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CIE 9709 2023 June Paper 63 Q8

A Level / CIE / S2

CIE 9709 2023 June Paper 63 Paper · Question 8

题目

Problem

A new light was installed on a certain footpath. A town councillor decided to use a hypothesis test to investigate whether the number of people using the path in the evening had increased.

Before the light was installed, the mean number of people using the path during any 2020-minute period during the evening was 1.011.01.

After the light was installed, the total number, nn, of people using the path during 33 randomly chosen 2020-minute periods during the evening was noted.

(a) Given that the value of nn was 66, use a Poisson distribution to carry out the test at the 5%5\% significance level.

(6)

(b) Later a similar test, at the 5%5\% significance level, was carried out using another 33 randomly chosen 2020-minute periods during the evening.

Find the probability of a Type I error.

(2)

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

(1)

(d) State, in context, what further information would be needed in order to find the probability of a Type II error. Do not carry out any further calculation.

(2)
题目中文翻译

某条人行小路上安装了一盏新灯。一名镇议员决定用假设检验调查晚上使用这条小路的人数是否有所增加。

安装这盏灯之前，晚上任意一段 2020 分钟的时间内，使用这条小路的人数的均值为 1.011.01。

安装这盏灯之后，从晚上的时间中随机选取 33 段各长 2020 分钟的时段，记录这些时段内使用这条小路的总人数 nn。

(a) 已知 nn 的值为 66，用泊松分布在 5%5\% 的显著性水平下进行检验。

(6)

(b) 后来，又从晚上的时间中随机选取 33 段各长 2020 分钟的时段，在 5%5\% 的显著性水平下进行了一次类似的检验。

求发生第一类错误的概率。

(2)

(c) 说明第一类错误在本题中的含义。

(1)

(d) 结合题意，写出为求发生第二类错误的概率还需要什么信息。不要作进一步计算。

(2)
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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以每20分钟的总体均数为参数，备择为增加。原假设下三段总人数服从Po3.03，使用至少6人的右尾概率。

答题过程

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Let λ\lambda be the population mean number of people using the path per 20-minute evening period after the light was installed.

H0:λ=1.01,H1:λ>1.01.H_0:\lambda=1.01,\qquad H_1:\lambda>1.01.

Under H0H_0, the total NN over three periods has distribution Po(3(1.01))=Po(3.03)\mathrm{Po}(3(1.01))=\mathrm{Po}(3.03).

For the observed total 6,

 P(N≥6∣H0)= 1−e−3.03(1+3.03+3.0322!) −e−3.03(3.0333!+3.0344!+3.0355!)= 0.0869727….\begin{align*} &\,\mathrm{P}(N\geq6\mid H_0)\\[4mm] =&\,1-e^{-3.03}\bigg(1+3.03+\frac{3.03^2}{2!}\bigg)\\[4mm] &\,-e^{-3.03}\bigg(\frac{3.03^3}{3!}+\frac{3.03^4}{4!}+\frac{3.03^5}{5!}\bigg)\\[4mm] =&\,0.0869727\ldots. \end{align*}

Since 0.0869727…>0.050.0869727\ldots>0.05, we do not reject H0H_0. There is insufficient evidence at the 5% significance level that the mean number of people using the path in the evening has increased.

(b)

解法一

思路

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比较相邻右尾概率，至少6人的概率超过5%，至少7人才进入拒绝域，原假设下这一概率为实际第一类错误概率。

答题过程

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Under H0H_0, N∼Po(3.03)N\sim\mathrm{Po}(3.03). From part (a),

P(N≥6)=0.0869727…>0.05.\mathrm{P}(N\geq6)=0.0869727\ldots>0.05.

But

P(N≥7)= 0.0869727…−e−3.033.0366!= 0.0350436…<0.05.\begin{align*} \mathrm{P}(N\geq7) =&\,0.0869727\ldots-e^{-3.03}\frac{3.03^6}{6!}\\[4mm] =&\,0.0350436\ldots<0.05. \end{align*}

Thus the rejection region is N≥7N\geq7, giving

P(Type I error)≈0.0350(3 s.f.).\mathrm{P}(\text{Type I error})\approx\boxed{0.0350}\quad\text{(3 s.f.).}

(c)

解法一

思路

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第一类错误为实际均数没有增加，却错误推断安装灯后晚间使用人数均数增加。

答题过程

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It means concluding that the population mean number using the path per 20-minute period in the evening has increased when it has actually remained at 1.01.

(d)

解法一

思路

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计算第二类错误须知道备择成立时的真实总体均数，需说明对应安装灯后晚间每20分钟的人数。

答题过程

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The true population mean number of people using the path per 20-minute evening period after the light was installed would be needed.