Chapter 4.6, Problem 22PFA

a.

To determine

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

Answer to Problem 22PFA

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 22PFA

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, substitutingx =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, substitutingx =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.