Chapter 4.6, Problem 6PPS

a.

To determine

To find: An equation of the regression line.

Answer to Problem 6PPS

The equation of regression line is $y=3.317x+16.166$ .

Explanation of Solution

Given Information: A table showing the sales at a clothing chain since 2008.

Calculation:

An equation of the best fit line is given by,

  $y=a+bx$ ...... (1)

• Here, a is the intercept and b is the slope.

The years start from 2008 and go on till 2014.

Assume the year 2008 as 1 and x be the number of years since 2008.

Finding $xy\text{ and }{x}^2$ ,

$x$	$y$	$xy$	$x^2$
1	6.84	6.84	1
2	7.6	15.2	4
3	10.9	32.7	9
4	115.4	461.6	16
5	17.6	88	25
6	21.2	127.2	36
7	26.5	185.5	49

Sum of every column is,

  $\begin{array}{c}{\displaystyle \sum x}=28\\ {\displaystyle \sum y}=206.4\\ {\displaystyle \sum x}y=917.04\\ {\displaystyle \sum x^2}=140\end{array}$

Calculate a as,

  $\begin{array}{c}a=\frac{{\displaystyle \sum y\cdot {{\displaystyle \sum x}}^2}-{\displaystyle \sum x\cdot {\displaystyle \sum x}y}}{n\cdot {\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ =\frac{206.04\times 140-28\times 917.04}{7\times 140-{28}^2}\\ =16.166\end{array}$

Calculate b as,

  $\begin{array}{c}b=\frac{n\cdot {\displaystyle \sum xy}-{\displaystyle \sum x\cdot {\displaystyle \sum y}}}{n\cdot {\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ =\frac{7\times 917.04-28\times 917.04}{7\times 140-{28}^2}\\ =3.317\end{array}$

Substitute the values of a and b in equation (1),

  $\begin{array}{c}y=a+bx\\ y=3.317x+16.166\end{array}$

• Here, x be the number of years since 2008.

Hence, the equation of regression line is,

  $y=3.317x+16.166$

b.

To determine

To graph: the residual graph.

Answer to Problem 6PPS

The linear regression model is not a good fit.

Explanation of Solution

Given Information: A table showing the sales at a clothing chain since 2008.

Calculation:

Find the predicted values and error,

$x$	$y$ (observed)	$y$ (predicted)	Residual (Observed-Predicted)
1	6.84	$\begin{array}{c}y=3.317x+16.166\\ =3.317\left(1\right)+16.166\\ =1.505\end{array}$	5.335
2	7.6	1.55	6.05
3	10.9	1.595	9.305
4	115.4	1.64	113.76
5	17.6	1.685	15.915
6	21.2	1.73	19.47
7	26.5	1.775	24.725

The residual graph is,

  

From the residual plot it is evident that all the residuals are positive and that the linear model is not that a good fit.