题目

Problem

The variables $x$ and $y$ have the following regression equations based on the same $12$ observations.

\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 $\sum x=25$

(4)

Given that

\sum xy=\frac{6961}{60}

(b) Find $S_{xy}$

(4)

解答

(a)(i)

解法一

思路

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

答题过程

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

\begin{align*} x=&\,1.2+0.2(1.4x+1.5)\\[3mm] x=&\,1.2+0.28x+0.3\\[3mm] 0.72x=&\,1.5\\[3mm] x=&\,\frac{25}{12}. \end{align*}

Then

\begin{align*} y =&\,1.4\left(\frac{25}{12}\right)+1.5\\[3mm] =&\,\frac{7}{5}\cdot\frac{25}{12}+\frac{3}{2}\\[3mm] =&\,\frac{35}{12}+\frac{18}{12}\\[3mm] =&\,\frac{53}{12}. \end{align*}

The point of intersection is

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

(a)(ii)

解法一

思路

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

答题过程

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

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

Since there are $12$ observations,

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

This shows that $\sum x=25$ .

(b)

解法一

思路

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

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

答题过程

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Since

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

we have

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

Therefore

\begin{align*} S_{xy} =&\,\sum xy-\frac{\sum x\sum y}{n}\\[3mm] =&\,\frac{6961}{60}-\frac{25(53)}{12}\\[3mm] =&\,\frac{6961}{60}-\frac{6625}{60}\\[3mm] =&\,\frac{336}{60}\\[3mm] =&\,5.6. \end{align*}

(c)

解法一

思路

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回归线斜率分别满足

b_{yx}=\frac{S_{xy}}{S_{xx}},\qquad b_{xy}=\frac{S_{xy}}{S_{yy}}.

先求 $S_{xx}$ 和 $S_{yy}$ ，再代入相关系数公式。

答题过程

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

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

Using $S_{xy}=5.6$ ,

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

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

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

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

Therefore

\begin{align*} r =&\,\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}\\[3mm] =&\,\frac{5.6}{\sqrt{4\cdot28}}\\[3mm] =&\,0.5291\ldots \end{align*}

The product moment correlation coefficient is

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

to $3$ significant figures.

IAL 2023 Oct Q6

题目

解答

(a)(i)

解法一

思路

答题过程

(a)(ii)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程