题目

Problem

A university awards its graduates a degree in one of three categories, Distinction, Merit or Pass.

Table 1 shows information about a random sample of 200 graduates from three departments, Arts, Humanities and Sciences.

Degree classification	Arts	Humanities	Sciences	Total
Distinction	22	32	38	92
Merit	15	30	13	58
Pass	18	15	17	50
Total	55	77	68	200

Xiu wants to carry out a test of independence between the category of degree and the department.

Table 2 shows some of the values of $\dfrac{(O-E)^2}{E}$ for this test.

Degree classification	Arts	Humanities	Sciences	Total
Distinction	0.43	0.33	1.44	2.20
Merit	0.06	2.63	2.29	4.98
Pass

(a) Complete Table 2.

(4)

(b) Hence, complete Xiu’s hypothesis test using a 5% level of significance. You should state the hypotheses, the degrees of freedom and the critical value used for this test.

(5)

(Total 9 marks)

题目中文翻译

一所大学向毕业生授予三个类别之一的学位：Distinction、Merit 或 Pass。

表 1 给出了来自 Arts、Humanities 和 Sciences 三个学系的 200 名毕业生随机样本的信息。

学位等级	Arts	Humanities	Sciences	合计
Distinction	22	32	38	92
Merit	15	30	13	58
Pass	18	15	17	50
合计	55	77	68	200

Xiu 希望检验学位类别与学系是否相互独立。

表 2 给出了这项检验中部分格子的 $\dfrac{(O-E)^2}{E}$ 值。

学位等级	Arts	Humanities	Sciences	合计
Distinction	0.43	0.33	1.44	2.20
Merit	0.06	2.63	2.29	4.98
Pass

(a) 补全表 2。

(b) 由此，在 5% 显著性水平下完成 Xiu 的假设检验。请写出假设、自由度和这项检验所用的临界值。

解答

(a)

解法一

思路

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表 2 缺少的是 Pass 这一行。先用“行总数乘列总数再除以总数”求三个格子的期望频数，然后逐格计算 $(O-E)^2/E$ ，最后与题目已经给出的两行合计相加，得到总检验统计量。

答题过程

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For the Pass row, the expected frequencies are

\begin{align*} E_{\mathrm{Arts}}=&\,\frac{50(55)}{200}=13.75,\\ E_{\mathrm{Humanities}}=&\,\frac{50(77)}{200}=19.25,\\ E_{\mathrm{Sciences}}=&\,\frac{50(68)}{200}=17. \end{align*}

The three contributions are

\begin{align*} \frac{(18-13.75)^2}{13.75}=&\,1.31,\\ \frac{(15-19.25)^2}{19.25}=&\,0.94,\\ \frac{(17-17)^2}{17}=&\,0. \end{align*}

Thus, the completed table is

Degree classification	Arts	Humanities	Sciences	Total
Distinction	0.43	0.33	1.44	2.20
Merit	0.06	2.63	2.29	4.98
Pass	1.31	0.94	0	2.25
Total				9.43

(b)

解法一

思路

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由 (a) 得检验统计量 $\chi^2=9.43$ 。这是 $3\times3$ 列联表的独立性检验，所以自由度为 $(3-1)(3-1)=4$ 。将统计量与 $5\%$ 上尾临界值比较，并以学位类别和学系为语境写结论。

答题过程

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The hypotheses are:

$H_0$ : degree category and department are independent.
$H_1$ : there is an association between degree category and department.

From part (a),

\chi^2=9.43.

The number of degrees of freedom is

\nu=(3-1)(3-1)=4.

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

\chi^2_4(0.05)=9.488.

Since

9.43<9.488,

the result is not significant, so $H_0$ is not rejected. There is insufficient evidence of an association between degree category and department.

IAL 2020 Oct S3 Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程