Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 2 / 6H / ii

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- Solution ii

Solution

For observed counts

O_{j}

and null expected counts

E_{j} = n p_{j}

, the Pearson chi-squared goodness-of-fit test uses

X^{2} = \sum_{j = 1}^{k} \frac{( O _{j} - E _{j} ) ^{2}}{E _{j}} .

When the null cell probabilities are specified and the expected counts grow, its limiting null distribution is

X^{2} d χ_{k - 1}^{2} .

Here

k = 6

, so the limit is

χ_{5}^{2}

.

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