Applied Political Research Session 13: Multiple Regression Analysis - - PowerPoint PPT Presentation

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POLI 443 Applied Political Research Session 13: Multiple Regression Analysis Lecturer: Prof. A. Essuman-Johnson , Dept. of Political Science Contact Information: aessuman-johnson@ug.edu.gh College of Education School of Continuing and Distance

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College of Education School of Continuing and Distance Education

2014/2015 – 2016/2017

POLI 443 Applied Political Research

Session 13: Multiple Regression Analysis

Lecturer: Prof. A. Essuman-Johnson, Dept. of Political Science Contact Information: aessuman-johnson@ug.edu.gh

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Multiple Regression Analysis

Introduction
Linear regression involves the use of one

independent variable to predict. However, political analysis usually involves more than one variable due to the complex nature of social and political

ccurrences and processes. Multiple regression

enable political scientists to do multivariate analysis. Researchers often examine relationships involving more than 2 variables. These variables are often found in questionnaires and interview research.

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Usually questions measuring several variables will be

included in one questionnaire from which we relate the respondents’ scores/answers on each variable. The most common procedure is to correlate several X variables with one Y variable e.g. there is a positive correlation between a person’s height and his/her ability to play basketball. Taller people tend to make more baskets. There is also a positive correlation such that the more people practice basketball the more baskets they tend to make.

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For maximum accuracy in predicting how well people

shoot baskets, we will consider both how tall they are and how much they practice. Here, there are two predictor variables – height and practice – that predict the criterion variable – basket shooting. Multiple regression allows the researcher to predict someone’s Y score by simultaneously considering his/her scores on all X variables. The general linear equation of Y on X is given as:

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Y = a + bX
Where Y is predicted on solving for the 2 unknown

variables a and b using the value of the independent variable X to get the actual value of the dependent Y variable.

Multiple regression does a similar prediction based
n solving unknown variables and 2 or 3

independent (X) variables. The general form of the linear multiple regression equation is given as follows:

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Y = bO + b1X1 + b2 X2 + Error Term
Where bO = MY – b1MX1 – b2 MX2
b1 = SDY/SDX1 * β1
b2 = SDY/SDX2 * β2
β1 = rY1 – (r12)( rY2)/1 – (r12)2
β2 = rY2 – (r12)( rY1)/1 – (r12)2
Note: M = Mean
SD = Standard deviation
r = Linear correlation

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Illustration

The following data was collected from 10 students in

Political Science to predict performance in POLI 403 using performance in POLI 303 (X1) and POLI 304 (X2).

POLI 403 (Y)

45 55 60 40 60 45 70 60 70 63

POLI 303 (X1)

40 60 65 50 70 65 58 68 79 80

POLI 304 (X2)

50 60 66 45 70 61 50 75 70 75

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a. Write out the multiple equation you would use to

estimate a student’s performance in POLI 403?

b. Compute the b coefficients for the data and form

the regression equation to predict students’ performance in POLI 403.

c. Estimate the marks of a student who scores 45 in

POLI 303 and 55 in POLI 304

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Step 1 Let POLI 403 = Y POLI 303 = X1 POLI 304 = X2 Step 2 From SPSS output extract correlations (X1X2) = r12 (X1Y) = rY1 (X2Y) = rY2

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Step 3
Extract means for Y, X1 X2 i.e. MY MX1 MX2
Step 4
Extract standard deviations for Y, X1 X2 i.e. SDY SDX1

SDX2

The SPSS output for the data is as follows:

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r12 = 0.691 MY = 56.7

SDY = 14.3218

rY1 = 0.849 MX1 = 66.5

SDX1 = 16.2745

rY2 = 0.791 MX2 = 62.4

SDX2 = 17.420

Step 5
To obtain b values, there is need to compute β values
β1 = rY1 – (r12)(rY2)/1-(r12)2
= 0.849 – (0.691)(0.791)/1 – (0.691)2
= 0.849 – 0.546581/1 – 0.477481
= 0.302419/0.522519
= 0.579

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β2 = rY2 – (r12)(rY1)/1 – (r12)2
= 0.791 – (0.691)(0.849)/1 – (0.691)2
= 0.791 – 0.586659/1 – 0.477481
= 0.204341/0.522519
= 0.391
b1 = SDY/SDX1 * β1
= 14.3218/16.2745 * 0.579
= 0.8800 *0.579
= 0.51

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b2 = 14.3218/17.420 * 0.391
= 0.822 * 0.391
= 0.321
bO = MY – b1MX1 – b2MX2
= 56.7-0.51(63.5) – 0.321(62.4)
= 56.7 – 32.385 -20.0304
= 56.7 -52.415
bO = 4.285

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The estimating equation for predicting Y i.e. marks

for POLI 403 is as follows:

Y = 4.29 + 0.51X1 + 0.321X2 + Error Term
When X1= 45 and X2 = 55
Y = 4.29 + 0.51(45) + 0.321(55)
= 4.29 +22.95 + 17.655
= 44.89

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Y = 44.9 + Error Term
A student who scores 45 in POLI 303 (X1) and 55 in

POLI 304 (X2) will probably score 45 in POLI 403 (Y).

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Exercise

In the example above use the following data to compute

the following

r12 = 0.591 MY = 46.7

SDY = 14.321

rY1 = 0.749 MX1 = 56.5

SDX1 = 16.274

rY2 = 0.891 MX2 = 72.4

SDX2 = 17.420

Write out the multiple equations you would use to

estimate a student’s performance in POLI 403.

Compute the b coefficients for the data and form the

regression equation to predict students’ performance in POLI 403.

Estimate the marks of a student who scores 45 in POLI

303 and 55 in POLI 304.

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