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S1 2019 Jan Q6

A Level / Edexcel / S1

IAL 2019 Jan Paper · Question 6

题目

Problem

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

xˉ=35.75yˉ=25.75σx=7.79σy=11.91∑xy=15837\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)(38,0) is possibly an outlier.

(ii) Suggest a possible reason for this result.

(2)

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

(c) Find the value of xˉ\bar x and the value of yˉ\bar y for these 15 students.

(3)

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

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

(3)

For these 15 students, Chetna calculates Sxx=965.6S_{xx}=965.6 and Syy=1561.7S_{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 yy on xx for these 15 students, giving your answer in the form y=a+bxy=a+bx

(4)

The product moment correlation coefficient between xx and yy 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)(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)
题目中文翻译

学校考试后，Chetna 研究班内 16 名学生的成绩。每名学生的试卷 1 分数 xx 与试卷 2 分数 yy 由题中统计量汇总。

(a) 评论学生在两份试卷上的成绩差异。

Chetna 把 16 名学生的成绩绘制成散点图。

(b) (i) 解释圈出的点 (38,0)(38,0) 为何可能是离群点；(ii) 提出出现此结果的一种可能原因。

Chetna 删去数据点 (38,0)(38,0)，研究其余 15 名学生的成绩。

(c) 求这 15 名学生的 xˉ\bar x 与 yˉ\bar y。

(d) (i) 解释为何 ∑xy\sum xy 仍为 15 837；(ii) 证明 Sxy=1169.8S_{xy}=1169.8。

已知这 15 名学生的 Sxx=965.6S_{xx}=965.6、Syy=1561.7S_{yy}=1561.7（精确至 1 位小数）。

(e) 计算这 15 名学生成绩的积矩相关系数。

(f) 求这 15 名学生的 yy 关于 xx 的回归直线方程，答案写成 y=a+bxy=a+bx。

16 名学生全部数据的积矩相关系数为 0.746。

(g) 解释 (e) 的计算如何支持 Chetna 在求线性回归方程前删去点 (38,0)(38,0) 的决定。

(h) 估计试卷 1 得 38 分的学生在试卷 2 的成绩。

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 σy>σx\sigma_y>\sigma_x.

(b)

(i)

思路

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(38,0)(38,0) 离其他点形成的整体趋势很远。

答题过程

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

(ii)

思路

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给出能够导致第二份试卷得 0 分的合理情境，例如缺考、漏阅或成绩记录错误。

答题过程

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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,

xˉ=16(35.75)−3815=53415=35.6.\begin{align*} \bar x=\frac{16(35.75)-38}{15}=\frac{534}{15}=35.6. \end{align*}

For paper 2,

yˉ=16(25.75)−015=41215=27.466….\begin{align*} \bar y=\frac{16(25.75)-0}{15}=\frac{412}{15}=27.466\ldots. \end{align*}

So

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

(d)

(i)

思路

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删去的点对 ∑xy\sum xy 的贡献为 0。

答题过程

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The omitted point has

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

So ∑xy\sum xy is still 1583715837.

(ii)

思路

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用更新后的 ∑x\sum x、∑y\sum y 与 n=15n=15 代入 SxyS_{xy} 公式。

答题过程

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For the remaining 15 students,

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

Therefore

Sxy= ∑xy−∑x∑yn= 15837−534(412)15= 1169.8.\begin{align*} S_{xy} =&\,\sum xy-\frac{\sum x\sum y}{n}\\ =&\,15837-\frac{534(412)}{15}\\ =&\,1169.8. \end{align*}

(e)

解法一

思路

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直接代入相关系数公式。

答题过程

展开 r= SxySxxSyy= 1169.8965.6(1561.7)= 0.9526….\begin{align*} r =&\,\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}\\ =&\,\frac{1169.8}{\sqrt{965.6(1561.7)}}\\ =&\,0.9526\ldots. \end{align*}

So

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

(f)

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}}，截距用 (xˉ,yˉ)(\bar x,\bar y) 求。

答题过程

展开 b=1169.8965.6=1.2114….\begin{align*} b=\frac{1169.8}{965.6}=1.2114\ldots. \end{align*}

Using a=yˉ−bxˉa=\bar y-b\bar x,

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

Therefore the regression line is

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

(g)

解法一

思路

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删掉该点后 rr 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 0.7460.746 to 0.9530.953.

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

(h)

解法一

思路

展开

把 x=38x=38 代入回归线。

答题过程

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

The estimated mark in paper 2 is approximately

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