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, $x$ in $10m^2$ , and the annual rent, $y$ in thousands of dollars, were recorded.

The data are summarised by the following statistics

\sum x=900\qquad \sum x^2=84818\qquad \sum y=183\qquad \sum y^2=3434

and the regression line of $y$ on $x$ has equation $y=6.066+0.136x$

(a) Use the regression line to estimate the annual rent in dollars for a shop with a floor area of $800m^2$

(2)

(b) Find $S_{yy}$ and $S_{xx}$

(3)

(4)

An 11th shop is added to the sample. The floor area is $900m^2$ and the annual rent is 15 000 dollars.

(d) Use the formula $S_{xy}=\sum(x-\bar x)(y-\bar y)$ to show that the value of $S_{xy}$ 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 $y$ on $x$ for the 11 shops.

(3)

The company is considering renting a larger shop with area of $3000m^2$

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

(1)

解答

(a)

解法一

思路

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$x$ 的单位是 $10m^2$ ，所以 $800m^2$ 对应 $x=80$ 。代入后 $y$ 的单位是 thousands of dollars。

答题过程

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For a floor area of $800m^2$ ,

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

Using the regression line,

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

Since $y$ is in thousands of dollars, the estimated annual rent is

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

(b)

解法一

思路

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分别使用 $S_{yy}=\sum y^2-\frac{(\sum y)^2}{n}$ 和 $S_{xx}=\sum x^2-\frac{(\sum x)^2}{n}$ 。

答题过程

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\begin{aligned} S_{yy} =&\,\sum y^2-\frac{(\sum y)^2}{n}\\ =&\,3434-\frac{183^2}{10}\\ =&\,85.1. \end{aligned}

Also,

\begin{aligned} S_{xx} =&\,\sum x^2-\frac{(\sum x)^2}{n}\\ =&\,84818-\frac{900^2}{10}\\ =&\,3818. \end{aligned}

(c)

解法一

思路

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回归线 $y$ on $x$ 的斜率 $b=\frac{S_{xy}}{S_{xx}}$ ，先求 $S_{xy}$ ，再求相关系数。

答题过程

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For the regression line of $y$ on $x$ ,

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

Hence

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

Now

\begin{aligned} r =&\,\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}\\ =&\,\frac{519.248}{\sqrt{3818(85.1)}}\\ =&\,0.910944\ldots. \end{aligned}

Therefore

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

答题过程

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For the original 10 shops,

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

For the 11 shops,

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

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

\begin{align*} S_{xy}=\sum(x-\bar x)(y-\bar y) \end{align*}

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

Therefore $S_{xy}$ is unchanged.

解法二

思路

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

\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： $n=10$ ， $\sum x=900$ ， $\sum y=183$ ，因此 $S_{xy} = \sum xy_{10} - \frac{900 \times 183}{10} = \sum xy_{10} - 16470$ 。
对于 11 个商店：新加入的数据为 $x_{11}=90$ ， $y_{11}=15$ （注意年租金 $\$ 15000 $对应$ y=15 $）。新样本数$ n’=11 $，$ \sum x’ = 900+90 = 990 $，$ \sum y’ = 183+15 = 198 $，新乘积和$ \sum xy_{11} = \sum xy_{10} + 90 \times 15 = \sum xy_{10} + 1350 $。将上述数值代入 11 个商店的$ S_{xy} $简化公式中展开计算，即可证明新旧$ S_{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 $S_{xy}$ :

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

For the original $10$ shops:

$n = 10$
$\sum x = 900$
$\sum y = 183$

Let $\sum xy_{10}$ be the sum of products for the original $10$ shops. The original $S_{xy}$ is:

\begin{align*} S_{xy,\text{old}} =&\,\, \sum xy_{10} - \frac{900 \times 183}{10}\\[3mm] =&\,\, \sum xy_{10} - 16470. \end{align*}

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

$n' = 11$
$\sum x' = 900 + 90 = 990$
$\sum y' = 183 + 15 = 198$
$\sum xy_{11} = \sum xy_{10} + 90 \times 15 = \sum xy_{10} + 1350$

The new $S_{xy}$ is:

\begin{align*} S_{xy,\text{new}} =&\,\, \sum xy_{11} - \frac{\sum x' \sum y'}{n'}\\[3mm] =&\,\, \left(\sum xy_{10} + 1350\right) - \frac{990 \times 198}{11}\\[3mm] =&\,\, \sum xy_{10} + 1350 - 90 \times 198\\[3mm] =&\,\, \sum xy_{10} + 1350 - 17820\\[3mm] =&\,\, \sum xy_{10} - 16470. \end{align*}

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

(e)

解法一

思路

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

答题过程

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The gradient remains

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

The new mean of $y$ is

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

The new mean of $x$ is still

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

The new mean of $y$ is $0.3$ less than the original mean of $y$ , while $\bar x$ and the gradient are unchanged.

So the intercept is also $0.3$ less than before:

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

Therefore the new regression line is

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

(f)

解法一

思路

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

答题过程

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For a floor area of $3000m^2$ ,

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

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

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

IAL 2020 Oct Q5

题目

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(a)

解法一

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(b)

解法一

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(c)

解法一

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(d)

解法一

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解法二

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(e)

解法一

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(f)

解法一

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