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IAL 2022 Jan Q2

A Level / Edexcel / S1

IAL 2022 Jan Paper · Question 2

题目

Problem

Tom’s car holds 5050 litres of petrol when the fuel tank is full. For each of 1010 journeys, each starting with 5050 litres of petrol in the fuel tank, Tom records the distance travelled, dd kilometres, and the amount of petrol used, pp litres. The summary statistics for the 1010 journeys are given below.

∑d=1029∑p=50.8∑dp=5240.8\sum d=1029\qquad \sum p=50.8\qquad \sum dp=5240.8 Sdd=344.9Spp=0.576S_{dd}=344.9\qquad S_{pp}=0.576

(a) Calculate the product moment correlation coefficient between dd and pp

(3)

The amount of petrol remaining in the fuel tank for each journey, ww litres, is recorded.

(b) (i) Write down an equation for ww in terms of pp

(ii) Hence, write down the value of the product moment correlation coefficient between ww and pp

(2)

(c) Write down the value of the product moment correlation coefficient between dd and ww

(1)

解答

(a)

解法一

思路

展开

先求 SdpS_{dp}，再代入相关系数公式。

答题过程

展开 Sdp= ∑dp−∑d∑pn= 5240.8−1029(50.8)10= 13.48.\begin{align*} S_{dp} =&\,\sum dp-\frac{\sum d\sum p}{n}\\[3mm] =&\,5240.8-\frac{1029(50.8)}{10}\\[3mm] =&\,13.48. \end{align*}

Thus

r= SdpSddSpp= 13.48344.9(0.576)= 0.956…\begin{align*} r =&\,\frac{S_{dp}}{\sqrt{S_{dd}S_{pp}}}\\[3mm] =&\,\frac{13.48}{\sqrt{344.9(0.576)}}\\[3mm] =&\,0.956\ldots \end{align*}

So

r=0.956.\begin{align*} r=0.956. \end{align*}

(b)

解法一

思路

展开

油箱满时有 5050 litres，剩余油量 = 50−50- 使用油量。由于 w=50−pw=50-p 是对 pp 的负线性变换，所以与 pp 的相关系数为 −1-1。

答题过程

展开 w=50−p.\begin{align*} w=50-p. \end{align*}

Therefore the product moment correlation coefficient between ww and pp is

−1.\begin{align*} -1. \end{align*}

(c)

解法一

思路

展开

ww 与 pp 完全负相关，所以 dd 与 ww 的相关系数是 dd 与 pp 的相关系数变号。

答题过程

展开

The product moment correlation coefficient between dd and ww is

−0.956.\begin{align*} -0.956. \end{align*}

解法二

思路

展开

线性编码相关系数符号法则。 乘积累积相关系数（PMCC）具有线性变换不变性，只受变换系数的正负符号影响。 定理：若 X′=aX+bX' = a X + b 且 Y′=cY+dY' = c Y + d（其中 a,c≠0a, c \neq 0），则新变量 X′X' 与 Y′Y' 的相关系数为：

r(X′,Y′)=sgn⁡(a)sgn⁡(c)r(X,Y)\begin{align*} r(X', Y') = \operatorname{sgn}(a) \operatorname{sgn}(c) r(X, Y) \end{align*}

本题中：

  • 距离 dd 未发生变换，即 d′=1⋅d+0d' = 1 \cdot d + 0，变换系数 a=1>0a = 1 > 0；
  • 剩余油量 ww 与使用油量 pp 关系为 w=−1⋅p+50w = -1 \cdot p + 50，变换系数 c=−1<0c = -1 < 0。 因此：
rdw=sgn⁡(1)sgn⁡(−1)rdp=1×(−1)×rdp=−rdp\begin{align*} r_{dw} = \operatorname{sgn}(1) \operatorname{sgn}(-1) r_{dp} = 1 \times (-1) \times r_{dp} = -r_{dp} \end{align*}

代入 rdp=0.956r_{dp} = 0.956 即可得到结果。该方法为定性结论提供了严密的统计学公式支持。

答题过程

展开

Consider the linear transformation from (d,p)(d, p) to (d,w)(d, w):

  • For dd: d=1⋅d+0d = 1 \cdot d + 0, where the coefficient of the variable is 1>01 > 0.
  • For ww: w=−1⋅p+50w = -1 \cdot p + 50, where the coefficient of pp is −1<0-1 < 0.

Using the property of the product moment correlation coefficient under linear coding:

rdw=  sgn⁡(1)sgn⁡(−1)rdp…=  (1)(−1)rdp=  −rdp.\begin{align*} r_{dw} =&\,\, \operatorname{sgn}(1) \operatorname{sgn}(-1) r_{dp}\dots\\[3mm] =&\,\, (1)(-1) r_{dp}\\[3mm] =&\,\, -r_{dp}. \end{align*}

From part (a), rdp=0.956r_{dp} = 0.956. Therefore:

rdw=  −0.956.\begin{align*} r_{dw} =&\,\, -0.956. \end{align*}