STATISTICS
chapter14.ppt

# chapter14.ppt - Chapter 14 Section 1 Evaluating the...

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Chapter 14 Section 1 Evaluating the Validity and Use of the Least-Squares Regression Model

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Chapter 14 – Section 1 In Chapter 4, we developed linear models that used an explanatory (or predictor) variable to explain the values of a response variable We computed the slope and intercept of the line of best fit model, but We did not have the tools to evaluate the validity or accuracy of the model. 1. The first requirement is that the response values (Y), on the average, are a linear function of the explanatory values (X). The relationship between Y and X is a straight line. 2.The second requirement is that X is a significant predictor of Y. In other words, the slope of the regression line can be positive or negative but cannot be equal to 0.
Chapter 14 – Section 1 The least-squares regression model is given by where Y is the response variable X is the explanatory variable ε is a random error (residual) which is Y – Yhat = [Y - (β 0 + β 1 X)] the actual observation subtract the model’s prediction. So we say Yhat = where Yhat is also called the "fit" and Y is the actual value from the actual data. X Y 1 0 X 1 0

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Chapter 14 – Section 1 The conditions on the error terms ε i are that The ε i are normally distributed with mean 0 and some standard deviation σ We can use a normal probability plot to verify normality of the residuals. Minitab can provide this plot easily. If the residuals are normally distributed, we can be reasonably confident that the relationship between Y and X is a straight line and not some sort of curve.
Example First, consider a situation in which the relationship between Y and X is a straight line (linear). Consider the data set found in K:\Courses\Daniels, John\STA 282\Lecture Data\ch14machinevalue.mtw In this example, the dependent variable is the sales price of a machine (\$)while the independent value is the age of the machine (years). The raw data and fitted line plot is as follows:

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Using Minitab age salesprice 7 6 5 4 3 2 120 110 100 90 80 70 60 50 40 S 6.22018 R-Sq 91.6% R-Sq(adj) 90.6% Fitted Line Plot salesprice = 141.4 - 13.70 age Certainly, it appears a linear relationship is appropriate. However, we need to examine the normality of the residuals to confirm this. In Minitab, open Stat>Regression>Regression. The response is sales price and the predictor is age. Under storage, click on residuals. The residuals from this model will be placed in a column on the worksheet.
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