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S1 2020 Oct Q5

A Level / Edexcel / S1

IAL 2020 Oct Paper · Question 5

题目

Problem

A large company rents shops in different parts of the country. A random sample of 10 shops was taken and the floor area, xx in 10 m210\,\mathrm{m}^2, and the annual rent, yy in thousands of dollars, were recorded.

The data are summarised by the following statistics

∑x= 900∑x2=84818∑y= 183∑y2=3434\begin{align*} \sum x=&\,900\qquad \sum x^2=84818\\ \sum y=&\,183\qquad \sum y^2=3434 \end{align*}

and the regression line of yy on xx has equation y=6.066+0.136xy=6.066+0.136x

(a) Use the regression line to estimate the annual rent in dollars for a shop with a floor area of 800 m2800\,\mathrm{m}^2

(2)

(b) Find SyyS_{yy} and SxxS_{xx}

(3)

(c) Find the product moment correlation coefficient between yy and xx.

(4)

An 11th shop is added to the sample. The floor area is 900 m2900\,\mathrm{m}^2 and the annual rent is 15 000 dollars.

(d) Use the formula Sxy=∑(x−xˉ)(y−yˉ)S_{xy}=\sum(x-\bar x)(y-\bar y) to show that the value of SxyS_{xy} for the 11 shops will be the same as it was for the original 10 shops.

(2)

(e) Find the new equation of the regression line of yy on xx for the 11 shops.

(3)

The company is considering renting a larger shop with area of 3000 m23000\,\mathrm{m}^2

(f) Comment on the suitability of using the new regression line to estimate the annual rent. Give a reason for your answer.

(1)
题目中文翻译

一家大型公司在全国不同地区租赁商店。现随机抽取 10 家商店,记录其楼面面积 xx(单位为 10 m210\,\mathrm{m}^2)及年租金 yy(单位为千美元)。

数据汇总如下:

∑x= 900∑x2=84818∑y= 183∑y2=3434\begin{align*} \sum x=&\,900\qquad \sum x^2=84818\\ \sum y=&\,183\qquad \sum y^2=3434 \end{align*}

且 yy 对 xx 的回归直线方程为 y=6.066+0.136xy=6.066+0.136x。

(a) 用该回归直线估计楼面面积为 800 m2800\,\mathrm{m}^2 的商店每年应付的租金(以美元为单位)。

(b) 求 SyyS_{yy} 与 SxxS_{xx}。

(c) 求 yy 与 xx 的积矩相关系数。

在样本中加入第 11 家商店。其楼面面积为 900 m2900\,\mathrm{m}^2,年租金为 15 000 美元。

(d) 使用公式 Sxy=∑(x−xˉ)(y−yˉ)S_{xy}=\sum(x-\bar x)(y-\bar y),证明 11 家商店的 SxyS_{xy} 与原来 10 家商店的 SxyS_{xy} 相同。

(e) 求 11 家商店的新回归直线方程(yy 对 xx)。

公司正在考虑租赁一家面积为 3000 m23000\,\mathrm{m}^2 的较大商店。

(f) 评论使用新回归直线估计年租金是否合适,并说明理由。

解答

(a)

解法一

思路

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xx 的单位是 10 m210\,\mathrm{m}^2,所以 800 m2800\,\mathrm{m}^2 对应 x=80x=80。代入后 yy 的单位是 thousands of dollars。

答题过程

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For a floor area of 800 m2800\,\mathrm{m}^2,

x=80.\begin{align*} x=80. \end{align*}

Using the regression line,

y=6.066+0.136(80)=16.946.\begin{align*} y=6.066+0.136(80)=16.946. \end{align*}

Since yy is in thousands of dollars, the estimated annual rent is

$16946.\begin{align*} \$16946. \end{align*}

(b)

解法一

思路

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分别使用 Syy=∑y2−(∑y)2nS_{yy}=\sum y^2-\frac{(\sum y)^2}{n} 和 Sxx=∑x2−(∑x)2nS_{xx}=\sum x^2-\frac{(\sum x)^2}{n}。

答题过程

展开 Syy= ∑y2−(∑y)2n= 3434−183210= 85.1.\begin{align*} S_{yy} =&\,\sum y^2-\frac{(\sum y)^2}{n}\\ =&\,3434-\frac{183^2}{10}\\ =&\,85.1. \end{align*}

Also,

Sxx= ∑x2−(∑x)2n= 84818−900210= 3818.\begin{align*} S_{xx} =&\,\sum x^2-\frac{(\sum x)^2}{n}\\ =&\,84818-\frac{900^2}{10}\\ =&\,3818. \end{align*}

(c)

解法一

思路

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回归线 yy on xx 的斜率 b=SxySxxb=\frac{S_{xy}}{S_{xx}},先求 SxyS_{xy},再求相关系数。

答题过程

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For the regression line of yy on xx,

b=SxySxx.\begin{align*} b=\frac{S_{xy}}{S_{xx}}. \end{align*}

Hence

Sxy=0.136(3818)=519.248.\begin{align*} S_{xy}=0.136(3818)=519.248. \end{align*}

Now

r= SxySxxSyy= 519.2483818(85.1)= 0.910944….\begin{align*} r =&\,\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}\\ =&\,\frac{519.248}{\sqrt{3818(85.1)}}\\ =&\,0.910944\ldots. \end{align*}

Therefore

r=0.911to 3 significant figures.\begin{align*} r=0.911\quad\text{to 3 significant figures}. \end{align*}

(d)

解法一

思路

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原本 xˉ=90010=90\bar x=\frac{900}{10}=90。新 shop 的 xx 也是 90,所以加入后新的 xˉ\bar x 仍是 90。新增项中的 x−xˉ=0x-\bar x=0,对 SxyS_{xy} 没有贡献。

答题过程

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For the original 10 shops,

xˉ=90010=90.\begin{align*} \bar x=\frac{900}{10}=90. \end{align*}

For the 11 shops,

xˉ=900+9011=90.\begin{align*} \bar x=\frac{900+90}{11}=90. \end{align*}

The new shop has x=90x=90, so its contribution to

Sxy=∑(x−xˉ)(y−yˉ)\begin{align*} S_{xy}=\sum(x-\bar x)(y-\bar y) \end{align*}

is

(90−90)(15−yˉ)=0.\begin{align*} (90-90)(15-\bar y)=0. \end{align*}

Therefore SxyS_{xy} is unchanged.

