# The Method of Least Squares 1

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Straight Lines and Linear Functions

 The Cartesian Coordinate System
 Straight Lines
 Linear Functions and Mathematical
Models
 Intersection of Straight Lines
 The Method of Least Squares
1.1
The Cartesian Coordinate System

y

P(x, y)
C(h, k)
k                 r

x
h
The Cartesian Coordinate System
 We can represent real numbers geometrically by points on
a real number, or coordinate, line:
 This line includes all real numbers.
 Exactly one point on the line is associated with each real
number, and vice-versa (one dimensional space).

Origin

Negative Direction                Positive Direction

–4    –3   –2    –1     0      1         2   3       4

 2                     3           p
The Cartesian Coordinate System
 The Cartesian coordinate system extends this concept to a
plane (two dimensional space) by adding a vertical axis.

4

3

2

1

–4   –3   –2   –1        1   2    3   4
–1

–2

–3

–4
The Cartesian Coordinate System
 The horizontal line is called the x-axis, and the vertical line is
called the y-axis.
y
4

3

2

1

x
–4   –3    –2   –1            1   2   3   4
–1

–2

–3

–4
The Cartesian Coordinate System
 The point where these two lines intersect is called the origin.

y
4

3

2

1   Origin
x
–4   –3   –2   –1            1    2   3   4
–1

–2

–3

–4
The Cartesian Coordinate System
 In the x-axis, positive numbers are to the right and negative
numbers are to the left of the origin.
y
4

3

2

Negative Direction    1           Positive Direction
x
–4   –3    –2   –1            1      2      3     4
–1

–2

–3

–4
The Cartesian Coordinate System
 In the y-axis, positive numbers are above and negative
numbers are below the origin.
y
4

Positive Direction
3

2

1

x
–4   –3   –2   –1                 1               2   3   4
–1
Negative Direction
–2

–3

–4
The Cartesian Coordinate System
 A point in the plane can now be represented uniquely in this
coordinate system by an ordered pair of numbers (x, y).
y
(– 2, 4)              4

3                          (4, 3)
2

1

x
–4     –3       –2   –1            1   2   3      4
–1
(3, –1)
(–1, – 2)        –2

–3

–4
The Cartesian Coordinate System
 The axes divide the plane into four quadrants as shown below.

y
4
(–, +)                            (+, +)
2

1

x
–4   –3   –2   –1            1     2   3   4
–1

(–, –)            –3              (+, –)
–4
The Distance Formula

 The distance between any two points in the plane can be
expressed in terms of the coordinates of the points.

Distance formula
 The distance d between two points P1(x1, y1)
and P2(x2, y2) in the plane is given by

d     x2  x1    y2  y1 
2                 2
Examples
 Find the distance between the points (– 4, 3) and (2, 6).
Solution
 Let P1(– 4, 3) and P2(2, 6) be points in the plane.
 We have
x1 = – 4       y1 = 3         x2 = 2         y2 = 6
 Using the distance formula, we have

d    x2  x1          y2  y1 
2                   2

    2  ( 4)   6  3
2          2

 62  32  45  3 5
Example 1, page 4
Examples
 Let P(x, y) denote a point lying on the circle with radius r
and center C(h, k). Find a relationship between x and y.
Solution
 By the definition of a circle, the distance between P(x, y)
and C(h, k) is r.
 With the distance formula               y
we get
P(x, y)
 x  h   y  k   r
2          2
C(h, k)
k             r

 Squaring both sides gives

 x  h   y  k   r2
2          2

x
h
Example 3, page 4
Equation of a Circle

 An equation of a circle with center C(h, k)
and radius r is given by

 x  h         y  k   r2
2            2
Examples
 Find an equation of the circle with radius 2 and center (–1, 3).
Solution
 We use the circle formula with r = 2, h = –1, and k = 3:

 x  h   y  k   r2
2                2
y

 x  ( 1)   y  3          22
2            2
(–1, 3)
3
 x  1         y  3  4
2            2
2

x
–1

Example 4, page 5
Examples
 Find an equation of the circle with radius 3 and center
located at the origin.
Solution
 We use the circle formula with r = 3, h = 0, and k = 0:
y
 x  h   y  k   r2
2         2

 x  0    y  0   32
2         2
3
x
x  y 9
2   2

Example 4, page 5
1.2
Straight Lines
Slope of a Vertical Line
 Let L denote the unique straight line that passes through
the two distinct points (x1, y1) and (x2, y2).
 If x1 = x2, then L is a vertical line, and the slope is
undefined.

