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{{Template:LDABOOK_SUB|Appendix A|Least Squares/Rank Regression Equations}}
{{Template:LDABOOK|Appendix A|Least Squares/Rank Regression Equations}}


==Rank Regression on Y==
==Rank Regression on Y==
Assume that a set of data pairs (x1, y1), (x2, y2), ... , (xN, yN), were obtained and plotted. Then, according to the least squares principle, which minimizes the vertical distance between the data points and the straight line fitted to the data, the best fitting straight line to these data is the straight line y = + x such that:
Assume that a set of data pairs (x1, y1), (x2, y2), ... , (xN, yN), were obtained and plotted. Then, according to the least squares principle, which minimizes the vertical distance between the data points and the straight line fitted to the data, the best fitting straight line to these data is the straight line y = a + bx such that:


::<math>\sum_{i=1}^N (\hat{a}+\hat{b} x_i - y_i)^2=min(a,b)\sum_{i=1}^N (a+b x_i-y_i)^2 </math>
::<math>\sum_{i=1}^N (\hat{a}+\hat{b} x_i - y_i)^2=min(a,b)\sum_{i=1}^N (a+b x_i-y_i)^2 \,\!</math>


and where and are the least squares estimates of ''a'' and ''b'', and ''N'' is the number of data points.
and where <math>\hat{a}\,\!</math> and <math>\hat{b}\,\!</math> are the least squares estimates of ''a'' and ''b'', and ''N'' is the number of data points.


To obtain <math>\hat{a}</math> and <math>\hat{b}</math>, let:
To obtain <math>\hat{a}\,\!</math> and <math>\hat{b}\,\!</math>, let:


::<math>F=\sum_{i=1}^N (a+bx_i-y_i)^2 </math>
::<math>F=\sum_{i=1}^N (a+bx_i-y_i)^2 \,\!</math>


Differentiating F with respect to a and b yields:
Differentiating F with respect to a and b yields:


::<math>\frac{\partial F}{\partial a}=2\sum_{i=1}^N (a+b x_i-y_i) </math> (1)
::<math>\frac{\partial F}{\partial a}=2\sum_{i=1}^N (a+b x_i-y_i) \,\!</math> (1)


:and:
:and:


::<math>\frac{\partial F}{\partial b}=2\sum_{i=1}^N (a+b x_i-y_i)x_i </math> (2)
::<math>\frac{\partial F}{\partial b}=2\sum_{i=1}^N (a+b x_i-y_i)x_i \,\!</math> (2)


Setting Eqns. (1) and (2) equal to zero yields:
Setting Eqns. (1) and (2) equal to zero yields:


::<math>\sum_{i=1}^N (a+b x_i-y_i)x_i=\sum_{i=1}^N(\hat{y}_i-y_i)=-\sum_{i=1}^N(y_i-\hat{y}_i)=0 </math>
::<math>\sum_{i=1}^N (a+b x_i-y_i)=\sum_{i=1}^N(\hat{y}_i-y_i)=-\sum_{i=1}^N(y_i-\hat{y}_i)=0 \,\!</math>


:and:
:and:


::<math>\sum_{i=1}^N (a+b x_i-y_i)x_i=\sum_{i=1}^N(\hat{y}_i-y_i)x_i=-\sum_{i=1}^N(y_i-\hat{y}_i)x_i =0</math>
::<math>\sum_{i=1}^N (a+b x_i-y_i)x_i=\sum_{i=1}^N(\hat{y}_i-y_i)x_i=-\sum_{i=1}^N(y_i-\hat{y}_i)x_i =0\,\!</math>


Solving the equations simultaneously yields:
Solving the equations simultaneously yields:


::<math>\hat{a}=\frac{\displaystyle \sum_{i=1}^N y_i}{N}-\hat{b}\frac{\displaystyle \sum_{i=1}{N} x_i}{N}=\bar{y}-\hat{b}\bar{x} </math> (3)
::<math>\hat{a}=\frac{\displaystyle \sum_{i=1}^N y_i}{N}-\hat{b}\frac{\displaystyle \sum_{i=1}^N x_i}{N}=\bar{y}-\hat{b}\bar{x} \,\!</math> (3)


:and:
:and:


