题目

Problem

Two different types of juice extractor, machine X and machine Y, are being compared. The manufacturer claims that machine Y extracts more juice per orange on average than machine X.

A random sample of 20 oranges is selected and each carefully measured. Oranges of a similar size and mass are then paired and numbered from 1 to 10. One orange from each pair is randomly allocated to machine X with the other allocated to machine Y. The amount of juice, in ml, extracted from each orange is recorded in the following table.

pair	1	2	3	4	5	6	7	8	9	10
machine X	65	73	58	61	72	79	64	65	69	71
machine Y	68	72	64	63	75	82	63	63	72	74

(a) Use a t-test to test the manufacturer’s claim at the $1\%$ significance level.

[7]

(b) State an assumption required for the test in 3(a) to be valid.

[1]

The manufacturer notices that the amount of juice extracted from the oranges in pair 4 were recorded incorrectly. In fact, the amount of juice extracted from machine X was $60\text{ ml}$ and from machine Y was $62\text{ ml}$ .

[2]

题目中文翻译

正在比较两种不同类型的榨汁机，machine X 和 machine Y。制造商声称 machine Y 平均每个橙子榨出的果汁量比 machine X 更多。

随机选取 20 个橙子，并仔细测量每个橙子。然后将大小和质量相近的橙子配对，并编号为 1 到 10。每一对中，一个橙子随机分配给 machine X，另一个分配给 machine Y。每个橙子榨出的果汁量（单位为 ml）记录在下表中。

配对编号	1	2	3	4	5	6	7	8	9	10
X 型机器	65	73	58	61	72	79	64	65	69	71
Y 型机器	68	72	64	63	75	82	63	63	72	74

(a) 使用 t 检验，在 $1\%$ 显著性水平下检验制造商的声称。

(7)

(b) 写出 3(a) 中检验有效所需的一个假设。

(1)

制造商注意到第 4 对橙子的榨汁量记录有误。实际上，machine X 榨出的果汁量为 $60\text{ ml}$ ，machine Y 榨出的果汁量为 $62\text{ ml}$ 。

(2)

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

展开

两台机器的数据按橙子配对，检验的是每对的差值。按厂家声明设右尾备择假设。

答题过程

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Let $D$ be the juice volume from machine Y minus that from machine X for a matched pair, and let $\mu_D$ be its population mean. Test $H_0\colon\mu_D=0$ against $H_1\colon\mu_D>0$ .

The differences are $3,-1,6,2,3,3,-1,-2,3,3$ . They give $\sum d=19$ , $\sum d^2=91$ and

\begin{align*} &\,\bar d=1.9,\\ &\,s_D^2=\frac{91-19^2/10}{9}=6.1. \end{align*}

The paired test statistic is

t=\frac{1.9}{\sqrt{6.1/10}}=2.43270\ldots.

The upper $1\%$ critical value with $9$ degrees of freedom is $2.821$ . Since $2.43<2.821$ , we fail to reject $H_0$ . There is insufficient evidence at the $1\%$ level that machine Y extracts more juice per orange on average than machine X.

(b)

解法一

思路

展开

配对检验使用的是差值样本，因此正态假设也针对差值。

答题过程

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The population distribution of paired differences in juice volume is assumed to be normal.

(c)

解法一

思路

展开

只需比较更正前后的第四对差值。配对检验用的全部差值若不变，检验结果就不变。

答题过程

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Pair 4 originally gave $63-61=2$ ml and now gives $62-60=2$ ml. Every paired difference is unchanged, so the sample mean and variance of the differences and the test statistic are unchanged. The conclusion remains the same.

CIE 9231 2025 Nov Paper 44 Q3

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