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CIE 9709 2023 March Paper 62 Q4

A Level / CIE / S2

CIE 9709 2023 March Paper 62 Paper · Question 4

题目

Problem

The number of accidents per 33-month period on a certain road has the distribution Po(λ)\mathrm{Po}(\lambda). In the past the value of λ\lambda has been 5.75.7. Following some changes to the road, the council carries out a hypothesis test to determine whether the value of λ\lambda has decreased. If there are fewer than 33 accidents in a randomly chosen 33-month period, the council will conclude that the value of λ\lambda has decreased.

(a) Find the probability of a Type I error.

(2)

(b) Find the probability of a Type II error if the mean number of accidents per 33-month period is now actually 0.90.9.

(3)
题目中文翻译

某条道路每 33 个月发生的事故次数服从 Po(λ)\mathrm{Po}(\lambda) 分布。过去 λ\lambda 的值为 5.75.7。道路作了一些改动后，地方议会进行假设检验，以判断 λ\lambda 的值是否下降。如果在随机选取的一段为期 33 个月的时间内发生的事故少于 33 次，议会就会得出 λ\lambda 的值已经下降的结论。

(a) 求发生第一类错误的概率。

(2)

(b) 如果现在每 33 个月的平均事故次数实际为 0.90.9，求发生第二类错误的概率。

(3)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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拒绝域为少于3次事故，第一类错误用原假设参数5.7计算落入该域的概率。

答题过程

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Under the null hypothesis, the three-month accident count A∼Po(5.7)A\sim\mathrm{Po}(5.7). The rejection region is A<3A<3, so

 P(Type I error)= e−5.7(1+5.7+5.722!)≈ 0.0768(3 s.f.).\begin{align*} &\,\mathrm{P}(\text{Type I error})\\[4mm] =&\,e^{-5.7}\bigg(1+5.7+\frac{5.7^2}{2!}\bigg)\\[4mm] \approx&\,\boxed{0.0768}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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真实参数为0.9时，第二类错误是至少3次事故而不拒绝原假设。

答题过程

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When the true mean is 0.9, A∼Po(0.9)A\sim\mathrm{Po}(0.9). A Type II error occurs when A≥3A\geq3, giving

 P(Type II error)= 1−e−0.9(1+0.9+0.922!)= 0.0628569…= 0.0629(3 s.f.).\begin{align*} &\,\mathrm{P}(\text{Type II error})\\[4mm] =&\,1-e^{-0.9}\bigg(1+0.9+\frac{0.9^2}{2!}\bigg)\\[4mm] =&\,0.0628569\ldots\\[4mm] =&\,\boxed{0.0629}\quad\text{(3 s.f.).} \end{align*}