Chapter 4.6, Problem 5PPS

To determine

To find: An equation of the regression line and the correlation coefficient.

Answer to Problem 5PPS

The equation of best line is $y=3.536x+16.143$ .

The correlation coefficient is 0.9007. The relation is strongly positively correlated.

Explanation of Solution

Given Information: A table showing the number of annual violin auditions held by a youth symphony each year since 2009.

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 2009 and go on till 2015.

Assume the year 2009 as 1 and x be the number of years since 2009.

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

$x$	$y$	$xy$	$x^2$
1	22	22	1
2	19	38	4
3	25	75	9
4	37	148	16
5	32	160	25
6	35	210	36
7	42	294	49

Sum of every column is,

  $\begin{array}{c}{\displaystyle \sum x}=28\\ {\displaystyle \sum y}=212\\ {\displaystyle \sum x}y=947\\ {\displaystyle \sum x^2}=140\end{array}$

Calculate a as,

  $\begin{array}{c}a=\frac{{\displaystyle \sum y\cdot {{\displaystyle \sum x}}^2}-{\displaystyle \sum x\cdot {\displaystyle \sum x}y}}{n\cdot {\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ =\frac{212\times 140-28\times 947}{7\times 140-{28}^2}\\ =16.143\end{array}$

Calculate b as,

  $\begin{array}{c}b=\frac{n\cdot {\displaystyle \sum xy}-{\displaystyle \sum x\cdot {\displaystyle \sum y}}}{n\cdot {\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ =\frac{7\times 947-28\times 212}{7\times 140-{28}^2}\\ =3.536\end{array}$

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

  $\begin{array}{c}y=a+bx\\ y=3.536x+16.143\end{array}$

• Here, x be the number of years since 2009.

Hence, the equation of best line is,

  $y=3.536x+16.143$

Find $xy\text{ , }{x}^2\text{ and }{y}^2$ ,

$x$	$y$	$xy$	$x^2$	$y^2$
1	22	22	1	484
2	19	38	4	361
3	25	75	9	625
4	37	148	16	1369
5	32	160	25	1024
6	35	210	36	1225
7	42	294	49	1764

Sum of every column is,

  $\begin{array}{c}{\displaystyle \sum x}=28\\ {\displaystyle \sum y}=212\\ {\displaystyle \sum x}y=947\\ {\displaystyle \sum x^2}=140\\ {\displaystyle \sum y^2}=6852\end{array}$

Calculate r as,

  $\begin{array}{c}r=\frac{n\cdot {\displaystyle \sum xy}-{\displaystyle \sum x\cdot {\displaystyle \sum y}}}{\sqrt{\left[n\cdot {\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2\right].\left[n\cdot {\displaystyle \sum y^2}-{\left({\displaystyle \sum y}\right)}^2\right]}}\\ =\frac{7\times 947-28\times 212}{\sqrt{\left[7\times 140-{28}^2\right]-\left[7\times 6852-{212}^2\right]}}\\ =0.9007\end{array}$

Therefore, the correlation coefficient is 0.9007

The correlation coefficient lies between $-1\text{ and }1$ , here since the value is close to 1 the relation is strongly positively correlated.