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Regression

Regression. Regression Specific statistical methods for finding the “line of best fit” for one response (dependent) numerical variable based on one or more explanatory (independent) variables. Curve Fitting vs. Regression. Regression

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Regression

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  1. Regression • Regression • Specific statistical methods for finding the “line of best fit” for one response (dependent) numerical variable based on one or more explanatory (independent) variables.

  2. Curve Fitting vs. Regression • Regression • Includes using statistical methods to assess the "goodness of fit" of the model.  (ex. Correlation Coefficient) Cause and effect Relationship between variables

  3. Regression: 3 Main Purposes • To describe (or model) • To predict (or estimate) • To control (or administer)

  4. Simple Linear Regression • Statistical method for finding • the “line of best fit” • for one response (dependent) numerical variable • based on one explanatory (independent) variable.  

  5. Least Squares Regression • GOAL - minimize the sum of the square of the errors of the data points. • This minimizes the Mean Square Error

  6. Steps to Reaching a Solution • Draw a scatterplot of the data. • Visually, consider the strength of the linear relationship. • If the relationship appears relatively strong, find the correlation coefficient as a numerical verification. • If the correlation is still relatively strong, then find the simple linear regression line.

  7. Least square method • Interpreting the result: y = a + bx • The valueofbis the slope • The value of ais the y-intercept Deviation method = deviations taken from actual mean values of x and y Y – Y = byx (X – X ) byx= ∑dxdy – (∑dx∑dy) ; ∑dx2 – (∑dx)2 byx= r σ y σx

  8. example • The hrd mgr of a company wants to find a measure which he can used to fix the monthly income of persons applying for a job in the production department. As an experimental project he collected data on 7 persons from that department referring to years of service and their monthly income. • Find regression equation of income on yrs of service. • What initial start would you recommend for a person applying for the job after having served in a similar capacity in another company for 13 yrs.

  9. SINGLE REGRESSION Problem 1 N = 1000 Average Salary = Rs. 22,500 σ Rs. 9050 Average training score = 400 points σ 107 points Co- efficient of correlation between salary & training 0.82 (ryx = 0.82) Forecast salary from training score Solution Let Salary = Y Let Training score = X . ˙ . Y = Avg salary = 22,500 Std deviation = σ Y = 9050 Avg Training score = X = 400 . ˙ . Std deviation = σ X = 107 The Regression, equation of Y on X Y – Y = byx (X – X ) . ˙ . Y – 22,500 = [ (r) (σ y) ][X – 400] (σ x) Regression 1

  10. . ˙ . Y – 22,500 = [ (0.82) (9050) ][X – 400] (107) . ˙ . Y – 22,500 = 69.36 (X – 400) . ˙ . Y = 69.36 X – 27744 + 22500 . ˙ . Y = 69.36 X – 5,244 If training score was 450 Then expected salary Y = 69.36 * 450 – 5244 = Rs 25,968 Regression 2

  11. MULTIPLE REGRESSION Problem 2 Following are the Intercorrelation , Means and Standard Deviations of Leader Competence (y), Emotional Awareness of Others (X2) and Interpersonal Skills Scores (X3). Using the data derive a Multiple Regression Equation. Regression 3

  12. Solution Leader Competence = X1 X1= 12 Emotional Awareness = X2X2 = 20 Interpersonal Skill = X3X3 = 18 r 12 = 0.75 σ 1= 3 r 13 = 0.85 σ 2 = 5 r 23 = 0.65 σ 3 = 4 The multiple regression equation is X1 = a + b12.3 (X2) + b13.2 (X3) To find b12.3 = σ 1[ r12 – (r13) (r23) ] = 3 [ 0.75 – 0.85 * 0.65 ] σ 21 – (r23)2 5 1 – (0.65)2 = 0.6 * 0.2 = 0.21 0.58 B13.2 = σ 1 [ r12 – (r12) (r23) ] = 3[ 0.85 – 0.75 * 0.65 ] σ 3 1 – (r23)2 4 1 – (0.65)2 = 0.75 * 0.36 = 0.47 0.58 Regression 4

  13. To find ‘a’ X1 = a + b12.3 (X2) + b13.2 (X3) . ˙ . 12 = a + 0.21 (20) + 0.47 (18) . ˙ . 12 – 4.2 – 8.46 = a . ˙ . A = - 0.66 The Regression Equation is X1 = a + b12.3 ( X2) + b13.2 ( X3) . ˙ . X1 = - 0.66 + 0.21 (X2) + 0.47 (X3) Regression 5

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