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IAL 2026 Jan S3 A Q3

A Level / Edexcel / S3

IAL 2026 Jan A Paper · Question 3

题目

Problem

A researcher investigates the results of candidates who took their driving test at one of three driving test centres.

A random sample of 620 candidates gave the following results.

ResultABCTotal
Pass9911068277
Fail108116119343
Total207226187620

(a) Test, at the 5% level of significance, whether there is an association between the results of candidates’ driving tests and the driving test centre. Show your working clearly. You should state your hypotheses, expected frequencies, test statistic and the critical value for this test.

(10)

The researcher decides to conduct a further investigation into the results of candidates’ driving tests.

(b) State which driving test centre you would recommend for further investigation. Give a reason for your answer.

(2)
(Total for Question 3 is 12 marks)
题目中文翻译

一位研究者调查了在三个驾驶考试中心之一参加驾驶考试的考生结果。

从 620 名考生中抽得一个随机样本,结果如下。

结果ABC总计
通过9911068277
未通过108116119343
总计207226187620

(a) 在 5% 的显著性水平下检验考生的驾驶考试结果与驾驶考试中心之间是否存在关联。请清楚写出你的过程。你应写出原假设、期望频数、检验统计量和临界值。

研究者决定对考生驾驶考试结果做进一步调查。

(b) 说明你会建议进一步调查哪个驾驶考试中心,并给出理由。

解答

(a)

解法一

思路

展开

这是一个 2×32\times3 列联表的卡方独立性检验。先在“考试中心与结果相互独立”的原假设下,用“行总计乘列总计再除以总人数”求每格期望频数;然后计算六格对 χ2\chi^2 的贡献,确定自由度、临界值并作出有语境的结论。

答题过程

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Let H0H_0 be the hypothesis that driving test centre and result are independent.

Let H1H_1 be the hypothesis that there is an association between driving test centre and result.

Under H0H_0, each expected frequency is calculated using

E=(row total)(column total)grand total.E=\frac{(\text{row total})(\text{column total})}{\text{grand total}}.

For example,

EPass,A=277×207620=92.482.E_{\text{Pass},A} =\frac{277\times207}{620} =92.482\ldots.

The complete table of expected frequencies, rounded to 2 decimal places, is

ResultABCTotal
Pass92.48100.9783.55277
Fail114.52125.03103.45343
Total207226187620

Now calculate each contribution (OE)2E\frac{(O-E)^2}{E}:

ResultABC
Pass0.45960.80752.8941
Fail0.37120.65212.3373

Therefore,

χ2=0.4596+0.8075+2.8941+0.3712+0.6521+2.3373=7.522(approximately).\begin{align*} \chi^2 =&\,0.4596+0.8075+2.8941\\ +&\,0.3712+0.6521+2.3373\\ =&\,7.522\quad\text{(approximately)}. \end{align*}

The degrees of freedom are

ν=(21)(31)=2.\nu=(2-1)(3-1)=2.

At the 5%5\% significance level, the critical value is

χ2,0.052=5.991,\chi^2_{2,0.05}=5.991,

so the critical region is χ25.991\chi^2\geq5.991.

Since

7.522>5.991,7.522>5.991,

the result is significant and H0H_0 is rejected. There is evidence, at the 5%5\% significance level, of an association between driving test centre and driving test result.

The official alternative calculation gives the same statistic:

χ2=O2E620=627.522620=7.522.\begin{align*} \chi^2 =&\,\sum\frac{O^2}{E}-620\\ =&\,627.522\ldots-620\\ =&\,7.522\ldots. \end{align*}

(b)

解法一

思路

展开

进一步调查应聚焦于对检验统计量贡献最大的中心。把同一中心“通过”和“未通过”两格的贡献相加,C 中心的总贡献约为 5.235.23,远高于 A、B,说明 C 中心的观察频数与独立性假设下的期望频数差异最大。

答题过程

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The contributions to the test statistic from each centre are

Centre A:0.4596+0.3712=0.8308,Centre B:0.8075+0.6521=1.4596,Centre C:2.8941+2.3373=5.2314.\begin{align*} \text{Centre A}:&\quad 0.4596+0.3712=0.8308,\\ \text{Centre B}:&\quad 0.8075+0.6521=1.4596,\\ \text{Centre C}:&\quad 2.8941+2.3373=5.2314. \end{align*}

Therefore, the researcher should investigate centre C\boxed{\text{centre C}}, because it makes the largest contribution to the χ2\chi^2 test statistic. In particular, its observed pass frequency, 68, is much lower than its expected frequency, approximately 83.55.