::<math>\hat{b}=\frac{\displaystyle \sum{i=1}^N x_i y_i-\frac{\displaystyle \sum_{i=1}^N x_i \sum_{i=1}^N y_i}{N}}{\displaystyle \sum_{i=1}^N x_i^2-\frac{\left(\displaystyle\sum_{i=1}^N x_i\right)^2}{N}}</math>(4)
::<math>\hat{b}=\frac{\displaystyle \sum_{i=1}^N x_i y_i-\frac{\displaystyle \sum_{i=1}^N x_i \sum_{i=1}^N y_i}{N}}{\displaystyle \sum_{i=1}^N x_i^2-\frac{\left(\displaystyle\sum_{i=1}^N x_i\right)^2}{N}}\,\!</math>(4)




==Rank Regression on X==
==Rank Regression on X==
Assume that a set of data pairs (x1, y1), (x2, y2), ... , (xN, yN) were obtained and plotted. Then, according to the least squares principle, which minimizes the horizontal distance between the data points and the straight line fitted to the data, the best fitting straight line to these data is the straight line x = + y such that:
Assume that a set of data pairs (x1, y1), (x2, y2), ... , (xN, yN) were obtained and plotted. Then, according to the least squares principle, which minimizes the horizontal distance between the data points and the straight line fitted to the data, the best fitting straight line to these data is the straight line x = a + by such that:


::<math>\displaystyle\sum_{i=1}^N(\hat{a}+\hat{b}y_i-x_i)^2=min(a,b)\displaystyle\sum_{i=1}^N (a+by_i-x_i)^2 </math>
::<math>\displaystyle\sum_{i=1}^N(\hat{a}+\hat{b}y_i-x_i)^2=min(a,b)\displaystyle\sum_{i=1}^N (a+by_i-x_i)^2 \,\!</math>


Again, <math>\hat{a}</math> and <math>\hat{b}</math> are the least squares estimates of a and b, and N is the number of data points.
Again, <math>\hat{a}\,\!</math> and <math>\hat{b}\,\!</math> are the least squares estimates of a and b, and N is the number of data points.


To obtain <math>\hat{a}</math> and <math>\hat{b}</math>, let:
To obtain <math>\hat{a}\,\!</math> and <math>\hat{b}\,\!</math>, let:


::<math>F=\displaystyle\sum_{i=1}^N(a+by_i-x_i)^2</math>
::<math>F=\displaystyle\sum_{i=1}^N(a+by_i-x_i)^2\,\!</math>


Differentiating F with respect to a and b yields:
Differentiating F with respect to a and b yields:


::<math>\frac{\partial F}{\partial a}=2\displaystyle\sum_{i=1}^N(a+by_i-x_i</math> (5)
::<math>\frac{\partial F}{\partial a}=2\displaystyle\sum_{i=1}^N(a+by_i-x_i)\,\!</math> (5)


:and:
:and:


::<math>\frac{\partial F}{\partial b}=2\displaystyle\sum_{i=1}^N(a+by_i-x_i)y_i</math>(6)
::<math>\frac{\partial F}{\partial b}=2\displaystyle\sum_{i=1}^N(a+by_i-x_i)y_i\,\!</math>(6)


Setting Eqns. (5) and (6) equal to zero yields:
Setting Eqns. (5) and (6) equal to zero yields:


::<math>\displaystyle\sum_{i=1}^N(a+by_i-x_i)=\displaystyle\sum_{i=1}^N(\widehat{x}_i-x_i)=-\displaystyle\sum_{i=1}^N(x_i-\widehat{x}_i)=0</math>
::<math>\displaystyle\sum_{i=1}^N(a+by_i-x_i)=\displaystyle\sum_{i=1}^N(\widehat{x}_i-x_i)=-\displaystyle\sum_{i=1}^N(x_i-\widehat{x}_i)=0\,\!</math>


:and:
:and:


::<math>\displaystyle\sum_{i=1}^N(a+by_i-x_i)y_i=\displaystyle\sum_{i=1}^N(\widehat{x}_i-x_i)y_i=-\displaystyle\sum_{i=1}^N(x_i-\widehat{x}_i)y_i=0</math>
::<math>\displaystyle\sum_{i=1}^N(a+by_i-x_i)y_i=\displaystyle\sum_{i=1}^N(\widehat{x}_i-x_i)y_i=-\displaystyle\sum_{i=1}^N(x_i-\widehat{x}_i)y_i=0\,\!</math>


Solving the above equations simultaneously yields:
Solving the above equations simultaneously yields:


