250 likes | 512 Views
LESSON 21: REGRESSION ANALYSIS. Outline Linear Regression The Method of Least Squares. Linear Regression. This lesson addresses the problem of finding a relationship between two population. The goal is to predict values for one population on the basis of observations taken from the other.
E N D
LESSON 21: REGRESSION ANALYSIS Outline • Linear Regression • The Method of Least Squares
Linear Regression • This lesson addresses the problem of finding a relationship between two population. The goal is to predict values for one population on the basis of observations taken from the other. • Engineers often encounter such problems whenever they need to fit a curve to experimental data. Recall that a scatter diagram is a graph that plots points representing relationship between two variables, an independent variable and a dependent variable . A curve can be fitted to the scatter plot using a regression analysis. • A linear regression assumes a linear relationship and fits a straight line. The method of least square is a method that finds the particular line.
Y Dependent variable X Independent variable Linear Regression The scatter diagram on this slide shows a linear relationship between two variables.
Regression equation: Y = a + bX Y Dependent variable X Independent variable Linear Regression A straight line of the form Y = a+bX nicely fits the points. Linear regression provides values of parameters a and b
Deviation, or error Regression equation: Y = a + bX Y Estimate of Y from regression equation { Actual value of Y Dependent variable Value of X used to estimate Y X Independent variable The Method of Least Squares The least square method finds a and b such that the sum of the squares of errors is minimum
The Method of Least Squares • Consider n observations • For the ith observation Xi, the predicted value • The least square method finds a and b to minimize the sum of the squares of deviations of the predicted values from the actual values
The Method of Least Squares • The following a and b minimize the sum of the squares of deviations
Linear Regression • The standard deviation of the individual Y observations: • The standard error of the Y estimate • Note: The divisor is n minus the number of regression coefficients.
Sales Advertising Month (000 units) (000 $) 1 264 2.5 2 116 1.3 3 165 1.4 4 101 1.0 5 209 2.0 Linear Regression Using Least Squares Future sales are unknown, but future advertising expenses are given by marketing plan. A known value of advertising expense is used to forecast sales. For such a forecast, we need the relationship between advertising and sales. Since sales depends on advertising, sales is the dependent variable and shown on the Y-axis. Advertising is the independent variable and shown on the X-axis.
Sales, Y Advertising, X Month (000 units) (000 $) XYX 2 1 264 2.5 2 116 1.3 3 165 1.4 4 101 1.0 5 209 2.0 Total Y= X = XY - nXY X 2 - nX 2 a = Y - bX b = Linear Regression Using Least Squares
Sales, Y Advertising, X Month (000 units) (000 $) XYX 2 1 264 2.5 660.0 6.25 2 116 1.3 150.8 1.69 3 165 1.4 231.0 1.96 4 101 1.0 101.0 1.00 5 209 2.0 418.0 4.00 Total 855 8.2 1560.8 14.90 Y= 171 X = 1.64 XY - nXY X 2 - nX 2 a = Y - bX b = Linear Regression Using Least Squares
Sales, Y Advertising, X Month (000 units) (000 $) XYX 2 1 264 2.5 660.0 6.25 2 116 1.3 150.8 1.69 3 165 1.4 231.0 1.96 4 101 1.0 101.0 1.00 5 209 2.0 418.0 4.00 Total 855 8.2 1560.8 14.90 Y = 171 X = 1.64 a = - 8.136 b = 109.229 Linear Regression Using Least Squares
300 — 250 — 200 — 150 — 100 — 50 Linear Regression Using Least Squares Y = - 8.136 +109.229(X) Interpretation: For each $1000 increase in advertising, sales increases by 109,229 units. Sales (000s) | | | | 1.0 1.5 2.0 2.5
Linear Regression Using Least Squares The regression equation can be used to forecast sales of Month 6 from a known value of advertising expenditure in Month 6. Forecast for Month 6: Let advertising expenditure = $1750 Y =
Linear Regression Using Least Squares Forecast for Month 6: Let advertising expenditure = $1750 Y =-8.136+109.229(1.75) = 183.015 thousand units
Example Example 1: The following sample observations have been obtained by a chemical engineer investigating the relationship between weight of final product Y (in pounds) and volume of raw materials X in gallons. Find a and b: X Y 14 68 23 105 9 40 17 79 10 51
Example X Y XY X2 14 68 23 105 9 40 17 79 10 51
Example Example 2: Consider Example 1. Compute the sample standard deviation for final product weight.
Example Example 3: Consider Example 1. Compute the standard error of estimate for final product weight.
READING AND EXERCISES Lesson 21 Reading: Section 4-1 pp. 90-102 Exercises: 4-1, 4-2