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IAL 2021 Oct S3 Q4

A Level / Edexcel / S3

IAL 2021 Oct Paper · Question 4

题目

Problem

A local village radio station, LSB, decides to survey adults in its broadcasting area about the programmes it produces.

LSB broadcasts to 4 villages A, B, C and D. The number of households in each of the villages is given below.

VillageNumber of households
A41
B164
C123
D82

LSB decides to take a stratified sample of 200 households.

(a) Explain how to select the households for this stratified sample.

(3)

One of the questions in the survey related to the age group of each member of the household and whether they listen to LSB. The data received are shown below.

Age group18–4950–69Older than 69
Listen to LSB13016265
Do not listen to LSB789862

The data are to be used to determine whether or not there is an association between the age group and whether they listen to LSB.

(b) Calculate the expected frequencies for the age group 50–69 that

(i) listen to LSB

(ii) do not listen to LSB

(2)

Given that for the other 4 classes (OE)2E=4.657\sum \dfrac{(O-E)^2}{E} = 4.657 to 3 decimal places,

(c) test at the 5% level of significance whether or not there is evidence of an association between age and listening to LSB. Show your working clearly, stating the degrees of freedom and the critical value.

(6)
(Total 11 marks)
题目中文翻译

当地一个村庄电台 LSB 决定就其制作的节目对其广播范围内的成年人进行调查。

LSB 向 A、B、C、D 四个村庄广播。各村的住户数量如下。

村庄住户数量
A41
B164
C123
D82

LSB 决定抽取一个 200 户的分层样本。

(a) 说明如何抽取这个分层样本。

调查中的一个问题与每个家庭成员的年龄组以及他们是否收听 LSB 有关。收集到的数据如下。

年龄组18–4950–6969 岁以上
收听 LSB13016265
不收听 LSB789862

这些数据将用于判断年龄组与是否收听 LSB 之间是否存在关联。

(b) 计算年龄组 50–69 的期望频数:

(i) 收听 LSB

(ii) 不收听 LSB

已知其余 4 个单元格的 (OE)2E=4.657\sum \dfrac{(O-E)^2}{E}=4.657,保留到 3 位小数,

(c) 在 5% 显著性水平下检验:年龄与是否收听 LSB 之间是否存在关联。清楚写出你的工作过程,说明自由度和临界值。

解答

(a)

解法一

思路

展开

以四个村庄作为四个层。总户数为 410410,样本量为 200200,所以各层按住户数比例分配;随后必须在每个村庄内部随机抽取相应数量的住户。

答题过程

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The total number of households is

41+164+123+82=410.41+164+123+82=410.

The required sample sizes are

VillageSample sizeA41410(200)=20B164410(200)=80C123410(200)=60D82410(200)=40\begin{array}{c|c} \text{Village}&\text{Sample size}\\ \hline A&\dfrac{41}{410}(200)=20\\ B&\dfrac{164}{410}(200)=80\\ C&\dfrac{123}{410}(200)=60\\ D&\dfrac{82}{410}(200)=40 \end{array}

Label the households in village A from 11 to 4141, in B from 11 to 164164, in C from 11 to 123123, and in D from 11 to 8282. Use random numbers to select, without replacement, 2020, 8080, 6060 and 4040 households respectively from villages A, B, C and D.

(b)(i)

解法一

思路

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独立性检验中,每个单元格的期望频数等于“所在行总数乘所在列总数,再除以总数”。收听 LSB 的行总数为 357357,年龄 50506969 的列总数为 260260,总人数为 595595

答题过程

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The relevant totals are

listen to LSB:  357,age 50–69:  260,grand total:  595.\begin{aligned} \text{listen to LSB: }&\ 357,\\ \text{age 50--69: }&\ 260,\\ \text{grand total: }&\ 595. \end{aligned}

Therefore, the expected frequency is

E=357(260)595=156.\begin{align*} E =&\,\frac{357(260)}{595}\\ =&\,\boxed{156}. \end{align*}

(b)(ii)

解法一

思路

展开

不收听 LSB 的行总数为 238238,其余计算与 (b)(i) 相同;同一年龄组的两个期望频数也应相加为列总数 260260

答题过程

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The expected frequency is

E=238(260)595=104.\begin{align*} E =&\,\frac{238(260)}{595}\\ =&\,\boxed{104}. \end{align*}

As a check, 156+104=260156+104=260, the total for the age group 50506969.

(c)

解法一

思路

展开

先提出年龄组与是否收听 LSB 相互独立的原假设。题目已给出其余四格对卡方统计量的总贡献,只需算出年龄 50506969 两格的贡献并加上 4.6574.657。这是 2×32\times3 列联表,所以自由度为 (21)(31)=2(2-1)(3-1)=2

答题过程

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The hypotheses are

H0: age group and whether an adult listens to LSB are independent,H1: there is an association between age group and whether an adult listens to LSB.\begin{aligned} H_0:&\ \text{age group and whether an adult listens to LSB}\\ &\ \text{are independent},\\ H_1:&\ \text{there is an association between age group}\\ &\ \text{and whether an adult listens to LSB}. \end{aligned}

For the two classes in the age group 50506969,

(162156)2156=0.230769,(98104)2104=0.346153\begin{align*} \frac{(162-156)^2}{156} =&\,0.230769\ldots,\\ \frac{(98-104)^2}{104} =&\,0.346153\ldots \end{align*}

Hence the test statistic is

χ2=4.657+0.230769+0.346153=5.233925.23.\begin{align*} \chi^2 =&\,4.657+0.230769\ldots\\ &\,\hspace{2pt}+0.346153\ldots\\ =&\,5.23392\ldots\\ \approx&\,5.23. \end{align*}

The number of degrees of freedom is

ν=(21)(31)=2.\nu=(2-1)(3-1)=2.

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

χ22(0.05)=5.991.\chi_2^2(0.05)=5.991.

Since

5.23392<5.991,5.23392\ldots<5.991,

H0H_0 is not rejected. There is insufficient evidence at the 5%5\% significance level of an association between age group and whether an adult listens to LSB.