::<math>\widehat{a}=\frac{\displaystyle\sum_{i=1}^N x_i}{N}-\widehat{b}\frac{\displaystyle\sum_{i=1}^N y_i}{N}=\bar{x}-\widehat{b}\bar{y}</math>(7)
::<math>\widehat{a}=\frac{\displaystyle\sum_{i=1}^N x_i}{N}-\widehat{b}\frac{\displaystyle\sum_{i=1}^N y_i}{N}=\bar{x}-\widehat{b}\bar{y}\,\!</math>(7)


:and:
:and:


::<math>\widehat{b}=\frac{\displaystyle\sum_{i=1}^N x_iy_i-\frac{\displaystyle\sum_{i=1}^N x_i\displaystyle\sum_{i=1}^N y_i}{N}}{\displaystyle\sum_{i=1}^N y_i^2-\frac{\left(\displaystyle\sum_{i=1}^N y_i\right)^2}{N}}</math>(8)
::<math>\widehat{b}=\frac{\displaystyle\sum_{i=1}^N x_iy_i-\frac{\displaystyle\sum_{i=1}^N x_i\displaystyle\sum_{i=1}^N y_i}{N}}{\displaystyle\sum_{i=1}^N y_i^2-\frac{\left(\displaystyle\sum_{i=1}^N y_i\right)^2}{N}}\,\!</math>(8)


Solving the equation of the line for y yields:
Solving the equation of the line for y yields:


::<math>y=-\frac{\hat{a}}{\hat{b}}+\frac{1}{\hat{b}} x </math>
::<math>y=-\frac{\hat{a}}{\hat{b}}+\frac{1}{\hat{b}} x \,\!</math>


== Example  ==
== Example  ==


Fit a least squares straight line using regression on X and regression on Y to the following data:  
Fit a least squares straight line using regression on X and regression on Y to the following data:  
<br>


{| align="center" border="0" cellspacing="5" width="300" cellpadding="5"
{| align="center" border="0" cellspacing="5" width="300" cellpadding="5"
Line 104: Line 102:
| align="left" valign="middle" | 10
| align="left" valign="middle" | 10
|}
|}
<br>


The first step is to generate the following table:  
The first step is to generate the following table:  


<br>
{| border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
 
{| align="center" border="1" cellpadding="5" cellspacing="1" width="350"
|-
|-
! scope="col" | '''<math>i</math>'''  
! scope="col" | '''<math>i\,\!</math>'''  
! scope="col" | '''<math>x_i</math>'''  
! scope="col" | '''<math>x_i\,\!</math>'''  
! scope="col" | '''<math>y_i</math>'''  
! scope="col" | '''<math>y_i\,\!</math>'''  
! scope="col" | '''<math>x_i^2</math>'''  
! scope="col" | '''<math>x_i^2\,\!</math>'''  
! scope="col" | '''<math>x_iy_i</math>'''  
! scope="col" | '''<math>x_iy_i\,\!</math>'''  
! scope="col" | '''<math>y_i^2</math>'''
! scope="col" | '''<math>y_i^2\,\!</math>'''
|-
|-
! scope="row" | 1  
! scope="row" | 1  
Line 176: Line 170:
| align="center" |100
| align="center" |100
|-
|-
! scope="row" | <math>\Sigma</math>  
! scope="row" | <math>\Sigma\,\!</math>  
| align="center" |56.5  
| align="center" |56.5  
| align="center" |41.5  
| align="center" |41.5  
Line 183: Line 177:
| align="center" |276.25
| align="center" |276.25
|}
|}
<br>


From the table then, and for rank regression on Y (RRY):
From the table then, and for rank regression on Y (RRY):


::<math>\widehat{b}=\frac{387.5-(56.5)(41.5)/8}{550.25-(56.5)^2/8}</math>  
::<math>\widehat{b}=\frac{387.5-(56.5)(41.5)/8}{550.25-(56.5)^2/8}\,\!</math>  
::<math>\widehat{b}=0.6243</math>
::<math>\widehat{b}=0.6243\,\!</math>


:and:
:and:


::<math>\widehat{a}=\frac{41.5}{8}-0.6243\frac{56.5}{8}</math>  
::<math>\widehat{a}=\frac{41.5}{8}-0.6243\frac{56.5}{8}\,\!</math>  
::<math>\widehat{a}=0.77836</math>
::<math>\widehat{a}=0.77836\,\!</math>


The least squares line is given by:  
The least squares line is given by:  


::<math>y=0.77836+0.6243x</math>
::<math>\begin{align}
y=0.77836+0.6243x
\end{align}\,\!</math>


The plotted line is shown in the next figure.  
The plotted line is shown in the next figure.  


