Elementary Statistics (Text Only)
Elementary Statistics (Text Only)
2nd Edition
ISBN: 9780077836351
Author: Author
Publisher: McGraw Hill
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Chapter 4, Problem 6CS

The relationship between inflation and unemployment is not very strong. However .if we are interested in predicting unemployment, we would probably want to predict next year’s unemployment from this year’s inflation we can construct equation to do this by matching each year? Inflation with the next year’s unemployment. As shown in the following table.

Chapter 4, Problem 6CS, The relationship between inflation and unemployment is not very strong. However .if we are

Compute the least-squares line for predicting next year’s unemployment from this year’s inflation

Expert Solution & Answer
Check Mark
To determine

To calculate:

To compute the least squares regression line for the given data set.

Answer to Problem 6CS

y=0.2251x+6.7017

Explanation of Solution

Given information:

The following table presents the inflation rate and unemployment rate, both in percent, for the years 1985-2012.

Year Inflation Unemployment
1985 3.8 7.0
1986 1.1 6.2
1987 4.4 5.5
1988 4.4 5.3
1989 4.6 5.6
1990 6.1 6.8
1991 3.1 7.5
1992 2.9 6.9
1993 2.7 6.1
1994 2.7 5.6
1995 2.5 5.4
1996 3.3 4.9
1997 1.7 4.5
1998 1.6 4.2
1999 2.7 4.0
2000 3.4 4.7
2001 1.6 5.8
2002 2.4 6.0
2003 1.9 5.5
2004 3.3 5.1
2005 3.4 4.6
2006 2.5 4.6
2007 4.1 5.8
2008 0.1 9.3
2009 2.7 9.6
2010 1.5 8.9
2011 3.0 8.1

Formula Used:

The equation for least-square regression line:

y=b0+b1x

Where b1=rsysx is the slope and b0=y¯b1x¯ is the y -intercept.

The correlation coefficient of a data is given by:

r=1n(xx ¯ )(yy ¯ )sxsy

Where,

x¯,y¯ represent the mean of x and y respectively. sx,sy represent the standard deviations of x and y. n represents the number of terms.

The standard deviations are given by:

sx=(xx ¯ )2n,sy=(yy ¯ )2n

The mean of x is given by:

x¯=xn

The mean of y is given by:

y¯=yn

Calculation:

The mean of x is given by:

x¯=xn= 3.8+1.1+4.4+4.4+4.6+6.1+3.1+2.9+2.7+2.7+2.5+3.3+1.7+1.6+ 2.7+3.4+1.6+2.4+1.9+3.3+3.4+2.5+4.1+0.1+2.7+1.5+3.027x¯=xn=77.5027=2.87037

The mean of y is given by:

y¯=yn= 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+8.127y¯=yn=163.527=6.05556

The data can be represented in tabular form as:

x y Elementary Statistics (Text Only), Chapter 4, Problem 6CS , additional homework tip  1 Elementary Statistics (Text Only), Chapter 4, Problem 6CS , additional homework tip  2 Elementary Statistics (Text Only), Chapter 4, Problem 6CS , additional homework tip  3 Elementary Statistics (Text Only), Chapter 4, Problem 6CS , additional homework tip  4
3.8 7.0 0.92963 0.86421 0.94444 0.89198
1.1 6.2 -1.77037 3.13421 0.14444 0.02086
4.4 5.5 1.52963 2.33977 -0.55556 0.30864
4.4 5.3 1.52963 2.33977 -0.75556 0.57086
4.6 5.6 1.72963 2.99162 -0.45556 0.20753
6.1 6.8 3.22963 10.43051 0.74444 0.55420
3.1 7.5 0.22963 0.05273 1.44444 2.08642
2.9 6.9 0.02963 0.00088 0.84444 0.71309
2.7 6.1 -0.17037 0.02903 0.04444 0.00198
2.7 5.6 -0.17037 0.02903 -0.45556 0.20753
2.5 5.4 -0.37037 0.13717 -0.65556 0.42975
3.3 4.9 0.42963 0.18458 -1.15556 1.33531
1.7 4.5 -1.17037 1.36977 -1.55556 2.41975
1.6 4.2 -1.27037 1.61384 -1.85556 3.44309
2.7 4.0 -0.17037 0.02903 -2.05556 4.22531
3.4 4.7 0.52963 0.28051 -1.35556 1.83753
1.6 5.8 -1.27037 1.61384 -0.25556 0.06531
2.4 6.0 -0.47037 0.22125 -0.05556 0.00309
1.9 5.5 -0.97037 0.94162 -0.55556 0.30864
3.3 5.1 0.42963 0.18458 -0.95556 0.91309
3.4 4.6 0.52963 0.28051 -1.45556 2.11864
2.5 4.6 -0.37037 0.13717 -1.45556 2.11864
4.1 5.8 1.22963 1.51199 -0.25556 0.06531
0.1 9.3 -2.77037 7.67495 3.24444 10.52642
2.7 9.6 -0.17037 0.02903 3.54444 12.56309
1.5 8.9 -1.37037 1.87791 2.84444 8.09086
3.0 8.1 0.12963 0.01680 2.04444 4.17975
x¯=2.87037 y¯=6.05556 (xx ¯ )2=40.31630 (yy ¯ )2=60.20667

