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IAL 2021 Jan S3 Q3

A Level / Edexcel / S3

IAL 2021 Jan Paper · Question 3

题目

Problem

The students in a group of schools can choose a club to join. There are 4 clubs available: Music, Art, Sports and Computers. The director collected information about the number of students in each club, using a random sample of 88 students from across the schools. The results are given in Table 1 below.

MusicArtSportsComputers
No. of students14282719

The director uses a chi-squared test to determine whether or not the students are uniformly distributed across the 4 clubs.

(a) (i) Find the expected frequencies he should use.

Given that the test statistic he calculated was 6.09 (to 3 significant figures)

(ii) use a 5% level of significance to complete the test. You should state the degrees of freedom and the critical value used.

(4)

The director wishes to examine the situation in more detail and takes a second random sample of 88 students. The director assumes that within each school, students select their clubs independently. The students come from 3 schools and the distribution of the students from each school amongst the clubs is given in Table 2 below.

SchoolMusicArtSportsComputers
School A31098
School B111135
School C11674

The director wishes to test for an association between a student’s school and the club they choose.

(b) State hypotheses suitable for such a test.

(1)

(c) Calculate the expected frequency for School C and the Computers club.

(1)

The director calculates the test statistic to be 7.29 (to 3 significant figures) with 4 degrees of freedom.

(d) Explain clearly why his test has 4 degrees of freedom.

(2)

(e) Complete the test using a 5% level of significance and stating clearly your critical value.

(2)
(Total 10 marks)
题目中文翻译

一群学校的学生可以选择加入一个社团。有 4 个可选社团：音乐、美术、体育和计算机。主任用来自各学校的 88 名学生随机样本收集了各社团的人数信息。结果如下表 1。

音乐美术体育计算机
学生人数14282719

主任使用卡方检验来判断学生是否在 4 个社团中均匀分布。

(a) (i) 求他应使用的期望频数。

已知他计算出的检验统计量为 6.09（保留 3 位有效数字）

(ii) 用 5% 显著性水平完成检验。你应写出自由度和所用临界值。

主任希望更详细地考察这一情况，并再次抽取 88 名学生的随机样本。主任假定在每所学校内，学生选择社团是相互独立的。学生来自 3 所学校，各校学生在各社团中的分布如下表 2。

学校音乐美术体育计算机
学校 A31098
学校 B111135
学校 C11674

主任希望检验学生所在学校与他们选择的社团之间是否存在关联。

(b) 写出适合这种检验的假设。

(c) 求学校 C 和计算机社团的期望频数。

主任计算得到检验统计量为 7.29（保留 3 位有效数字），自由度为 4。

(d) 清楚解释为什么这个检验有 4 个自由度。

(e) 用 5% 显著性水平完成检验，并清楚写出临界值。

解答

(a)(i)

解法一

思路

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若学生在四个社团中均匀分布，每个社团的期望人数相同，因此把总人数 8888 平均分成四份。

答题过程

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Under a uniform distribution, the expected frequency for each club is

884=22.\frac{88}{4}=22.

Therefore, the expected frequencies are

22,22,22,22.\boxed{22, 22, 22, 22}.

(a)(ii)

解法一

思路

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这是四个类别、没有估计参数的拟合优度检验，所以自由度为 4−1=34-1=3。把已给的检验统计量与 5%5\% 上尾临界值比较；统计量没有落入拒绝域，便不能否定均匀分布模型。

答题过程

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The hypotheses are:

  • H0H_0: the students are uniformly distributed across the four clubs.
  • H1H_1: the students are not uniformly distributed across the four clubs.

The number of degrees of freedom is

4−1=3.4-1=3.

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

χ32(0.05)=7.815.\chi^2_3(0.05)=7.815.

Since

6.09<7.815,6.09<7.815,

the result is not significant, so H0H_0 is not rejected. There is insufficient evidence to suggest that the students are not uniformly distributed across the four clubs.

(b)

解法一

思路

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独立性检验的原假设应表示“学校与社团选择相互独立”，备择假设则表示两者有关联。假设必须同时带有题目语境。

答题过程

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  • H0H_0: a student’s school and choice of club are independent.
  • H1H_1: there is an association between a student’s school and choice of club.

(c)

解法一

思路

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列联表中某格的期望频数等于“所在行总数乘以所在列总数，再除以总人数”。学校 C 的行总数是 2828，计算机社团的列总数是 1717。

答题过程

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The expected frequency for School C and the Computers club is

E= (28)(17)88= 5.409…\begin{align*} E=&\,\frac{(28)(17)}{88}\\ =&\,5.409\ldots \end{align*}

Therefore, the expected frequency is

5.41\boxed{5.41}

to 3 significant figures.

(d)

解法一

思路

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原来的 3×43\times4 列联表中，学校 C 与音乐社团一格的期望频数小于 55，不满足卡方近似的使用条件。把音乐列与另一社团列合并后，表格变成 3×33\times3，所以自由度是 (3−1)(3−1)=4(3-1)(3-1)=4。

答题过程

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The expected frequency for School C and the Music club is

E= (28)(15)88= 4.77…<5.\begin{align*} E=&\,\frac{(28)(15)}{88}\\ =&\,4.77\ldots<5. \end{align*}

Therefore, the Music column must be pooled with another club column. This produces a 3×33\times3 contingency table, so the number of degrees of freedom is

(3−1)(3−1)=4.\boxed{(3-1)(3-1)=4}.

(e)

解法一

思路

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使用题目所给的 44 个自由度，将检验统计量 7.297.29 与 5%5\% 上尾临界值比较。统计量没有进入拒绝域，因此没有足够证据认为学校与社团选择有关联。

答题过程

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For 4 degrees of freedom at the 5%5\% significance level, the critical value is

χ42(0.05)=9.488.\chi^2_4(0.05)=9.488.

Since

7.29<9.488,7.29<9.488,

the result is not significant, so H0H_0 is not rejected. There is insufficient evidence of an association between a student’s school and the club chosen.