题目

Problem

The number of accidents per year on a certain road has the distribution $\mathrm{Po}(\lambda)$ . In the past the value of $\lambda$ was $3.3$ . Recently, a new speed limit was imposed and the council wishes to test whether the value of $\lambda$ has decreased. The council notes the total number, $X$ , of accidents during two randomly chosen years after the speed limit was introduced and it carries out a test at the $5\%$ significance level.

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

(4)

(b) Given that $X=2$ , carry out the test.

(3)

(c) The council decides to carry out another similar test at the $5\%$ significance level using the same hypotheses and two different randomly chosen years.

Given that the true value of $\lambda$ is $0.6$ , calculate the probability of a Type II error.

(3)

(d) Using $\lambda=0.6$ and a suitable approximating distribution, find the probability that there will be more than $10$ accidents in $30$ years.

(4)

题目中文翻译

某条道路每年发生的事故次数服从 $\mathrm{Po}(\lambda)$ 分布。过去 $\lambda$ 的值为 $3.3$ 。最近实施了新的限速规定，地方议会想检验 $\lambda$ 的值是否下降。议会从实施限速后的年份中随机选取两年，记录这两年的事故总次数 $X$ ，并在 $5\%$ 的显著性水平下进行检验。

(a) 计算发生第一类错误的概率。

(4)

(b) 已知 $X=2$ ，进行该检验。

(3)

(c) 地方议会决定采用相同的假设，另随机选取两年，在 $5\%$ 的显著性水平下再进行一次类似的检验。

已知 $\lambda$ 的真值为 $0.6$ ，计算发生第二类错误的概率。

(3)

(d) 使用 $\lambda=0.6$ 和适当的近似分布，求 $30$ 年内发生超过 $10$ 次事故的概率。

(4)

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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先把两年总次数在原假设下的参数设为6.6。找最大的左尾拒绝域使概率不超过5%，该概率就是第一类错误概率。

答题过程

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Under the null hypothesis of an annual mean of 3.3 accidents, the two-year total has distribution

X\sim\mathrm{Po}(6.6).

The lower-tail probabilities are

\begin{align*} \mathrm{P}(X\leq2)=&\,e^{-6.6}\bigg(1+6.6+\frac{6.6^2}{2!}\bigg)\\[4mm] =&\,0.0399676\ldots<0.05, \end{align*}

whereas

\begin{align*} \mathrm{P}(X\leq3)=&\,e^{-6.6}(1+6.6)\\[4mm] &\,+e^{-6.6}\bigg(\frac{6.6^2}{2!}+\frac{6.6^3}{3!}\bigg)\\[4mm] =&\,0.105151\ldots>0.05. \end{align*}

Thus the rejection region is $X\leq2$ . The probability of a Type I error is

\boxed{0.0400}\quad\text{(3 s.f.).}

(b)

解法一

思路

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明确以每年的参数建立假设。观测值2落在前一问确定的拒绝域，因此有证据表明事故均数下降。

答题过程

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Using $\lambda$ for the annual mean number of accidents,

H_0:\lambda=3.3,\qquad H_1:\lambda<3.3.

The observed value $X=2$ lies in the rejection region $X\leq2$ . Equivalently, its lower-tail probability under $H_0$ is

0.0399676\ldots<0.05.

We reject $H_0$ . There is sufficient evidence at the 5% significance level that the annual mean number of accidents has decreased.

(c)

解法一

思路

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真实年度参数为0.6时，两年总数的参数为1.2。第二类错误为未落入拒绝域，即总数至少3。

答题过程

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When $\lambda=0.6$ , $X\sim\mathrm{Po}(1.2)$ . A Type II error occurs when $X>2$ , so

\begin{align*} &\,\mathrm{P}(\text{Type II error})\\[4mm] =&\,1-e^{-1.2}\bigg(1+1.2+\frac{1.2^2}{2!}\bigg)\\[4mm] =&\,0.120513\ldots\\[4mm] =&\,\boxed{0.121}\quad\text{(3 s.f.).} \end{align*}

(d)

解法一

思路

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30年总次数的泊松参数为18，采用题目要求的正态近似，并在超过10的边界作连续性修正。

答题过程

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Let $Y$ be the total number of accidents in 30 years. Then

\begin{align*} Y\sim&\,\mathrm{Po}(30(0.6))=\mathrm{Po}(18)\\[4mm] \approx&\,\mathrm{N}(18,18). \end{align*}

Using a continuity correction,

\begin{align*} \mathrm{P}(Y>10)\approx&\,1-\Phi\bigg(\frac{10.5-18}{\sqrt{18}}\bigg)\\[4mm] =&\,0.961450\ldots\\[4mm] =&\,\boxed{0.961}\quad\text{(3 s.f.).} \end{align*}

CIE 9709 2024 November Paper 61 Q7

题目

解答

(a)

解法一

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答题过程

(b)

解法一

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答题过程

(c)

解法一

思路

答题过程

(d)

解法一

思路

答题过程