题目
Problem
A biologist is studying bears.
The biologist records the length, d d d cm, and the girth, g g g cm, of 8 8 8 bears.
The biologist summarises the data as follows
∑ d = 1456.8 , ∑ g = 713.2 , ∑ d g = 141978.84 , ∑ g 2 = 72675.98 \sum d=1456.8,\quad \sum g=713.2,\quad
\sum dg=141978.84,\quad \sum g^2=72675.98 ∑ d = 1456.8 , ∑ g = 713.2 , ∑ d g = 141978.84 , ∑ g 2 = 72675.98
S d d = 16769.78. S_{dd}=16769.78. S dd = 16769.78.
(a) Calculate the exact value of S d g S_{dg} S d g and the exact value of S g g S_{gg} S g g .
(3)
(b) Calculate the value of the product moment correlation coefficient between d d d and g g g .
(2)
(c) Show that the equation of the regression line of g g g on d d d can be written as
g = − 42.3 + 0.722 d g=-42.3+0.722d g = − 42.3 + 0.722 d
where the values of the intercept and gradient are given to 3 3 3 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 2.5 2.5 2.5 metres,
(ii) explain why an estimate for the girth of a bear with a length of 0.5 0.5 0.5 metres is not reliable.
(2)
Using the regression line from part (c), the biologist estimates that for each x x x cm increase in the length of a bear there will be a 17.3 17.3 17.3 cm increase in the girth.
(f) Find the value of x x x .
(2)
题目中文翻译
一名生物学家研究熊,并记录 8 8 8 只熊的身长 d d d cm 和胸围 g g g cm;题目给出相应的汇总统计量。
(a) 计算 S d g S_{dg} S d g 和 S g g S_{gg} S g g 的精确值。
(b) 计算 d d d 与 g g g 的积矩相关系数。
(c) 证明 g g g 关于 d d d 的回归直线可写成 g = − 42.3 + 0.722 d g=-42.3+0.722d g = − 42.3 + 0.722 d ,其中截距和斜率均取 3 3 3 位有效数字。
(d) 结合情境解释回归直线斜率的意义。
利用 (c) 的回归直线:
(e) (i) 估计身长为 2.5 2.5 2.5 m 的熊的胸围;
(ii) 解释为什么对身长为 0.5 0.5 0.5 m 的熊所作的胸围估计不可靠。
利用 (c) 的回归直线,生物学家估计熊的身长每增加 x x x cm,胸围会增加 17.3 17.3 17.3 cm。
(f) 求 x x x 。
解答
(a)
解法一
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用
S d g = ∑ d g − ∑ d ∑ g n , \begin{align*}
S_{dg}=\sum dg-\frac{\sum d\sum g}{n},
\end{align*} S d g = ∑ d g − n ∑ d ∑ g ,
以及
S g g = ∑ g 2 − ( ∑ g ) 2 n . \begin{align*}
S_{gg}=\sum g^2-\frac{(\sum g)^2}{n}.
\end{align*} S g g = ∑ g 2 − n ( ∑ g ) 2 .
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S d g = 141978.84 − 1456.8 ( 713.2 ) 8 = 12105.12. \begin{align*}
S_{dg}
=&\,141978.84-\frac{1456.8(713.2)}{8}\\
=&\,12105.12.
\end{align*} S d g = = 141978.84 − 8 1456.8 ( 713.2 ) 12105.12.
Also,
S g g = 72675.98 − 713.2 2 8 = 9094.2. \begin{align*}
S_{gg}
=&\,72675.98-\frac{713.2^2}{8}\\
=&\,9094.2.
\end{align*} S g g = = 72675.98 − 8 713. 2 2 9094.2.
(b)
解法一
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PMCC 的公式是
r = S d g S d d S g g . \begin{align*}
r=\frac{S_{dg}}{\sqrt{S_{dd}S_{gg}}}.
\end{align*} r = S dd S g g S d g .
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r = S d g S d d S g g = 12105.12 16769.78 ( 9094.2 ) = 0.9802 … . \begin{align*}
r
=&\,\frac{S_{dg}}{\sqrt{S_{dd}S_{gg}}}\\
=&\,\frac{12105.12}{\sqrt{16769.78(9094.2)}}\\
=&\,0.9802\ldots.
