Skip to content
CalcGospel 國際數學圖譜
返回

IAL 2022 Jan S3 UNUSED Q4

A Level / Edexcel / S3

IAL 2022 Jan Unused Paper · Question 4

题目

Problem

A manager at a large estate agency believes that the type of property affects the time taken to sell it.

A random sample of 125 properties sold is shown in the table.

Type of propertyBungalowFlatHouseTotal
Sold within three months7294682
Sold in more than three months9191543
Total164861125

Test, at the 5% level of significance, whether there is evidence for an association between the type of property and the time taken to sell it. You should state your hypotheses, expected frequencies, test statistic and the critical value used for this test.

(10)
(Total 10 marks)
题目中文翻译

一家大型房产中介的经理认为,房产类型会影响出售所需时间。

下表给出随机抽取的 125 套已售房产样本。

房产类型平房公寓独立屋合计
在三个月内售出7294682
超过三个月才售出9191543
合计164861125

在 5% 显著性水平下检验:房产类型与出售所需时间之间是否存在关联。 你应写出原假设、期望频数、检验统计量和该检验所用的临界值。

解答

解法一

思路

展开

这是 2×32\times3 列联表的卡方独立性检验。先写“房产类型”与“售出时间”是否独立的假设,再用行总计乘列总计除以总数求六个期望频数,计算检验统计量,并与自由度为 (21)(31)=2(2-1)(3-1)=2 的临界值比较。

答题过程

展开

The hypotheses are

H0: Property type and time taken to sell are independent,H1: Property type and time taken to sell are associated.\begin{aligned} H_0:&\ \text{Property type and time taken to sell are independent,}\\ H_1:&\ \text{Property type and time taken to sell are associated.} \end{aligned}

Each expected frequency is calculated using

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

For example,

Ewithin 3 months, bungalow=82(16)125=10.496.E_{\text{within 3 months, bungalow}} =\frac{82(16)}{125}=10.496.

The expected frequencies are

BungalowFlatHouseTotal
Sold within three months10.49631.48840.01682
Sold after three months5.50416.51220.98443
Total164861125

Therefore,

χ2=(OE)2E=(710.496)210.496+(2931.488)231.488+(4640.016)240.016+(95.504)25.504+(1916.512)216.512+(1520.984)220.984=6.557796.56.\begin{align*} \chi^2 =&\,\sum\frac{(O-E)^2}{E}\\ =&\,\frac{(7-10.496)^2}{10.496}\\ &\,\hspace{2pt}+\frac{(29-31.488)^2}{31.488}\\ &\,\hspace{4pt}+\frac{(46-40.016)^2}{40.016}\\ &\,\hspace{6pt}+\frac{(9-5.504)^2}{5.504}\\ &\,\hspace{8pt}+\frac{(19-16.512)^2}{16.512}\\ &\,\hspace{10pt}+\frac{(15-20.984)^2}{20.984}\\ =&\,6.55779\ldots\\ \approx&\,6.56. \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

6.55779>5.991,6.55779\ldots>5.991,

the test statistic lies in the critical region, so H0H_0 is rejected. There is sufficient evidence at the 5%5\% significance level to suggest an association between property type and the time taken to sell the property.