Skip to content
CalcGospel 國際數學圖譜
返回

S2 2025 Oct 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。

(1)

(b) 识别抽样单位。

(1)

(c) 给出普查而非抽样调查的一个优点和一个缺点。

(2)

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。

(5)

(第 2 题共 9 分)

解答

(a)

解法一

思路

展开

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

展开

A suitable sampling frame is a complete list of all customers who eat at the restaurant during the four weeks.

(b)

解法一

思路

展开

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

展开

The sampling units are the individual customers who eat at the restaurant.

(c)

解法一

思路

展开

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

展开

An advantage of taking a census is that the results are likely to be more accurate.

A disadvantage is that collecting and analysing the data is more time-consuming.

(d)

解法一

思路

展开

令 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 X∼B(50,0.3)X\sim B(50,0.3);观察到 20 人,计算 P(X≥20)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

展开

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.\begin{align*} &\,H_0:p=0.3,\\ &\,H_1:p>0.3. \end{align*}

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

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

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

 P(X≥20) =1−P(X≤19) ≈1−0.9152 =0.0848.\begin{align*} &\,P(X\geq20)\\ &\,=1-P(X\leq19)\\ &\,\approx1-0.9152\\ &\,=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

展开

也可以使用临界区域法。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(X≥20)P(X\geq20) 与 P(X≥21)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

展开

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

 P(X≥20) ≈0.0848>0.05\begin{align*} &\,P(X\geq20)\\ &\,\approx0.0848>0.05 \end{align*}

and

 P(X≥21) =1−P(X≤20) ≈0.0478<0.05.\begin{align*} &\,P(X\geq21)\\ &\,=1-P(X\leq20)\\ &\,\approx0.0478<0.05. \end{align*}

Therefore, the critical region is

X≥21.\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.