题目

Problem

Students studying geography at a college choose either human geography or physical geography. A teacher is investigating the times taken by students to complete assignments of equal difficulty in human geography and physical geography. He selects a random sample of $7$ human geography students and records the times taken, $x$ minutes, to complete an assignment. He also selects a random sample of $9$ physical geography students and records the times taken, $y$ minutes, to complete an assignment. The results are summarised as follows.

\begin{align*} \sum x=&\,445,\qquad\sum x^2=29144,\\ \sum y=&\,310,\qquad\sum y^2=11622. \end{align*}

You should assume that the underlying distributions are normal and have equal population variances.

Find a $95\%$ confidence interval for the difference in population mean times taken to complete assignments in human geography and physical geography.

(6)

题目中文翻译

某学院学习地理的学生选择人文地理或自然地理。一位教师正在调查学生完成人文地理和自然地理中难度相同的作业所需的时间。他随机抽取 $7$ 名人文地理学生，记录每人完成一份作业所需的时间 $x$ 分钟；另随机抽取 $9$ 名自然地理学生，记录每人完成一份作业所需的时间 $y$ 分钟。结果汇总如下。

\begin{align*} \sum x=&\,445,\qquad\sum x^2=29144,\\ \sum y=&\,310,\qquad\sum y^2=11622. \end{align*}

应假设相应的总体分布均为正态分布，且总体方差相等。

求完成人文地理和自然地理作业所需时间的总体均值之差的 $95\%$ 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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总体正态且方差相等。用两组的平方离差和合并方差，再构造小样本 $t$ 区间。

答题过程

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The sample means are $445/7$ and $310/9$ . The sums of squared deviations are

\begin{align*} S_{xx}=&\,29144-\frac{445^2}{7}=\frac{5983}{7},\\ S_{yy}=&\,11622-\frac{310^2}{9}=\frac{8498}{9}. \end{align*}

The pooled variance estimate is

\begin{align*} &\,s_p^2=\frac{(29144-445^2/7)+(11622-310^2/9)}{7+9-2}\\ =&\,128.49546\ldots. \end{align*}

Using $t_{0.975,14}=2.145$ , the $95\%$ interval for the human-minus-physical population mean difference is

\begin{align*} &\,\frac{445}{7}-\frac{310}{9}\\ &\,\ \pm\ 2.145\sqrt{s_p^2(1/7+1/9)}\\ =&\,(16.87346\ldots,41.38050\ldots). \end{align*}

Thus the interval is $(16.9,41.4)$ minutes.

CIE 9231 2026 June Paper 41 Q3

题目

解答

解法一

思路

答题过程