题目

Problem

Rhiannon is investigating the mass of individual bananas sold in two supermarkets. Rhiannon claims that the mean mass of bananas sold by supermarket $X$ is less than the mean mass of bananas sold by supermarket $Y$ .

She takes random samples of $44$ bananas from supermarket $X$ and $52$ bananas from supermarket $Y$ . She records their masses, $x\text{ kg}$ and $y\text{ kg}$ , respectively. Her results are summarised as follows.

\begin{align*} \sum x=&\,6.63,\qquad\sum x^2=1.21,\\ \sum y=&\,8.56,\qquad\sum y^2=1.65. \end{align*}

You should not assume that the population variances of the masses of bananas sold by each supermarket are equal.

(a) Test Rhiannon’s claim at the $10\%$ significance level.

(8)

(b) Find a $90\%$ confidence interval for the difference between the population mean masses of bananas in supermarkets $X$ and $Y$ .

(3)

题目中文翻译

Rhiannon 正在调查两家超市所售单根香蕉的质量。Rhiannon 声称，超市 $X$ 所售香蕉的平均质量小于超市 $Y$ 所售香蕉的平均质量。

她从超市 $X$ 随机抽取 $44$ 根香蕉，从超市 $Y$ 随机抽取 $52$ 根香蕉，分别记录它们的质量为 $x\text{ kg}$ 和 $y\text{ kg}$ 。结果汇总如下。

\begin{align*} \sum x=&\,6.63,\qquad\sum x^2=1.21,\\ \sum y=&\,8.56,\qquad\sum y^2=1.65. \end{align*}

不应假设两家超市所售香蕉质量的总体方差相等。

(a) 在 $10\%$ 的显著性水平下检验 Rhiannon 的声称。

(8)

(b) 求超市 $X$ 和超市 $Y$ 所售香蕉质量的总体均值之差的 $90\%$ 置信区间。

(3)

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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两个样本量都较大，且总体方差不假设相同。分别估计方差，对均值差做左尾正态近似检验。

答题过程

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Test $H_0\colon\mu_X=\mu_Y$ against $H_1\colon\mu_X<\mu_Y$ . The means are $6.63/44$ and $8.56/52$ . The unbiased variance estimates are

\begin{align*} &\,s_X^2=\frac{1.21-6.63^2/44}{43},\\ &\,s_Y^2=\frac{1.65-8.56^2/52}{51}. \end{align*}

Thus $s_X^2=0.00490650\ldots$ and $s_Y^2=0.00472337\ldots$ . For the independent samples,

\begin{align*} \widehat{\operatorname{Var}}(\bar X-\bar Y) =&\,\frac{s_X^2}{44}+\frac{s_Y^2}{52}\\ =&\,0.00020234559\ldots. \end{align*}

The estimated standard error is

\sqrt{s_X^2/44+s_Y^2/52}=0.01422482\ldots.

Thus the statistic is $(6.63/44-8.56/52)/0.01422482\ldots=-0.97952\ldots$ . The lower $10\%$ critical value is $-1.282$ . Since the statistic is higher, we fail to reject $H_0$ . There is insufficient evidence at the $10\%$ level that bananas from supermarket X have a lower population mean mass.

(b)

解法一

思路

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沿用 (a) 的独立均值差标准误。用双侧 $90\%$ 临界值构造区间。

答题过程

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The interval for $\mu_X-\mu_Y$ is

\begin{align*} &\,\Big(\frac{6.63}{44}-\frac{8.56}{52}\Big)\\ &\,\ \pm\ 1.645\sqrt{s_X^2/44+s_Y^2/52}\\ =&\,(-0.0373334\ldots,0.00946626\ldots). \end{align*}

Thus the $90\%$ confidence interval is $(-0.0373,0.00947)$ kg.

CIE 9231 2026 June Paper 43 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