题目
Problem
Statistical models are a cheap and quick way to make predictions about real-world situations.
(a) Give one other reason why statistical models are used.
(1)
Madison wants to develop a model to describe the relationship between the average daily temperature, t ∘ C t\,^\circ\text{C} t ∘ C d m 3 d\,\text{m}^3 d m 3 12 12 12 w = 11.5 d w=11.5d w = 11.5 d
S t t = 26.43 , S t w = − 91.55 , ∑ w = 339.25 , ∑ t = 9.1 , ∑ w 2 = 10 036.45. \begin{align*}
S_{tt}=&\,26.43,\qquad S_{tw}=-91.55,\\[2mm]
\sum w=&\,339.25,\qquad \sum t=9.1,\\[2mm]
\sum w^2=&\,10\,036.45.
\end{align*} S tt = ∑ w = ∑ w 2 = 26.43 , S tw = − 91.55 , 339.25 , ∑ t = 9.1 , 10 036.45. (b) Show that S w w = 445.57 S_{ww}=445.57 S w w = 445.57 2 2 2
(1)
(c) Find the value of S t d S_{td} S t d S d d S_{dd} S dd
(3)
(d) Find the product moment correlation coefficient between d d d t t t
(2)
(e) Give an interpretation, in context, of your product moment correlation coefficient.
(1)
(f) Show that the equation of the regression line of w w w t t t
w = − 3.46 t + 30.9 w=-3.46t+30.9 w = − 3.46 t + 30.9 where the values of the intercept and the gradient are given to 3 3 3
(3)
(g) Write down an equation of the regression line of d d d t t t
(1)
(h) Using your equation in part (g)
(i) estimate the daily gas consumption in winter when the temperature is 2 ∘ C 2\,^\circ\text{C} 2 ∘ C
(ii) interpret the effect an increase of 1 ∘ C 1\,^\circ\text{C} 1 ∘ C
(2)
解答
(a)
解法一
思路
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题目已经说了 statistical models 可以 cheap、quick、make predictions，所以答案不能只重复这些。可以说模型帮助理解、简化真实世界，或展示变量之间的关系。
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Statistical models can be used to show relationships between variables.
(b)
解法一
思路
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用
S w w = ∑ w 2 − ( ∑ w ) 2 n . \begin{align*}
S_{ww}=\sum w^2-\frac{(\sum w)^2}{n}.
\end{align*} S w w = ∑ w 2 − n ( ∑ w ) 2 . 题目给了 ∑ w 2 \sum w^2 ∑ w 2 ∑ w \sum w ∑ w n = 12 n=12 n = 12
答题过程
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S w w = ∑ w 2 − ( ∑ w ) 2 n = 10036.45 − ( 339.25 ) 2 12 = 10036.45 − 9590.878 … = 445.571 … . \begin{align*}
S_{ww}
=&\,\sum w^2-\frac{(\sum w)^2}{n}\\[3mm]
=&\,10036.45-\frac{(339.25)^2}{12}\\[3mm]
=&\,10036.45-9590.878\ldots\\[3mm]
=&\,445.571\ldots.
\end{align*} S w w = = = = ∑ w 2 − n ( ∑ w ) 2 10036.45 − 12 ( 339.25 ) 2 10036.45 − 9590.878 … 445.571 … . So
S w w = 445.57 \begin{align*}
S_{ww}=445.57
\end{align*} S w w = 445.57 to 2 2 2
(c)
解法一
思路
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因为 w = 11.5 d w=11.5d w = 11.5 d d = w 11.5 d=\dfrac{w}{11.5} d = 11.5 w S S S 11.5 11.5 11.5 11.5 11.5 11.5 S S S 11.5 11.5 11.5 11.5 2 11.5^2 11. 5 2
答题过程
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Since
w = 11.5 d , \begin{align*}
w=11.5d,
\end{align*} w = 11.5 d , we have
d = w 11.5 . \begin{align*}
d=\frac{w}{11.5}.
\end{align*} d = 11.5 w . Therefore
S t d = S t w 11.5 = − 91.55 11.5 = − 7.9608 … . \begin{align*}
S_{td}
=&\,\frac{S_{tw}}{11.5}\\[3mm]
=&\,\frac{-91.55}{11.5}\\[3mm]
=&\,-7.9608\ldots.
