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IAL 2018 Oct Q1

A Level / Edexcel / S1

IAL 2018 Oct Paper · Question 1

题目

Problem

The heights above sea level (hh hundred metres) and the temperatures (t∘Ct^\circ C) at 12 randomly selected places in France, at 7 am on July 31st, were recorded. The data are summarised as follows

∑h=112∑t=136∑t2=1828Sht=−236Shh=297\sum h=112\qquad \sum t=136\qquad \sum t^2=1828\qquad S_{ht}=-236\qquad S_{hh}=297

(a) Find the value of SttS_{tt}

(2)

(b) Calculate the product moment correlation coefficient for these data.

(2)

(c) Interpret the relationship between tt and hh.

(1)

(d) Find an equation of the regression line of tt on hh.

(3)

At 7 am on July 31st Yinka is on holiday in South Africa. He uses the regression equation to estimate the temperature when the height above sea level is 500 m.

(e) Find the estimated temperature Yinka calculates.

(2)

(f) Comment on the validity of your answer in part (e).

(1)

解答

解法一

思路

展开

先求 SttS_{tt}，再求相关系数和回归线。注意 hh 的单位是 hundred metres，所以 500 m 对应 h=5h=5。

答题过程

展开 Stt=∑t2−(∑t)2n=1828−136212=286.667….S_{tt}=\sum t^2-\frac{(\sum t)^2}{n} =1828-\frac{136^2}{12} =286.667\ldots.

So

Stt=287to 3 significant figures.\begin{align*} S_{tt}=287\quad\text{to 3 significant figures}. \end{align*}

The product moment correlation coefficient is

r= ShtShhStt= −236297(286.667…)= −0.809….\begin{aligned} r =&\,\frac{S_{ht}}{\sqrt{S_{hh}S_{tt}}}\\ =&\,\frac{-236}{\sqrt{297(286.667\ldots)}}\\ =&\,-0.809\ldots. \end{aligned}

Thus

r=−0.809.\begin{align*} r=-0.809. \end{align*}

This shows strong negative correlation: as height above sea level increases, temperature tends to decrease.

For the regression line of tt on hh,

b=ShtShh=−236297=−0.7946….\begin{align*} b=\frac{S_{ht}}{S_{hh}}=\frac{-236}{297}=-0.7946\ldots. \end{align*}

Also,

hˉ=11212=9.333…,tˉ=13612=11.333….\bar h=\frac{112}{12}=9.333\ldots,\qquad \bar t=\frac{136}{12}=11.333\ldots.

So

a=tˉ−bhˉ=11.333…−(−0.7946…)(9.333…)=18.75….a=\bar t-b\bar h =11.333\ldots-(-0.7946\ldots)(9.333\ldots) =18.75\ldots.

The regression line is

t=18.8−0.795h.\begin{align*} t=18.8-0.795h. \end{align*}

For 500 m above sea level,

h=5.\begin{align*} h=5. \end{align*}

So

t=18.75…−0.7946…(5)=14.78….\begin{align*} t=18.75\ldots-0.7946\ldots(5)=14.78\ldots. \end{align*}

The estimated temperature is approximately

14.8∘C.\begin{align*} 14.8^\circ C. \end{align*}

This estimate is not very valid, because the regression line is based on places in France, but Yinka is in South Africa.