y
L

(x1, y1)

(x2, y2)

x
Slope of a Nonvertical Line
 If (x1, y1) and (x2, y2) are two distinct points on a
nonvertical line L, then the slope m of L is given by
y y2  y1
m           
x x2  x1

y
L
(x2, y2)

y2 – y1 = y
(x1, y1)
x2 – x1 = x

x
Slope of a Nonvertical Line

 If m > 0, the line slants upward from left to right.

y

m=1                 L

y = 1

x = 1
x
Slope of a Nonvertical Line

 If m > 0, the line slants upward from left to right.

y
L
m=2

y = 2

x = 1
x
Slope of a Nonvertical Line

 If m < 0, the line slants downward from left to right.

y

m = –1
x = 1

y = –1

x
L
Slope of a Nonvertical Line

 If m < 0, the line slants downward from left to right.

y

m = –2
x = 1

y = –2

x

L
Examples

 Sketch the straight line that passes through the point
(2, 5) and has slope – 4/3.
Solution
1. Plot the point (2, 5).             y
6
2. A slope of – 4/3 means                          x = 3
5
that if x increases by 3,              (2, 5)
4
y decreases by 4.                                            y = – 4
3
3. Plot the resulting            2
point (5, 1).                 1                              (5, 1)
4. Draw a line through                                                       x
1    2   3   4    5       6
the two points.                                                       L
Examples

 Find the slope m of the line that goes through the points
(–1, 1) and (5, 3).
Solution
 Choose (x1, y1) to be (–1, 1) and (x2, y2) to be (5, 3).
 With x1 = –1, y1 = 1, x2 = 5, y2 = 3, we find

y2  y1   3 1    2 1
m                  
x2  x1 5  ( 1) 6 3

Example 2, page 11
Examples

 Find the slope m of the line that goes through the points
(–2, 5) and (3, 5).
Solution
 Choose (x1, y1) to be (–2, 5) and (x2, y2) to be (3, 5).
 With x1 = –2, y1 = 5, x2 = 3, y2 = 5, we find

y2  y1   55     0
m                  0
x2  x1 3  ( 2) 5

Example 3, page 11
Examples

 Find the slope m of the line that goes through the points
(–2, 5) and (3, 5).
Solution
 The slope of a horizontal line is zero:

y
6
(–2, 5)                        (3, 5)
L
4
3                        m=0
2
1
x
–2   –1           1   2   3   4
Example 3, page 11
Parallel Lines

 Two distinct lines are parallel if and only if their
slopes are equal or their slopes are undefined.
Example
 Let L1 be a line that passes through the points (–2, 9) and
(1, 3), and let L2 be the line that passes through the points
(– 4, 10) and (3, – 4).
 Determine whether L1 and L2 are parallel.
Solution
 The slope m1 of L1 is given by
39
m1             2
1  ( 2)
 The slope m2 of L2 is given by
4  10
m2             2
3  ( 4)
 Since m1 = m2, the lines L1 and L2 are in fact parallel.
Example 4, page 12
Equations of Lines

 Let L be a straight line
parallel to the y-axis.
y
 Then L crosses the x-axis at              L
some point (a, 0) , with the
(a, y )
x-coordinate given by x = a,
where a is a real number.
 Any other point on L has
the form (a, y), where y
is an appropriate number.                 (a, 0)
x
 The vertical line L can
therefore be described as
x=a
Equations of Lines
 Let L be a nonvertical line with a slope m.
 Let (x1, y1) be a fixed point lying on L, and let (x, y) be a
variable point on L distinct from (x1, y1).
 Using the slope formula by letting (x, y) = (x2, y2), we get
y  y1
m
x  x1
 Multiplying both sides by x – x1 we get

y  y1  m( x  x1 )
Point-Slope Form

 An equation of the line that has slope m and
passes through point (x1, y1) is given by
y  y1  m( x  x1 )
Examples