[[Image:LdaappendixA.1.gif|thumb|center|500px]]  
[[Image:LdaappendixA_1_new.png|center|500px]]  


For rank regression on X (RRX) using the same table yields:  
For rank regression on X (RRX) using the same table yields:  


::<math>\widehat{b}=\frac{387.5-(56.5)(41.5)/8}{276.25-(41.5)^2/8}</math>  
::<math>\widehat{b}=\frac{387.5-(56.5)(41.5)/8}{276.25-(41.5)^2/8}\,\!</math>  
::<math>\widehat{a}=-0.97002</math>
::<math>\widehat{b}=1.5484\,\!</math>


:and:
:and:


::<math>\widehat{a}=\frac{56.5}{8}-1.5484\frac{41.5}{8}</math>  
::<math>\widehat{a}=\frac{56.5}{8}-1.5484\frac{41.5}{8}\,\!</math>  
::\widehat{a}=-0.97002
::<math>\widehat{a}=-0.97002\,\!</math>


The least squares line is given by:  
The least squares line is given by:  


::<math>y=-\frac{(-0.97002)}{1.5484}+\frac{1}{1.5484}\cdot x</math>  
::<math>y=-\frac{(-0.97002)}{1.5484}+\frac{1}{1.5484}\cdot x\,\!</math>  
::<math>y=0.62645+0.64581\cdot x</math>
::<math>y=0.62645+0.64581\cdot x\,\!</math>


The plotted line is shown in the next figure.  
The plotted line is shown in the next figure.  


[[Image:LdaappendixA.2.gif|thumb|center|500px]]  
[[Image:LdaappendixA_2_new.png|thumb|center|500px]]  


Note that the regression on Y is not necessarily the same as the regression on X. The only time when the two regressions are the same (i.e. will yield the same equation for a line) is when the data lie perfectly on a line.  
Note that the regression on Y is not necessarily the same as the regression on X. The only time when the two regressions are the same (i.e., will yield the same equation for a line) is when the data lie perfectly on a line.  


The correlation coefficient is given by:  
The correlation coefficient is given by:  


::<math>\hat{\rho}=\frac{\displaystyle\sum_{i=1}^N x_iy_i-\frac{\displaystyle\sum_{i=1}^N x_i\displaystyle\sum_{i=1}^N y_i}{N}}{\sqrt{\left(\displaystyle\sum_{i=1}^N x_i^2-\frac{(\displaystyle\sum_{i=1}^N x_i)^2}{N}\right)\left(\displaystyle\sum_{i=1}^N y_i^2-\frac{(\displaystyle\sum_{i=1}^N y_i)^2}{N}\right)}}</math>
::<math>\hat{\rho}=\frac{\displaystyle\sum_{i=1}^N x_iy_i-\frac{\displaystyle\sum_{i=1}^N x_i\displaystyle\sum_{i=1}^N y_i}{N}}{\sqrt{\left(\displaystyle\sum_{i=1}^N x_i^2-\frac{(\displaystyle\sum_{i=1}^N x_i)^2}{N}\right)\left(\displaystyle\sum_{i=1}^N y_i^2-\frac{(\displaystyle\sum_{i=1}^N y_i)^2}{N}\right)}}\,\!</math>


::<math>\widehat{\rho}=\frac{387.5-(56.5)(41.5)/8}{[(550.25-(56.5)^2/8)(276.25-(41.5)^2/8)]^{\frac{1}{2}}}</math>
::<math>\widehat{\rho}=\frac{387.5-(56.5)(41.5)/8}{[(550.25-(56.5)^2/8)(276.25-(41.5)^2/8)]^{\frac{1}{2}}}\,\!</math>


::<math>\widehat{\rho}=0.98321</math>
::<math>\widehat{\rho}=0.98321\,\!</math>


<br>
<br>

Latest revision as of 19:15, 15 September 2023

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Chapter Appendix A: Least Squares/Rank Regression Equations


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Chapter Appendix A  
Least Squares/Rank Regression Equations  

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More Resources:
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Rank Regression on Y

Assume that a set of data pairs (x1, y1), (x2, y2), ... , (xN, yN), were obtained and plotted. Then, according to the least squares principle, which minimizes the vertical distance between the data points and the straight line fitted to the data, the best fitting straight line to these data is the straight line y = a + bx such that:

[math]\displaystyle{ \sum_{i=1}^N (\hat{a}+\hat{b} x_i - y_i)^2=min(a,b)\sum_{i=1}^N (a+b x_i-y_i)^2 \,\! }[/math]

and where [math]\displaystyle{ \hat{a}\,\! }[/math] and [math]\displaystyle{ \hat{b}\,\! }[/math] are the least squares estimates of a and b, and N is the number of data points.

