题目
Problem
The heights above sea level (h h h t ∘ C t^\circ C t ∘ C
∑ h = 112 ∑ t = 136 ∑ t 2 = 1828 S h t = − 236 S h h = 297 \sum h=112\qquad \sum t=136\qquad \sum t^2=1828\qquad S_{ht}=-236\qquad S_{hh}=297 ∑ h = 112 ∑ t = 136 ∑ t 2 = 1828 S h t = − 236 S hh = 297 (a) Find the value of S t t S_{tt} S tt
(2)
(b) Calculate the product moment correlation coefficient for these data.
(2)
(c) Interpret the relationship between t t t h h h
(1)
(d) Find an equation of the regression line of t t t h h h
(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)
中文翻译
在 7 月 31 日上午 7 时，随机记录了法国 12 个地点的海拔高度（h h h t ∘ C t^\circ C t ∘ C
(a) 求 S t t S_{tt} S tt
(b) 计算这些数据的积矩相关系数。
(c) 解释 t t t h h h
(d) 求 t t t h h h
同一时刻，Yinka 正在南非度假。他用该回归方程估计海拔 500 m 处的气温。
(e) 求 Yinka 算得的估计气温。
(f) 评论 (e) 中答案的有效性。
解答
(a)
思路
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利用 S t t = ∑ t 2 − ( ∑ t ) 2 n S_{tt}=\sum t^2-\dfrac{(\sum t)^2}{n} S tt = ∑ t 2 − n ( ∑ t ) 2
答题过程
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S t t = 1828 − 136 2 12 = 286.667 … S_{tt}=1828-\frac{136^2}{12}=286.667\ldots S tt = 1828 − 12 13 6 2 = 286.667 … S t t = 287 (3 s.f.) . S_{tt}=287\quad\text{(3 s.f.)}. S tt = 287 (3 s.f.) .
(b)
思路
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把 S h t S_{ht} S h t S h h S_{hh} S hh S t t S_{tt} S tt
答题过程
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r = S h t S h h S t t = − 236 297 ( 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} r = S hh S tt S h t = 297 ( 286.667 … ) − 236 = − 0.809 … r = − 0.809. r=-0.809. r = − 0.809.
(c)
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 height above sea level increases, temperature tends to decrease.
(d)
思路
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回归直线写作 t = a + b h t=a+bh t = a + bh b = S h t S h h b=\dfrac{S_{ht}}{S_{hh}} b = S hh S h t ( h ˉ , t ˉ ) (\bar h,\bar t) ( h ˉ , t ˉ ) a a a
答题过程
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b = S h t S h h = − 236 297 = − 0.7946 … b=\frac{S_{ht}}{S_{hh}}=\frac{-236}{297}=-0.7946\ldots b = S hh S h t = 297 − 236 = − 0.7946 … h ˉ = 112 12 = 9.333 … , t ˉ = 136 12 = 11.333 … \bar h=\frac{112}{12}=9.333\ldots,\qquad
\bar t=\frac{136}{12}=11.333\ldots h ˉ = 12 112 = 9.333 … , t ˉ = 12 136 = 11.333 … a = t ˉ − b h ˉ = 11.333 … − ( − 0.7946 … ) ( 9.333 … ) = 18.7497 … \begin{aligned}
a&=\bar t-b\bar h\\
&=11.333\ldots-(-0.7946\ldots)(9.333\ldots)\\
&=18.7497\ldots
\end{aligned} a = t ˉ − b h ˉ = 11.333 … − ( − 0.7946 … ) ( 9.333 … ) = 18.7497 … t = 18.7 − 0.795 h t=18.7-0.795h t = 18.7 − 0.795 h (or an equivalent equation using more accurate coefficients).
(e)
思路
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h h h h = 5 h=5 h = 5
答题过程
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t = 18.7497 … − 0.7946 … ( 5 ) = 14.776 … \begin{aligned}
t&=18.7497\ldots-0.7946\ldots(5)\\
&=14.776\ldots
\end{aligned} t = 18.7497 … − 0.7946 … ( 5 ) = 14.776 … 14.8 ∘ C . 14.8^\circ C. 14. 8 ∘ C .
(f)
思路
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判断资料的地区与估计对象是否相符。
答题过程
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The estimate is not very valid because the regression line is based on places in France, whereas Yinka is in South Africa.