 Find an equation of the line that passes through the point
(1, 3) and has slope 2.
Solution
 Use the point-slope form
y  y1  m( x  x1 )
 Substituting for point (1, 3) and slope m = 2, we obtain

y  3  2( x  1)
 Simplifying we get
2x  y  1  0

Example 5, page 13
Examples
 Find an equation of the line that passes through the points
(–3, 2) and (4, –1).
Solution
 The slope is given by
y y     1  2     3
m 2 1                
x2  x1 4  (3)    7
 Substituting in the point-slope form for point (4, –1) and
slope m = – 3/7, we obtain
3
y  1   ( x  4)
7
7 y  7  3x  12

3x  7 y  5  0
Example 6, page 14
Perpendicular Lines

 If L1 and L2 are two distinct nonvertical lines that
have slopes m1 and m2, respectively, then L1 is
perpendicular to L2 (written L1 ┴ L2) if and only if

1
m1  
m2
Example
 Find the equation of the line L1 that passes through the
point (3, 1) and is perpendicular to the line L2 described by
y  3  2( x  1)
Solution
 L2 is described in point-slope form, so its slope is m2 = 2.
 Since the lines are perpendicular, the slope of L1 must be
m1 = –1/2
 Using the point-slope form of the equation for L1 we obtain

1
y  1   ( x  3)
2
2 y  2  x  3
x  2y 5  0
Example 7, page 14
Crossing the Axis
 A straight line L that is neither horizontal nor vertical
cuts the x-axis and the y-axis at, say, points (a, 0) and
(0, b), respectively.
 The numbers a and b are called the x-intercept and
y-intercept, respectively, of L.

y
y-intercept
(0, b)

x-intercept

x
(a, 0)
L
Slope-Intercept Form

 An equation of the line that has slope m and
intersects the y-axis at the point (0, b) is given by
y = mx + b
Examples

 Find the equation of the line that has slope 3 and
y-intercept of – 4.
Solution
 We substitute m = 3 and b = – 4 into y = mx + b and get

y = 3x – 4

Example 8, page 15
Examples

 Determine the slope and y-intercept of the line whose
equation is 3x – 4y = 8.
Solution
 Rewrite the given equation in the slope-intercept form.

3x  4 y  8
4 y  8  3 x
3
y  x2
4

 Comparing to y = mx + b, we find that m = ¾ and b = – 2.
 So, the slope is ¾ and the y-intercept is – 2.

Example 9, page 15
Applied Example
 Suppose an art object purchased for \$50,000 is expected to
appreciate in value at a constant rate of \$5000 per year for
the next 5 years.
 Write an equation predicting the value of the art object for
any given year.
 What will be its value 3 years after the purchase?
Solution
 Let     x = time (in years) since the object was purchased
y = value of object (in dollars)
 Then, y = 50,000 when x = 0, so the y-intercept is b = 50,000.
 Every year the value rises by 5000, so the slope is m = 5000.
 Thus, the equation must be y = 5000x + 50,000.
 After 3 years the value of the object will be \$65,000:
y = 5000(3) + 50,000 = 65,000
Applied Example 11, page 16
General Form of a Linear Equation

 The equation
Ax + By + C = 0
where A, B, and C are constants and A and B
are not both zero, is called the general form
of a linear equation in the variables x and y.
General Form of a Linear Equation

 An equation of a straight line is a linear
equation; conversely, every linear equation
represents a straight line.
Example

 Sketch the straight line represented by the equation
3x – 4y – 12 = 0
Solution
 Since every straight line is uniquely determined by two
distinct points, we need find only two such points through
which the line passes in order to sketch it.
 For convenience, let’s compute the x- and y-intercepts:
✦ Setting y = 0, we find x = 4; so the x-intercept is 4.
✦ Setting x = 0, we find y = –3; so the y-intercept is –3.
 Thus, the line goes through the points (4, 0) and (0, –3).

