题目

Problem

Zimo wants to apply for a job with a large international company.

All applicants must take an aptitude test to see if they are suitable for the job.

There are two centres where Zimo can take the test, centre A and centre B.

Zimo found out that last month

90 people took the test at centre A and 60 passed
110 people took the test at centre B and 80 passed

Test, at the 5% level of significance, whether or not the aptitude test result is independent of the test centre. Show your working clearly. You must state your hypotheses, expected frequencies, test statistic and critical value.

(8)

(Total for Question 1 is 8 marks)

题目中文翻译

Zimo 想申请一份大型国际公司的工作。

所有应聘者都必须参加一项能力测试，以判断是否适合这份工作。

Zimo 可以在两个中心参加测试，A 中心和 B 中心。

Zimo 发现上个月：

A 中心有 90 人参加测试，其中 60 人通过
B 中心有 110 人参加测试，其中 80 人通过

在 5% 的显著性水平下，检验该能力测试结果是否与考试中心相互独立。请清楚写出你的过程。你必须写出原假设、期望频数、检验统计量和临界值。

解答

解法一

思路

展开

把题目资料整理成“中心 × 测试结果”的 $2\times2$ 列联表，在结果与中心相互独立的原假设下求四格期望频数。然后计算卡方统计量，与自由度 1、显著性水平 $5\%$ 的临界值比较，并给出包含“测试结果”和“考试中心”的结论。

答题过程

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Let $H_0$ be the hypothesis that aptitude test result is independent of test centre.

Let $H_1$ be the hypothesis that aptitude test result is not independent of test centre.

The observed frequencies are

	Pass	Fail	Total
Centre A	60	30	90
Centre B	80	30	110
Total	140	60	200

Under $H_0$ ,

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

For example,

E_{A,\text{Pass}} =\frac{90\times140}{200} =63.

The expected frequencies are therefore

	Pass	Fail	Total
Centre A	63	27	90
Centre B	77	33	110
Total	140	60	200

The four contributions to the test statistic are

Cell	$\dfrac{(O-E)^2}{E}$
Centre A, Pass	0.1429
Centre A, Fail	0.3333
Centre B, Pass	0.1169
Centre B, Fail	0.2727

Hence,

\begin{align*} \chi^2 =&\,0.1429+0.3333\\ +&\,0.1169+0.2727\\ =&\,0.8658\ldots\\ =&\,0.866\quad\text{(3 s.f.)}. \end{align*}

The degrees of freedom are

\nu=(2-1)(2-1)=1.

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

\chi^2_{1,0.05}=3.841.

Since

0.866<3.841,

the result is not significant and $H_0$ is not rejected. There is insufficient evidence, at the $5\%$ significance level, that aptitude test result differs by test centre.

The official alternative form of the test statistic gives the same result:

\begin{align*} \chi^2 =&\,\sum\frac{O^2}{E}-200\\ =&\,\frac{60^2}{63}+\frac{80^2}{77}\\ +&\,\frac{30^2}{27}+\frac{30^2}{33}-200\\ =&\,0.8658\ldots. \end{align*}

IAL 2026 Jan S3 Q1

题目

解答

解法一

思路

答题过程