# Regression  Regression  Specific statistical methods for finding the “line of best fit” for one response (dependent) numerical variable based on one.

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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  Includes using statistical methods to assess the "goodness of fit" of the model. (ex. Correlation Coefficient) Cause and effect Relationship between variables

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

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

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

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.

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

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. Yrs of service 117958610 Income (rs) 108659711

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

. ˙. 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

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. X1X2X3 X11.000.750.85 X20.751.000.65 X30.850.651.00 Means12.0020.0018.00 S.D.3.005.004.00 Regression 3

Solution Leader Competence = X 1 X 1 = 12 Emotional Awareness = X 2 X 2 = 20 Interpersonal Skill = X 3 X 3 = 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 X 1 = a + b 12.3 (X 2 ) + b 13.2 (X 3 ) To find b 12.3 = σ 1 [ r 12 – (r 13 ) (r 23 ) ] = 3 [ 0.75 – 0.85 * 0.65 ] σ 2 1 – (r 23 ) 2 5 1 – (0.65) 2 = 0.6 * 0.2 = 0.21 0.58 B 13.2 = σ 1 [ r 12 – (r 12 ) (r 23 ) ] = 3 [ 0.85 – 0.75 * 0.65 ] σ 3 1 – (r 23 ) 2 4 1 – (0.65) 2 = 0.75 * 0.36 = 0.47 0.58 Regression 4

To find ‘a’ X 1 = a + b 12.3 (X 2 ) + b 13.2 (X 3 ). ˙. 12 = a + 0.21 (20) + 0.47 (18). ˙. 12 – 4.2 – 8.46 = a. ˙. A = - 0.66 The Regression Equation is X 1 = a + b 12.3 ( X 2 ) + b 13.2 ( X 3 ). ˙. X 1 = - 0.66 + 0.21 (X 2 ) + 0.47 (X 3 ) Regression 5

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