题目
Problem
Two economics students, Andi and Behrouz, are studying some data relating to unemployment, x % x\% x % y % y\% y % y y y x x x
y = 3.684 − 0.3242 x y=3.684-0.3242x y = 3.684 − 0.3242 x and
∑ y = 23.7 ∑ y 2 = 42.63 ∑ x 2 = 756.81 n = 16 \sum y=23.7\qquad \sum y^2=42.63\qquad \sum x^2=756.81\qquad n=16 ∑ y = 23.7 ∑ y 2 = 42.63 ∑ x 2 = 756.81 n = 16 (a) Show that S y y = 7.524375 S_{yy}=7.524375 S y y = 7.524375
(1)
(b) Find S x x S_{xx} S xx
(4)
(c) Find the product moment correlation coefficient between x x x y y y
(3)
Behrouz claims that, assuming the model is valid, the data show that when unemployment is 2% wages increase at over 3%
(d) Explain how Behrouz could have come to this conclusion.
(1)
Andi uses the formula
range = mean ± 3 × standard deviation \text{range}=\text{mean}\pm3\times\text{standard deviation} range = mean ± 3 × standard deviation to estimate the range of values for x x x
(e) Find estimates of the minimum value and the maximum value of x x x
(3)
(f) Comment, giving a reason, on the reliability of Behrouz’s claim.
(2)
Andi suggests using the regression line with equation y = 3.684 − 0.3242 x y=3.684-0.3242x y = 3.684 − 0.3242 x
(g) Comment, giving a reason, on Andi’s suggestion.
(2)
解答
(a)
解法一
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这是 show that，所以要写出 S y y = ∑ y 2 − ( ∑ y ) 2 n S_{yy}=\sum y^2-\frac{(\sum y)^2}{n} S y y = ∑ y 2 − n ( ∑ y ) 2
答题过程
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S y y = ∑ y 2 − ( ∑ y ) 2 n = 42.63 − 23.7 2 16 = 42.63 − 561.69 16 = 7.524375. \begin{aligned}
S_{yy}
=&\,\sum y^2-\frac{(\sum y)^2}{n}\\
=&\,42.63-\frac{23.7^2}{16}\\
=&\,42.63-\frac{561.69}{16}\\
=&\,7.524375.
\end{aligned} S y y = = = = ∑ y 2 − n ( ∑ y ) 2 42.63 − 16 23. 7 2 42.63 − 16 561.69 7.524375. Hence S y y = 7.524375 S_{yy}=7.524375 S y y = 7.524375
(b)
解法一
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回归线 y = a + b x y=a+bx y = a + b x ( x ˉ , y ˉ ) (\bar x,\bar y) ( x ˉ , y ˉ ) y ˉ \bar y y ˉ x ˉ \bar x x ˉ ∑ x \sum x ∑ x S x x S_{xx} S xx
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First,
y ˉ = 23.7 16 = 1.48125. \begin{align*}
\bar y=\frac{23.7}{16}=1.48125.
\end{align*} y ˉ = 16 23.7 = 1.48125. The regression line passes through ( x ˉ , y ˉ ) (\bar x,\bar y) ( x ˉ , y ˉ )
1.48125 = 3.684 − 0.3242 x ˉ . \begin{align*}
1.48125=3.684-0.3242\bar x.
\end{align*} 1.48125 = 3.684 − 0.3242 x ˉ . Thus
0.3242 x ˉ = 3.684 − 1.48125 = 2.20275 , \begin{align*}
0.3242\bar x=3.684-1.48125=2.20275,
\end{align*} 0.3242 x ˉ = 3.684 − 1.48125 = 2.20275 , and
x ˉ = 2.20275 0.3242 = 6.7944 … . \begin{align*}
\bar x=\frac{2.20275}{0.3242}=6.7944\ldots.
\end{align*} x ˉ = 0.3242 2.20275 = 6.7944 … . So
∑ x = 16 ( 6.7944 … ) = 108.7106 … . \begin{align*}
\sum x=16(6.7944\ldots)=108.7106\ldots.
