题目

Problem

A researcher is studying household waste recycling in two different cities, city $A$ and city $B$ . She randomly selects $8$ households in city $A$ and $9$ households in city $B$ . The researcher records the percentage of waste that each household recycles. Her results are as follows.

City $A$ :

\begin{align*} &\,25\quad49\quad41\quad38\\ &\,31\quad21\quad25\quad40 \end{align*}

City $B$ :

\begin{align*} &\,18\quad35\quad30\quad22\quad13\\ &\,23\quad24\quad36\quad24 \end{align*}

The population variances of the percentages of waste recycled in each city are assumed to be equal.

(a) Carry out a $t$ -test at the $10\%$ significance level to investigate whether there is a difference in the average percentage of waste that is recycled in city $A$ and city $B$ .

(9)

(b) State another assumption that is necessary for the test in 6(a) to be valid.

(1)

题目中文翻译

一位研究人员正在研究两个不同城市 $A$ 和 $B$ 的家庭垃圾回收情况。她在城市 $A$ 随机抽取 $8$ 户家庭，在城市 $B$ 随机抽取 $9$ 户家庭，记录每户家庭回收的垃圾占其垃圾总量的百分比。结果如下。

城市 $A$ ：

\begin{align*} &\,25\quad49\quad41\quad38\\ &\,31\quad21\quad25\quad40 \end{align*}

城市 $B$ ：

\begin{align*} &\,18\quad35\quad30\quad22\quad13\\ &\,23\quad24\quad36\quad24 \end{align*}

假设两个城市垃圾回收百分比的总体方差相等。

(a) 在 $10\%$ 的显著性水平下进行 $t$ 检验，考察城市 $A$ 和城市 $B$ 的平均垃圾回收百分比是否存在差异。

(9)

(b) 写出使 6(a) 中的检验有效所必需的另一项假设。

(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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两个样本独立，题给总体方差相同。用合并方差检验两个总体均值是否不同，因此做双侧检验。

答题过程

展开

Let $\mu_A,\mu_B$ be the population mean recycling percentages. Test $H_0\colon\mu_A=\mu_B$ against $H_1\colon\mu_A\ne\mu_B$ . The sample means are $270/8=33.75$ and $225/9=25$ . The sums of squared deviations are

\begin{align*} S_{AA}=&\,9778-\frac{270^2}{8}=665.5,\\ S_{BB}=&\,6079-\frac{225^2}{9}=454.0. \end{align*}

The pooled variance estimate is

\begin{align*} &\,s_p^2=\frac{(9778-270^2/8)+(6079-225^2/9)}{8+9-2}\\ =&\,74.63333\ldots. \end{align*}

The test statistic is

t=\frac{33.75-25}{\sqrt{s_p^2(1/8+1/9)}}=2.08441\ldots.

For $15$ degrees of freedom, the two-sided $10\%$ critical values are $\pm1.753$ . Since $|t|>1.753$ , we reject $H_0$ . There is sufficient evidence at the $10\%$ level that the population mean recycling percentages differ between the cities.

(b)

解法一

思路

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除了等方差，还要对两个城市各自的总体分布作正态假设。

答题过程

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The population distribution of household recycling percentages is assumed to be normal in each city.

CIE 9231 2026 June Paper 44 Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程