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IAL 2024 Oct Q2

A Level / Edexcel / S1

IAL 2024 Oct Paper · Question 2

题目

Problem

A biologist records the length, yy cm, and the weight, ww kg, of 5050 rabbits.

The following summary statistics are obtained.

∑y=2015,∑y2=81938.5,∑w=125,\sum y=2015,\quad \sum y^2=81938.5,\quad \sum w=125, Sww=72.25,Syw=219.55.S_{ww}=72.25,\qquad S_{yw}=219.55.

(a) (i) Show that Syy=734S_{yy}=734.

(ii) Calculate the product moment correlation coefficient for these data. Give your answer to 33 decimal places.

(3)

(b) Interpret the product moment correlation coefficient in context.

(1)

(c) State, giving a reason, whether the value of the product moment correlation coefficient is consistent with the use of a linear regression model.

(1)

(d) Find the equation of the regression line of ww on yy, giving your answer in the form w=a+byw=a+by.

(4)

A rabbit named Jeff has length 4545 cm.

(e) Use your regression equation to estimate the weight of Jeff.

(2)

解答

(a)(i)

解法一

思路

展开

用

Syy=∑y2−(∑y)2n.\begin{align*} S_{yy}=\sum y^2-\frac{(\sum y)^2}{n}. \end{align*}

题目要求 show that，所以要写出代入式和计算结果。

答题过程

展开 Syy= ∑y2−(∑y)2n= 81938.5−2015250= 81938.5−81204.5= 734.\begin{align*} S_{yy} =&\,\sum y^2-\frac{(\sum y)^2}{n}\\[3mm] =&\,81938.5-\frac{2015^2}{50}\\[3mm] =&\,81938.5-81204.5\\[3mm] =&\,734. \end{align*}

(a)(ii)

解法一

思路

展开

PMCC 的公式是

r=SywSyySww.\begin{align*} r=\frac{S_{yw}}{\sqrt{S_{yy}S_{ww}}}. \end{align*}

答题过程

展开 r= SywSyySww= 219.55734(72.25)= 0.95338….\begin{align*} r =&\,\frac{S_{yw}}{\sqrt{S_{yy}S_{ww}}}\\[3mm] =&\,\frac{219.55}{\sqrt{734(72.25)}}\\[3mm] =&\,0.95338\ldots. \end{align*}

So, to 33 decimal places,

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

(b)

解法一

思路

展开

rr 接近 11，表示长度和重量之间有强正相关。解释时要放回语境：兔子越长，通常越重。

答题过程

展开

There is a strong positive correlation between length and weight. In general, the longer the rabbit, the greater its weight.

(c)

解法一

思路

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r=0.953r=0.953 很接近 11，说明数据点应当相当接近一条正斜率直线，所以适合线性回归模型。

答题过程

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Yes. The value of rr is close to 11, so the data are consistent with a linear regression model.

(d)

解法一

思路

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回归线 w=a+byw=a+by 中，

b=SywSyy.\begin{align*} b=\frac{S_{yw}}{S_{yy}}. \end{align*}

然后用均值点 (yˉ,wˉ)(\bar{y},\bar{w}) 在回归线上，求截距 aa。

答题过程

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The gradient of the regression line of ww on yy is

b= SywSyy= 219.55734= 0.29911….\begin{align*} b =&\,\frac{S_{yw}}{S_{yy}}\\[3mm] =&\,\frac{219.55}{734}\\[3mm] =&\,0.29911\ldots. \end{align*}

Also,

yˉ=201550=40.3,wˉ=12550=2.5.\bar{y}=\frac{2015}{50}=40.3,\qquad \bar{w}=\frac{125}{50}=2.5.

Using a=wˉ−byˉa=\bar{w}-b\bar{y},

a= 2.5−(0.29911…)(40.3)= −9.554….\begin{align*} a =&\,2.5-(0.29911\ldots)(40.3)\\[3mm] =&\,-9.554\ldots. \end{align*}

Therefore the regression equation is

w=−9.55+0.299y.\begin{align*} w=-9.55+0.299y. \end{align*}

(e)

解法一

思路

展开

把 Jeff 的长度 y=45y=45 代入上一小题的回归方程。

答题过程

展开

When y=45y=45,

w= −9.55+0.299(45)= 3.905.\begin{align*} w =&\,-9.55+0.299(45)\\[3mm] =&\,3.905. \end{align*}

The estimated weight of Jeff is

3.91 kg.\begin{align*} 3.91\text{ kg}. \end{align*}