Elementary Statistics 2nd Edition
Elementary Statistics 2nd Edition
2nd Edition
ISBN: 9781259724275
Author: William Navidi, Barry Monk
Publisher: McGraw-Hill Education
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Chapter 4, Problem 2CS

The following table, Reproduce from chapter introduction. Presents the inflation rate and unemployment rate, both in percent, for the years 1985-2012

Chapter 4, Problem 2CS, The following table, Reproduce from chapter introduction. Presents the inflation rate and

We will investigate some methods for predicting unemployment first. We will try to predict the unemployment rate from the inflation rate.

Compute the least-squares line for predicting unemployment from 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 2CS

  y=0.199x+6.6594

Explanation of Solution

Given information:

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

    YearInflationUnemployment
    19853.87.2
    19861.17.0
    19874.46.2
    19884.45.5
    19894.65.3
    19906.15.6
    19913.16.8
    19922.97.5
    19932.76.9
    19942.76.1
    19952.55.6
    19963.35.4
    19971.74.9
    19981.64.5
    19992.74.2
    20003.44.0
    20011.64.7
    20022.45.8
    20031.96.0
    20043.35.5
    20053.45.1
    20062.54.6
    20074.14.6
    20080.15.8
    20092.79.3
    20101.59.6
    20113.08.9
    20121.78.1

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( x x ¯ )( y y ¯ )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= ( x x ¯ ) 2 n,sy= ( y y ¯ ) 2 n

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+3.0+1.728x¯=xn=79.2028=2.82857

The mean of y is given by:

  y¯=yn= 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+8.128y¯=yn=170.728=6.09643

The data can be represented in tabular form as:

    x y Elementary Statistics 2nd Edition, Chapter 4, Problem 2CS , additional homework tip  1Elementary Statistics 2nd Edition, Chapter 4, Problem 2CS , additional homework tip  2Elementary Statistics 2nd Edition, Chapter 4, Problem 2CS , additional homework tip  3Elementary Statistics 2nd Edition, Chapter 4, Problem 2CS , additional homework tip  4
    3.87.20.971430.943671.103571.21787
    1.17.0-1.728572.987960.903570.81644
    4.46.21.571432.469390.103570.01073
    4.45.51.571432.46939-0.596430.35573
    4.65.31.771433.13796-0.796430.63430
    6.15.63.2714310.70224-0.496430.24644
    3.16.80.271430.073670.703570.49501
    2.97.50.071430.005101.403571.97001
    2.76.9-0.128570.016530.803570.64573
    2.76.1-0.128570.016530.003570.00001
    2.55.6-0.328570.10796-0.496430.24644
    3.35.40.471430.22224-0.696430.48501
    1.74.9-1.128571.27367-1.196431.43144
    1.64.5-1.228571.50939-1.596432.54858
    2.74.2-0.128570.01653-1.896433.59644
    3.44.00.571430.32653-2.096434.39501
    1.64.7-1.228571.50939-1.396431.95001
    2.45.8-0.428570.18367-0.296430.08787
    1.96.0-0.928570.86224-0.096430.00930
    3.35.50.471430.22224-0.596430.35573
    3.45.10.571430.32653-0.996430.99287
    2.54.6-0.328570.10796-1.496432.23930
    4.14.61.271431.61653-1.496432.23930
    0.15.8-2.728577.44510-0.296430.08787
    2.79.3-0.128570.016533.2035710.26287
    1.59.6-1.328571.765103.5035712.27501
    3.08.90.171430.029392.803577.86001
    1.78.1-1.128571.273672.003574.01430
    x¯=2.82857y¯=6.09643 ( x x ¯ )2=41.63714 ( y y ¯ )2=61.46964

Hence, the standard deviation is given by:

  sx= ( x x ¯ ) 2 nsx= 41.63714 28sx=1.48704

And,

  sy= ( y y ¯ ) 2 nsy= 61.46964 28sy=2.19534

Consider, r=1n( x x ¯ )( y y ¯ )sxsy

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

    x y Elementary Statistics 2nd Edition, Chapter 4, Problem 2CS , additional homework tip  5Elementary Statistics 2nd Edition, Chapter 4, Problem 2CS , additional homework tip  6Elementary Statistics 2nd Edition, Chapter 4, Problem 2CS , additional homework tip  7
    3.87.20.971431.103571.07204
    1.17.0-1.728570.90357-1.56189
    4.46.21.571430.103570.16276
    4.45.51.57143-0.59643-0.93724
    4.65.31.77143-0.79643-1.41082
    6.15.63.27143-0.49643-1.62403
    3.16.80.271430.703570.19097
    2.97.50.071431.403570.10026
    2.76.9-0.128570.80357-0.10332
    2.76.1-0.128570.00357-0.00046
    2.55.6-0.32857-0.496430.16311
    3.35.40.47143-0.69643-0.32832
    1.74.9-1.12857-1.196431.35026
    1.64.5-1.22857-1.596431.96133
    2.74.2-0.12857-1.896430.24383
    3.44.00.57143-2.09643-1.19796
    1.64.7-1.22857-1.396431.71561
    2.45.8-0.42857-0.296430.12704
    1.96.0-0.92857-0.096430.08954
    3.35.50.47143-0.59643-0.28117
    3.45.10.57143-0.99643-0.56939
    2.54.6-0.32857-1.496430.49168
    4.14.61.27143-1.49643-1.90260
    0.15.8-2.72857-0.296430.80883
    2.79.3-0.128573.20357-0.41189
    1.59.6-1.328573.50357-4.65474
    3.08.90.171432.803570.48061
    1.78.1-1.128572.00357-2.26117
    x¯=2.82857y¯=6.09643( x x ¯ )( y y ¯ )=8.28714

Plugging the values in the formula,

  r=1n ( x x ¯ )( y y ¯ )sxsyr=1288.28714( 1.48704 )( 2.19534 )r=0.16381

Plugging the values to obtain b1 ,

  b1=rsysxb1=1288.28714( 1.48704 )( 2.19534 )( 2.19534 )( 1.48704 )b1=0.1990

Plugging the values to obtain b0 ,

  b0=y¯b1x¯b0=(6.09643)(0.1990)(2.82857)b0=6.6594

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

  y=b0+b1xy=(6.6594)+(0.1990)xy=0.199x+6.6594

Therefore, the least squares regression line for the given data set is y=0.199x+6.6594

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Chapter 4 Solutions

Elementary Statistics 2nd Edition

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