Question

Refer to Figure 13 of this section. The least-squares linefor these data is the line y=mxthat fits the databest, in that the sum of the squares of the vertical distancesbetween the line and the data points is minimal.We want to minimize the sum

${\mathbf{(}\mathbf{m}{\mathbf{x}}_{\mathbf{1}}\mathbf{-}{\mathbf{y}}_{\mathbf{1}}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathbf{m}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}{\mathbf{y}}_{\mathbf{2}}\mathbf{)}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{(}\mathbf{m}{\mathbf{x}}_{\mathbf{5}}\mathbf{-}{\mathbf{y}}_{\mathbf{5}}\mathbf{)}}^{\mathbf{2}}$

In vector notation, to minimize the sum means to find the scalar msuch that${||m\stackrel{r}{x}-\stackrel{r}{y}||}^{\mathbf{2}}$ is minimal. Arguing geometrically, explain how you can find m. Use the accompanying sketch, which is not drawn to scale.

Find mnumerically, and explain the relationship between mand the correlation coefficient r. You may find the following information helpful:

$\stackrel{\mathbf{\perp }}{\mathbf{x}}\mathbf{.}\stackrel{\mathbf{\perp }}{\mathbf{y}}\mathbf{=}\mathbf{4182}\mathbf{.}\mathbf{9}\mathbf{,}||\stackrel{\perp }{x}||\mathbf{\approx }\mathbf{198}\mathbf{.}\mathbf{53}\mathbf{,}||\stackrel{\perp }{y}||\mathbf{\approx }\mathbf{21}\mathbf{.}\mathbf{539}$

To check whether your solution mis reasonable, draw the line y=mxin Figure 13. (A more thorough discussion of least-squares approximations will follow in Section 5.4.)

Short answer

The value of$m=0.11$ , the value of r is very close to 1. So, for the line$y=0.11x$ , all the point lies almost on the line.

Step-by-step solution

Find the value of m

The objective is to find m such that${||m\stackrel{r}{x}-\stackrel{r}{y}||}^2$ is a minimum.

Consider the following expression.

$$\begin{gathered} {||m\stackrel{r}{x}-\stackrel{r}{y}||}^2=\left(m\stackrel{r}{x}-\stackrel{r}{y}\right).m\stackrel{r}{x}-\stackrel{r}{y} \\ =m^2\left(\stackrel{r}{x}.\stackrel{r}{x}\right)-2m\left(\stackrel{r}{x}.\stackrel{r}{y}\right)+\left(\stackrel{r}{y}.\stackrel{r}{y}\right) \\ =m^2\left(||\stackrel{r}{x}||.||\stackrel{r}{x}||\cos \left(0\right)\right)-2m\left(\stackrel{r}{x}.\stackrel{r}{y}\right)+\left(||\stackrel{r}{y}||.||\stackrel{r}{y}||\cos \left(0\right)\right),\left(\theta =0\right) \\ =m^2{||\stackrel{r}{x}||}^2-2m\left(\stackrel{r}{x}.\stackrel{r}{y}\right)+{||\stackrel{r}{y}||}^2 \end{gathered}$$

Substitute, $\stackrel{\perp }{\mathrm{x}}.\stackrel{\perp }{\mathrm{y}}=4182.9,||\stackrel{\perp }{\mathrm{x}}||\approx 198.53,||\stackrel{\perp }{\mathrm{y}}||\approx 21.539$in the above equation we get

$$\begin{gathered} {||m\stackrel{r}{x}-\stackrel{r}{y}||}^2=m^2{||\stackrel{r}{x}||}^2-2m\left(\stackrel{r}{x}.\stackrel{r}{y}\right)+{||\stackrel{r}{y}||}^2 \\ \Rightarrow m^2{\left(198.53\right)}^2-2m\left(4128.9\right)+{\left(21.539\right)}^2 \\ \Rightarrow 3941.1609m^2-8365.8m+463.928521 \end{gathered}$$

Now put the last equation equal to zero we get

$$\begin{gathered} {\left(3941.1609m^2-8365.8m+463.928521\right)}^{\text{'}}=0 \\ 2\left(39414.1609\right)m^{2-1}-1\left(8365.8\right)m^{1-1}+0=0 \\ 78828.321m-8365.8=0 \\ m\approx 0.11 \end{gathered}$$

Consider the diagram below

Consider the graph of the line $y=0.11x$below

Hence, the solution$m\approx 0.11$ is reasonable, from the graph$y=mx$.

Find r

Find the value of correlation coefficient r by using the formula $r=\cos \theta =\frac{\overrightarrow{x}.\overrightarrow{y}}{||\overrightarrow{x}||.||\overrightarrow{y}||}$, where $\overrightarrow{x}.\overrightarrow{y}=4182.9,||\overrightarrow{x}||=198.53and||\overrightarrow{y}||=21.539$.

$$\begin{gathered} r=\cos \theta =\frac{4182.9}{198.53.21.539} \\ \approx 0.9782 \end{gathered}$$

Which is very close to 1. So, for the line y=0.11x , all the point lie almost on the line.