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CIE 9709 2025 November Paper 63 Q2

A Level / CIE / S2

CIE 9709 2025 Nov Paper 63 Paper · Question 2

题目

Problem

The mean mass of packets of Trueleaf tea is supposed to be 500500 grams. An inspector wishes to test whether this value is correct. He weighs 6060 randomly chosen packets and notes the mass, xx grams, of each packet. The results are summarised as follows.

n=60∑x=29 970∑x2=14 970 300n=60\qquad \sum x=29\,970\qquad \sum x^2=14\,970\,300

Test, at the 5%5\% significance level, whether the population mean mass is 500500 grams.

(8)
题目中文翻译

Trueleaf 茶每包的平均质量应为 500500 克。一名检验员想检验这一数值是否正确。他随机抽取 6060 包,称量并记录每包的质量 xx(以克为单位)。结果汇总如下。

n=60∑x=29 970∑x2=14 970 300n=60\qquad \sum x=29\,970\qquad \sum x^2=14\,970\,300

在 5%5\% 的显著性水平下检验总体平均质量是否为 500500 克。

(8)
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

解法一

思路

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先从汇总数据估计总体均值和无偏方差;样本量较大,可对样本均值用正态近似。检验的是均值是否正确,采用双侧备择。

答题过程

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The unbiased estimates are

x‾=2997060=499.5 g,\overline x=\frac{29970}{60}=499.5\ \text{g},

and

s2= 159(14970300−29970260)= 28559≈4.83051 g2.\begin{align*} s^2=&\,\frac1{59}\bigg(14970300-\frac{29970^2}{60}\bigg)\\[4mm] =&\,\frac{285}{59}\approx4.83051\ \text{g}^2. \end{align*}

Let μ\mu be the population mean mass of the tea packets. Test

H0:μ=500,H1:μ≠500.H_0:\mu=500,\qquad H_1:\mu\ne500.

For the large random sample, the Central Limit Theorem gives an approximately normal sample mean. Using s2s^2 to estimate the population variance,

z=499.5−500(285/59)/60≈−1.762.z=\frac{499.5-500}{\sqrt{(285/59)/60}}\approx-1.762.

At the 5%5\% significance level, the two-tailed critical values are ±1.96\pm1.96. Since ∣z∣<1.96|z|<1.96, do not reject H0H_0.

There is insufficient evidence at the 5%5\% level that the population mean mass differs from 500 g500\ \text{g}.