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IAL 2024 May Q4

A Level / Edexcel / S1

IAL 2024 May Paper · Question 4

题目

Problem

A biologist is studying bears. The biologist records the length, dd cm, and the girth, gg cm, of 88 bears. The biologist summarises the data as follows

∑d=1456.8,∑g=713.2,∑dg=141978.84,∑g2=72675.98\sum d=1456.8,\quad \sum g=713.2,\quad \sum dg=141978.84,\quad \sum g^2=72675.98 Sdd=16769.78.S_{dd}=16769.78.

(a) Calculate the exact value of SdgS_{dg} and the exact value of SggS_{gg}.

(3)

(b) Calculate the value of the product moment correlation coefficient between dd and gg.

(2)

(c) Show that the equation of the regression line of gg on dd can be written as

g=−42.3+0.722dg=-42.3+0.722d

where the values of the intercept and gradient are given to 33 significant figures.

(3)

(d) Give an interpretation, in context, of the gradient of the regression line.

(1)

Using the equation of the regression line given in part (c)

(e) (i) estimate the girth of a bear with a length of 2.52.5 metres,

(ii) explain why an estimate for the girth of a bear with a length of 0.50.5 metres is not reliable.

(2)

Using the regression line from part (c), the biologist estimates that for each xx cm increase in the length of a bear there will be a 17.317.3 cm increase in the girth.

(f) Find the value of xx.

(2)

解答

(a)

解法一

思路

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用

Sdg=∑dg−∑d∑gn,\begin{align*} S_{dg}=\sum dg-\frac{\sum d\sum g}{n}, \end{align*}

以及

Sgg=∑g2−(∑g)2n.\begin{align*} S_{gg}=\sum g^2-\frac{(\sum g)^2}{n}. \end{align*}

答题过程

展开 Sdg= 141978.84−1456.8(713.2)8= 12105.12.\begin{align*} S_{dg} =&\,141978.84-\frac{1456.8(713.2)}{8}\\[3mm] =&\,12105.12. \end{align*}

Also,

Sgg= 72675.98−713.228= 9094.2.\begin{align*} S_{gg} =&\,72675.98-\frac{713.2^2}{8}\\[3mm] =&\,9094.2. \end{align*}

(b)

解法一

思路

展开

PMCC 的公式是

r=SdgSddSgg.\begin{align*} r=\frac{S_{dg}}{\sqrt{S_{dd}S_{gg}}}. \end{align*}

答题过程

展开 r= SdgSddSgg= 12105.1216769.78(9094.2)= 0.9802….\begin{align*} r =&\,\frac{S_{dg}}{\sqrt{S_{dd}S_{gg}}}\\[3mm] =&\,\frac{12105.12}{\sqrt{16769.78(9094.2)}}\\[3mm] =&\,0.9802\ldots. \end{align*}

Therefore

r=0.980\begin{align*} r=0.980 \end{align*}

to 33 significant figures.

(c)

解法一

思路

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回归线 g=a+bdg=a+bd 中，

b=SdgSdd.\begin{align*} b=\frac{S_{dg}}{S_{dd}}. \end{align*}

再用均值点 (dˉ,gˉ)(\bar d,\bar g) 求截距。

答题过程

展开

The gradient is

b= SdgSdd= 12105.1216769.78= 0.7218….\begin{align*} b =&\,\frac{S_{dg}}{S_{dd}}\\[3mm] =&\,\frac{12105.12}{16769.78}\\[3mm] =&\,0.7218\ldots. \end{align*}

Also,

dˉ=1456.88=182.1,gˉ=713.28=89.15.\bar d=\frac{1456.8}{8}=182.1,\qquad \bar g=\frac{713.2}{8}=89.15.

So

a= gˉ−bdˉ= 89.15−(0.7218…)(182.1)= −42.297….\begin{align*} a =&\,\bar g-b\bar d\\[3mm] =&\,89.15-(0.7218\ldots)(182.1)\\[3mm] =&\,-42.297\ldots. \end{align*}

Therefore, to 33 significant figures,

g=−42.3+0.722d.\begin{align*} g=-42.3+0.722d. \end{align*}

(d)

解法一

思路

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斜率表示 dd 每增加 11 cm，预测的 gg 增加多少 cm。

答题过程

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For each 11 cm increase in the length of a bear, the girth is predicted to increase by about 0.7220.722 cm.

