题目

Problem

A company manufactures copper pipes. The pipes are produced by two different machines, $A$ and $B$ . An inspector claims that the mean diameter of the pipes produced by machine $A$ is greater than the mean diameter of the pipes produced by machine $B$ . He takes a random sample of 12 pipes produced by machine $A$ and measures their diameters, $x$ cm. His results are summarised as follows.

\sum x = 6.24 \qquad \sum x^2 = 3.26

He also takes a random sample of 10 pipes produced by machine $B$ and measures their diameters in cm. His results are as follows.

\begin{gathered} 0.48,\ 0.53,\ 0.47,\ 0.54,\ 0.54,\\ 0.55,\ 0.46,\ 0.55,\ 0.50,\ 0.48. \end{gathered}

The diameters of the pipes produced by each machine are assumed to be normally distributed with equal population variances.

Test at the 2.5% significance level whether the data supports the inspector’s claim.

[9]

题目中文翻译

一家公司生产铜管。铜管由两台不同的机器 $A$ 和 $B$ 生产。一名检验员声称，机器 $A$ 生产的铜管平均直径大于机器 $B$ 生产的铜管平均直径。他随机抽取 12 根由机器 $A$ 生产的铜管，并测量它们的直径，记为 $x$ cm。结果汇总如下。

\sum x = 6.24 \qquad \sum x^2 = 3.26

他还随机抽取 10 根由机器 $B$ 生产的铜管，并测量它们的直径，单位为 cm。结果如下。

\begin{gathered} 0.48,\ 0.53,\ 0.47,\ 0.54,\ 0.54,\\ 0.55,\ 0.46,\ 0.55,\ 0.50,\ 0.48. \end{gathered}

假设每台机器生产的铜管直径都服从正态分布，并且总体方差相等。

在 2.5% 显著性水平下检验数据是否支持该检验员的说法。

(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

展开

两个独立小样本来自等方差的正态总体。合并组内离差后，检验机器 $A$ 的平均直径是否较大。

答题过程

展开

Let $\mu_A$ and $\mu_B$ be the population mean diameters. Test $H_0\colon\mu_A=\mu_B$ against $H_1\colon\mu_A>\mu_B$ . For machine B, the measurements give $\sum y=5.10$ and $\sum y^2=2.6124$ . The sample means are $6.24/12=0.52$ and $5.10/10=0.51$ . The unbiased variance estimates are

\begin{align*} s_A^2=&\,\frac{3.26-6.24^2/12}{11}=\frac{19}{13750},\\ s_B^2=&\,\frac{2.6124-5.10^2/10}{9}=\frac{19}{15000}. \end{align*}

The pooled variance is

\begin{align*} &\,s_p^2\\ =&\,\frac{(3.26-6.24^2/12)+(2.6124-5.10^2/10)}{12+10-2}\\ =&\,0.00133. \end{align*}

The test statistic is

t=\frac{0.52-0.51}{\sqrt{0.00133(1/12+1/10)}}=0.64040\ldots.

The upper $2.5\%$ critical value with $20$ degrees of freedom is $2.086$ . Since the statistic is smaller, we fail to reject $H_0$ . There is insufficient evidence at the $2.5\%$ level that machine A produces pipes with a greater population mean diameter.

CIE 9231 2022 Nov Paper 43 Q6

题目

解答

解法一

思路

答题过程