\end{align*} S t d = = = 11.5 S tw 11.5 − 91.55 − 7.9608 … . Also,
S d d = S w w ( 11.5 ) 2 = 445.571 … 132.25 = 3.3691 … . \begin{align*}
S_{dd}
=&\,\frac{S_{ww}}{(11.5)^2}\\[3mm]
=&\,\frac{445.571\ldots}{132.25}\\[3mm]
=&\,3.3691\ldots.
\end{align*} S dd = = = ( 11.5 ) 2 S w w 132.25 445.571 … 3.3691 … . So
S t d = − 7.96 , S d d = 3.37 \begin{align*}
S_{td}=-7.96,\qquad S_{dd}=3.37
\end{align*} S t d = − 7.96 , S dd = 3.37 to 3 3 3
解法二
思路
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公式推导法。直接利用样本共变异数和样本方差的定义式，通过代数代换，严格证明编码转换的关系：
S t d = ∑ ( t − t ˉ ) ( d − d ˉ ) = ∑ ( t − t ˉ ) ( w 11.5 − w ˉ 11.5 ) = 1 11.5 S t w \begin{align*}
S_{td} = \sum(t-\bar{t})(d-\bar{d}) = \sum(t-\bar{t})\left(\frac{w}{11.5}-\frac{\bar{w}}{11.5}\right) = \frac{1}{11.5}S_{tw}
\end{align*} S t d = ∑ ( t − t ˉ ) ( d − d ˉ ) = ∑ ( t − t ˉ ) ( 11.5 w − 11.5 w ˉ ) = 11.5 1 S tw 和
S d d = ∑ ( d − d ˉ ) 2 = ∑ ( w 11.5 − w ˉ 11.5 ) 2 = 1 11.5 2 S w w \begin{align*}
S_{dd} = \sum(d-\bar{d})^2 = \sum\left(\frac{w}{11.5}-\frac{\bar{w}}{11.5}\right)^2 = \frac{1}{11.5^2}S_{ww}
\end{align*} S dd = ∑ ( d − d ˉ ) 2 = ∑ ( 11.5 w − 11.5 w ˉ ) 2 = 11. 5 2 1 S w w 最后代入已知的 S t w S_{tw} S tw S w w S_{ww} S w w
答题过程
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By definition, the coded value is d = w 11.5 d = \frac{w}{11.5} d = 11.5 w d d d d ˉ = w ˉ 11.5 \bar{d} = \frac{\bar{w}}{11.5} d ˉ = 11.5 w ˉ
For S t d S_{td} S t d
S t d = ∑ ( t − t ˉ ) ( d − d ˉ ) = ∑ ( t − t ˉ ) ( w 11.5 − w ˉ 11.5 ) = 1 11.5 ∑ ( t − t ˉ ) ( w − w ˉ ) = S t w 11.5 = − 91.55 11.5 = − 7.9608 … \begin{align*}
S_{td} =&\,\, \sum(t-\bar{t})(d-\bar{d})\\[3mm]
=&\,\, \sum(t-\bar{t})\left(\frac{w}{11.5} - \frac{\bar{w}}{11.5}\right)\\[3mm]
=&\,\, \frac{1}{11.5}\sum(t-\bar{t})(w-\bar{w})\\[3mm]
=&\,\, \frac{S_{tw}}{11.5}\\[3mm]
=&\,\, \frac{-91.55}{11.5}\\[3mm]
=&\,\, -7.9608\ldots
\end{align*} S t d = = = = = = ∑ ( t − t ˉ ) ( d − d ˉ ) ∑ ( t − t ˉ ) ( 11.5 w − 11.5 w ˉ ) 11.5 1 ∑ ( t − t ˉ ) ( w − w ˉ ) 11.5 S tw 11.5 − 91.55 − 7.9608 … For S d d S_{dd} S dd
S d d = ∑ ( d − d ˉ ) 2 = ∑ ( w 11.5 − w ˉ 11.5 ) 2 = 1 11.5 2 ∑ ( w − w ˉ ) 2 = S w w 11.5 2 = 445.571 … 132.25 = 3.3691 … \begin{align*}
S_{dd} =&\,\, \sum(d-\bar{d})^2\\[3mm]
=&\,\, \sum\left(\frac{w}{11.5} - \frac{\bar{w}}{11.5}\right)^2\\[3mm]
=&\,\, \frac{1}{11.5^2}\sum(w-\bar{w})^2\\[3mm]
=&\,\, \frac{S_{ww}}{11.5^2}\\[3mm]
=&\,\, \frac{445.571\ldots}{132.25}\\[3mm]
=&\,\, 3.3691\ldots
\end{align*} S dd = = = = = = ∑ ( d − d ˉ ) 2 ∑ ( 11.5 w − 11.5 w ˉ ) 2 11. 5 2 1 ∑ ( w − w ˉ ) 2 11. 5 2 S w w 132.25 445.571 … 3.3691 … Therefore,
S t d = − 7.96 , S d d = 3.37 \begin{align*}
S_{td}=-7.96,\qquad S_{dd}=3.37
\end{align*} S t d = − 7.96 , S dd = 3.37 to 3 3 3
(d)
解法一
思路
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PMCC 公式是
r = S t d S t t S d d . \begin{align*}
r=\frac{S_{td}}{\sqrt{S_{tt}S_{dd}}}.
