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CIE 9709 2025 June Paper 62 Q5

A Level / CIE / S2

CIE 9709 2025 June Paper 62 Paper · Question 5

题目

Problem

The amount of time, in minutes, spent by a customer on one visit to a certain shop is modelled by the random variable X∼N(μ,σ2)X\sim\mathrm{N}(\mu,\sigma^2). In the past, the values of μ\mu and σ\sigma were 10.510.5 and 3.83.8 respectively. The shop has recently moved to a new location, and the manager hopes that the new value of μ\mu will be greater than 10.510.5. He takes a random sample of 1010 customers and notes the time they each spend in the shop. He then calculates the sample mean x‾\overline{x} for these 1010 times.

Using a hypothesis test at the 5%5\% significance level, the manager finds that there is sufficient evidence to conclude that the new value of μ\mu is greater than 10.510.5.

Stating a necessary assumption, find the smallest possible value of x‾\overline{x}.

(4)
题目中文翻译

用随机变量 X∼N(μ,σ2)X\sim\mathrm{N}(\mu,\sigma^2) 对顾客一次到某商店购物所停留的时间建模，时间以分钟为单位。过去 μ\mu 和 σ\sigma 的值分别为 10.510.5 和 3.83.8。这家店最近搬到了新地点，店长希望新的 μ\mu 值大于 10.510.5。他随机抽取 1010 名顾客，记录每人在店内停留的时间，并计算这 1010 个时间的样本均值 x‾\overline{x}。

店长在 5%5\% 的显著性水平下进行假设检验，发现有充分证据得出新的 μ\mu 值大于 10.510.5 的结论。

写出一个必要的假设，并求 x‾\overline{x} 的最小可能值。

(4)
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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总体正态已经给定，额外假设是标准差仍为 3.83.8。由右尾 5%5\% 拒绝边界反求样本均值。

答题过程

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Assume that the population standard deviation is unchanged at 3.83.8 minutes. Test H0:μ=10.5H_0:\mu=10.5 against H1:μ>10.5H_1:\mu>10.5.

Under H0H_0,

X‾∼N(10.5,3.8210).\overline X\sim\mathrm N\bigg(10.5,\frac{3.8^2}{10}\bigg).

At the lower boundary of the upper-tail critical region,

c−10.53.8/10= Φ−1(0.95)≈ 1.645,c= 10.5+1.6453.810≈ 12.477.\begin{align*} \frac{c-10.5}{3.8/\sqrt{10}}=&\,\Phi^{-1}(0.95)\\[4mm] \approx&\,1.645,\\[4mm] c=&\,10.5+1.645\frac{3.8}{\sqrt{10}}\\[4mm] \approx&\,12.477. \end{align*}

Thus the smallest sample mean at the critical boundary is 12.5 minutes\boxed{12.5\text{ minutes}} to 33 significant figures; values above the unrounded boundary lead to rejection.