IAI Actuarial Core Principles · Actuarial Statistics · Hypothesis testing and goodness of fit
In a chi-squared test of independence on a 3×3 table, several expected frequencies are below 5. Which action is the standard remedy?
Combine adjacent categories so that expected frequencies reach about 5 or more. The chi-squared approximation is unreliable with small expected counts, and merging categories restores validity, whereas the other options have no statistical basis.
- ACombine adjacent categories so that expected frequencies are at least 5Correct
- BIncrease the degrees of freedom to 9
- CUse the t-test for the difference of means instead
- DApply a Yates continuity correction to every cell
- Multiply the statistic by the number of small cells
Explanation
The chi-squared approximation is poor when expected counts are small, so categories are merged (where sensible) to lift expected counts to about 5 or more. Yates correction applies to 2×2 tables only, and the other choices have no justification.
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