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

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


  1. 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 Education 2014/2015 – 2016/2017

  2. 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 occurrences 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. Slide 2

  3. • 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. Slide 3

  4. • 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: Slide 4

  5. • 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 on solving unknown variables and 2 or 3 independent (X) variables. The general form of the linear multiple regression equation is given as follows: Slide 5

  6. • Y = b O + b 1 X 1 + b 2 X 2 + Error Term • Where b O = M Y – b 1 M X1 – b 2 M X2 • b 1 = SD Y /SD X1 * β 1 • b 2 = SD Y /SD X2 * β 2 • β 1 = r Y1 – (r 12 )( r Y2 )/1 – (r 12 ) 2 • β 2 = r Y2 – (r 12 )( r Y1 )/1 – (r 12 ) 2 • Note: M = Mean • SD = Standard deviation • r = Linear correlation Slide 6

  7. Illustration • The following data was collected from 10 students in Political Science to predict performance in POLI 403 using performance in POLI 303 (X 1 ) and POLI 304 (X 2 ). • POLI 403 (Y) 45 55 60 40 60 45 70 60 70 63 • POLI 303 (X 1 ) 40 60 65 50 70 65 58 68 79 80 • POLI 304 (X 2 ) 50 60 66 45 70 61 50 75 70 75 Slide 7

  8. 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 Slide 8

  9. Step 1 Let POLI 403 = Y POLI 303 = X 1 POLI 304 = X 2 Step 2 From SPSS output extract correlations (X 1 X 2 ) = r 12 (X 1 Y) = r Y1 (X 2 Y) = r Y2 • Slide 9

  10. • Step 3 • Extract means for Y, X 1 X 2 i.e. M Y M X1 M X2 • Step 4 • Extract standard deviations for Y, X 1 X 2 i.e. SD Y SD X1 SD X2 • The SPSS output for the data is as follows: Slide 10

  11. • r 12 = 0.691 M Y = 56.7 SD Y = 14.3218 • r Y1 = 0.849 M X1 = 66.5 SD X1 = 16.2745 • r Y2 = 0.791 M X2 = 62.4 SD X2 = 17.420 • Step 5 • To obtain b values, there is need to compute β values • β 1 = r Y1 – (r 12 )(r Y2 )/1-(r 12 ) 2 • = 0.849 – (0.691)(0.791)/1 – (0.691) 2 • = 0.849 – 0.546581/1 – 0.477481 • = 0.302419/0.522519 • = 0.579 Slide 11

  12. • β 2 = r Y2 – (r 12 )(r Y1 )/1 – (r 12 ) 2 • = 0.791 – (0.691)(0.849)/1 – (0.691) 2 • = 0.791 – 0.586659/1 – 0.477481 • = 0.204341/0.522519 • = 0.391 • b 1 = SD Y /SD X1 * β 1 • = 14.3218/16.2745 * 0.579 • = 0.8800 *0.579 • = 0.51 Slide 12

  13. • b 2 = 14.3218/17.420 * 0.391 • = 0.822 * 0.391 • = 0.321 • b O = M Y – b 1 M X1 – b 2 M X2 • = 56.7-0.51(63.5) – 0.321(62.4) • = 56.7 – 32.385 -20.0304 • = 56.7 -52.415 • b O = 4.285 Slide 13

  14. 4 • The estimating equation for predicting Y i.e. marks for POLI 403 is as follows: • Y = 4.29 + 0.51X 1 + 0.321X 2 + Error Term • When X 1 = 45 and X 2 = 55 • Y = 4.29 + 0.51(45) + 0.321(55) • = 4.29 +22.95 + 17.655 • = 44.89 Slide 14

  15. • Y = 44.9 + Error Term • A student who scores 45 in POLI 303 (X 1 ) and 55 in POLI 304 (X 2 ) will probably score 45 in POLI 403 (Y). Slide 15

  16. Exercise • In the example above use the following data to compute the following • r 12 = 0.591 M Y = 46.7 SD Y = 14.321 • r Y1 = 0.749 M X1 = 56.5 SD X1 = 16.274 • r Y2 = 0.891 M X2 = 72.4 SD X2 = 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. Slide 16

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