题目

Problem

A French test and a Spanish test were sat by $11$ students. The table below shows their marks.

Student	A	B	C	D	E	F	G	H	I	J	K
French mark $(f)$	$24$	$30$	$32$	$32$	$36$	$36$	$40$	$44$	$50$	$60$	$68$
Spanish mark $(s)$	$16$	$90$	$24$	$28$	$32$	$36$	$38$	$44$	$48$	$48$	$68$

Greg says that if these points were plotted on a scatter diagram, then the point $(30,90)$ would be an outlier because $90$ is an outlier for the Spanish marks.

An outlier is defined as a value that is

\text{greater than }Q_3+1.5(Q_3-Q_1)

\text{smaller than }Q_1-1.5(Q_3-Q_1).

(a) Show that $90$ is an outlier for the Spanish marks.

(3)

Ignoring the point $(30,90)$ , Greg calculated the following summary statistics.

\sum f=422,\qquad \sum s=382,\qquad S_{ff}=1667.6,\qquad S_{fs}=1735.6.

(b) Use these summary statistics to show that the equation of the least squares regression line of $s$ on $f$ for the remaining $10$ students is

s=-5.72+1.04f

where the values of the intercept and gradient are given to $3$ significant figures. You must show your working.

(3)

(1)

Two further students sat the French test but missed the Spanish test.

(d) Using the equation given in part (b), estimate

(i) a Spanish mark for the student who scored $55$ marks in their French test,

(ii) a Spanish mark for the student who scored $18$ marks in their French test.

(3)

(e) State, giving a reason, which of the two estimates found in part (d) would be the more reliable estimate.

(2)

题目中文翻译

$11$ 名学生参加了法语和西班牙语测试，分数如表所示。

Greg 认为，若把这些点画在散点图上，点 $(30,90)$ 会是离群点，因为 $90$ 是西班牙语分数中的离群值。题目给出了离群值的定义。

(a) 证明 $90$ 是西班牙语分数中的离群值。

忽略点 $(30,90)$ 后，Greg 得到题目所列的汇总统计量。

(b) 利用这些统计量证明其余 $10$ 名学生的 $s$ 关于 $f$ 的最小二乘回归直线为 $s=-5.72+1.04f$ ，截距和斜率均取 $3$ 位有效数字。必须写出过程。

另有两名学生参加了法语测试，但缺席西班牙语测试。

(d) 利用 (b) 的方程，分别估计法语得分为 $55$ 和 $18$ 的学生的西班牙语分数。

(e) 说明 (d) 中哪个估计更可靠，并给出理由。

解答

(a)

解法一

思路

展开

把 Spanish marks 排序，找上下四分位数，再算上 outlier boundary。只要 $90$ 超过上界，就说明它是 outlier。

答题过程

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For the Spanish marks,

\begin{align*} Q_1=28,\qquad Q_3=48. \end{align*}

The upper outlier boundary is

\begin{align*} Q_3+1.5(Q_3-Q_1) =&\,48+1.5(48-28)\\ =&\,48+30\\ =&\,78. \end{align*}

Since

\begin{align*} 90>78, \end{align*}

$90$ is an outlier for the Spanish marks.

(b)

解法一

思路

展开

回归线 $s=a+bf$ 中，

\begin{align*} b=\frac{S_{fs}}{S_{ff}}. \end{align*}

然后用均值点求 $a$ 。

答题过程

展开

The gradient is

\begin{align*} b =&\,\frac{S_{fs}}{S_{ff}}\\ =&\,\frac{1735.6}{1667.6}\\ =&\,1.0407\ldots. \end{align*}

For the remaining $10$ students,

\bar f=\frac{422}{10}=42.2,\qquad \bar s=\frac{382}{10}=38.2.

\begin{align*} a =&\,\bar s-b\bar f\\ =&\,38.2-(1.0407\ldots)(42.2)\\ =&\,-5.719\ldots. \end{align*}

Therefore, to $3$ significant figures,

\begin{align*} s=-5.72+1.04f. \end{align*}

(c)

解法一

思路

展开

斜率 $1.04$ 表示 French mark 每增加 $1$ 分，预测 Spanish mark 平均增加 $1.04$ 分。

答题过程

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For each extra mark in the French test, the Spanish test mark is predicted to increase by about $1.04$ marks.

(d)(i)

解法一

思路

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把 $f=55$ 代入回归方程。

答题过程

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When $f=55$ ,

\begin{align*} s =&\,-5.72+1.04(55)\\ =&\,51.48. \end{align*}

The estimate is

\begin{align*} 51.5. \end{align*}

(d)(ii)

解法一

思路

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把 $f=18$ 代入同一个回归方程。

答题过程

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When $f=18$ ,

\begin{align*} s =&\,-5.72+1.04(18)\\ =&\,13.0. \end{align*}

The estimate is

\begin{align*} 13. \end{align*}

(e)

解法一

思路

展开

剩余 $10$ 个学生的 French marks 范围是 $24$ 到 $68$ 。 $55$ 在范围内，是 interpolation； $18$ 在范围外，是 extrapolation。因此 $55$ 对应的估计更可靠。

答题过程

展开

The estimate for the student who scored $55$ in the French test is more reliable.

This is because $55$ is within the range of the French marks used to form the regression line, whereas $18$ is outside this range.

S1 2024 Jan Q4

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)(i)

解法一

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(d)(ii)

解法一

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(e)

解法一

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答题过程