解法二

思路

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计算简化公式法。 除了使用 SxyS_{xy} 的定义式进行证明外,我们也可以使用其计算简化公式:

Sxy=∑xy−∑x∑yn\begin{align*} S_{xy} = \sum xy - \frac{\sum x \sum y}{n} \end{align*}

我们可以对比原 10 个商店与加入第 11 个商店后新系统的这三部分变化:

  • 对于 10 个商店:n=10n=10,∑x=900\sum x=900,∑y=183\sum y=183,因此 Sxy=∑xy10−900×18310=∑xy10−16470S_{xy} = \sum xy_{10} - \frac{900 \times 183}{10} = \sum xy_{10} - 16470。
  • 对于 11 个商店:新加入的数据为 x11=90x_{11}=90,y11=15y_{11}=15(注意年租金 15 000 美元对应 y=15y=15)。 新样本数 n′=11n'=11,∑x′=900+90=990\sum x' = 900+90 = 990,∑y′=183+15=198\sum y' = 183+15 = 198,新乘积和 ∑xy11=∑xy10+90×15=∑xy10+1350\sum xy_{11} = \sum xy_{10} + 90 \times 15 = \sum xy_{10} + 1350。 将上述数值代入 11 个商店的 SxyS_{xy} 简化公式中展开计算,即可证明新旧 SxyS_{xy} 恒等。许多同学更习惯使用计算简化公式,此方法提供了一个完全可行且严密的代数证明视角。

答题过程

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Use the computational formula for SxyS_{xy}:

Sxy=∑xy−∑x∑yn.\begin{align*} S_{xy} = \sum xy - \frac{\sum x \sum y}{n}. \end{align*}

For the original 1010 shops:

  • n=10n = 10
  • ∑x=900\sum x = 900
  • ∑y=183\sum y = 183

Let ∑xy10\sum xy_{10} be the sum of products for the original 1010 shops. The original SxyS_{xy} is:

Sxy,old= ∑xy10−900×18310= ∑xy10−16470.\begin{align*} S_{xy,\text{old}} =&\,\sum xy_{10}-\frac{900\times183}{10}\\ =&\,\sum xy_{10}-16470. \end{align*}

For the 1111 shops, the new shop has x11=90x_{11} = 90 and y11=15y_{11} = 15 (since rent is in thousands of dollars). The new summary statistics are:

  • n′=11n' = 11
  • ∑x′=900+90=990\sum x' = 900 + 90 = 990
  • ∑y′=183+15=198\sum y' = 183 + 15 = 198
  • ∑xy11=∑xy10+90×15=∑xy10+1350\sum xy_{11} = \sum xy_{10} + 90 \times 15 = \sum xy_{10} + 1350

The new SxyS_{xy} is:

Sxy,new= ∑xy11−∑x′∑y′n′= (∑xy10+1350)−990×19811= ∑xy10+1350−90×198= ∑xy10+1350−17820= ∑xy10−16470.\begin{align*} S_{xy,\text{new}} =&\,\sum xy_{11}-\frac{\sum x'\sum y'}{n'}\\ =&\,\left(\sum xy_{10}+1350\right)-\frac{990\times198}{11}\\ =&\,\sum xy_{10}+1350-90\times198\\ =&\,\sum xy_{10}+1350-17820\\ =&\,\sum xy_{10}-16470. \end{align*}

Since Sxy,new=Sxy,oldS_{xy,\text{new}} = S_{xy,\text{old}}, the value of SxyS_{xy} remains unchanged.

(e)

解法一

思路

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由于 SxyS_{xy} 不变,且新 shop 的 xx 等于平均 xx,SxxS_{xx} 也不变,所以斜率仍是 0.136。新的 yˉ\bar y 用 11 个 shops 计算。

答题过程

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The gradient remains

b=0.136.\begin{align*} b=0.136. \end{align*}

The new mean of yy is

yˉ=183+1511=18.\begin{align*} \bar y=\frac{183+15}{11}=18. \end{align*}

The new mean of xx is still

xˉ=90.\begin{align*} \bar x=90. \end{align*}

The new mean of yy is 0.30.3 less than the original mean of yy, while xˉ\bar x and the gradient are unchanged.

So the intercept is also 0.30.3 less than before:

a=6.066−0.3=5.766.\begin{align*} a=6.066-0.3=5.766. \end{align*}

Therefore the new regression line is

y=5.766+0.136x.\begin{align*} y=5.766+0.136x. \end{align*}

(f)

解法一

思路

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3000 m23000\,\mathrm{m}^2 对应 x=300x=300,远大于样本平均 x=90x=90,属于外推,因此不适合。

答题过程

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For a floor area of 3000 m23000\,\mathrm{m}^2,

x=300.\begin{align*} x=300. \end{align*}

This is far outside the range of the data, compared with the mean value xˉ=90\bar x=90.

Therefore using the regression line is not suitable, because it would involve extrapolation.