Chapter 1, Problem 148RE

a

To determine

To find: A linear model for the data.

Answer to Problem 148RE

The regression line is $y=-0.00889x+4.08558.$

Explanation of Solution

Given: The following ordered pairs ( x,y ) represents Olympic year x and winning time y (in minutes) in the men’s 400-meter freestyle swimming event.

  $\left[\begin{array}{l}\left(1968,4.150\right),\left(1972,4.005\right),\left(1976,3.866\right),\left(1980,3.855\right),\\ \left(1984,3.854\right),\left(1988,3.783\right),\left(1992,3.750\right),\left(1996,3.800\right),\\ \left(2000,3.677\right),\left(2004,3.718\right),\left(2008,3.698\right),\left(2012,3.669\right)\end{array}\right]$

Calculation:

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 =8 corresponding to year 1968 then the above ordered pairs can be written as below table.

Year x	8	12	16	20	24	28	32	36	40	44	48	52
Winning Time t	4.150	4.005	3.866	3.855	3.854	3.783	3.750	3.800	3.677	3.718	3.698	3.669

Using linear regression calculator on above result data table, linear regression line is,

  $\begin{array}{l}\text{Sum}\text{of}x=360.\\ \text{Sum}\text{}\text{of}y=45.825.\\ \text{Mean}x=30.\\ \text{Mean}y=\mathrm{3.8188.}\\ \text{Sum}\text{of}\text{squares}(S{S}_x)=2288.\\ \text{Sum}\text{of}\text{products}(SP)=-\mathrm{20.35.}\\ \text{Regression}\text{equation}y=ax+b.\\ a=SP/S{S}_x=-20.35/2288=-0.00889.\\ b=M_y-a{M}_x=3.82-(-0.02\times 30)=4.08558.\\ y=-0.00889x+4.08558.\end{array}$

The regression coefficient is -0.9013.

b

To determine

To graph: A scatter plot of the data and the model in the same window.

Explanation of Solution

Given: The following ordered pairs ( x,y ) represents Olympic year x and winning time y (in minutes) in the men’s 400-meter freestyle swimming event.

  $\left[\begin{array}{l}\left(1968,4.150\right),\left(1972,4.005\right),\left(1976,3.866\right),\left(1980,3.855\right),\\ \left(1984,3.854\right),\left(1988,3.783\right),\left(1992,3.750\right),\left(1996,3.800\right),\\ \left(2000,3.677\right),\left(2004,3.718\right),\left(2008,3.698\right),\left(2012,3.669\right)\end{array}\right]$

Calculation:

Let x =8 corresponding to year 1968 then the above ordered pairs can be written as below table.

Year x	8	12	16	20	24	28	32	36	40	44	48	52
Winning Time t	4.150	4.005	3.866	3.855	3.854	3.783	3.750	3.800	3.677	3.718	3.698	3.669

Graph:

Scatter plot for the above table data is shown below.

  

Conclusion: The regression model is the good fit for data.

c

To determine

To find: the good fit for the data.

Answer to Problem 148RE

The regression model is the good fit for data.

Explanation of Solution

Given: The following ordered pairs ( x,y ) represents Olympic year x and winning time y (in minutes) in the men’s 400-meter freestyle swimming event.

  $\left[\begin{array}{l}\left(1968,4.150\right),\left(1972,4.005\right),\left(1976,3.866\right),\left(1980,3.855\right),\\ \left(1984,3.854\right),\left(1988,3.783\right),\left(1992,3.750\right),\left(1996,3.800\right),\\ \left(2000,3.677\right),\left(2004,3.718\right),\left(2008,3.698\right),\left(2012,3.669\right)\end{array}\right]$

Calculation:

The regression coefficient is -0.9013.

So, regression model is the good fit for data because regression coefficient is -0.9013. which is close to -1.

Thus, the regression model is the good fit for data.

d.

To determine

To find: the model appropriate for predicting the winning times in future Olympic or not.

Answer to Problem 148RE

Yes, this model appropriate for predicting the winning times in future Olympic.

Explanation of Solution

Given: The following ordered pairs ( x,y ) represents Olympic year x and winning time y (in minutes) in the men’s 400-meter freestyle swimming event.

  $\left[\begin{array}{l}\left(1968,4.150\right),\left(1972,4.005\right),\left(1976,3.866\right),\left(1980,3.855\right),\\ \left(1984,3.854\right),\left(1988,3.783\right),\left(1992,3.750\right),\left(1996,3.800\right),\\ \left(2000,3.677\right),\left(2004,3.718\right),\left(2008,3.698\right),\left(2012,3.669\right)\end{array}\right]$

Calculation:

Regression line is $y=-0.00889x+4.08558.$

The regression coefficient is -0.9013.

Regression model is the good fit for data because regression coefficient is -0.9013. It is close to -1.

Since Regression model is the good fit for data because regression coefficient is -0.9013. it is close to -1. So, this model appropriate for predicting the winning times in future Olympic.