题目

Problem

A survey was carried out with students that had studied Maths, Physics and Chemistry at a college between 2016 and 2020. The students were divided into two groups A and B.

(a) Explain how a sample could be obtained from this population using quota sampling.

(2)

The students were asked which of the three subjects they enjoyed the most. The results of the survey are shown in the table.

	Maths	Physics	Chemistry	Total
Group A	16	10	13	39
Group B	38	13	10	61
Total	54	23	23	100

(b) Test, at the 5% level of significance, whether the subject enjoyed the most is independent of group. You should state your hypotheses, expected frequencies, test statistic and the critical value used for this test.

(8)

The Headteacher discovered later that the results were actually based on a random sample of 200 students but had been recorded in the table as percentages.

(c) For the test in part (b), state with reasons the effect, if any, that this information would have on (i) the null and alternative hypotheses, (ii) the critical value, (iii) the value of the test statistic, (iv) the conclusion of the test.

(4)

(Total for Question 4 is 14 marks)

题目中文翻译

在 2016 到 2020 年间，某学院对学过数学、物理和化学的学生进行了调查。这些学生被分为 A 组和 B 组。

(a) 说明如何使用配额抽样从该总体中抽样。

向学生询问他们最喜欢的三门学科中的哪一门。调查结果如下表。

	数学	物理	化学	总计
A 组	16	10	13	39
B 组	38	13	10	61
总计	54	23	23	100

(b) 在 5% 显著性水平下检验“最喜欢的学科”是否与组别独立。你应写出原假设、期望频数、检验统计量和所用临界值。

后来校长发现，这些结果实际上来自一个 200 人的随机样本，但在表中被记录成百分比。

解答

(a)

解法一

思路

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配额抽样先按总体中的相关类别划分组别，并按各组所占比例规定需要抽取的人数；之后在每组中以非随机方式寻找参与者，直到各组配额填满。两个关键点是“按组设配额”和“组内非随机选取”。

答题过程

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Divide the population into suitable groups, for example by group A or B and year of attendance, and set a quota for each group in proportion to its size in the population. Select students non-randomly from each group until every quota has been filled.

(b)

解法一

思路

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这是 $2\times3$ 列联表的卡方独立性检验。先用“行总计乘列总计再除以总数”求六个期望频数，再计算 $\chi^2=\sum\frac{(O-E)^2}{E}$ 。自由度是 $(2-1)(3-1)=2$ ，最后与 $5\%$ 上尾临界值比较。

答题过程

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The hypotheses are

\begin{aligned} H_0:&\ \text{The subject enjoyed most is independent of group,}\\ H_1:&\ \text{The subject enjoyed most is not independent of group.} \end{aligned}

Each expected frequency is calculated using

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

For example,

E_{A,\text{Maths}}=\frac{39(54)}{100}=21.06.

The complete table of expected frequencies is

	Maths	Physics	Chemistry	Total
Group A	21.06	8.97	8.97	39
Group B	32.94	14.03	14.03	61
Total	54	23	23	100

Therefore,

\begin{align*} \chi^2 =&\,\sum\frac{(O-E)^2}{E}\\ =&\,\frac{(16-21.06)^2}{21.06}\\ &\,\hspace{2pt}+\frac{(10-8.97)^2}{8.97}\\ &\,\hspace{4pt}+\frac{(13-8.97)^2}{8.97}\\ &\,\hspace{6pt}+\frac{(38-32.94)^2}{32.94}\\ &\,\hspace{8pt}+\frac{(13-14.03)^2}{14.03}\\ &\,\hspace{10pt}+\frac{(10-14.03)^2}{14.03}\\ =&\,5.15508\ldots\\ \approx&\,5.16. \end{align*}

The number of degrees of freedom is

\nu=(2-1)(3-1)=2.

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

\chi^2_2(0.05)=5.991.

Since

5.15508\ldots<5.991,

the test statistic does not lie in the critical region, so $H_0$ is not rejected. There is insufficient evidence at the $5\%$ significance level to suggest that the subject enjoyed most is not independent of group.

(c)(i)

解法一

思路

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新信息只改变实际频数的规模，没有改变所检验的两个分类变量。因此原假设与备择假设都不变。

答题过程

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There is no change to the hypotheses because the test is still assessing whether the subject enjoyed most is independent of group.

(c)(ii)

解法一

思路

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表格的行数和列数没有改变，因此自由度仍为 $2$ ；显著性水平也仍为 $5\%$ ，所以临界值不变。

答题过程

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There is no change to the critical value because the degrees of freedom remain

(2-1)(3-1)=2.

The critical value is still $5.991$ .

(c)(iii)

解法一

思路

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原表记录的是百分比，而实际样本量为 $200$ ，所以每个观测频数和期望频数都应乘以 $2$ 。代入卡方统计量的单格贡献可见，每一项都会变成原来的 $2$ 倍，因此整个统计量也翻倍。

答题过程

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Both the observed and expected frequencies are doubled. For each cell,

\frac{(2O-2E)^2}{2E} =2\frac{(O-E)^2}{E}.

Therefore, the test statistic doubles:

\chi^2=2(5.15508\ldots)=10.3102\ldots

(c)(iv)

解法一

思路

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承接 (c)(ii) 与 (c)(iii)，临界值仍为 $5.991$ ，但检验统计量现在约为 $10.31$ ，已经进入临界域，所以检验结论与 (b) 相反。

答题过程

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Since

10.3102\ldots>5.991,

the test statistic now lies in the critical region. Therefore, $H_0$ is rejected. There is sufficient evidence at the $5\%$ significance level to suggest that the subject enjoyed most is not independent of group.

IAL 2022 Jan S3 Q4

题目

解答

(a)

解法一

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(b)

解法一

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(c)(i)

解法一

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(c)(ii)

解法一

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(c)(iii)

解法一

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(c)(iv)

解法一

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