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IAL 2025 Oct S2 A Q2

A Level / Edexcel / S2

IAL 2025 Oct A Paper · Question 2

题目

Problem

Bill owns a restaurant.

Over the next four weeks Bill decides to carry out a sample survey to obtain the customers’ opinions.

(a) Suggest a suitable sampling frame for the sample survey.

(1)

(b) Identify the sampling units.

(1)

(c) Give one advantage and one disadvantage of taking a census rather than a sample survey.

(2)

Bill believes that only 30% of customers would like a greater choice on the menu.

He takes a random sample of 50 customers and finds that 20 of them would like a greater choice on the menu.

(d) Test, at the 5% significance level, whether or not the percentage of customers who would like a greater choice on the menu is more than Bill believes. State your hypotheses clearly.

(5)

(Total for Question 2 is 9 marks)

题目中文翻译

Bill 拥有一家餐厅。

在接下来的四周里,Bill 决定进行抽样调查以获取顾客的意见。

(a) 建议一个适合此抽样调查的抽样框架。

(b) 识别抽样单位。

(c) 给出普查而非抽样调查的一个优点和一个缺点。

Bill 认为只有 30% 的顾客希望菜单有更多选择。

他随机抽取了 50 名顾客样本,发现其中 20 人希望菜单有更多选择。

(d) 在 5% 的显著性水平下,检验希望菜单有更多选择的顾客比例是否大于 Bill 的预期。清楚地陈述你的假设。

(第 2 题共 9 分)

解答

(a)

解法一

思路

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抽样框应是能够识别目标总体中每个抽样单位的完整名单。这里的总体是在这四周内到餐厅用餐的所有顾客,因此应使用这些顾客的完整记录,而不能只列部分顾客或顾客意见。

答题过程

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A suitable sampling frame is

a complete list of all customers who eatat the restaurant during the four weeks.\boxed{ \begin{gathered} \text{a complete list of all customers who eat}\\ \text{at the restaurant during the four weeks} \end{gathered} }.

(b)

解法一

思路

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抽样单位是实际被选择和调查的基本对象,而不是顾客给出的意见。因此,本调查中的抽样单位是每一名到餐厅用餐的顾客。

答题过程

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The sampling units are

the individual customerswho eat at the restaurant.\boxed{ \begin{gathered} \text{the individual customers}\\ \text{who eat at the restaurant} \end{gathered} }.

(c)

解法一

思路

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普查涵盖总体中的每一名顾客,因此通常能减少抽样误差,结果更准确;但需要收集并分析更多数据,所以更加耗时。题目各要求一个优点和缺点,清楚标明即可。

答题过程

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An advantage of taking a census is that

the results are likelyto be more accurate.\boxed{ \begin{gathered} \text{the results are likely}\\ \text{to be more accurate} \end{gathered} }.

A disadvantage is that

collecting and analysing the datais more time-consuming.\boxed{ \begin{gathered} \text{collecting and analysing the data}\\ \text{is more time-consuming} \end{gathered} }.

(d)

解法一

思路

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pp 为希望菜单有更多选择的顾客比例。题目检验该比例是否大于 0.3,所以使用右尾二项检验。在原假设下 XB(50,0.3)X\sim B(50,0.3);观察到 20 人,计算 P(X20)P(X\geq20) 并与 0.05 比较,最后写出结合餐厅情境的结论。

答题过程

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Let pp be the proportion of customers who would like a greater choice on the menu. The hypotheses are

H0:p=0.3,H1:p>0.3.H_0:p=0.3, \qquad H_1:p>0.3.

Let XX be the number of such customers in the sample. Under H0H_0,

XB(50,0.3).X\sim\operatorname{B}(50,0.3).

The observed value is X=20X=20, so the upper-tail probability is

P(X20)=1P(X19)=10.9152=0.0848.\begin{align*} P(X\geq20) =&\,1-P(X\leq19)\\ =&\,1-0.9152\\ =&\,0.0848. \end{align*}

Since

0.0848>0.05,0.0848>0.05,

H0H_0 is not rejected. There is insufficient evidence at the 5%5\% significance level to suggest that more than 30%30\% of the restaurant’s customers would like a greater choice on the menu.

解法二

思路

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官方评分资料也接受临界区域法。从较大的整数开始寻找右尾概率不超过 0.05 的最小边界;比较 P(X20)P(X\geq20)P(X21)P(X\geq21) 后得到临界区域,再检查观察值 20 是否落入其中。

答题过程

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Under H0H_0, XB(50,0.3)X\sim\operatorname{B}(50,0.3). The adjacent upper-tail probabilities are

P(X20)=0.0848>0.05P(X\geq20)=0.0848>0.05

and

P(X21)=1P(X20)=0.0478<0.05.P(X\geq21)=1-P(X\leq20)=0.0478<0.05.

Therefore, the critical region is

X21.\boxed{X\geq21}.

The observed value X=20X=20 is not in the critical region, so H0H_0 is not rejected. There is insufficient evidence at the 5%5\% significance level to suggest that more than 30%30\% of the restaurant’s customers would like a greater choice on the menu.