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CIE 9709 2024 June Paper 61 Q4

A Level / CIE / S2

CIE 9709 2024 June Paper 61 Paper · Question 4

题目

Problem

(a) A random sample of 88 boxes of cereal from a certain supplier was taken. Each box was weighed and the masses in grams were as follows.

261249259252255256258254261\qquad 249\qquad 259\qquad 252\qquad 255\qquad 256\qquad 258\qquad 254

Find unbiased estimates of the population mean and variance.

(3)

(b) The supplier claims that the mean mass of boxes of cereal is 253 g253\,\mathrm{g}. A quality control officer suspects that the mean mass is actually more than 253 g253\,\mathrm{g}. In order to test this claim, he weighs a random sample of 100100 boxes of cereal and finds that the total mass is 25 360 g25\,360\,\mathrm{g}.

(i) Given that the population standard deviation of the masses is 3.5 g3.5\,\mathrm{g}, test at the 5%5\% significance level whether the population mean mass is more than 253 g253\,\mathrm{g}.

(5)

An employee says, ‘This test is invalid because it uses the normal distribution, but we do not know whether the masses of the boxes are normally distributed.’

(ii) Explain briefly whether this statement is true or not.

(1)
题目中文翻译

(a) 从某供应商提供的盒装谷物中随机抽取 88 盒。逐盒称量，质量以克为单位，结果如下。

261249259252255256258254261\qquad 249\qquad 259\qquad 252\qquad 255\qquad 256\qquad 258\qquad 254

求总体均值和方差的无偏估计值。

(3)

(b) 供应商声称盒装谷物的平均质量为 253 g253\,\mathrm{g}。一名质量检验员怀疑实际的平均质量超过 253 g253\,\mathrm{g}。为了检验这一说法，他随机抽取 100100 盒谷物进行称量，发现总质量为 25 360 g25\,360\,\mathrm{g}。

(i) 已知质量的总体标准差为 3.5 g3.5\,\mathrm{g}，在 5%5\% 的显著性水平下检验总体平均质量是否超过 253 g253\,\mathrm{g}。

(5)

一名员工说：“这个检验无效，因为它使用了正态分布，但我们不知道盒装谷物的质量是否服从正态分布。”

(ii) 简要说明这句话是否正确。

(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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先计算数据总和与平方和，再用样本均值及分母为7的无偏方差公式。

答题过程

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From the eight masses,

∑x=2044,∑x2=522348.\sum x=2044,\qquad \sum x^2=522348.

The unbiased estimate of the population mean is

xˉ=20448=255.5 g.\bar x=\frac{2044}{8}=\boxed{255.5\text{ g}}.

The unbiased estimate of the population variance is

s2= 522348−20442/87=1067≈ 15.1 g2(3 s.f.).\begin{align*} s^2=&\,\frac{522348-2044^2/8}{7}=\frac{106}{7}\\[4mm] \approx&\,\boxed{15.1\text{ g}^2}\quad\text{(3 s.f.).} \end{align*}

(b)(i)

解法一

思路

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本问采用另一个100盒样本及已知总体标准差，不使用前一问的估计方差。进行总体均值的右尾正态检验。

答题过程

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Let μ\mu be the population mean mass of a cereal box, in grams.

H0:μ=253,H1:μ>253.H_0:\mu=253,\qquad H_1:\mu>253.

For the sample of 100 boxes,

xˉ=25360100=253.6.\bar x=\frac{25360}{100}=253.6.

The test statistic is

z=253.6−2533.5/100=1.714285….z=\frac{253.6-253}{3.5/\sqrt{100}} =1.714285\ldots.

The upper-tail 5% critical value is 1.6451.645. Since 1.714285…>1.6451.714285\ldots>1.645, we reject H0H_0.

There is sufficient evidence at the 5% significance level that the population mean mass is greater than 253 g.

(b)(ii)

解法一

思路

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样本量100足够大，中心极限定理使样本均值近似正态；不需要原始质量分布为正态。

答题过程

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The statement is not true. The sample size of 100 is large, so the Central Limit Theorem gives an approximately normal distribution for the sample mean even if individual box masses are not normally distributed.