Chapter 4.6, Problem 4PPS

To determine

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

Answer to Problem 4PPS

The equation of best line is $y=-2.752x+102.533$ .

The correlation coefficient is -0.6071. The relation is strongly correlated.

Explanation of Solution

Given Information: A table showing the ranks of the ten tallest buildings in which Shai has been.

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.

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

$x$	$y$	$xy$	$x^2$
1	101	101	1
2	88	176	4
3	110	330	9
4	88	352	16
5	88	440	25
6	80	480	36
7	69	483	49
8	102	816	64
9	78	702	81
10	70	700	100

Sum of every column is,

  $\begin{array}{c}{\displaystyle \sum x}=55\\ {\displaystyle \sum y}=874\\ {\displaystyle \sum x}y=4580\\ {\displaystyle \sum x^2}=385\end{array}$

Calculate a as,

  $\begin{array}{c}a=\frac{{\displaystyle \sum y\cdot {{\displaystyle \sum x}}^2}-{\displaystyle \sum x\cdot {\displaystyle \sum x}y}}{n\cdot {\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ =\frac{874\times 385-55\times 4580}{10\times 385-{55}^2}\\ =102.533\end{array}$

Calculate b as,

  $\begin{array}{c}b=\frac{n\cdot {\displaystyle \sum xy}-{\displaystyle \sum x\cdot {\displaystyle \sum y}}}{n\cdot {\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ =\frac{10\times 4580-55\times 874}{10\times 385-{55}^2}\\ =-2.752\end{array}$

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

  $\begin{array}{c}y=a+bx\\ y=-2.752x+102.533\end{array}$

Hence, the equation of the regression line is,

  $y=-2.752x+102.533$

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

$x$	$y$	$xy$	$x^2$	$y^2$
1	101	101	1	10201
2	88	176	4	7744
3	110	330	9	12100
4	88	352	16	7744
5	88	440	25	7744
6	80	480	36	6400
7	69	483	49	4761
8	102	816	64	10404
9	78	702	81	6084
10	70	700	100	4900

Sum of every column is,

  $\begin{array}{c}{\displaystyle \sum x}=55\\ {\displaystyle \sum y}=874\\ {\displaystyle \sum x}y=4580\\ {\displaystyle \sum x^2}=385\\ {\displaystyle \sum y^2}=78082\end{array}$

Calculate r as,

  $\begin{array}{c}r=\frac{n\cdot {\displaystyle \sum xy}-{\displaystyle \sum x\cdot {\displaystyle \sum y}}}{\sqrt{\left[n\cdot {\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2\right].\left[n\cdot {\displaystyle \sum y^2}-{\left({\displaystyle \sum y}\right)}^2\right]}}\\ =\frac{10\times 4580-55\times 874}{\sqrt{\left[10\times 4580-{55}^2\right]-\left[10\times 78082-{874}^2\right]}}\\ =-0.6071\end{array}$

Therefore, the correlation coefficient is -0.6071

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