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 2.5. He takes a random sample of 120 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 metre	0	1	2	3	4	5	6	$\geq 7$
Observed frequency	12	20	36	32	13	6	1	0
Expected frequency	9.85	24.63	30.78	25.65	$p$	8.02	3.34	$q$

(a) Find the values of $p$ and $q$ , correct to 2 decimal places.

[2]

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

[6]

题目中文翻译

一名科学家正在研究某地区一种特定蝴蝶的数量。他声称，每平方米发现的这种蝴蝶数量可以用均值为 2.5 的 Poisson 分布来建模。他随机抽取 120 个区域，每个区域面积为一平方米，并统计每个区域中这种蝴蝶的数量。下表给出了观察频数，以及使用该科学家的 Poisson 分布得到的部分期望频数。

每平方米数量	0	1	2	3	4	5	6	$\geq 7$
观察频数	12	20	36	32	13	6	1	0
期望频数	9.85	24.63	30.78	25.65	$p$	8.02	3.34	$q$

(a) 求 $p$ 和 $q$ 的值，精确到 2 位小数。

(2)

(b) 在 10% 显著性水平下进行拟合优度检验，检验这名科学家的声称。

(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

展开

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

展开

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

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

and

\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 $p=16.03$ and $q=1.70$ to two decimal places.

(b)

解法一

思路

展开

合并 $6$ 与至少 $7$ 两格，使合并格的期望频数大于 $5$ 。均值已给定，所以自由度为格数减一。

答题过程

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Test $H_0$ that $\operatorname{Poisson}(2.5)$ fits the butterfly counts per square metre against $H_1$ that it does not. Combining the last two categories gives observed count $1$ and expected count $5.04252\ldots$ . There are seven categories, all with expected counts above $5$ . Using unrounded expected counts,

\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 $6$ degrees of freedom the upper $10\%$ critical value is $10.645$ . The statistic is smaller, so we fail to reject $H_0$ . There is insufficient evidence at the $10\%$ level that the scientist’s Poisson model fails to fit the counts.

CIE 9231 2022 June Paper 42 Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程