题目

Problem

The manager of a hardware store is interested in whether there is a difference in the amount spent per customer on weekdays ( $x$ ) compared to weekends ( $y$ ). Random samples of 120 customers on weekdays and 80 customers on weekends are taken and the amount spent by each customer is recorded. The results are summarised as follows.

\begin{align*} &\,\sum x = 10\,470 \qquad \sum (x - \bar{x})^2 = 12\,283\\ &\,\sum y = 6560 \qquad \sum (y - \bar{y})^2 = 13\,520 \end{align*}

Test at the 1% significance level whether there is a difference in the mean amount spent per customer on weekdays compared to weekends. You should not assume that the population variances of the amounts spent on weekdays and weekends are equal.

[7]

题目中文翻译

一家五金店的经理想研究工作日每位顾客的消费金额（ $x$ ）与周末每位顾客的消费金额（ $y$ ）是否存在差异。随机抽取 120 名工作日顾客和 80 名周末顾客，并记录每位顾客的消费金额。结果概括如下。

\begin{align*} &\,\sum x = 10\,470 \qquad \sum (x - \bar{x})^2 = 12\,283\\ &\,\sum y = 6560 \qquad \sum (y - \bar{y})^2 = 13\,520 \end{align*}

在 1% 显著性水平下检验工作日与周末每位顾客的总体平均消费金额是否存在差异。你不应假设工作日和周末消费金额的总体方差相等。

(7)

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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两个样本量都大，且不假设方差相等。分别估计方差并用均值差的正态近似，做双侧检验。

答题过程

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Let $\mu_x$ and $\mu_y$ be the population mean spending per customer on weekdays and weekends. Test $H_0\colon\mu_x=\mu_y$ against $H_1\colon\mu_x\ne\mu_y$ . The sample means are $10470/120=87.25$ and $6560/80=82$ . The unbiased variance estimates are $12283/119$ and $13520/79$ . The estimated variance of the difference of independent sample means is

\frac{12283/119}{120}+\frac{13520/79}{80}=2.999394567\ldots.

The large-sample statistic is

\begin{align*} &\,z=\frac{87.25-82}{\sqrt{(12283/119)/120+(13520/79)/80}}\\ =&\,3.03139\ldots. \end{align*}

The two-sided $1\%$ critical values are $\pm2.576$ . Since $|z|>2.576$ , we reject $H_0$ . There is sufficient evidence at the $1\%$ level that mean spending per customer differs between weekdays and weekends.

CIE 9231 2025 June Paper 41 Q1

题目

解答

解法一

思路

答题过程