Chapter 4, Problem 12CS

To determine

To calculate: The correlation coefficient between this year’s unemployment and next year’s unemployment.

Answer to Problem 12CS

0.3275

Explanation of Solution

Given information:

A model in which previous values of a variable are used to predict future values of the same variable is called an autoregressive model. The following table presents the data needed to fit this model.

Year	This Year’sUnemployment	Next Year’sUnemployment
1985	7.2	7.0
1986	7.0	6.2
1987	6.2	5.5
1988	5.5	5.3
1989	5.3	5.6
1990	5.6	6.8
1991	6.8	7.5
1992	7.5	6.9
1993	6.9	6.1
1994	6.1	5.6
1995	5.6	5.4
1996	5.4	4.9
1997	4.9	4.5
1998	4.5	4.2
1999	4.2	4.0
2000	4.0	4.7
2001	4.7	5.8
2002	5.8	6.0
2003	6.0	5.5
2004	5.5	5.1
2005	5.1	4.6
2006	4.6	4.6
2007	4.6	5.8
2008	5.8	9.3
2009	9.3	9.6
2010	9.6	8.9
2011	8.9	8.1

Formula Used:

The correlation coefficient of a data is given by:

  $r=\frac{1}{n}\frac{{\displaystyle \sum \left(x-\overline{x}\right)\left(y-\overline{y}\right)}}{s_x{s}_y}$

Where,

  $\overline{x},\overline{y}$ represent the mean of x and y respectively. $s_x,s_y$ represent the standard deviations of x and y . n represents the number of terms.

The standard deviations are given by:

  $s_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}},s_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}$

The mean of x is given by:

  $\overline{x}=\frac{{\displaystyle \sum x}}{n}$

The mean of y is given by:

  $\overline{y}=\frac{{\displaystyle \sum y}}{n}$

Calculation:

The mean of x is given by:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x}}{n}=\frac{\begin{array}{l}7.2+7.0+6.2+5.5+5.3+5.6+6.8+7.5+6.9+6.1+5.6+5.4+4.9+4.5+\\ 4.2+4.0+4.7+5.8+6.0+5.5+5.1+4.6+4.6+5.8+9.3+9.6+8.9\end{array}}{27}\\ \overline{x}=\frac{{\displaystyle \sum x}}{n}=\frac{162.60}{27}=6.02222\end{array}$

The mean of y is given by:

  $\begin{array}{l}\overline{y}=\frac{{\displaystyle \sum y}}{n}=\frac{\begin{array}{l}7.0+6.2+5.5+5.3+5.6+6.8+7.5+6.9+6.1+5.6+5.4+4.9+4.5+\\ 4.2+4.0+4.7+5.8+6.0+5.5+5.1+4.6+4.6+5.8+9.3+9.6+8.9+8.1\end{array}}{27}\\ \overline{y}=\frac{{\displaystyle \sum y}}{n}=\frac{163.5}{27}=6.05556\end{array}$

The data can be represented in tabular form as:

