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CIE 9709 2022 November Paper 62 Q4

A Level / CIE / S2

CIE 9709 2022 Nov Paper 62 Paper · Question 4

题目

Problem

The number of faults in cloth made on a certain machine has a Poisson distribution with mean 2.42.4 per 10 m210\,\mathrm{m}^2. An adjustment is made to the machine. It is required to test at the 5%5\% significance level whether the mean number of faults has decreased. A randomly selected 30 m230\,\mathrm{m}^2 of cloth is checked and the number of faults is found.

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

(1)

(b) Find the probability of a Type I error.

(3)

Exactly 33 faults are found in the randomly selected 30 m230\,\mathrm{m}^2 of cloth.

(c) Carry out the test at the 5%5\% significance level.

(2)

Later a similar test was carried out at the 5%5\% significance level, using another randomly selected 30 m230\,\mathrm{m}^2 of cloth.

(d) Given that the number of faults actually has a Poisson distribution with mean 0.50.5 per 10 m210\,\mathrm{m}^2, find the probability of a Type II error.

(2)
题目中文翻译

某台机器织出的布料上的疵点数服从泊松分布,均值为每 10 m210\,\mathrm{m}^2 有 2.42.4 个疵点。对该机器作了调整,需要在 5%5\% 的显著性水平下检验疵点数的均值是否下降。随机选取 30 m230\,\mathrm{m}^2 的布料进行检查,数出其中的疵点数。

(a) 写出适合该检验的原假设和备择假设。

(1)

(b) 求发生第一类错误的概率。

(3)

在随机选取的这 30 m230\,\mathrm{m}^2 的布料中,恰好发现 33 个疵点。

(c) 在 5%5\% 的显著性水平下进行检验。

(2)

后来,另随机选取 30 m230\,\mathrm{m}^2 的布料,在 5%5\% 的显著性水平下进行了一次类似的检验。

(d) 已知疵点数实际服从均值为每 10 m210\,\mathrm{m}^2 有 0.50.5 个疵点的泊松分布,求发生第二类错误的概率。

(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

展开

以每10平方米平均疵点数为参数,备择为减少。

答题过程

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Let λ\lambda be the population mean number of faults per 10 m210\text{ m}^2 after the adjustment.

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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原假设下30平方米参数为7.2。比较相邻左尾概率确定拒绝域,实际第一类错误概率就是该域在原假设下的概率。

答题过程

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Let FF be the number of faults in 30 m230\text{ m}^2. Under H0H_0, F∼Po(7.2)F\sim\mathrm{Po}(7.2).

P(F≤2)= e−7.2(1+7.2+7.222!)= 0.0254735…<0.05,\begin{align*} \mathrm{P}(F\leq2)=&\,e^{-7.2}\bigg(1+7.2+\frac{7.2^2}{2!}\bigg)\\[4mm] =&\,0.0254735\ldots<0.05, \end{align*}

whereas

P(F≤3)= e−7.2(1+7.2) +e−7.2(7.222!+7.233!)= 0.0719171…>0.05.\begin{align*} \mathrm{P}(F\leq3)=&\,e^{-7.2}(1+7.2)\\[4mm] &\,+e^{-7.2}\bigg(\frac{7.2^2}{2!}+\frac{7.2^3}{3!}\bigg)\\[4mm] =&\,0.0719171\ldots>0.05. \end{align*}

Thus the rejection region is F≤2F\leq2, and

P(Type I error)≈0.0255(3 s.f.).\mathrm{P}(\text{Type I error})\approx\boxed{0.0255}\quad\text{(3 s.f.).}

(c)

解法一

思路

展开

3个疵点不在拒绝域,因此不足以认为总体平均疵点数下降。

答题过程

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The observed count of 3 is outside the rejection region F≤2F\leq2, so H0H_0 is not rejected. There is insufficient evidence at the 5% significance level that the population mean number of faults has decreased.

(d)

解法一

思路

展开

真实参数每10平方米0.5时,30平方米参数为1.5。第二类错误为至少3个疵点而未拒绝。

答题过程

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When the true rate is 0.5 faults per 10 m210\text{ m}^2, F∼Po(1.5)F\sim\mathrm{Po}(1.5). A Type II error occurs when F≥3F\geq3, giving

 P(Type II error)= 1−e−1.5(1+1.5+1.522!)≈ 0.191(3 s.f.).\begin{align*} &\,\mathrm{P}(\text{Type II error})\\[4mm] =&\,1-e^{-1.5}\bigg(1+1.5+\frac{1.5^2}{2!}\bigg)\\[4mm] \approx&\,\boxed{0.191}\quad\text{(3 s.f.).} \end{align*}