Example 12, page 17
Example

 Sketch the straight line represented by the equation
3x – 4y – 12 = 0
Solution
 Graph the line going through the points (4, 0) and (0, –3).

y                               L
1
(4, 0)
x
1      2   3    4   5   6
–1
–2
–3   (0, – 3)
–4

Example 12, page 17
Equations of Straight Lines

Vertical line:   x=a
Horizontal line:    y=b
Point-slope form:    y – y1 = m(x – x1)
Slope-intercept form:    y = mx + b
General Form:      Ax + By + C = 0
1.3
Linear Functions and Mathematical Models

Real - world      Formulate       Mathematical
problem                           model

Test                                  Solve

-
Solution of real                   Solution of
world Problem       Interpret   mathematical model
Mathematical Modeling
 Mathematics can be used to solve real-world problems.
 Regardless of the field from which the real-world problem
is drawn, the problem is analyzed using a process called
mathematical modeling.
 The four steps in this process are:

Real-world          Formulate
Mathematical   model
problem

Test                                      Solve

Solution of real-                      Solution of
world Problem          Interpret   mathematical model
Functions

 A function f is a rule that assigns to each value of x one and
only one value of y.
 The value y is normally denoted by f(x), emphasizing the
dependency of y on x.
Example
 Let x and y denote the radius and area of a circle,
respectively.
 From elementary geometry we have

y = px2
 This equation defines y as a function of x, since for each
admissible value of x there corresponds precisely one
number y = px2 giving the area of the circle.
 The area function may be written as

f(x) = px2
 To compute the area of a circle with a radius of 5 inches, we
simply replace x in the equation by the number 5:
f(5) = p52)= 25p
Domain and Range

 Suppose we are given the function y = f(x).
 The variable x is referred to as the independent variable,
and the variable y is called the dependent variable.
 The set of all the possible values of x is called the domain of
the function f.
 The set of all the values of f(x) resulting from all the possible
values of x in its domain is called the range of f.
 The output f(x) associated with an input x is unique:
✦ Each x must correspond to one and only one value of f(x).
Linear Function

 The function f defined by

f ( x)  mx  b
where m and b are constants, is called a linear
function.
Applied Example: U.S. Health-Care Expenditures
 Because the over-65 population will be growing more rapidly
in the next few decades, health-care spending is expected to
increase significantly in the coming decades.
 The following table gives the projected U.S. health-care
expenditures (in trillions of dollars) from 2005 through 2010:

Year              2005   2006   2007    2008   2009   2010
Expenditure       2.00   2.17   2.34    2.50   2.69   2.90

 A mathematical model giving the approximate U.S. health-
care expenditures over the period in question is given by
S (t )  0.178t  1.989
where t is measured in years, with t = 0 corresponding to 2005.
Applied Example 1, page 29
Applied Example: U.S. Health-Care Expenditures
 We have                      S (t )  0.178t  1.989
Year             2005   2006   2007    2008   2009   2010
Expenditure      2.00   2.17   2.34    2.50   2.69   2.90

a. Sketch the graph of the function S and the given data on
the same set of axes.
b. Assuming that the trend continues, how much will U.S.
health-care expenditures be in 2011?
c. What is the projected rate of increase of U.S. health-
care expenditures over the period in question?

Applied Example 1, page 29
Applied Example: U.S. Health-Care Expenditures
 We have                           S (t )  0.178t  1.989
Year                2005    2006    2007      2008       2009       2010
Expenditure         2.00    2.17    2.34       2.50          2.69       2.90

Solution
a. The graph of the given data and of the function S is:
S(t)
3.0
S (t )  0.178t  1.989
2.8
2.6
2.4
2.2
2.0
t
1        2        3          4             5
Applied Example 1, page 29
Applied Example: U.S. Health-Care Expenditures
 We have                         S (t )  0.178t  1.989
Year               2005   2006    2007    2008   2009   2010
Expenditure        2.00   2.17    2.34    2.50   2.69   2.90

Solution
b. The projected U.S. health-care expenditures in 2011 is
S (6)  0.178(6)  1.989  3.057
or approximately \$3.06 trillion.

Applied Example 1, page 29
Applied Example: U.S. Health-Care Expenditures
 We have                      S (t )  0.178t  1.989
Year             2005   2006   2007    2008   2009   2010
Expenditure      2.00   2.17   2.34    2.50   2.69   2.90

Solution
c. The function S is linear, so the rate of increase of the
U.S. health-care expenditures is given by the slope of
the straight line represented by S, which is
approximately \$0.18 trillion per year.

