Chapter 4, Problem 3CS

To determine

To predict the unemployment in a year when inflation is 3%

Answer to Problem 3CS

The unemployment rate is 5.97% when the inflation rate is 3%

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,

  $\gamma =\frac{{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}}{n-1}=\frac{-4.4169802}{28-1}=\frac{-4.4169802}{27}=-0.163591\approx -0.16$

  $\Rightarrow b_1=\gamma \frac{s_y}{s_x}=\left(-0.16\right)\left(\frac{1.5115}{1.2798}\right)=-0.1932$

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

Hence least squares regression line is,

  $\hat{y}=6.5458-0.1932x$

Now substituting $x=3$ in the least squares regression line, we get

  $\begin{array}{l}\hat{y}=6.5458-0.1932\left(3\right)\\ =6.5458-0.5796\\ =5.9662\\ \approx 5.97\end{array}$