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CIE 9231 2026 June Paper 42 Q2

A Level / CIE / FS

CIE 9231 2026 June Paper 42 Paper · Question 2

题目

Problem

Three different bus companies, AA, BB and CC, record the reliability of their service. Each bus journey is recorded as being on time, late or cancelled. A reliable bus service is considered to be one that is on time. The records from a random sample of buses for each company are taken. These records are summarised in the table.

company AAcompany BBcompany CCtotal
on time7222049
late14111237
cancelled46414
total253936100

(a) Test at the 5%5\% significance level whether the reliability of the service is independent of the bus company.

(8)

(b) Further evidence indicates that one company may be particularly unreliable compared with the other two companies. Use your calculations in 2(a) to suggest which company this is likely to be.

(2)
题目中文翻译

三家不同的公交公司 AA、BB 和 CC 记录各自服务的可靠性。每趟公交行程被记录为准点、晚点或取消。准点的公交服务被视为可靠。对每家公司随机抽取公交车辆的记录,汇总如下表。

公司 AA公司 BB公司 CC合计
准点7222049
晚点14111237
取消46414
合计253936100

(a) 在 5%5\% 的显著性水平下,检验服务的可靠性是否与公交公司相互独立。

(8)

(b) 进一步的证据表明,与另外两家公司相比,其中一家公司可能特别不可靠。利用你在 2(a) 中的计算,指出最可能是哪家公司。

(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

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先按独立假设算九格期望频数。因为可靠指准时,将晚点和取消合成同一个不准时类别。

答题过程

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Test H0H_0 that service reliability is independent of bus company against H1H_1 that it is associated with bus company. The original expected counts, using row total times column total divided by 100100, are

ArrivalCompany ACompany BCompany C
On time12.2519.1117.64
Late9.2514.4313.32
Cancelled3.505.465.04

The expected count 3.503.50 is below 55. Combine late and cancelled into the not-on-time category. Its observed counts are 18,17,1618,17,16, and its expected counts are 12.75,19.89,18.3612.75,19.89,18.36. All pooled expected frequencies exceed 55. The statistic is

 χ2=∑(O−E)2E= (7−12.25)212.25+ (22−19.11)219.11+⋯= 5.8878….\begin{align*} &\,\chi^2=\sum\frac{(O-E)^2}{E}\\ =&\,\frac{(7-12.25)^2}{12.25}\\ +&\,\frac{(22-19.11)^2}{19.11}+\cdots\\ =&\,5.8878\ldots. \end{align*}

There are (2−1)(3−1)=2(2-1)(3-1)=2 degrees of freedom. The upper 5%5\% critical value is 5.9915.991. Since the statistic is smaller, we fail to reject H0H_0. There is insufficient evidence at the 5%5\% level that reliability is associated with bus company.

(b)

解法一

思路

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比较各公司的不准时班次与独立假设下的期望值。找出实际频数高于期望频数的公司。

答题过程

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Company A appears least reliable. It had 1818 late or cancelled journeys against an expected 12.7512.75. For companies B and C, the observed numbers of late or cancelled journeys were below their expected counts.