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CIE 9709 2023 June Paper 61 Q7

A Level / CIE / S2

CIE 9709 2023 June Paper 61 Paper · Question 7

题目

Problem

The number of accidents per week at a certain factory has a Poisson distribution. In the past the mean has been 1.91.9 accidents per week. Last year, the manager gave all his employees a new booklet on safety. He decides to test, at the 5%5\% significance level, whether the mean number of accidents has been reduced. He notes the number of accidents during 44 randomly chosen weeks this year.

(a) State suitable null and alternative hypotheses for the test.

(1)

(b) Find the critical region for the test and state the probability of a Type I error.

(6)

(c) State what is meant by a Type I error in this context.

(1)

(d) During the 44 randomly chosen weeks there are a total of 33 accidents.

State the conclusion that the manager should reach. Give a reason for your answer.

(2)

(e) Assuming that the mean remains 1.91.9 accidents per week, use a suitable approximation to calculate the probability that there will be more than 100100 accidents during a 5252-week period.

(4)
题目中文翻译

某工厂每周发生的事故次数服从泊松分布。过去平均每周发生 1.91.9 次事故。去年，经理给所有员工发放了一本新的安全手册。他决定在 5%5\% 的显著性水平下检验平均事故次数是否减少。他从今年随机选取 44 周，记录这几周的事故次数。

(a) 写出适合该检验的原假设和备择假设。

(1)

(b) 求该检验的拒绝域，并写出发生第一类错误的概率。

(6)

(c) 说明第一类错误在本题中的含义。

(1)

(d) 在随机选取的这 44 周内，共发生了 33 次事故。

写出经理应得出的结论，并给出理由。

(2)

(e) 假设平均每周仍发生 1.91.9 次事故，用适当的近似计算一段为期 5252 周的时间内发生超过 100100 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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用每周总体平均事故次数为参数，备择为下降。

答题过程

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Let λ\lambda be the population mean number of accidents per week this year.

H0:λ=1.9,H1:λ<1.9.\boxed{H_0:\lambda=1.9,\qquad H_1:\lambda<1.9}.

(b)

解法一

思路

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四周总数在原假设下服从Po7.6。比较相邻左尾累计概率与5%确定拒绝域，并给出实际第一类错误概率。

答题过程

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Let TT be the total number of accidents in the four weeks. Under H0H_0,

T∼Po(4(1.9))=Po(7.6).T\sim\mathrm{Po}(4(1.9))=\mathrm{Po}(7.6).

We have

P(T≤2)= e−7.6(1+7.6+7.622!)= 0.0187569…<0.05,\begin{align*} \mathrm{P}(T\leq2)=&\,e^{-7.6}\bigg(1+7.6+\frac{7.6^2}{2!}\bigg)\\[4mm] =&\,0.0187569\ldots<0.05, \end{align*}

but

P(T≤3)= e−7.6(1+7.6) +e−7.6(7.622!+7.633!)= 0.0553713…>0.05.\begin{align*} \mathrm{P}(T\leq3)=&\,e^{-7.6}(1+7.6)\\[4mm] &\,+e^{-7.6}\bigg(\frac{7.6^2}{2!}+\frac{7.6^3}{3!}\bigg)\\[4mm] =&\,0.0553713\ldots>0.05. \end{align*}

Hence the critical region is T≤2\boxed{T\leq2}, and

P(Type I error)≈0.0188(3 s.f.).\mathrm{P}(\text{Type I error})\approx\boxed{0.0188}\quad\text{(3 s.f.).}

(c)

解法一

思路

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在此情境中第一类错误是实际没有下降却错误推断事故均数下降。

答题过程

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A Type I error is concluding that the population mean number of accidents per week has decreased when it has actually remained at 1.9.

(d)

解法一

思路

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观测总数3不在不超过2的拒绝域，因此不能认为均数下降。

答题过程

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The observed total of 3 is outside the critical region T≤2T\leq2, so H0H_0 is not rejected. There is insufficient evidence at the 5% significance level that the mean number of accidents per week has decreased.

(e)

解法一

思路

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52周总事故参数为98.8，可用同均值方差正态近似；超过100的连续性修正边界为100.5。

答题过程

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Let AA be the number of accidents in 52 weeks. Then

A∼ Po(52(1.9))=Po(98.8)≈ N(98.8,98.8).\begin{align*} A\sim&\,\mathrm{Po}(52(1.9))=\mathrm{Po}(98.8)\\[4mm] \approx&\,\mathrm{N}(98.8,98.8). \end{align*}

With a continuity correction,

P(A>100)≈ 1−Φ(100.5−98.898.8)= 0.432100…= 0.432(3 s.f.).\begin{align*} \mathrm{P}(A>100)\approx&\,1-\Phi\bigg(\frac{100.5-98.8}{\sqrt{98.8}}\bigg)\\[4mm] =&\,0.432100\ldots\\[4mm] =&\,\boxed{0.432}\quad\text{(3 s.f.).} \end{align*}