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S1 2023 Oct Q6

A Level / Edexcel / S1

IAL 2023 Oct Paper · Question 6

题目

Problem

The variables xx and yy have the following regression equations based on the same 1212 observations.

Regression equationy on xy=1.4x+1.5x on yx=1.2+0.2y\begin{array}{c|c} \text{Regression equation}&\\ \hline y\text{ on }x&y=1.4x+1.5\\ x\text{ on }y&x=1.2+0.2y \end{array}

(a) (i) Find the point of intersection of these lines.

(ii) Hence show that ∑x=25\sum x=25

(4)

Given that

∑xy=696160\sum xy=\frac{6961}{60}

(b) Find SxyS_{xy}

(4)

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

(4)
题目中文翻译

变量 xx 与 yy 的下列两条回归方程基于同一组 1212 个观测值：

回归方程y 关于 xy=1.4x+1.5x 关于 yx=1.2+0.2y\begin{array}{c|c} \text{回归方程}&\\ \hline y\text{ 关于 }x&y=1.4x+1.5\\ x\text{ 关于 }y&x=1.2+0.2y \end{array}

(a)

(i) 求这两条直线的交点。

(ii) 由此证明 ∑x=25\sum x=25。

已知

∑xy=696160.\sum xy=\frac{6961}{60}.

(b) 求 SxyS_{xy}。

(c) 求 xx 与 yy 的积矩相关系数。

解答

(a)(i)

解法一

思路

展开

两条回归线的交点是 (xˉ,yˉ)(\bar{x},\bar{y})。先联立两条直线求交点。

答题过程

展开

Substitute y=1.4x+1.5y=1.4x+1.5 into x=1.2+0.2yx=1.2+0.2y:

x= 1.2+0.2(1.4x+1.5)x= 1.2+0.28x+0.30.72x= 1.5x= 2512.\begin{align*} x=&\,1.2+0.2(1.4x+1.5)\\ x=&\,1.2+0.28x+0.3\\ 0.72x=&\,1.5\\ x=&\,\frac{25}{12}. \end{align*}

Then

y= 1.4(2512)+1.5= 75⋅2512+32= 3512+1812= 5312.\begin{align*} y =&\,1.4\left(\frac{25}{12}\right)+1.5\\ =&\,\frac{7}{5}\cdot\frac{25}{12}+\frac{3}{2}\\ =&\,\frac{35}{12}+\frac{18}{12}\\ =&\,\frac{53}{12}. \end{align*}

The point of intersection is

(2512,5312).\begin{align*} \left(\frac{25}{12},\frac{53}{12}\right). \end{align*}

(a)(ii)

解法一

思路

展开

回归线交于 (xˉ,yˉ)(\bar{x},\bar{y})，所以 xˉ=2512\bar{x}=\dfrac{25}{12}。共有 1212 组数据，因此 ∑x=12xˉ\sum x=12\bar{x}。

答题过程

展开

The regression lines intersect at (xˉ,yˉ)(\bar{x},\bar{y}).

So

xˉ=2512.\begin{align*} \bar{x}=\frac{25}{12}. \end{align*}

Since there are 1212 observations,

∑x=12xˉ=12⋅2512=25.\begin{align*} \sum x=12\bar{x}=12\cdot\frac{25}{12}=25. \end{align*}

This shows that ∑x=25\sum x=25.

(b)

解法一

思路

展开

先由交点得到 yˉ=5312\bar{y}=\dfrac{53}{12}，所以 ∑y=53\sum y=53。再使用

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

答题过程

展开

Since

yˉ=5312,\begin{align*} \bar{y}=\frac{53}{12}, \end{align*}

we have

∑y=12⋅5312=53.\begin{align*} \sum y=12\cdot\frac{53}{12}=53. \end{align*}

Therefore

Sxy= ∑xy−∑x∑yn= 696160−25(53)12= 696160−662560= 33660= 5.6.\begin{align*} S_{xy} =&\,\sum xy-\frac{\sum x\sum y}{n}\\ =&\,\frac{6961}{60}-\frac{25(53)}{12}\\ =&\,\frac{6961}{60}-\frac{6625}{60}\\ =&\,\frac{336}{60}\\ =&\,5.6. \end{align*}

(c)

解法一

思路

展开

回归线斜率分别满足

byx=SxySxx,bxy=SxySyy.b_{yx}=\frac{S_{xy}}{S_{xx}},\qquad b_{xy}=\frac{S_{xy}}{S_{yy}}.

先求 SxxS_{xx} 和 SyyS_{yy}，再代入相关系数公式。

答题过程

展开

For the regression line of yy on xx,

SxySxx=1.4.\begin{align*} \frac{S_{xy}}{S_{xx}}=1.4. \end{align*}

Using Sxy=5.6S_{xy}=5.6,

Sxx=5.61.4=4.\begin{align*} S_{xx}=\frac{5.6}{1.4}=4. \end{align*}

For the regression line of xx on yy,

SxySyy=0.2.\begin{align*} \frac{S_{xy}}{S_{yy}}=0.2. \end{align*}

So

Syy=5.60.2=28.\begin{align*} S_{yy}=\frac{5.6}{0.2}=28. \end{align*}

Therefore

r= SxySxxSyy= 5.64⋅28= 0.5291…\begin{align*} r =&\,\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}\\ =&\,\frac{5.6}{\sqrt{4\cdot28}}\\ =&\,0.5291\ldots \end{align*}

The product moment correlation coefficient is

0.529\begin{align*} 0.529 \end{align*}

to 33 significant figures.

解法二

思路

展开

两条回归线斜率的乘积等于 r2r^2。本题两条斜率都为正，因此相关系数取正根。

答题过程

展开 r2= (1.4)(0.2)=0.28.\begin{align*} r^2=&\,(1.4)(0.2)=0.28. \end{align*}

Both regression coefficients are positive, so r>0r>0. Hence

r= 0.28= 0.529(3 s.f.).\begin{align*} r=&\,\sqrt{0.28}\\ =&\,0.529\quad\text{(3 s.f.)}. \end{align*}