题目

Problem

A factory produces small bottles of natural spring water. Two different machines, $X$ and $Y$ , are used to fill empty bottles with the water. A quality control engineer checks the volumes of water in the bottles filled by each of the machines. He chooses a random sample of 60 bottles filled by machine $X$ and a random sample of 75 bottles filled by machine $Y$ . The volumes of water, $x$ and $y$ respectively, in millilitres, are summarised as follows.

\begin{gathered} \sum x = 6345,\qquad \sum (x-\bar{x})^2 = 243.8,\\ \sum y = 7614,\qquad \sum (y-\bar{y})^2 = 384.9. \end{gathered}

$\bar{x}$ and $\bar{y}$ are the sample means of the volume of water in the bottles filled by machines $X$ and $Y$ respectively.

Find a 95% confidence interval for the difference between the mean volume of water in bottles filled by machine $X$ and the mean volume of water in bottles filled by machine $Y$ .

[6]

题目中文翻译

一家工厂生产小瓶天然泉水。有两台不同的机器 $X$ 和 $Y$ 用来向空瓶中灌水。质量控制工程师检查由两台机器灌装的瓶中水的体积。他从机器 $X$ 灌装的瓶中随机抽取 60 瓶，从机器 $Y$ 灌装的瓶中随机抽取 75 瓶。水的体积分别记为 $x$ 和 $y$ ，单位为毫升，汇总如下。

\begin{gathered} \sum x = 6345,\qquad \sum (x-\bar{x})^2 = 243.8,\\ \sum y = 7614,\qquad \sum (y-\bar{y})^2 = 384.9. \end{gathered}

$\bar{x}$ 和 $\bar{y}$ 分别是机器 $X$ 与机器 $Y$ 灌装的瓶中水体积的样本均值。

求机器 $X$ 灌装瓶中水的平均体积与机器 $Y$ 灌装瓶中水的平均体积之差的 95% 置信区间。

(6)

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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两组样本量都大。用各组平方离差和求无偏方差，再相加得到独立均值差的方差估计。

答题过程

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The sample means are $\bar x=6345/60=105.75$ and $\bar y=7614/75=101.52$ . The variance estimates are $s_X^2=243.8/59$ and $s_Y^2=384.9/74$ . Using the large-sample normal approximation, the estimated standard error of the difference is

\begin{align*} &\,\sqrt{\frac{243.8/59}{60}+\frac{384.9/74}{75}}\\ =&\,0.371781\ldots. \end{align*}

The $95\%$ confidence interval for $\mu_X-\mu_Y$ is

\begin{align*} &\,4.23\pm1.96(0.371781\ldots)\\ =&\,(3.50130\ldots,4.95869\ldots). \end{align*}

Thus it is $(3.50,4.96)$ ml to three significant figures.

CIE 9231 2023 Nov Paper 42 Q1

题目

解答

解法一

思路

答题过程