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CIE 9231 2025 Nov Paper 42 Q5

A Level / CIE / FS

CIE 9231 2025 Nov Paper 42 Paper · Question 5

题目

Problem

An engineer is comparing the tensile strengths of steel rods made from two machines, A and B. The engineer randomly selects 8 rods from machine A and 6 rods from machine B. The tensile strengths, in appropriate units, are given in the following table.

Machine A402403415412409407406410
Machine B401398395397410405

You should assume that the two distributions are normal and have the same population variance.

Use a t-test at the 5% significance level to test whether there is any difference in the mean tensile strengths of steel rods from the two machines.

[9]
题目中文翻译

一名工程师正在比较由两台机器 A 和 B 制造的钢棒的抗拉强度。该工程师从机器 A 随机选取 8 根钢棒,从机器 B 随机选取 6 根钢棒。以适当单位表示的抗拉强度如下表所示。

机器 A402403415412409407406410
机器 B401398395397410405

你应假设这两个分布均为正态分布,并且具有相同的总体方差。

在 5% 显著性水平下使用 t 检验,检验来自两台机器的钢棒平均抗拉强度是否存在差异。

(9)
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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两个样本独立,题目假设总体方差相等。先合并组内平方离差,再用双侧两样本 tt 检验。

答题过程

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Test H0 ⁣:μA=μBH_0\colon\mu_A=\mu_B against H1 ⁣:μA≠μBH_1\colon\mu_A\ne\mu_B, where μA\mu_A and μB\mu_B are the population mean tensile strengths. The sample means are 3264/8=4083264/8=408 and 2406/6=4012406/6=401. The two within-sample sums of squares are

 1331848−326428=136, 964964−240626=158.\begin{align*} &\,1331848-\frac{3264^2}{8}=136,\\ &\,964964-\frac{2406^2}{6}=158. \end{align*}

The pooled variance is sp2=(136+158)/(8+6−2)=24.5s_p^2=(136+158)/(8+6-2)=24.5. The statistic is

t=408−40124.5(1/8+1/6)=2.61861….t=\frac{408-401}{\sqrt{24.5(1/8+1/6)}}=2.61861\ldots.

For 1212 degrees of freedom the two-sided 5%5\% critical values are ±2.179\pm2.179. Since ∣t∣>2.179|t|>2.179, we reject H0H_0. There is sufficient evidence at the 5%5\% level that the two machines produce rods with different mean tensile strengths.