Chapter 4, Problem 2CS

To determine

To find the least square regression line for the given data

Answer to Problem 2CS

Least square regression line is $\hat{y}=6.5458-0.1932x$

Explanation of Solution

Given:

The inflation rate and the unemployment rate, both in percent, for the years 1988-2015 is as shown below.

Year	Inflation	Unemployment
1988	4.4	5.5
1989	4.6	5.3
1990	6.1	5.6
1991	3.1	6.8
1992	2.9	7.5
1993	2.7	6.9
1994	2.7	6.1
1995	2.5	5.6
1996	3.3	5.4
1997	1.7	4.9
1998	1.6	4.5
1999	2.7	4.2
2000	3.4	4.0
2001	1.6	4.7
2002	2.4	5.8
2003	1.9	6.0
2004	3.3	5.5
2005	3.4	5.1
2006	2.5	4.6
2007	4.1	4.6
2008	0.1	5.8
2009	2.7	9.3
2010	1.5	9.6
2011	3.0	8.9
2012	1.7	8.1
2013	1.5	7.4
2014	0.8	6.2
2015	0.7	5.3

Concept used:

Given ordered pairs $\left(x,y\right)$ with sample means $\overline{x}\text{and}\overline{y}$ , sample standard deviations $s_x\text{and}{s}_y$ and the correlation coefficient $\gamma$ , the equation of the least squares regression line for predicting $y$ from $x$ is,

  $\hat{y}=b_0+b_{1}x$ , where $b_1=\gamma \frac{s_y}{s_x}\text{is}\text{the}\text{slpoe}\text{and}{b}_0=\overline{y}-b_1\overline{x}\text{is}\text{the}y-\text{intercept}\text{.}$

First we find the sample means $\overline{x}\text{and}\overline{y}$ .

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x}}{n}=\frac{72.9}{28}=2.603571\\ \overline{y}=\frac{{\displaystyle \sum y}}{n}=\frac{169.2}{28}=6.042857\end{array}$

Now to find standard deviations, we construct the table as shown below:

$x$	$y$	${\left(x-\overline{x}\right)}^2$	${\left(y-\overline{y}\right)}^2$
4.4	5.5	3.227157	0.294693
4.6	5.3	3.985728	0.551836
6.1	5.6	12.225015	0.196122
3.1	6.8	0.246441	0.573265
2.9	7.5	0.087870	2.123265
2.7	6.9	0.009298	0.734694
2.7	6.1	0.009298	0.003265
2.5	5.6	0.010726	0.196122
3.3	5.4	0.485013	0.413265
1.7	4.9	0.816440	1.306122
1.6	4.5	1.007154	2.380407
2.7	4.2	0.009298	3.396121
3.4	4.0	0.634299	4.173264
1.6	4.7	1.007154	1.803264
2.4	5.8	0.041441	0.058979
1.9	6.0	0.495012	0.001836
3.3	5.5	0.485013	0.294693
3.4	5.1	0.634299	0.888979
2.5	4.6	0.010726	2.081836
4.1	4.6	2.239299	2.081836
0.1	5.8	6.267867	0.058979
2.7	9.3	0.009298	10.608980
1.5	9.6	1.217868	12.653266
3.0	8.9	0.157155	8.163266
1.7	8.1	0.816440	4.231837
1.5	7.4	1.217868	1.841837
0.8	6.2	3.252868	0.024693
0.7	5.3	3.623582	0.551836
		${\displaystyle \sum {\left(x-\overline{x}\right)}^2}=$ 44.229617	${\displaystyle \sum {\left(y-\overline{y}\right)}^2}=$ 61.688558

Therefore, standard deviations are,

  $\begin{array}{l}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}=\sqrt{\frac{44.229627}{28-1}}=\sqrt{\frac{44.229617}{27}}=\sqrt{1.6381339}=1.2798\\ s_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n-1}}=\sqrt{\frac{61.688558}{28-1}}=\sqrt{\frac{61.688558}{27}}=\sqrt{2.2847614}=1.5115\end{array}$

Now to find correlation coefficient, we construct the table as shown below:

$x$	$y$	$\frac{x-\overline{x}}{s_x}$	$\frac{y-\overline{y}}{s_y}$	$\left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)$
4.4	5.5	1.4036794	$-$ 0.3591511	$-$ 0.5041330
4.6	5.3	1.5599538	$-$ 0.4914700	$-$ 0.7666704
6.1	5.6	2.7320120	$-$ 0.2929917	$-$ 0.8004568
3.1	6.8	0.3878957	0.5009216	0.1943053
2.9	7.5	0.2316213	0.9640377	0.2232916
2.7	6.9	0.0753469	0.5670810	0.0427277
2.7	6.1	0.0753469	0.0378054	0.0028485
2.5	5.6	$-$ 0.0809274	$-$ 0.2929917	0.0237110
3.3	5.4	0.5441701	$-$ 0.4253106	$-$ 0.2314413
1.7	4.9	$-$ 0.7060251	$-$ 0.7561078	0.5338310
1.6	4.5	$-$ 0.7841623	$-$ 1.0207456	0.8004302
2.7	4.2	0.0753469	$-$ 1.2192239	$-$ 0.0918647
3.4	4.0	0.6223073	$-$ 1.3515428	$-$ 0.8410749
1.6	4.7	$-$ 0.7841623	$-$ 0.8884267	0.6966707
2.4	5.8	$-$ 0.1590646	$-$ 0.1606728	0.0255573
1.9	6.0	$-$ 0.5497507	$-$ 0.0283539	0.0155875
3.3	5.5	0.5441701	$-$ 0.3591511	$-$ 0.1954392
3.4	5.1	0.6223073	$-$ 0.6237889	$-$ 0.3881883
2.5	4.6	$-$ 0.0809274	$-$ 0.9545861	0.0772521
4.1	4.6	1.1692678	$-$ 0.9545861	$-$ 1.1161667
0.1	5.8	$-$ 1.9562205	$-$ 0.1606728	0.3143114
2.7	9.3	0.0753469	2.1549077	0.1623656
1.5	9.6	$-$ 0.8622995	2.3533860	$-$ 2.0293235
3.0	8.9	0.3097585	1.8902699	0.5855271
1.7	8.1	$-$ 0.7060251	1.3609943	$-$ 0.9608961
1.5	7.4	$-$ 0.8622995	0.8978782	$-$ 0.7742399
0.8	6.2	$-$ 1.4092600	0.1039649	$-$ 0.1465135
0.7	5.3	$-$ 1.4873972	$-$ 0.4914700	0.7310111
				${\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}$ = $-$ 4.4169802

Therefore,

  $\gamma =\frac{{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}}{n-1}=\frac{-4.4169802}{28-1}=\frac{-4.4169802}{27}=-0.163591\approx -0.164$

  $\Rightarrow b_1=\gamma \frac{s_y}{s_x}=\left(-0.164\right)\left(\frac{1.5115}{1.2798}\right)=-0.1932$

  $\begin{array}{l}\&b_0=\overline{y}-b_1\overline{x}=6.042857-\left(-0.1932\right)2.603571\\ =6.042857+0.503033\\ =6.54589\end{array}$

Henceleast squares regression line is,

  $\hat{y}=6.5458-0.1932x$