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CIE 9709 2026 June Paper 62 Q3

A Level / CIE / S2

CIE 9709 2026 June Paper 62 Paper · Question 3

题目

Problem

The number of vehicles, XX, passing a certain junction between 1.00 pm and 1.01 pm is modelled by a Poisson distribution. In the past, the population mean of XX was 2.42.4. After a change is made to the junction, the value of XX is noted on a randomly chosen day. A hypothesis test at the 5%5\% significance level is carried out to determine whether the population mean of XX has increased.

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

(1)

(b) Find the smallest value of XX that would result in rejecting the null hypothesis.

(4)

(c) State the value of P(Type I error)\mathrm{P}(\text{Type I error}).

(1)
题目中文翻译

用泊松分布对下午 1:00 至 1:01 之间通过某路口的车辆数 XX 建模。过去,XX 的总体均值为 2.42.4。路口改造后,随机选取一天,记录 XX 的值。在 5%5\% 的显著性水平下进行假设检验,以判断 XX 的总体均值是否有所增加。

(a) 写出该检验适用的原假设与备择假设。

(1)

(b) 求使原假设被拒绝的 XX 的最小值。

(4)

(c) 写出 P(第一类错误)\mathrm{P}(\text{第一类错误}) 的值。

(1)
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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假设针对总体每分钟车流量的均值,增加对应右尾检验。

答题过程

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Let λ\lambda be the population mean number of vehicles passing the junction in the stated one-minute interval. Suitable hypotheses are

H0:λ=2.4,H1:λ>2.4.\boxed{H_0:\lambda=2.4,\qquad H_1:\lambda>2.4}.

(b)

解法一

思路

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拒绝域是右尾,所以要算“至少某个车数”的累计概率,而不是该车数的单点概率。检查相邻两个候选边界来证明最小值。

答题过程

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Under H0H_0, X∼Po(2.4)X\sim\mathrm{Po}(2.4). The critical region has the form X≥cX\ge c, where its probability under H0H_0 must be at most 0.050.05.

For c=5c=5,

P(X≥5)= 1−e−2.4∑r=042.4rr!= 1−e−2.4(1+2.4) −e−2.4(2.422!+2.433!+2.444!)≈ 0.0958686>0.05.\begin{align*} \mathrm P(X\ge5) =&\,1-\mathrm e^{-2.4}\sum_{r=0}^4\frac{2.4^r}{r!}\\[4mm] =&\,1-\mathrm e^{-2.4}(1+2.4)\\[4mm] &\,-\mathrm e^{-2.4}\bigg(\frac{2.4^2}{2!}+\frac{2.4^3}{3!}+\frac{2.4^4}{4!}\bigg)\\[4mm] \approx&\,0.0958686>0.05. \end{align*}

For c=6c=6,

P(X≥6)= P(X≥5)−e−2.42.455!≈ 0.0958686−0.0601961≈ 0.0356725<0.05.\begin{align*} \mathrm P(X\ge6) =&\,\mathrm P(X\ge5)-\mathrm e^{-2.4}\frac{2.4^5}{5!}\\[4mm] \approx&\,0.0958686-0.0601961\\[4mm] \approx&\,0.0356725<0.05. \end{align*}

Since upper-tail probabilities decrease as cc increases, the smallest observed value that leads to rejection is X=6\boxed{X=6}, with critical region X≥6X\ge6.

(c)

解法一

思路

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第一类错误是原假设真实时却落入拒绝域,所以实际显著性水平等于上一问右尾概率,并不恰好等于名义上的 5%5\%。

答题过程

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A Type I error occurs when H0H_0 is true but is rejected. Thus

P(Type I error)= P(X≥6∣λ=2.4)≈ 0.0357.\begin{align*} \mathrm P(\text{Type I error})=&\,\mathrm P(X\ge6\mid\lambda=2.4)\\[4mm] \approx&\,\boxed{0.0357}. \end{align*}