题目

Problem

A statistics course was taught by two professors. Students attend either all lectures given by professor A or all lectures given by professor B.

Students who took this course all sat the same examination at the end of the course and their mark, $x$ , was recorded.

The results of the examination for a random sample of students taught by each professor are summarised in the table below.

	$n$	$\sum x$	$\sum x^2$	Unbiased estimate of the mean	Unbiased estimate of the variance
Students taught by professor A	50	3610	260955.6	72.2	6.4
Students taught by professor B	50	3085	190457.2	$m$	$v$

(a) Find the value of $m$ and the value of $v$ .

(3)

The head of department claims that the mean mark of the students taught by professor A is more than 10 marks greater than the mean mark of the students taught by professor B.

(b) Use a suitable test, at the 5% level of significance, to investigate the claim made by the head of department.

(7)

(2)

(Total for Question 5 is 12 marks)

题目中文翻译

某统计课程由两位教授授课。学生要么全部听教授 A 的课，要么全部听教授 B 的课。

参加该课程的学生都在课程结束时参加同一场考试，并记录了他们的分数 $x$ 。

下表汇总了由两位教授分别教学的一个随机样本学生的考试结果。

	$n$	$\sum x$	$\sum x^2$	均值的无偏估计	方差的无偏估计
教授 A 的学生	50	3610	260955.6	72.2	6.4
教授 B 的学生	50	3085	190457.2	$m$	$v$

(a) 求 $m$ 和 $v$ 。

(3)

系主任声称，教授 A 所教学生的平均分比教授 B 所教学生的平均分高 10 分以上。

(b) 使用适当的检验，在 5% 显著性水平下检验系主任的说法。

(7)

(2)

解答

(a)

解法一

思路

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先用 B 组的总分除以样本量求样本均值 $m$ ，再用无偏样本方差公式计算 $v$ 。题目给出的 $v$ 是总体方差的无偏估计，因此分母应为 $n-1$ 。

答题过程

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For the students taught by professor B,

\begin{align*} m =&\,\frac{\sum x}{n}\\ =&\,\frac{3085}{50}\\ =&\,\boxed{61.7}. \end{align*}

The unbiased estimate of the variance is

\begin{align*} v =&\,\frac{\sum x^2-nm^2}{n-1}\\ =&\,\frac{190457.2-50(61.7)^2}{49}\\ =&\,\boxed{2.3}. \end{align*}

(b)

解法一

思路

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声称的是 A 组总体均值比 B 组总体均值高出超过 10 分，所以进行右尾检验。两组样本都很大，可用两样本均值差的正态近似检验，并以两组无偏样本方差估计总体方差。

答题过程

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Let $\mu_A$ and $\mu_B$ be the population mean marks for students taught by professors A and B respectively. The hypotheses are

\begin{align*} H_0:&\ \mu_A-\mu_B=10,\\ H_1:&\ \mu_A-\mu_B>10. \end{align*}

The standard error is

\begin{align*} \operatorname{SE}(\overline X_A-\overline X_B) =&\,\sqrt{\frac{6.4}{50}+\frac{2.3}{50}}\\ =&\,0.4171\ldots. \end{align*}

Therefore, the test statistic is

\begin{align*} z =&\,\frac{(72.2-61.7)-10} {\sqrt{\frac{6.4}{50}+\frac{2.3}{50}}}\\ =&\,1.1986\ldots\\ =&\,1.20. \end{align*}

For a one-tailed test at the $5\%$ significance level, the critical value is $1.6449$ , so the rejection region is

z\geqslant1.6449.

Since $1.20<1.6449$ , $H_0$ is not rejected. There is insufficient evidence at the $5\%$ significance level to support the head of department’s claim that professor A’s students have a mean mark more than 10 marks greater than professor B’s students.

(c)

解法一

思路

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检验中以两组样本方差代替相应总体方差，并依靠大样本的中心极限定理把两个样本均值的抽样分布近似为正态分布。这两点分别构成所需假设。

答题过程

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The assumptions are:

For both samples, the sample variance is equal to the corresponding population variance, so $s_A^2=\sigma_A^2$ and $s_B^2=\sigma_B^2$ .
Both samples are sufficiently large for the central limit theorem to apply, so the sampling distributions of both sample means are approximately normal.

S3 2026 Jan Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程