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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.

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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

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

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

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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

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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.

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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

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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 Income (rs)

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SINGLE REGRESSION Problem 1 N = 1000 Average Salary = Rs. 22,500 σ Rs 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

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

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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 X X X Means S.D Regression 3

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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 ) – (0.65) 2 = 0.6 * 0.2 = B 13.2 = σ 1 [ r 12 – (r 12 ) (r 23 ) ] = 3 [ 0.85 – 0.75 * 0.65 ] σ 3 1 – (r 23 ) – (0.65) 2 = 0.75 * 0.36 = Regression 4

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

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