To obtain [math]\displaystyle{ \hat{a}\,\! }[/math] and [math]\displaystyle{ \hat{b}\,\! }[/math], let:

[math]\displaystyle{ F=\sum_{i=1}^N (a+bx_i-y_i)^2 \,\! }[/math]

Differentiating F with respect to a and b yields:

[math]\displaystyle{ \frac{\partial F}{\partial a}=2\sum_{i=1}^N (a+b x_i-y_i) \,\! }[/math] (1)
and:
[math]\displaystyle{ \frac{\partial F}{\partial b}=2\sum_{i=1}^N (a+b x_i-y_i)x_i \,\! }[/math] (2)

Setting Eqns. (1) and (2) equal to zero yields:

[math]\displaystyle{ \sum_{i=1}^N (a+b x_i-y_i)=\sum_{i=1}^N(\hat{y}_i-y_i)=-\sum_{i=1}^N(y_i-\hat{y}_i)=0 \,\! }[/math]
and:
[math]\displaystyle{ \sum_{i=1}^N (a+b x_i-y_i)x_i=\sum_{i=1}^N(\hat{y}_i-y_i)x_i=-\sum_{i=1}^N(y_i-\hat{y}_i)x_i =0\,\! }[/math]

Solving the equations simultaneously yields:

[math]\displaystyle{ \hat{a}=\frac{\displaystyle \sum_{i=1}^N y_i}{N}-\hat{b}\frac{\displaystyle \sum_{i=1}^N x_i}{N}=\bar{y}-\hat{b}\bar{x} \,\! }[/math] (3)
and:
[math]\displaystyle{ \hat{b}=\frac{\displaystyle \sum_{i=1}^N x_i y_i-\frac{\displaystyle \sum_{i=1}^N x_i \sum_{i=1}^N y_i}{N}}{\displaystyle \sum_{i=1}^N x_i^2-\frac{\left(\displaystyle\sum_{i=1}^N x_i\right)^2}{N}}\,\! }[/math](4)


Rank Regression on X

Assume that a set of data pairs (x1, y1), (x2, y2), ... , (xN, yN) were obtained and plotted. Then, according to the least squares principle, which minimizes the horizontal distance between the data points and the straight line fitted to the data, the best fitting straight line to these data is the straight line x = a + by such that:

[math]\displaystyle{ \displaystyle\sum_{i=1}^N(\hat{a}+\hat{b}y_i-x_i)^2=min(a,b)\displaystyle\sum_{i=1}^N (a+by_i-x_i)^2 \,\! }[/math]

Again, [math]\displaystyle{ \hat{a}\,\! }[/math] and [math]\displaystyle{ \hat{b}\,\! }[/math] are the least squares estimates of a and b, and N is the number of data points.

To obtain [math]\displaystyle{ \hat{a}\,\! }[/math] and [math]\displaystyle{ \hat{b}\,\! }[/math], let:

[math]\displaystyle{ F=\displaystyle\sum_{i=1}^N(a+by_i-x_i)^2\,\! }[/math]

Differentiating F with respect to a and b yields:

[math]\displaystyle{ \frac{\partial F}{\partial a}=2\displaystyle\sum_{i=1}^N(a+by_i-x_i)\,\! }[/math] (5)
and:
[math]\displaystyle{ \frac{\partial F}{\partial b}=2\displaystyle\sum_{i=1}^N(a+by_i-x_i)y_i\,\! }[/math](6)

Setting Eqns. (5) and (6) equal to zero yields:

[math]\displaystyle{ \displaystyle\sum_{i=1}^N(a+by_i-x_i)=\displaystyle\sum_{i=1}^N(\widehat{x}_i-x_i)=-\displaystyle\sum_{i=1}^N(x_i-\widehat{x}_i)=0\,\! }[/math]
and:
[math]\displaystyle{ \displaystyle\sum_{i=1}^N(a+by_i-x_i)y_i=\displaystyle\sum_{i=1}^N(\widehat{x}_i-x_i)y_i=-\displaystyle\sum_{i=1}^N(x_i-\widehat{x}_i)y_i=0\,\! }[/math]

Solving the above equations simultaneously yields:

