Chapter 4.6, Problem 22PPS

a.

To determine

Find an equation for the regression line .Let x represents the number of years since 2006. ..

Answer to Problem 22PPS

The equation of regression line is $y=1.3745x+39.516$ .

Explanation of Solution

Given information: The table shows the average cost of a 10-person Thanksgiving

dinner.

Years	2006	2008	2010	2012	2014
Cost($)	38.10	44.61	43.47	43.48	49.41

Calculation:

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 2006.So the given table becomes.

Years x	0	2	4	6	8
Cost($) y	38.10	44.61	43.47	43.48	49.41

Use a graphing calculator for above table,

Equation of regression line is,

  $\begin{array}{l}\text{Sum}\text{of}x=20.\\ \text{Sum}\text{}\text{of}y=225.07.\\ \text{Mean}x=4.\\ \text{Mean}y=\mathrm{45.014.}\\ \text{Sum}\text{of}\text{squares}(S{S}_x)=40.\\ \text{Sum}\text{of}\text{products}(SP)=\mathrm{54.98.}\\ \text{Regression}\text{equation}y=ax+b.\\ a=SP/S{S}_x=54.98/40=\mathrm{1.3745.}\\ b=M_y-a{M}_x=45.01-(1.37\times 4)=\mathrm{39.516.}\\ y=1.3745x+\mathrm{39.516.}\end{array}$

The equation of regression line is $y=1.3745x+39.516$ .

b.

To determine

Which of the following statements are true.

A −The correlation coefficient indicates that’s the best-fit line does not model the data well.

B-The average cost in 2009 was about $43.64.

C-The average cost in 2026 was about $75.25.

Answer to Problem 22PPS

A- False.

B-True.

C-False.

Explanation of Solution

Given information: The table shows the average cost of a 10-person Thanksgiving

dinner.

Years	2006	2008	2010	2012	2014
Cost($)	38.10	44.61	43.47	43.48	49.41

Calculation:

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 2006.So the given table becomes.

Years x	0	2	4	6	8
Cost($) y	38.10	44.61	43.47	43.48	49.41

Use a graphing calculator for above table ,

Equation of regression line is,

  $\begin{array}{l}\text{Sum}\text{of}x=20.\\ \text{Sum}\text{}\text{of}y=225.07.\\ \text{Mean}x=4.\\ \text{Mean}y=\mathrm{45.014.}\\ \text{Sum}\text{of}\text{squares}(S{S}_x)=40.\\ \text{Sum}\text{of}\text{products}(SP)=\mathrm{54.98.}\\ \text{Regression}\text{equation}y=ax+b.\\ a=SP/S{S}_x=54.98/40=\mathrm{1.3745.}\\ b=M_y-a{M}_x=45.01-(1.37\times 4)=\mathrm{39.516.}\\ y=1.3745x+\mathrm{39.516.}\end{array}$

The equation of regression line is $y=1.3745x+39.516$ .

Use a graphing calculator for above table,

The correlation − coefficient is 0.83534650926641.

The correlation coefficient is 0.845 which is close to 1 indicates that’s the best-fit line model the data well.

For calculating average cost in 2009, substituting x =2009-2006=3 in equation (1),

  $\begin{array}{l}y=1.3745x+\mathrm{39.516.}\\ x=3.\\ y=1.3745\times 3+\mathrm{39.516.}\\ y=4.1235+39.516=43.6395=\mathrm{43.64.}\end{array}$

The average cost in 2009 was about $43.64. So B-statement is True.

For calculating average cost in 2026, substituting x =2026-2006=20 in equation (1),

  $\begin{array}{l}y=1.3745x+\mathrm{39.516.}\\ x=20.\\ y=1.3745\times 20+\mathrm{39.516.}\\ y=27.49+39.516=67.006=\mathrm{67.01.}\end{array}$

The average cost in 2026 was about $67.01Which is not $75.25. So C-statement is False.