题目

Problem

Following some school examinations, Chetna is studying the results of the 16 students in her class. The mark for paper 1, $x$ , and the mark for paper 2, $y$ , for each student are summarised in the following statistics.

\bar x=35.75\qquad \bar y=25.75\qquad \sigma_x=7.79\qquad \sigma_y=11.91\qquad \sum xy=15837

(a) Comment on the differences between the marks of the students on paper 1 and paper 2

(2)

Chetna decides to examine these data in more detail and plots the marks for each of the 16 students on the scatter diagram opposite.

(b) (i) Explain why the circled point $(38,0)$ is possibly an outlier.

(ii) Suggest a possible reason for this result.

(2)

Chetna decides to omit the data point $(38,0)$ and examine the other 15 students’ marks.

(3)

(d) (i) explain why $\sum xy$ is still 15 837

(ii) show that $S_{xy}=1169.8$

(3)

For these 15 students, Chetna calculates $S_{xx}=965.6$ and $S_{yy}=1561.7$ correct to 1 decimal place.

(e) Calculate the product moment correlation coefficient for these 15 students.

(2)

(f) Calculate the equation of the line of regression of $y$ on $x$ for these 15 students, giving your answer in the form $y=a+bx$

(4)

The product moment correlation coefficient between $x$ and $y$ for all 16 students is 0.746

(g) Explain how your calculation in part (e) supports Chetna’s decision to omit the point $(38,0)$ before calculating the equation of the linear regression line.

(1)

(h) Estimate the mark in the second paper for a student who scored 38 marks in the first paper.

(1)

解答

(a)

解法一

思路

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比较平均数和标准差。Paper 2 的平均分较低，但离散程度较大。

答题过程

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The mean mark on paper 2 is lower than the mean mark on paper 1.

The marks on paper 2 are more spread out than the marks on paper 1, since $\sigma_y>\sigma_x$ .

(b)

解法一

思路

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$(38,0)$ 离其他点和整体趋势都很远。实际原因可能是缺考或分数录入错误。

答题过程

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The point $(38,0)$ is possibly an outlier because it does not follow the pattern of the other points.

A possible reason is that the student was absent for paper 2.

(c)

解法一

思路

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先从 16 个学生的总和中减去被删掉的点，再除以 15。

答题过程

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For paper 1,

\begin{align*} \bar x=\frac{16(35.75)-38}{15}=\frac{534}{15}=35.6. \end{align*}

For paper 2,

\begin{align*} \bar y=\frac{16(25.75)-0}{15}=\frac{412}{15}=27.466\ldots. \end{align*}

\begin{align*} \bar x=35.6,\qquad \bar y=27.5. \end{align*}

(d)

解法一

思路

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删掉的点是 $(38,0)$ ，它对 $\sum xy$ 的贡献是 $38\times0=0$ ，所以 $\sum xy$ 不变。然后用 $S_{xy}=\sum xy-\frac{\sum x\sum y}{n}$ 。

答题过程

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The omitted point has

\begin{align*} xy=38(0)=0. \end{align*}

So $\sum xy$ is still $15837$ .

For the remaining 15 students,

\begin{align*} \sum x=534,\qquad \sum y=412. \end{align*}

Therefore

\begin{aligned} S_{xy} =&\,\sum xy-\frac{\sum x\sum y}{n}\\ =&\,15837-\frac{534(412)}{15}\\ =&\,1169.8. \end{aligned}

(e)

解法一

思路

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直接代入相关系数公式。

答题过程

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\begin{aligned} r =&\,\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}\\ =&\,\frac{1169.8}{\sqrt{965.6(1561.7)}}\\ =&\,0.9526\ldots. \end{aligned}

\begin{align*} r=0.953. \end{align*}

(f)

解法一

思路

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回归线 $y$ on $x$ 的斜率 $b=\frac{S_{xy}}{S_{xx}}$ ，截距用 $(\bar x,\bar y)$ 求。

答题过程

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\begin{align*} b=\frac{1169.8}{965.6}=1.2114\ldots. \end{align*}

Using $a=\bar y-b\bar x$ ,

\begin{align*} a=27.466\ldots-(1.2114\ldots)(35.6)=-15.66\ldots. \end{align*}

Therefore the regression line is

\begin{align*} y=-15.7+1.21x. \end{align*}

(g)

解法一

思路

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删掉该点后 $r$ 从 0.746 增加到 0.953，说明剩下的数据更接近直线。

答题过程

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After omitting the point, the product moment correlation coefficient increases from $0.746$ to $0.953$ .

This suggests the remaining points have a stronger linear relationship, supporting Chetna’s decision.

(h)

解法一

思路

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把 $x=38$ 代入回归线。

答题过程

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\begin{align*} y=-15.66\ldots+1.2114\ldots(38)=30.37\ldots. \end{align*}

The estimated mark in paper 2 is approximately

\begin{align*} 30. \end{align*}

IAL 2019 Jan Q6

题目

解答

(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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(e)

解法一

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(f)

解法一

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(g)

解法一

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(h)

解法一

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