\end{align*} r = S tt S dd S t d . 用上一小题的 S t d S_{td} S t d S d d S_{dd} S dd S t t = 26.43 S_{tt}=26.43 S tt = 26.43
答题过程
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r = S t d S t t S d d = − 7.9608 … 26.43 ( 3.3691 … ) = − 0.8436 … . \begin{align*}
r
=&\,\frac{S_{td}}{\sqrt{S_{tt}S_{dd}}}\\[3mm]
=&\,\frac{-7.9608\ldots}{\sqrt{26.43(3.3691\ldots)}}\\[3mm]
=&\,-0.8436\ldots.
\end{align*} r = = = S tt S dd S t d 26.43 ( 3.3691 … ) − 7.9608 … − 0.8436 … . Therefore
r = − 0.844 \begin{align*}
r=-0.844
\end{align*} r = − 0.844 to 3 3 3
(e)
解法一
思路
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r r r − 1 -1 − 1
答题过程
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There is a strong negative correlation.
As the average daily temperature increases, the daily gas consumption in winter tends to decrease.
(f)
解法一
思路
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回归线 w w w t t t
b = S t w S t t . \begin{align*}
b=\frac{S_{tw}}{S_{tt}}.
\end{align*} b = S tt S tw . 截距用
a = w ˉ − b t ˉ . \begin{align*}
a=\bar{w}-b\bar{t}.
\end{align*} a = w ˉ − b t ˉ . 其中 w ˉ = ∑ w 12 \bar{w}=\dfrac{\sum w}{12} w ˉ = 12 ∑ w t ˉ = ∑ t 12 \bar{t}=\dfrac{\sum t}{12} t ˉ = 12 ∑ t
答题过程
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The gradient is
b = S t w S t t = − 91.55 26.43 = − 3.464 … . \begin{align*}
b
=&\,\frac{S_{tw}}{S_{tt}}\\[3mm]
=&\,\frac{-91.55}{26.43}\\[3mm]
=&\,-3.464\ldots.
\end{align*} b = = = S tt S tw 26.43 − 91.55 − 3.464 … . Also,
w ˉ = 339.25 12 , t ˉ = 9.1 12 . \begin{align*}
\bar{w}=\frac{339.25}{12},\qquad \bar{t}=\frac{9.1}{12}.
\end{align*} w ˉ = 12 339.25 , t ˉ = 12 9.1 . So the intercept is
a = w ˉ − b t ˉ = 339.25 12 − ( − 3.464 … ) 9.1 12 = 30.897 … . \begin{align*}
a
=&\,\bar{w}-b\bar{t}\\[3mm]
=&\,\frac{339.25}{12}
-(-3.464\ldots)\frac{9.1}{12}\\[3mm]
=&\,30.897\ldots.
\end{align*} a = = = w ˉ − b t ˉ 12 339.25 − ( − 3.464 … ) 12 9.1 30.897 … . Therefore the regression line is
w = − 3.464 … t + 30.897 … = − 3.46 t + 30.9 \begin{align*}
w=&\,-3.464\ldots t+30.897\ldots\\[3mm]
=&\,-3.46t+30.9
\end{align*} w = = − 3.464 … t + 30.897 … − 3.46 t + 30.9 to 3 3 3
(g)
解法一
思路
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因为 w = 11.5 d w=11.5d w = 11.5 d d = w 11.5 d=\dfrac{w}{11.5} d = 11.5 w 11.5 11.5 11.5
答题过程
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Since
w = 11.5 d , \begin{align*}
w=11.5d,
\end{align*} w = 11.5 d , we divide the regression equation by 11.5 11.5 11.5
d = − 3.46 t + 30.9 11.5 = − 0.301 t + 2.69. \begin{align*}
d
=&\,\frac{-3.46t+30.9}{11.5}\\[3mm]
=&\,-0.301t+2.69.