Applied Example 1, page 29
Cost, Revenue, and Profit Functions

 Let x denote the number of units of a product
manufactured or sold.
 Then, the total cost function is
C(x) = Total cost of manufacturing x units of the
product
 The revenue function is
R(x) = Total revenue realized from the sale of x
units of the product
 The profit function is
P(x) = Total profit realized from manufacturing
and selling x units of the product
Applied Example: Profit Function
 Puritron, a manufacturer of water filters, has a monthly
fixed cost of \$20,000, a production cost of \$20 per unit,
and a selling price of \$30 per unit.
 Find the cost function, the revenue function, and the
profit function for Puritron.
Solution
 Let x denote the number of units produced and sold.
 Then,
C ( x)  20 x  20,000
R( x)  30 x
P( x )  R( x )  C ( x )
 30 x  (20 x  20,000)
 10 x  20,000
Applied Example 2, page 31
1.4
Intersections of Straight Lines
Finding the Point of Intersection
 Suppose we are given two straight lines L1 and L2 with
equations
y = m1x + b1 and     y = m2x + b2
(where m1, b1, m2, and b2 are constants) that intersect at
the point P(x0, y0).
 The point P(x0, y0) lies on the line L1 and so satisfies the
equation y = m1x + b1.
 The point P(x0, y0) also lies on the line L2 and so satisfies
y = m2x + b2 as well.
 Therefore, to find the point of intersection P(x0, y0) of the
lines L1 and L2, we solve for x and y the system composed
of the two equations
y = m1x + b1 and y = m2x + b2
Example
 Find the point of intersection of the straight lines that have
equations
y=x+1     and      y = – 2x + 4
Solution
 Substituting the value y as given in the first equation into
the second equation, we obtain
x  1  2 x  4
3x  3
x 1
 Substituting this value of x into either one of the given
equations yields y = 2.
 Therefore, the required point of intersection is (1, 2).

Example 1, page 40
Example
 Find the point of intersection of the straight lines that have
equations
y=x+1         and          y = – 2x + 4
Solution
 The graph shows the point of intersection (1, 2) of the two
lines:
y
5                              L1
4

3

2            (1, 2)
1

x
–1            1       2         3   4     5
L2
Example 1, page 40
Applied Example: Break-Even Level
 Prescott manufactures its products at a cost of \$4 per unit
and sells them for \$10 per unit.
 If the firm’s fixed cost is \$12,000 per month, determine the
firm’s break-even point.
Solution
 The revenue function R and the cost function C are given
respectively by
R( x)  10 x     and C( x)  4 x  12,000.
 Setting R(x) = C(x), we obtain

10 x  4 x  12,000
6 x  12,000
x  2000
Applied Example 2, page 41
Applied Example: Break-Even Level
 Prescott manufactures its products at a cost of \$4 per unit
and sells them for \$10 per unit.
 If the firm’s fixed cost is \$12,000 per month, determine the
firm’s break-even point.
Solution
 Substituting x = 2000 into R(x) = 10x gives

R(2000)  10(2000)  20,000
 So, Prescott’s break-even point is 2000 units of the product,
resulting in a break-even revenue of \$20,000 per month.

Applied Example 2, page 41
Applied Example: Market Equilibrium
 The management of ThermoMaster, which manufactures
an indoor-outdoor thermometer at its Mexico subsidiary,
has determined that the demand equation for its product is
5x  3 p  30  0
where p is the price of a thermometer in dollars and x is the
quantity demanded in units of a thousand.
 The supply equation of these thermometers is

52 x  30 p  45  0
where x (in thousands) is the quantity that ThermoMaster
will make available in the market at p dollars each.
 Find the equilibrium quantity and price.