[math]\displaystyle{ \widehat{a}=\frac{\displaystyle\sum_{i=1}^N x_i}{N}-\widehat{b}\frac{\displaystyle\sum_{i=1}^N y_i}{N}=\bar{x}-\widehat{b}\bar{y}\,\! }[/math](7)
and:
[math]\displaystyle{ \widehat{b}=\frac{\displaystyle\sum_{i=1}^N x_iy_i-\frac{\displaystyle\sum_{i=1}^N x_i\displaystyle\sum_{i=1}^N y_i}{N}}{\displaystyle\sum_{i=1}^N y_i^2-\frac{\left(\displaystyle\sum_{i=1}^N y_i\right)^2}{N}}\,\! }[/math](8)

Solving the equation of the line for y yields:

[math]\displaystyle{ y=-\frac{\hat{a}}{\hat{b}}+\frac{1}{\hat{b}} x \,\! }[/math]

Example

Fit a least squares straight line using regression on X and regression on Y to the following data:

x 1 2.5 4 6 8 9 11 15
y 1.5 2 4 4 5 7 8 10

The first step is to generate the following table:

[math]\displaystyle{ i\,\! }[/math] [math]\displaystyle{ x_i\,\! }[/math] [math]\displaystyle{ y_i\,\! }[/math] [math]\displaystyle{ x_i^2\,\! }[/math] [math]\displaystyle{ x_iy_i\,\! }[/math] [math]\displaystyle{ y_i^2\,\! }[/math]
1 1 1.5 1 1.5 2.25
2 2.5 2 6.25 5 4
3 4 4 16 16 16
4 6 4 36 24 16
5 8 5 64 40 25
6 9 7 81 63 49
7 11 8 121 88 64
8 15 10 225 150 100
[math]\displaystyle{ \Sigma\,\! }[/math] 56.5 41.5 550.25 387.5 276.25

From the table then, and for rank regression on Y (RRY):

[math]\displaystyle{ \widehat{b}=\frac{387.5-(56.5)(41.5)/8}{550.25-(56.5)^2/8}\,\! }[/math]
[math]\displaystyle{ \widehat{b}=0.6243\,\! }[/math]
and:
[math]\displaystyle{ \widehat{a}=\frac{41.5}{8}-0.6243\frac{56.5}{8}\,\! }[/math]
[math]\displaystyle{ \widehat{a}=0.77836\,\! }[/math]

The least squares line is given by:

[math]\displaystyle{ \begin{align} y=0.77836+0.6243x \end{align}\,\! }[/math]

The plotted line is shown in the next figure.

LdaappendixA 1 new.png

For rank regression on X (RRX) using the same table yields:

[math]\displaystyle{ \widehat{b}=\frac{387.5-(56.5)(41.5)/8}{276.25-(41.5)^2/8}\,\! }[/math]
[math]\displaystyle{ \widehat{b}=1.5484\,\! }[/math]
and:
[math]\displaystyle{ \widehat{a}=\frac{56.5}{8}-1.5484\frac{41.5}{8}\,\! }[/math]
[math]\displaystyle{ \widehat{a}=-0.97002\,\! }[/math]

The least squares line is given by:

[math]\displaystyle{ y=-\frac{(-0.97002)}{1.5484}+\frac{1}{1.5484}\cdot x\,\! }[/math]
[math]\displaystyle{ y=0.62645+0.64581\cdot x\,\! }[/math]

The plotted line is shown in the next figure.

LdaappendixA 2 new.png

Note that the regression on Y is not necessarily the same as the regression on X. The only time when the two regressions are the same (i.e., will yield the same equation for a line) is when the data lie perfectly on a line.

The correlation coefficient is given by:

[math]\displaystyle{ \hat{\rho}=\frac{\displaystyle\sum_{i=1}^N x_iy_i-\frac{\displaystyle\sum_{i=1}^N x_i\displaystyle\sum_{i=1}^N y_i}{N}}{\sqrt{\left(\displaystyle\sum_{i=1}^N x_i^2-\frac{(\displaystyle\sum_{i=1}^N x_i)^2}{N}\right)\left(\displaystyle\sum_{i=1}^N y_i^2-\frac{(\displaystyle\sum_{i=1}^N y_i)^2}{N}\right)}}\,\! }[/math]
[math]\displaystyle{ \widehat{\rho}=\frac{387.5-(56.5)(41.5)/8}{[(550.25-(56.5)^2/8)(276.25-(41.5)^2/8)]^{\frac{1}{2}}}\,\! }[/math]
[math]\displaystyle{ \widehat{\rho}=0.98321\,\! }[/math]