\end{align*} ∑ x = 16 ( 6.7944 … ) = 108.7106 … . Now
S x x = ∑ x 2 − ( ∑ x ) 2 n = 756.81 − ( 108.7106 … ) 2 16 = 18.1843 … . \begin{aligned}
S_{xx}
=&\,\sum x^2-\frac{(\sum x)^2}{n}\\
=&\,756.81-\frac{(108.7106\ldots)^2}{16}\\
=&\,18.1843\ldots.
\end{aligned} S xx = = = ∑ x 2 − n ( ∑ x ) 2 756.81 − 16 ( 108.7106 … ) 2 18.1843 … . Therefore
S x x = 18.2 to 3 significant figures . \begin{align*}
S_{xx}=18.2\quad\text{to 3 significant figures}.
\end{align*} S xx = 18.2 to 3 significant figures .
(c)
解法一
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对于 y y y x x x b = S x y S x x b=\frac{S_{xy}}{S_{xx}} b = S xx S x y S x y S_{xy} S x y
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For the regression line of y y y x x x
b = S x y S x x . \begin{align*}
b=\frac{S_{xy}}{S_{xx}}.
\end{align*} b = S xx S x y . Here b = − 0.3242 b=-0.3242 b = − 0.3242
S x y = ( − 0.3242 ) ( 18.1843 … ) = − 5.8953 … . \begin{align*}
S_{xy}=(-0.3242)(18.1843\ldots)=-5.8953\ldots.
\end{align*} S x y = ( − 0.3242 ) ( 18.1843 … ) = − 5.8953 … . The product moment correlation coefficient is
r = S x y S x x S y y . \begin{align*}
r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}.
\end{align*} r = S xx S y y S x y . Thus
r = − 5.8953 … ( 18.1843 … ) ( 7.524375 ) = − 0.5039 … . \begin{aligned}
r
=&\,\frac{-5.8953\ldots}
{\sqrt{(18.1843\ldots)(7.524375)}}\\
=&\,-0.5039\ldots.
\end{aligned} r = = ( 18.1843 … ) ( 7.524375 ) − 5.8953 … − 0.5039 … . Therefore
r = − 0.504 to 3 significant figures . \begin{align*}
r=-0.504\quad\text{to 3 significant figures}.
\end{align*} r = − 0.504 to 3 significant figures .
(d)
解法一
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把 x = 2 x=2 x = 2 y y y
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Substituting x = 2 x=2 x = 2
y = 3.684 − 0.3242 ( 2 ) = 3.0356. \begin{align*}
y=3.684-0.3242(2)=3.0356.
\end{align*} y = 3.684 − 0.3242 ( 2 ) = 3.0356. Since 3.0356 > 3 3.0356>3 3.0356 > 3 3 % 3\% 3%
(e)
解法一
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先求 x x x S x x / n \sqrt{S_{xx}/n} S xx / n
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The mean of x x x
x ˉ = 6.7944 … . \begin{align*}
\bar x=6.7944\ldots.
\end{align*} x ˉ = 6.7944 … . The standard deviation of x x x
S x x n = 18.1843 … 16 = 1.0660 … . \sqrt{\frac{S_{xx}}{n}}
=\sqrt{\frac{18.1843\ldots}{16}}
=1.0660\ldots. n S xx = 16 18.1843 … = 1.0660 … . Using Andi’s formula,
estimated range = 6.7944 … ± 3 ( 1.0660 … ) = 6.7944 … ± 3.1981 … . \begin{aligned}
\text{estimated range}
=&\,6.7944\ldots\pm3(1.0660\ldots)\\
=&\,6.7944\ldots\pm3.1981\ldots.
\end{aligned} estimated range = = 6.7944 … ± 3 ( 1.0660 … ) 6.7944 … ± 3.1981 … . Therefore the estimated minimum and maximum values are
3.6 and 10.0. \begin{align*}
3.6\quad\text{and}\quad10.0.
\end{align*} 3.6 and 10.0.
(f)
解法一
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根据 (e)，x = 2 x=2 x = 2
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The value x = 2 x=2 x = 2 3.6 ≤ x ≤ 10.0 3.6\leq x\leq10.0 3.6 ≤ x ≤ 10.0
Therefore Behrouz’s claim is based on extrapolation, so it is unreliable.
(g)
解法一
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题目给的是 y y y x x x x x x y y y y y y x x x x x x y y y
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Andi should use the regression line of x x x y y y
The given regression line is the regression line of y y y x x x
