha^k n_1 \cdot n_k} = {\alpha n \choose \alpha n_1 \cdot \alpha n_k}$$ and expect to get the same results. Test Assumptions It's possible your division of 3 is caused by this being an average of three observations. In this case, though, there's a problem with assuming that the test is relevant here: A common rule is 5 or more in all cells of a 2-by-2 table