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CIE 9231 2024 June Paper 43 Q3

A Level / CIE / FS

CIE 9231 2024 June Paper 43 Paper · Question 3

题目

Problem

There are three bus companies in a city. The council is investigating whether the buses reliably arrive at their destination on time. The results from random samples of buses from each company are summarised in the following table.

Bus company
ABCTotal
ArrivalEarly22221054
ArrivalOn time305242124
ArrivalLate28261872
Total8010070250

Test, at the 5% significance level, whether the reliability of buses is independent of bus company.

[7]
题目中文翻译

某城市有三家公交公司。市政部门正在调查公交车是否能可靠地准时到达目的地。从每家公司公交车的随机样本得到的结果概括如下表。

公交公司
ABC合计
到达情况提前22221054
到达情况准时305242124
到达情况晚点28261872
合计8010070250

在 5% 显著性水平下,检验公交车的可靠性是否与公交公司相互独立。

(7)
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

解法一

思路

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在独立假设下,从每个到达类别的总数和每家公司的总数求期望频数。将九格的偏差合成一个卡方统计量。

答题过程

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Test H0H_0 that arrival category is independent of bus company against H1H_1 that the variables are associated. The expected count for an early bus from company A is 54(80)/250=17.2854(80)/250=17.28. The full expected table is

ArrivalABC
Early17.2821.6015.12
On time39.6849.6034.72
Late23.0428.8020.16

All expected counts exceed 55. The contributions (O−E)2/E(O-E)^2/E, in the same cell order, are

ArrivalABC
Early1.2892590.0074071.733757
On time2.3614520.1161291.526452
Late1.0677780.2722220.231429

Summing the unrounded contributions gives

 χ2=∑(O−E)2E= (22−17.28)217.28+(22−21.60)221.60+⋯= 8.60588….\begin{align*} &\,\chi^2=\sum\frac{(O-E)^2}{E}\\ =&\,\frac{(22-17.28)^2}{17.28}+\frac{(22-21.60)^2}{21.60}+\cdots\\ =&\,8.60588\ldots. \end{align*}

The degrees of freedom are (3−1)(3−1)=4(3-1)(3-1)=4. The upper 5%5\% critical value is 9.4889.488. Since the statistic is smaller, we fail to reject H0H_0. There is insufficient evidence at the 5%5\% level that arrival reliability depends on bus company.