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CIE 9231 2022 June Paper 41 Q4

A Level / CIE / FS

CIE 9231 2022 June Paper 41 Paper · Question 4

题目

Problem

A scientist is investigating the numbers of a particular type of butterfly in a certain region. He claims that the numbers of these butterflies found per square metre can be modelled by a Poisson distribution with mean 2.52.5. He takes a random sample of 120120 areas, each of one square metre, and counts the number of these butterflies in each of these areas. The following table shows the observed frequencies together with some of the expected frequencies using the scientist’s Poisson distribution.

Number per square metre0123456≥7\ge7
Observed frequency1220363213610
Expected frequency9.8524.6330.7825.65pp8.023.34qq

(a) Find the values of pp and qq, correct to 22 decimal places.

(2)

(b) Carry out a goodness of fit test, at the 10%10\% significance level, to test the scientist’s claim.

(6)
题目中文翻译

一位科学家正在调查某地区某种蝴蝶的数量。他声称,每平方米发现的这种蝴蝶的数量可以用均值为 2.52.5 的泊松分布建模。他随机抽取 120120 块面积均为一平方米的区域,数出每块区域内这种蝴蝶的数量。下表列出了观测频数,以及利用这位科学家的泊松分布计算出的部分期望频数。

每平方米的数量0123456≥7\ge7
观测频数1220363213610
期望频数9.8524.6330.7825.65pp8.023.34qq

(a) 求 pp 和 qq 的值,精确到小数点后 22 位。

(2)

(b) 在 10%10\% 的显著性水平下进行拟合优度检验,检验这位科学家的声称。

(6)
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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恰好四只的期望频数用泊松概率公式。至少七只的期望频数用补事件。

答题过程

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For N∼Poisson⁡(2.5)N\sim\operatorname{Poisson}(2.5), the expected counts are

p=120e−2.52.544!=16.03222…p=120\mathrm e^{-2.5}\frac{2.5^4}{4!}=16.03222\ldots

and

 q=120(1−e−2.5∑j=062.5jj!)= 1.70247….\begin{align*} &\,q=120\Big(1-\mathrm e^{-2.5}\sum_{j=0}^6\frac{2.5^j}{j!}\Big)\\ =&\,1.70247\ldots. \end{align*}

Thus p=16.03p=16.03 and q=1.70q=1.70 to two decimal places.

(b)

解法一

思路

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合并 66 与至少 77 两格,使合并格的期望频数大于 55。均值已给定,所以自由度为格数减一。

答题过程

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Test H0H_0 that Poisson⁡(2.5)\operatorname{Poisson}(2.5) fits the butterfly counts per square metre against H1H_1 that it does not. Combining the last two categories gives observed count 11 and expected count 5.04252…5.04252\ldots. There are seven categories, all with expected counts above 55. Using unrounded expected counts,

 χ2=∑(O−E)2E= (12−9.85019…)29.85019…+ (20−24.62549…)224.62549…+ ⋯=8.11514….\begin{align*} &\,\chi^2=\sum\frac{(O-E)^2}{E}\\ =&\,\frac{(12-9.85019\ldots)^2}{9.85019\ldots}\\ +&\,\frac{(20-24.62549\ldots)^2}{24.62549\ldots}\\ +&\,\cdots=8.11514\ldots. \end{align*}

With 66 degrees of freedom the upper 10%10\% critical value is 10.64510.645. The statistic is smaller, so we fail to reject H0H_0. There is insufficient evidence at the 10%10\% level that the scientist’s Poisson model fails to fit the counts.