Applied Example 6, page 44
Applied Example: Market Equilibrium
Solution
 We need to solve the system of equations
5x  3 p  30  0
52 x  30 p  45  0
for x and p.
 Let’s solve the first equation for p in terms of x:

5x  3 p  30  0
3 p  5x  30
5
p   x  10
3

Applied Example 6, page 44
Applied Example: Market Equilibrium
Solution
 We need to solve the system of equations
5x  3 p  30  0
52 x  30 p  45  0
for x and p.
 Now we substitute the value of p into the second equation:
 5        
52 x  30   x  10   45  0
 3        
52 x  50x  300  45  0
102x  255  0
102 x  255
255 5
x      
Applied Example 6, page 44                          102 2
Applied Example: Market Equilibrium
Solution
 We need to solve the system of equations
5x  3 p  30  0
52 x  30 p  45  0
for x and p.
 Finally, we substitute the value x = 5/2 into the first equation
5 5
p      10
3 2

25
   10
6
35
     5.83
Applied Example 6, page 44          6
Applied Example: Market Equilibrium
Solution
 We conclude that the equilibrium quantity is 2500 units and
the equilibrium price is \$5.83 per thermometer.

Applied Example 6, page 44
1.5
The Method of Least Squares
The Method of Least Squares
 In this section, we describe a general method known as the
method for least squares for determining a straight line
that, in a sense, best fits a set of data points when the
points are scattered about a straight line.
The Method of Least Squares
 Suppose we are given five data points
P1(x1, y1), P2(x2, y2), P3(x3, y3), P4(x4, y4), and P5(x5, y5)
describing the relationship between two variables x and y.
 By plotting these data points, we obtain a scatter diagram:

y
P5
10                 P3

P2
P4
5

P1

x
5              10
The Method of Least Squares
 Suppose we try to fit a straight line L to the data points
P1, P2, P3, P4, and P5.
 The line will miss these points by the amounts
d1, d2, d3, d4, and d5 respectively.

y                  d5    L
10

d3
d4
5       d2

d1

x
5         10
The Method of Least Squares
 The principle of least squares states that the straight line L
that fits the data points best is the one chosen by requiring
that the sum of the squares of d1, d2, d3, d4, and d5, that is
d12  d 2  d 32  d 4  d 52
2            2

be made as small as possible.

y                      d5       L
10

d3
d4
5       d2

d1

x
5               10
The Method of Least Squares

 Suppose we are given n data points:
P1(x1, y1), P2(x2, y2), P3(x3, y3), . . . , Pn(xn, yn)
 Then, the least-squares (regression) line for the data
is given by the linear equation
y = f(x) = mx + b
where the constants m and b satisfy the equations
( x1  x2  x3    xn )m  nb  y1  y2  y3    yn
and
( x12  x2  x3    xn )m  ( x1  x2  x3    xn )b
2    2           2

 y1 x1  y2 x2  y3 x3    yn xn
simultaneously.
 These last two equations are called normal equations.
Example
 Find the equation of the least-squares line for the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
Solution
 Here, we have n = 5 and
x1 = 1      x2 = 2      x3 = 3       x4 = 4     x5 = 5
y1 = 1      y2 = 3      y3 = 4       y4 = 3     y5 = 6
 Before using the
x       y        x2       xy
equations it is convenient
1        1        1        1
to summarize these data
2        3        4        6
in the form of a table:
3        4        9        12
4        3       16        12
5        6       25        30
15       17      55        61
Example 1, page 53
Example
 Find the equation of the least-squares line for the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
Solution
 Here, we have n = 5 and
x1 = 1      x2 = 2      x3 = 3       x4 = 4     x5 = 5
y1 = 1      y2 = 3      y3 = 4       y4 = 3     y5 = 6
 Using the table to substitute in the second equation we get
( x12  x2  x3  ...  xn )m  ( x1  x2  x3  ...  xn )b
2    2          2

 y1 x1  y2 x2  y3 x3  ...  yn xn
55m  15b  61

Example 1, page 53
Example
 Find the equation of the least-squares line for the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
Solution
 Here, we have n = 5 and
x1 = 1      x2 = 2      x3 = 3       x4 = 4     x5 = 5
y1 = 1      y2 = 3      y3 = 4       y4 = 3     y5 = 6
 Using the table to substitute in the first equation we get

( x1  x2  x3  ...  xn )m  nb  y1  y2  y3  ...  yn

15m  5b  17

Example 1, page 53
Example
 Find the equation of the least-squares line for the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
Solution
 Now we need to solve the simultaneous equations
15m  5b  17
55m  15b  61
 Solving the first equation for b gives

