IAI Actuarial Core Principles · Actuarial Statistics · Hypothesis testing and goodness of fit
A test of independence is carried out on a table with 4 rows and 5 columns. How many degrees of freedom does the chi-squared test statistic have under the null hypothesis, when no parameters other than the marginal probabilities are estimated?
The statistic has 12 degrees of freedom. For an r by c table the degrees of freedom are (r−1)(c−1), so with 4 rows and 5 columns this is 3 times 4, which equals 12.
- A20
- B19
- C12Correct
- D9
- 11
Explanation
Degrees of freedom = (r−1)(c−1) = 3 × 4 = 12. Using rc−1 = 19 would apply to a goodness-of-fit test on a single set of 20 cells with no estimated parameters, and r+c−2 = 7 is not correct either. 9 is r+c, and 11 is rc−r−c+... mistaken.
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