Chapter 4, Problem 10CS

If we are going to use data from this year to predict unemployment next year’s unemployment? A model like this, in which previous values of a variable are used to predict future values of the same variables. Is called an autoregressive model. The following table presents the needed to fit this model.

Compute the least-square line for predicting next year’s unemployment from this year’s unemployment.

To determine

To calculate:

To compute the least squares regression line for predicting next year’s unemployment from this year’s unemployment.

Answer to Problem 10CS

$y=0.1448x+5.1838$

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 equation for least-square regression line:

$y=b_0+b_{1}x$

Where $b_1=r\frac{s_y}{s_x}$ is the slope and $b_0=\overline{y}-b_1\overline{x}$ is the y -intercept.

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

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,

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

Plugging the values to obtain b₁,

$\begin{array}{l}{b}_1=r\frac{s_y}{s_x}\\ b_1=\frac{1}{27}\cdot \frac{44.60992}{\left(\sqrt{11.41373}\right)\left(\sqrt{2.22988}\right)}\frac{\left(\sqrt{2.22988}\right)}{\left(\sqrt{11.41373}\right)}\\ b_1=0.1448\end{array}$

Plugging the values to obtain b₀,

$\begin{array}{l}{b}_0=\overline{y}-b_1\overline{x}\\ b_0=\left(6.05556\right)-\left(0.1448\right)\left(6.02222\right)\\ b_0=5.1838\end{array}$

Hence, the least-square regression line is given by:

$\begin{array}{l}y=b_0+b_{1}x\\ y=\left(5.1838\right)+\left(0.1448\right)x\\ y=0.1448x+5.1838\end{array}$

Therefore, the least squares regression line for the given data set is $y=0.1448x+5.1838$