15m  5b  17
5b  15m  17
17
b  3m 
5
Example 1, page 53
Example
 Find the equation of the least-squares line for the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
Solution
 Now we need to solve the simultaneous equations
15m  5b  17
55m  15b  61
 Substituting b into the second equation gives
55m  15b  61
      17 
55m  15  3m    61
       5
55m  45m  51  61
10m  10
Example 1, page 53                 m 1
Example
 Find the equation of the least-squares line for the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
Solution
 Now we need to solve the simultaneous equations
15m  5b  17
55m  15b  61
 Finally, substituting the value m = 1 into the first equation
17
b  3m 
5
17
 3(1) 
5
Example 1, page 53                    0.4
Example
 Find the equation of the least-squares line for the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
Solution
 Now we need to solve the simultaneous equations
15m  5b  17
55m  15b  61
 Thus, we find that m = 1 and b = 0.4.
 Therefore, the required least-squares line is

y  mx  b
y  x  0.4

Example 1, page 53
Example
 Find the equation of the least-squares line for the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
Solution
 Below is the graph of the required least-squares line
y = x + 0.4
y
6                           L

5
4
3
2
1
x
1   2   3   4   5
Example 1, page 53
Applied Example: U.S. Health-Care Expenditures
 Because the over-65 population will be growing more
rapidly in the next few decades, health-care spending is
expected to increase significantly in the coming decades.
 The following table gives the U.S. health expenditures
(in trillions of dollars) from 2005 through 2010:

Year, t           0      1      2      3      4      5
Expenditure, y   2.00   2.17   2.34   2.50   2.69   2.90

 Find a function giving the U.S. health-care spending
between 2005 and 2010, using the least-squares technique.

Applied Example 3, page 55
Applied Example: U.S. Health-Care Expenditures
Solution
 The calculations required for obtaining the normal
equations are summarized in the following table:
t       y        t2        ty
0     2.00        0         0
1     2.17        1       2.17
2     2.34        4       4.68
3     2.50        9       7.50
4     2.69       16      10.76
5     2.90       25      14.50
15    14.60       55      39.61

 Use the table to obtain the second normal equation:
(t12  t2  t3  ...  tn )m  (t1  t2  t3  ...  tn )b
2    2          2

 y1t1  y2t2  y3t3  ...  yntn

Applied Example 3, page 55         55m  15b  39.61
Applied Example: U.S. Health-Care Expenditures
Solution
 The calculations required for obtaining the normal
equations are summarized in the following table:
t        y        t2        ty
0      2.00        0         0
1      2.17        1       2.17
2      2.34        4       4.68
3      2.50        9       7.50
4      2.69       16      10.76
5      2.90       25      14.50
15     14.60       55      39.61

 Use the table to obtain the first normal equation:

(t1  t2  t3  ...  tn )m  nb  y1  y2  y3  ...  yn
15m  6b  14.60
Applied Example 3, page 55
Applied Example: U.S. Health-Care Expenditures
Solution
 Now we solve the simultaneous equations

15m  6b  14.60
55m  15b  39.61
 Solving the first equation for b gives

15m  6b  14.60
6b  15m  14.60
b  2.5m  2.4333

Applied Example 3, page 55
Applied Example: U.S. Health-Care Expenditures
Solution
 Now we solve the simultaneous equations

15m  6b  14.60
55m  15b  39.61
 Substituting b into the second equation gives

55m  15b  39.61
55m  15  2.5m  2.4333  39.61
55m  37.5m  36.4995  39.61
17.5m  3.1105
m  0.1777

Applied Example 3, page 55
Applied Example: U.S. Health-Care Expenditures
Solution
 Now we solve the simultaneous equations

15m  6b  14.60
55m  15b  39.61
 Finally, substituting the value m ≈ 0.1777 into the first
equation that we already solved gives
b  2.5m  2.4333

 2.5(0.1777)  2.4333

 1.9891

Applied Example 3, page 55
Applied Example: U.S. Health-Care Expenditures
Solution
 Now we solve the simultaneous equations

15m  6b  14.60
55m  15b  39.61
 Thus, we find that m ≈ 0.1777 and b ≈ 1.9891.
 Therefore, the required least-squares function is

S (t )  y  1.178t  1.989

Applied Example 3, page 55
End of
Chapter

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