题目

Problem

An inspector is checking the lengths of metal rods produced by two machines, X and Y. These rods should be of the same length, but the inspector suspects that those made by machine X are shorter, on average, than those made by machine Y. The inspector chooses a random sample of 80 rods made by machine X and a random sample of 60 rods made by machine Y. The lengths of these rods are $x$ cm and $y$ cm respectively. Her results are summarised as follows.

\begin{align*} \sum x=&\,164.0,\qquad \sum x^2=338.1,\\ \sum y=&\,124.8,\qquad \sum y^2=261.1. \end{align*}

(a) Test at the 10% significance level whether the data supports the inspector’s suspicion.

[8]

(b) Give a reason why it is not necessary to make any assumption about the distributions of the lengths of the rods.

[1]

题目中文翻译

一名检查员正在检查两台机器 X 和 Y 生产的金属棒长度。这些金属棒应当长度相同，但检查员怀疑机器 X 生产的金属棒平均而言比机器 Y 生产的更短。检查员随机抽取 80 根机器 X 生产的金属棒和 60 根机器 Y 生产的金属棒。这些金属棒的长度分别为 $x$ cm 和 $y$ cm。她的结果概括如下。

\begin{align*} \sum x=&\,164.0,\qquad \sum x^2=338.1,\\ \sum y=&\,124.8,\qquad \sum y^2=261.1. \end{align*}

(a) 在 10% 显著性水平下，检验数据是否支持检查员的怀疑。

(8)

(b) 说明为什么不需要对金属棒长度的分布作任何假设。

(1)

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

(a)

解法一

思路

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用两个大样本分别估计方差。机器 X 的平均长度较短是左尾备择假设。

答题过程

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Let $\mu_X,\mu_Y$ be the population mean rod lengths. Test $H_0\colon\mu_X=\mu_Y$ against $H_1\colon\mu_X<\mu_Y$ . The means are $164.0/80=2.05$ and $124.8/60=2.08$ . The unbiased variance estimates are

\begin{align*} s_X^2=&\,\frac{338.1-164.0^2/80}{79}=\frac{19}{790},\\ s_Y^2=&\,\frac{261.1-124.8^2/60}{59}=\frac{379}{14750}. \end{align*}

The estimated variance of the difference of sample means is

\begin{align*} \widehat{\operatorname{Var}}(\bar X-\bar Y) =&\,\frac{s_X^2}{80}+\frac{s_Y^2}{60}\\ =&\,\frac{203839}{279660000}\\ =&\,0.000728881498\ldots. \end{align*}

The large-sample statistic is

z=\frac{2.05-2.08}{\sqrt{s_X^2/80+s_Y^2/60}}=-1.11120\ldots.

The lower $10\%$ critical value is $-1.282$ . Since $z>-1.282$ , we fail to reject $H_0$ . There is insufficient evidence at the $10\%$ level that machine X makes rods of a shorter population mean length.

(b)

解法一

思路

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样本量很大。中心极限定理给出样本均值的近似分布。

答题过程

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Both samples are large. By the central limit theorem, the sample means, and hence their difference, are approximately normal even if the original length distributions are not normal.

CIE 9231 2023 June Paper 43 Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程