(e)(i)

解法一

思路

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注意单位：2.52.5 metres 是 250250 cm。代入回归方程。

答题过程

展开

2.52.5 metres is 250250 cm.

Using

g=−42.3+0.722d,\begin{align*} g=-42.3+0.722d, \end{align*}

with d=250d=250,

g= −42.3+0.722(250)= 138.2.\begin{align*} g =&\,-42.3+0.722(250)\\[3mm] =&\,138.2. \end{align*}

The estimated girth is

138 cm\begin{align*} 138\text{ cm} \end{align*}

to 33 significant figures.

(e)(ii)

解法一

思路

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0.50.5 metres 是 5050 cm。代入会得到负 girth，这在语境中不可能，所以不可靠。

答题过程

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0.50.5 metres is 5050 cm.

Substituting d=50d=50 gives

g=−42.3+0.722(50)=−6.2.\begin{align*} g=-42.3+0.722(50)=-6.2. \end{align*}

This is a negative girth, which is not possible, so the estimate is not reliable.

解法二

思路

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数据外推法（Extrapolation）。 观察样本数据发现，0.50.5 米即 5050 cm 的身长，远远超出了实验中 88 只熊的实际身长观测范围（根据 (c) 小题可知，身长平均数 dˉ≈182\bar{d} \approx 182 cm，数据点大致分布在 150∼210150 \sim 210 cm 之间）。 利用回归方程对观测范围外的数据进行预测称为外推（Extrapolation），由于无法保证该线性关系在范围外依然成立，因此该估计结果是极不可靠的。这是统计学中判断预测可靠性的一大核心判定依据。

答题过程

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0.50.5 metres is 5050 cm.

This value of d=50d = 50 cm lies far outside the range of the observed lengths in the sample data (the mean length is dˉ=182.1\bar{d} = 182.1 cm).

Using a regression line to predict values outside the range of the original data is called extrapolation. Since we cannot assume the linear relationship continues to hold outside the observed range, the estimate is not reliable.

(f)

解法一

思路

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斜率是每 11 cm length 增加带来的 girth 增加。因此 xx cm length 增加对应的 girth 增加是 0.722x0.722x。

答题过程

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Using the gradient,

0.722x=17.3.\begin{align*} 0.722x=17.3. \end{align*}

Therefore

x= 17.30.722= 23.96….\begin{align*} x =&\,\frac{17.3}{0.722}\\[3mm] =&\,23.96\ldots. \end{align*}

So

x=24\begin{align*} x=24 \end{align*}

to 22 significant figures.

解法二

思路

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精确斜率计算法。不使用 (c) 中已经四舍五入到三位有效数字的斜率 0.7220.722，而是使用在 (a) 和 (c) 步骤中计算出来的未舍入精确斜率值：

b=SdgSdd=12105.1216769.78≈0.721841\begin{align*} b = \frac{S_{dg}}{S_{dd}} = \frac{12105.12}{16769.78} \approx 0.721841 \end{align*}

我们列出方程：

b⋅x=17.3\begin{align*} b \cdot x = 17.3 \end{align*}

求解 xx。这样可以有效避免中间步骤的舍入误差累积，得到更精确的结果。

答题过程

展开

Use the unrounded gradient from part (c):

b=SdgSdd=12105.1216769.78≈0.721841.\begin{align*} b = \frac{S_{dg}}{S_{dd}} = \frac{12105.12}{16769.78} \approx 0.721841. \end{align*}

We have:

b⋅x=17.3.\begin{align*} b \cdot x = 17.3. \end{align*}

Therefore:

x=  17.30.721841…=  23.966…\begin{align*} x =&\,\, \frac{17.3}{0.721841\ldots}\\[3mm] =&\,\, 23.966\ldots \end{align*}

So

x=24\begin{align*} x = 24 \end{align*}

to 22 significant figures.