x	y
7.2	7.0	1.17778	1.38716	0.94444	0.89198
7.0	6.2	4.12963	17.05384	0.14444	0.02086
6.2	5.5	3.32963	11.08643	-0.55556	0.30864
5.5	5.3	2.62963	6.91495	-0.75556	0.57086
5.3	5.6	2.42963	5.90310	-0.45556	0.20753
5.6	6.8	2.72963	7.45088	0.74444	0.55420
6.8	7.5	3.92963	15.44199	1.44444	2.08642
7.5	6.9	4.62963	21.43347	0.84444	0.71309
6.9	6.1	4.02963	16.23791	0.04444	0.00198
6.1	5.6	3.22963	10.43051	-0.45556	0.20753
5.6	5.4	2.72963	7.45088	-0.65556	0.42975
5.4	4.9	2.52963	6.39903	-1.15556	1.33531
4.9	4.5	2.02963	4.11940	-1.55556	2.41975
4.5	4.2	1.62963	2.65569	-1.85556	3.44309
4.2	4.0	1.32963	1.76791	-2.05556	4.22531
4.0	4.7	1.12963	1.27606	-1.35556	1.83753
4.7	5.8	1.82963	3.34754	-0.25556	0.06531
5.8	6.0	2.92963	8.58273	-0.05556	0.00309
6.0	5.5	3.12963	9.79458	-0.55556	0.30864
5.5	5.1	2.62963	6.91495	-0.95556	0.91309
5.1	4.6	2.22963	4.97125	-1.45556	2.11864
4.6	4.6	1.72963	2.99162	-1.45556	2.11864
4.6	5.8	1.72963	2.99162	-0.25556	0.06531
5.8	9.3	2.92963	8.58273	3.24444	10.52642
9.3	9.6	6.42963	41.34014	3.54444	12.56309
9.6	8.9	6.72963	45.28791	2.84444	8.09086
8.9	8.1	6.02963	36.35643	2.04444	4.17975
$\overline{x}=2.87037$	$\overline{y}=6.05556$		$\begin{array}{l}{\displaystyle \sum {\left(x-\overline{x}\right)}^2}=\\ 308.17073\end{array}$		$\begin{array}{l}{\displaystyle \sum {\left(y-\overline{y}\right)}^2}=\\ 60.20667\end{array}$

Hence, the standard deviation is given by:

  $\begin{array}{l}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}}\\ s_x=\sqrt{\frac{308.17073}{27}}\\ s_x=\sqrt{11.41373}\end{array}$

And,

  $\begin{array}{l}{s}_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}\\ s_y=\sqrt{\frac{60.20667}{27}}\\ s_y=\sqrt{2.22988}\end{array}$

Consider,

Hence, the table for calculating coefficient of correlation is given by:

x	y
7.2	7.0	1.17778	0.94444	1.11235
7.0	6.2	4.12963	0.14444	0.59650
6.2	5.5	3.32963	-0.55556	-1.84979
5.5	5.3	2.62963	-0.75556	-1.98683
5.3	5.6	2.42963	-0.45556	-1.10683
5.6	6.8	2.72963	0.74444	2.03206
6.8	7.5	3.92963	1.44444	5.67613
7.5	6.9	4.62963	0.84444	3.90947
6.9	6.1	4.02963	0.04444	0.17909
6.1	5.6	3.22963	-0.45556	-1.47128
5.6	5.4	2.72963	-0.65556	-1.78942
5.4	4.9	2.52963	-1.15556	-2.92313
4.9	4.5	2.02963	-1.55556	-3.15720
4.5	4.2	1.62963	-1.85556	-3.02387
4.2	4.0	1.32963	-2.05556	-2.73313
4.0	4.7	1.12963	-1.35556	-1.53128
4.7	5.8	1.82963	-0.25556	-0.46757
5.8	6.0	2.92963	-0.05556	-0.16276
6.0	5.5	3.12963	-0.55556	-1.73868
5.5	5.1	2.62963	-0.95556	-2.51276
5.1	4.6	2.22963	-1.45556	-3.24535
4.6	4.6	1.72963	-1.45556	-2.51757
4.6	5.8	1.72963	-0.25556	-0.44202
5.8	9.3	2.92963	3.24444	9.50502
9.3	9.6	6.42963	3.54444	22.78947
9.6	8.9	6.72963	2.84444	19.14206
8.9	8.1	6.02963	2.04444	12.32724
$\overline{x}=2.87037$	$\overline{y}=6.05556$			$\begin{array}{l}{\displaystyle \sum \left(x-\overline{x}\right)\left(y-\overline{y}\right)}=\\ 44.60992\end{array}$

Plugging the values in the formula,

  $\begin{array}{l}r=\frac{1}{n}\frac{{\displaystyle \sum \left(x-\overline{x}\right)\left(y-\overline{y}\right)}}{s_x{s}_y}\\ r=\frac{1}{27}\cdot \frac{44.60992}{\left(\sqrt{11.41373}\right)\left(\sqrt{2.22988}\right)}\\ r=0.32750\end{array}$

Therefore, the correlation coefficient for the given data is 0.3275