Chapter 4.6, Problem 14PPS

a.

To determine

Find an equation for the regression line.

Answer to Problem 14PPS

Equation for the regression line $y=0.44571x+\mathrm{9.42857.}$

Explanation of Solution

Given information: A count was kept on the number of paintings at an auction each year.

Years Painted	1990	1995	2000	2005	2010	2025
Paintings Sold	8	5	25	21	9	22

Formula used:

The equation for the regression line is given below.

  $\begin{array}{c}y=ax+b.\\ \text{Where},a=\frac{n{\displaystyle \sum xy-({\displaystyle \sum x)({\displaystyle \sum y)}}}}{n{\displaystyle \sum x^2-({\displaystyle \sum x{)}^2}}}.\\ b=\frac{{\displaystyle \sum y-a{\displaystyle \sum x}}}{n}.\\ \text{Where, }\text{n=number}\text{of}\text{observations}.\end{array}$

Calculation:

Let x be the number of years since 1990.

Years Painted x	0	5	10	15	20	25
Paintings Sold y	8	5	25	21	9	22

  $\begin{array}{l}\text{Sum}\text{of}x=75.\\ \text{Sum}\text{}\text{of}y=90.\\ \text{Mean}x=\mathrm{12.5.}\\ \text{Mean}y=15.\\ \text{Sum}\text{of}\text{squares}(S{S}_x)=437.5.\\ \text{Sum}\text{of}\text{products}(SP)=195.\\ \text{Regression}\text{equation}y=ax+b.\\ a=SP/S{S}_x=195/437.5=0.44571.\\ b=M_y-a{M}_x=15-(0.45\times 12.5)=9.42857.\\ y=0.44571x+\mathrm{9.42857.}\end{array}$

b.

To determine

According to the equation, find the number of paintings was sold in 2001.

Answer to Problem 14PPS

Number of paintings in 2001 is $y\approx 14.$

Explanation of Solution

Given information: A count was kept on the number of paintings at an auction each

year.

Years Painted	1990	1995	2000	2005	2010	2025
Paintings Sold	8	5	25	21	9	22

Formula used:

The equation for the regression line is given below.

  $\begin{array}{c}y=ax+b.\\ \text{Where},a=\frac{n{\displaystyle \sum xy-({\displaystyle \sum x)({\displaystyle \sum y)}}}}{n{\displaystyle \sum x^2-({\displaystyle \sum x{)}^2}}}.\\ b=\frac{{\displaystyle \sum y-a{\displaystyle \sum x}}}{n}.\\ \text{Where, }\text{n=number}\text{of}\text{observations}.\end{array}$

Calculation:

Let x be the number of years since 1990.

Years Painted x	0	5	10	15	20	25
Paintings Sold y	8	5	25	21	9	22

  $\begin{array}{l}\text{Sum}\text{of}x=75.\\ \text{Sum}\text{}\text{of}y=90.\\ \text{Mean}x=\mathrm{12.5.}\\ \text{Mean}y=15.\\ \text{Sum}\text{of}\text{squares}(S{S}_x)=437.5.\\ \text{Sum}\text{of}\text{products}(SP)=195.\\ \text{Regression}\text{equation}y=ax+b.\\ a=SP/S{S}_x=195/437.5=0.44571.\\ b=M_y-a{M}_x=15-(0.45\times 12.5)=9.42857.\\ y=0.44571x+\mathrm{9.42857.}\end{array}$

To estimate the number of paintings sold in 2001, evaluate the regression equation for x = 11 because 2001 is 11 years after 1990.

  $\begin{array}{l}y=0.44571x+9.42857.\\ y=0.44571\times 11+9.42857.\\ y=4.90281+9.42857.\\ y=14.33138.\\ y\approx 14.\end{array}$

c.

To determine

To find, is the liner regression equation an accurate modal of data.

Answer to Problem 14PPS

No, the liner regression equation does not give an accurate modal of the number of painting or data.

Explanation of Solution

Given information: A count was kept on the number of paintings at an auction each

year.

Years Painted	1990	1995	2000	2005	2010	2025
Paintings Sold	8	5	25	21	9	22

Formula used:

The equation for the regression line is given below.

  $\begin{array}{c}y=ax+b.\\ \text{Where},a=\frac{n{\displaystyle \sum xy-({\displaystyle \sum x)({\displaystyle \sum y)}}}}{n{\displaystyle \sum x^2-({\displaystyle \sum x{)}^2}}}.\\ b=\frac{{\displaystyle \sum y-a{\displaystyle \sum x}}}{n}.\\ \text{Where, }\text{n=number}\text{of}\text{observations}.\end{array}$

Let x be the number of years since 1990.

Years Painted x	0	5	10	15	20	25
Paintings Sold y	8	5	25	21	9	22

  $\begin{array}{l}\text{Sum}\text{of}x=75.\\ \text{Sum}\text{}\text{of}y=90.\\ \text{Mean}x=\mathrm{12.5.}\\ \text{Mean}y=15.\\ \text{Sum}\text{of}\text{squares}(S{S}_x)=437.5.\\ \text{Sum}\text{of}\text{products}(SP)=195.\\ \text{Regression}\text{equation}y=ax+b.\\ a=SP/S{S}_x=195/437.5=0.44571.\\ b=M_y-a{M}_x=15-(0.45\times 12.5)=9.42857.\\ y=0.44571x+\mathrm{9.42857.}\end{array}$

To estimate the number of paintings sold in 2001, evaluate the regression equation for x = 11 because 2001 is 11 years after 1990.

  $\begin{array}{l}y=0.44571x+9.42857.\\ y=0.44571\times 11+9.42857.\\ y=4.90281+9.42857.\\ y=14.33138.\\ y\approx 14.\end{array}$

No, the liner regression equation does not give an accurate modal of the number of painting. The points show no linear correlation; therefore a line cannot accurately portray the data.