Hence, the standard deviation is given by:

sx=(xx ¯ )2nsx=40.3163027sx=1.49320

And,

sy=(yy ¯ )2nsy=60.2066727sy=2.22988

Consider, r=1n(xx ¯ )(yy ¯ )sxsy

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

x y Elementary Statistics (Text Only), Chapter 4, Problem 6CS , additional homework tip  5 Elementary Statistics (Text Only), Chapter 4, Problem 6CS , additional homework tip  6 Elementary Statistics (Text Only), Chapter 4, Problem 6CS , additional homework tip  7
3.8 7.0 0.92963 0.94444 0.87798
1.1 6.2 -1.77037 0.14444 -0.25572
4.4 5.5 1.52963 -0.55556 -0.84979
4.4 5.3 1.52963 -0.75556 -1.15572
4.6 5.6 1.72963 -0.45556 -0.78794
6.1 6.8 3.22963 0.74444 2.40428
3.1 7.5 0.22963 1.44444 0.33169
2.9 6.9 0.02963 0.84444 0.02502
2.7 6.1 -0.17037 0.04444 -0.00757
2.7 5.6 -0.17037 -0.45556 0.07761
2.5 5.4 -0.37037 -0.65556 0.24280
3.3 4.9 0.42963 -1.15556 -0.49646
1.7 4.5 -1.17037 -1.55556 1.82058
1.6 4.2 -1.27037 -1.85556 2.35724
2.7 4.0 -0.17037 -2.05556 0.35021
3.4 4.7 0.52963 -1.35556 -0.71794
1.6 5.8 -1.27037 -0.25556 0.32465
2.4 6.0 -0.47037 -0.05556 0.02613
1.9 5.5 -0.97037 -0.55556 0.53909
3.3 5.1 0.42963 -0.95556 -0.41053
3.4 4.6 0.52963 -1.45556 -0.77091
2.5 4.6 -0.37037 -1.45556 0.53909
4.1 5.8 1.22963 -0.25556 -0.31424
0.1 9.3 -2.77037 3.24444 -8.98831
2.7 9.6 -0.17037 3.54444 -0.60387
1.5 8.9 -1.37037 2.84444 -3.89794
3.0 8.1 0.12963 2.04444 0.26502
x¯=2.87037 y¯=6.05556 (xx ¯ )(yy ¯ )=9.07556

Plugging the values in the formula,

r=1n(xx ¯ )(yy ¯ )sxsyr=1279.07556( 1.49320 )( 2.22988 )r=0.18421

Plugging the values to obtain b1,

b1=rsysxb1=1279.07556( 1.49320 )( 2.22988 )( 2.22988 )( 1.49320 )b1=0.2251

Plugging the values to obtain b0,

b0=y¯b1x¯b0=(6.05556)(0.2251)(2.87037)b0=6.7017

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

y=b0+b1xy=(6.7017)+(0.2251)xy=0.2251x+6.7017

Therefore, the least squares regression line for the given data set is y=0.2251x+6.7017

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Chapter 4 Solutions

Elementary Statistics (Text Only)

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