题目
Problem
Two students, Olive and Shan, collect data on the weight, w w w t t t 15 15 15
Olive summarised the data as follows
S t t = 5.3173 ∑ w 2 = 6089.12 ∑ t w = 2304.53 S_{tt}=5.3173\qquad
\sum w^2=6089.12\qquad
\sum tw=2304.53 S tt = 5.3173 ∑ w 2 = 6089.12 ∑ tw = 2304.53 ∑ w = 297.8 ∑ t = 114.8 \sum w=297.8\qquad
\sum t=114.8 ∑ w = 297.8 ∑ t = 114.8 (a) Calculate the value of S t w S_{tw} S tw S w w S_{ww} S w w
(3)
(b) Calculate the value of the product moment correlation coefficient between w w w t t t
(2)
(c) Show that the equation of the regression line of w w w t t t
w = − 16.7 + 4.77 t w=-16.7+4.77t w = − 16.7 + 4.77 t (3)
(d) Give an interpretation of the gradient of the regression line.
(1)
(e) Explain why it would not be appropriate to use the regression line in part (c) to estimate the weight of a mouse with a tail length of 2 2 2
(2)
Shan decided to code the data using x = t − 6 x=t-6 x = t − 6 y = w 2 − 5 y=\dfrac{w}{2}-5 y = 2 w − 5
(f) Write down the value of the product moment correlation coefficient between x x x y y y
(1)
(g) Write down an equation of the regression line of y y y x x x
(1)
解答
(a)
解法一
思路
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直接用
S t w = ∑ t w − ∑ t ∑ w n , S w w = ∑ w 2 − ( ∑ w ) 2 n . S_{tw}=\sum tw-\frac{\sum t\sum w}{n},\qquad
S_{ww}=\sum w^2-\frac{(\sum w)^2}{n}. S tw = ∑ tw − n ∑ t ∑ w , S w w = ∑ w 2 − n ( ∑ w ) 2 . 答题过程
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S t w = 2304.53 − 114.8 ( 297.8 ) 15 = 25.367 … \begin{align*}
S_{tw}
=&\,2304.53-\frac{114.8(297.8)}{15}\\[3mm]
=&\,25.367\ldots
\end{align*} S tw = = 2304.53 − 15 114.8 ( 297.8 ) 25.367 … So
S t w = 25.4 \begin{align*}
S_{tw}=25.4
\end{align*} S tw = 25.4 to 3 3 3
Also,
S w w = 6089.12 − 297.8 2 15 = 176.797 … \begin{align*}
S_{ww}
=&\,6089.12-\frac{297.8^2}{15}\\[3mm]
=&\,176.797\ldots
\end{align*} S w w = = 6089.12 − 15 297. 8 2 176.797 … So
S w w = 177. \begin{align*}
S_{ww}=177.
\end{align*} S w w = 177. (b)
解法一
思路
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把 S t w S_{tw} S tw S t t S_{tt} S tt S w w S_{ww} S w w
答题过程
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r = S t w S t t S w w = 25.367 … 5.3173 ( 176.797 … ) = 0.82735 … \begin{align*}
r
=&\,\frac{S_{tw}}{\sqrt{S_{tt}S_{ww}}}\\[3mm]
=&\,\frac{25.367\ldots}{\sqrt{5.3173(176.797\ldots)}}\\[3mm]
=&\,0.82735\ldots
\end{align*} r = = = S tt S w w S tw 5.3173 ( 176.797 … ) 25.367 … 0.82735 … Therefore
r = 0.827 \begin{align*}
r=0.827
\end{align*} r = 0.827 to 3 3 3
(c)
解法一
思路
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回归线 w w w t t t S t w S t t \dfrac{S_{tw}}{S_{tt}} S tt S tw w ˉ − b t ˉ \bar{w}-b\bar{t} w ˉ − b t ˉ
答题过程
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The gradient is
b = S t w S t t = 25.367 … 5.3173 = 4.77065 … \begin{align*}
b
=&\,\frac{S_{tw}}{S_{tt}}\\[3mm]
=&\,\frac{25.367\ldots}{5.3173}\\[3mm]
=&\,4.77065\ldots
\end{align*} b = = = S tt S tw 5.3173 25.367 … 4.77065 … The intercept is
a = w ˉ − b t ˉ = 297.8 15 − 4.77065 … ( 114.8 15 ) = − 16.658 … \begin{align*}
a
=&\,\bar{w}-b\bar{t}\\[3mm]
=&\,\frac{297.8}{15}-4.77065\ldots\left(\frac{114.8}{15}\right)\\[3mm]
=&\,-16.658\ldots
\end{align*} a = = = w ˉ − b t ˉ 15 297.8 − 4.77065 … ( 15 114.8 ) − 16.658 … Therefore the regression line is
w = − 16.7 + 4.77 t . \begin{align*}
w=-16.7+4.77t.
\end{align*} w = − 16.7 + 4.77 t . (d)
解法一
思路
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斜率要在题目语境中解释：tail length 每增加 1 1 1
答题过程
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For each increase of 1 1 1 4.77 4.77 4.77
(e)
解法一
思路
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把 t = 2 t=2 t = 2
答题过程
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Using t = 2 t=2 t = 2
w = − 16.7 + 4.77 ( 2 ) = − 7.16. \begin{align*}
w
=&\,-16.7+4.77(2)\\[3mm]
=&\,-7.16.
\end{align*} w = = − 16.7 + 4.77 ( 2 ) − 7.16. This gives a negative weight, which is impossible for a mouse.
Therefore it is not appropriate to use the regression line for a tail length of 2 2 2
解法二
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数据外推法（Extrapolation）。
观察样本数据可以发现，尾部长 t = 2 t = 2 t = 2 15 15 15 t ˉ ≈ 7.65 \bar{t} \approx 7.65 t ˉ ≈ 7.65
答题过程
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The value t = 2 t = 2 t = 2 t ˉ = 7.65 \bar{t} = 7.65 t ˉ = 7.65
Predicting values outside the range of the original data using a regression line is extrapolation. Since we cannot assume the linear relationship continues to hold outside the observed range, it is not appropriate to use the regression line for a tail length of 2 2 2
(f)
解法一
思路
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正的线性编码不会改变相关系数。这里 x = t − 6 x=t-6 x = t − 6 y = w 2 − 5 y=\dfrac{w}{2}-5 y = 2 w − 5
答题过程
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The product moment correlation coefficient is unchanged by positive linear coding.
Therefore
r = 0.827. \begin{align*}
r=0.827.
\end{align*} r = 0.827. (g)
解法一
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用 t = x + 6 t=x+6 t = x + 6 w = 2 ( y + 5 ) w=2(y+5) w = 2 ( y + 5 )
答题过程
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Since
t = x + 6 , w = 2 ( y + 5 ) , \begin{align*}
t=x+6,\qquad w=2(y+5),
\end{align*} t = x + 6 , w = 2 ( y + 5 ) , substitute into
w = − 16.7 + 4.77 t . \begin{align*}
w=-16.7+4.77t.
\end{align*} w = − 16.7 + 4.77 t . This gives
2 ( y + 5 ) = − 16.7 + 4.77 ( x + 6 ) . \begin{align*}
2(y+5)=-16.7+4.77(x+6).
\end{align*} 2 ( y + 5 ) = − 16.7 + 4.77 ( x + 6 ) . 