题目

Problem

A researcher is interested in whether there is a difference between two schools in students’ aptitude for English. She randomly chooses 10 students from school X and 8 students of a similar age from school Y to take a written English test. The scores for the students from school X ( $x$ ) and school Y ( $y$ ) are summarised as follows.

\begin{align*} &\,\sum x = 612 \qquad \sum x^2 = 40\,104\\ &\,\sum y = 444 \qquad \sum y^2 = 27\,460 \end{align*}

You should assume that the two distributions are normal and have the same population variance.

(a) Find a 95% confidence interval for the difference in the mean scores for students from school X and students from school Y in the written English test.

[6]

(b) Use the confidence interval you found in part (a) to explain why there is insufficient evidence at a 5% significance level to suggest that the English scores of students from school X and students from school Y are different.

[1]

题目中文翻译

一位研究人员想研究两所学校学生的英语能力是否存在差异。她从学校 X 随机选取 10 名学生，并从学校 Y 随机选取 8 名年龄相近的学生，让他们参加一场英语笔试。学校 X 学生的分数（ $x$ ）和学校 Y 学生的分数（ $y$ ）概括如下。

\begin{align*} &\,\sum x = 612 \qquad \sum x^2 = 40\,104\\ &\,\sum y = 444 \qquad \sum y^2 = 27\,460 \end{align*}

你应假设两个分布均为正态分布，且具有相同的总体方差。

(a) 求学校 X 学生与学校 Y 学生在英语笔试中总体平均分之差的 95% 置信区间。

(6)

(b) 使用你在 part (a) 中求得的置信区间，解释为什么在 5% 显著性水平下，没有足够证据表明学校 X 和学校 Y 学生的英语分数不同。

(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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题目给出两个正态总体且方差相同。先合并组内方差，再用两样本 $t$ 区间。

答题过程

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The sample means are $612/10=61.2$ and $444/8=55.5$ . The within-sample sums of squares about the sample means are

\begin{align*} \sum(x-\bar x)^2=&\,40104-\frac{612^2}{10}=2649.6,\\ \sum(y-\bar y)^2=&\,27460-\frac{444^2}{8}=2818. \end{align*}

The pooled variance estimate is

s_p^2=\frac{(40104-612^2/10)+(27460-444^2/8)}{10+8-2}=341.725.

The $95\%$ interval for $\mu_X-\mu_Y$ uses $t_{0.975,16}=2.120$ . It is

\begin{align*} &\,5.7\pm2.120\sqrt{341.725(1/10+1/8)}\\ =&\,(-12.8894\ldots,24.2894\ldots). \end{align*}

Thus the confidence interval is $(-12.9,24.3)$ .

(b)

解法一

思路

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将零差值与置信区间比较。 $95\%$ 双侧区间与同假设下 $5\%$ 双侧检验对应。

答题过程

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The confidence interval contains $0$ , so a zero difference in population mean scores is compatible with the data at this level. The corresponding two-sided $5\%$ test does not reject equality. Thus there is insufficient evidence that the schools have different population mean English scores.

CIE 9231 2025 June Paper 42 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