题目

Problem

A random sample of $40$ observations of a random variable $X$ and a random sample of $60$ observations of a random variable $Y$ are taken. The results are summarised as follows.

\begin{align*} \sum x=&\,67.1\qquad\sum(x-\overline{x})^2=934.3\\[2mm] \sum y=&\,84.1\qquad\sum(y-\overline{y})^2=756.2 \end{align*}

Find a $95\%$ confidence interval for the difference between the population means of $X$ and $Y$ .

(6)

题目中文翻译

从随机变量 $X$ 中随机抽取 $40$ 个观测值，从随机变量 $Y$ 中随机抽取 $60$ 个观测值。结果汇总如下。

\begin{align*} \sum x=&\,67.1\qquad\sum(x-\overline{x})^2=934.3\\[2mm] \sum y=&\,84.1\qquad\sum(y-\overline{y})^2=756.2 \end{align*}

求 $X$ 和 $Y$ 的总体均值之差的 $95\%$ 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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两个样本量都较大。分别估计总体方差，再用独立均值差的正态近似构造区间。

答题过程

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The sample means are $67.1/40$ and $84.1/60$ . The variance estimates are $934.3/39$ and $756.2/59$ . For independent samples, the estimated variance of the difference of means is

\begin{align*} \widehat{\operatorname{Var}}(\bar X-\bar Y) =&\,\frac{934.3/39}{40}+\frac{756.2/59}{60}\\ =&\,0.81252607\ldots. \end{align*}

The estimated standard error is

\sqrt{\frac{934.3/39}{40}+\frac{756.2/59}{60}}=0.901402\ldots.

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

\begin{align*} &\,\Big(\frac{67.1}{40}-\frac{84.1}{60}\Big)\\ &\,\ \pm\ 1.96(0.901402\ldots)\\ =&\,(-1.49091\ldots,2.04258\ldots). \end{align*}

Thus it is $(-1.49,2.04)$ .

CIE 9231 2026 June Paper 44 Q1

题目

解答

解法一

思路

答题过程