\end{align*} d = = 11.5 − 3.46 t + 30.9 − 0.301 t + 2.69. An equation of the regression line of d d d t t t
d = − 0.301 t + 2.69. \begin{align*}
d=-0.301t+2.69.
\end{align*} d = − 0.301 t + 2.69.
解法二
思路
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公式基本计算法。不通过 w w w t t t S t d S_{td} S t d
b d = S t d S t t , a d = d ˉ − b d t ˉ \begin{align*}
b_d = \frac{S_{td}}{S_{tt}},\qquad a_d = \bar{d} - b_d\bar{t}
\end{align*} b d = S tt S t d , a d = d ˉ − b d t ˉ 重新从头计算回归系数的斜率与截距，并验证结果。
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Calculate the gradient b d b_d b d d d d t t t
b d = S t d S t t = − 7.9608 … 26.43 = − 0.3012 … \begin{align*}
b_d =&\,\, \frac{S_{td}}{S_{tt}}\\[3mm]
=&\,\, \frac{-7.9608\ldots}{26.43}\\[3mm]
=&\,\, -0.3012\ldots
\end{align*} b d = = = S tt S t d 26.43 − 7.9608 … − 0.3012 … Calculate the sample means of d d d t t t
d ˉ = w ˉ 11.5 = 339.25 / 12 11.5 ≈ 2.4583 … \begin{align*}
\bar{d} = \frac{\bar{w}}{11.5} = \frac{339.25/12}{11.5} \approx 2.4583\ldots
\end{align*} d ˉ = 11.5 w ˉ = 11.5 339.25/12 ≈ 2.4583 … t ˉ = 9.1 12 ≈ 0.7583 … \begin{align*}
\bar{t} = \frac{9.1}{12} \approx 0.7583\ldots
\end{align*} t ˉ = 12 9.1 ≈ 0.7583 … Calculate the intercept a d a_d a d
a d = d ˉ − b d t ˉ = 2.4583 … − ( − 0.3012 … ) ( 0.7583 … ) = 2.4583 … + 0.2284 … = 2.6867 … \begin{align*}
a_d =&\,\, \bar{d} - b_d\bar{t}\\[3mm]
=&\,\, 2.4583\ldots - (-0.3012\ldots)(0.7583\ldots)\\[3mm]
=&\,\, 2.4583\ldots + 0.2284\ldots\\[3mm]
=&\,\, 2.6867\ldots
\end{align*} a d = = = = d ˉ − b d t ˉ 2.4583 … − ( − 0.3012 … ) ( 0.7583 … ) 2.4583 … + 0.2284 … 2.6867 … So the regression line equation is:
d = − 0.3012 … t + 2.6867 … = − 0.301 t + 2.69 \begin{align*}
d =&\,\, -0.3012\ldots t + 2.6867\ldots\\[3mm]
=&\,\, -0.301t + 2.69
\end{align*} d = = − 0.3012 … t + 2.6867 … − 0.301 t + 2.69 to 3 3 3
(h)(i)
解法一
思路
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把 t = 2 t=2 t = 2 m 3 \text{m}^3 m 3
答题过程
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Using
d = − 0.301 t + 2.69 , \begin{align*}
d=-0.301t+2.69,
\end{align*} d = − 0.301 t + 2.69 , when t = 2 t=2 t = 2
d = − 0.301 ( 2 ) + 2.69 = 2.088 … . \begin{align*}
d
=&\,-0.301(2)+2.69\\[3mm]
=&\,2.088\ldots.
\end{align*} d = = − 0.301 ( 2 ) + 2.69 2.088 … . The estimated daily gas consumption is
2.09 m 3 . \begin{align*}
2.09\text{ m}^3.
\end{align*} 2.09 m 3 .
(h)(ii)
解法一
思路
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回归线斜率约为 − 0.301 -0.301 − 0.301 1 ∘ C 1\,^\circ\text{C} 1 ∘ C 0.301 m 3 0.301\text{ m}^3 0.301 m 3
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For each increase of 1 ∘ C 1\,^\circ\text{C} 1 ∘ C
0.301 m 3 . \begin{align*}
0.301\text{ m}^3.
\end{align*} 0.301 m 3 . 