\end{align*} r = = = S dd S g g S d g 16769.78 ( 9094.2 ) 12105.12 0.9802 … .
Therefore
r = 0.980 \begin{align*}
r=0.980
\end{align*} r = 0.980
to 3 3 3 significant figures.
(c)
解法一
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回归线 g = a + b d g=a+bd g = a + b d 中,
b = S d g S d d . \begin{align*}
b=\frac{S_{dg}}{S_{dd}}.
\end{align*} b = S dd S d g .
再用均值点 ( d ˉ , g ˉ ) (\bar d,\bar g) ( d ˉ , g ˉ ) 求截距。
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The gradient is
b = S d g S d d = 12105.12 16769.78 = 0.7218 … . \begin{align*}
b
=&\,\frac{S_{dg}}{S_{dd}}\\
=&\,\frac{12105.12}{16769.78}\\
=&\,0.7218\ldots.
\end{align*} b = = = S dd S d g 16769.78 12105.12 0.7218 … .
Also,
d ˉ = 1456.8 8 = 182.1 , g ˉ = 713.2 8 = 89.15. \bar d=\frac{1456.8}{8}=182.1,\qquad
\bar g=\frac{713.2}{8}=89.15. d ˉ = 8 1456.8 = 182.1 , g ˉ = 8 713.2 = 89.15.
So
a = g ˉ − b d ˉ = 89.15 − ( 0.7218 … ) ( 182.1 ) = − 42.297 … . \begin{align*}
a
=&\,\bar g-b\bar d\\
=&\,89.15-(0.7218\ldots)(182.1)\\
=&\,-42.297\ldots.
\end{align*} a = = = g ˉ − b d ˉ 89.15 − ( 0.7218 … ) ( 182.1 ) − 42.297 … .
Therefore, to 3 3 3 significant figures,
g = − 42.3 + 0.722 d . \begin{align*}
g=-42.3+0.722d.
\end{align*} g = − 42.3 + 0.722 d .
(d)
解法一
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斜率表示 d d d 每增加 1 1 1 cm,预测的 g g g 增加多少 cm。
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For each 1 1 1 cm increase in the length of a bear, the girth is predicted to increase by about 0.722 0.722 0.722 cm.
(e)(i)
解法一
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注意单位:2.5 2.5 2.5 metres 是 250 250 250 cm。代入回归方程。
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2.5 2.5 2.5 metres is 250 250 250 cm.
Using
g = − 42.3 + 0.722 d , \begin{align*}
g=-42.3+0.722d,
\end{align*} g = − 42.3 + 0.722 d ,
with d = 250 d=250 d = 250 ,
g = − 42.3 + 0.722 ( 250 ) = 138.2. \begin{align*}
g
=&\,-42.3+0.722(250)\\
=&\,138.2.
\end{align*} g = = − 42.3 + 0.722 ( 250 ) 138.2.
The estimated girth is
138 cm \begin{align*}
138\text{ cm}
\end{align*} 138 cm
to 3 3 3 significant figures.
(e)(ii)
解法一
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0.5 0.5 0.5 metres 是 50 50 50 cm。代入会得到负 girth,这在语境中不可能,所以不可靠。
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0.5 0.5 0.5 metres is 50 50 50 cm.
Substituting d = 50 d=50 d = 50 gives
g = − 42.3 + 0.722 ( 50 ) = − 6.2. \begin{align*}
g=-42.3+0.722(50)=-6.2.
\end{align*} g = − 42.3 + 0.722 ( 50 ) = − 6.2.
This is a negative girth, which is not possible, so the estimate is not reliable.
(f)
解法一
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斜率是每 1 1 1 cm length 增加带来的 girth 增加。因此 x x x cm length 增加对应的 girth 增加是 0.722 x 0.722x 0.722 x 。
答题过程
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Using the gradient,
0.722 x = 17.3. \begin{align*}
0.722x=17.3.
\end{align*} 0.722 x = 17.3.
Therefore
x = 17.3 0.722 = 23.96 … . \begin{align*}
x
=&\,\frac{17.3}{0.722}\\
=&\,23.96\ldots.
\end{align*} x = = 0.722 17.3 23.96 … .
So
x = 24 \begin{align*}
x=24
\end{align*} x = 24
to 2 2 2 significant figures.