6.0 Lesson Plan. Answer Questions. Regression. Transformation. Extrapolation. Residuals

Transcription

1 6.0 Lesson Plan Answer Questions Regression Transformation Extrapolation Residuals 1

2 Information about TAs Lab grader: Pontus, Hwk grader: Rachel, Quiz (Tuesday): Matt, Quiz (Thursday): Blake, All the grades are uploaded by Kai, The students can pick up the graded hwk, lab report and quiz in the filing cabinet in the SECC (Old chem, Rm 211). 2

3 6.1 More About Regression Recall the regression assumptions: 1. Each point (X i, Y i ) in the scatterplot satisfies: Y i = a + bx i + ɛ i where the ɛ i have a normal distribution with mean zero and (usually) unknown standard deviation. 2. The errors ɛ i have nothing to do with one another. A large error does not tend to be followed by another large error, for example. 3. The X i values are measured without error. (Thus all the error occurs in the vertical direction, and we do not need to minimize perpendicular distance to the line.) 3

4 A biologist wants to predict brain weight from body weight, based on a sample of 62 mammals. A portion of the data are shown below: bodywt brainwt log(bodywt) log(brainwt) arctic fox owl monkey cow grey wolf roe deer vervet Will this give an ecological correlation? 4

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6 The regression equation is Y = X The correlation is the square root of.87266, or.9344, but it is heavily influenced by a few outliers. (Note: The correlation is the positive square root because the line has positive slope.) The standard deviation of the residuals is Residue is the typical distance of a point to the line (in the vertical direction). Under the Parameter Estimates portion of the printout, the last column tells whether the intercept and slope are significantly different from zero. Small numbers indicate significant differences; values less than.05 are usually taken to indicate real differences from zero, as opposed to chance errors. 6

7 The root mean square (RMSE) is the standard deviation of the vertical distances between each point and the estimated line. It is an estimate of the standard deviation of the vertical distances between the observations and the true line. Formally, where RMSE = 1 n [(Y 1 (â + ˆbX 1 )) (Y n (â + ˆbX n )) 2 ] â = Ȳ ˆb 1 n X; ˆb = n i=1 X iy i XȲ n i=1 X2 i X. 2 1 n Note that â + ˆbX i is the mean of the Y -value at X i 7

8 The regression line predicts the average value for the Y values at a given X. In practice, one wants to predict the individual value for a particular value of X. For example, if Beavis weighs 50 kilograms, then how much would his brain weigh? The prediction (in grams, without transformation) is Y = â + ˆbX = = But this is just the average for all mammals who weigh as much as Beavis does. 8

9 The individual value is less exact than the average value. To predict the average value, the only source of uncertainty is the exact location of the regression line (i.e., â and ˆb are estimates of the true intercept and slope). In order to predict Beavis s brainweight, the uncertainty about Beavis s deviation from the average is added to the uncertainty about the location of the line. For example, if Beavis weighs 50 kilograms, then his brain should weigh grams + ɛ. Assuming the regression model is correct, then ɛ have a normal distribution with mean zero and standard deviation

10 6.2 Transformations The scatterplot of the brainweight against bodyweight showed the the line was probably controlled by a few large values. (These are sometimes called high-leverage points.) Even worse, the scatterplot did not resemble the cigar-shaped point cloud that supports the regression assumptions listed before. In cases like this, one can consider making a transformation of the response variable or the explanatory variable or both. For this data, consider taking the logarithm (base 10) of the brainweight and body weight. The following scatterplot is much better. 10

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12 Taking the log shows that the outliers are not surprising. The regression equation is now: log Y = log X Now 91.23% of the variation in brain weight is explained by body weight. Both the intercept and the slope are highly significant. The estimated standard deviation of ɛ is.317; this is the typical vertical distance between a point and the line. Making transformations is an art. Here the analysis suggests that log Y = log X.763 = Y = 8.1 X.763. So there is a power-law relationship between brain mass and body mass. (Note: 8.1 = ) 12

13 6.3 Extrapolation Predicting Y values for X values outside the range of X values observed in the data is extrapolation. This is risky, because you have no evidence that the linear relationship you have seen in the scatterplot continues to hold in the new X region. Extrapolated values can be entirely wrong. For example, it is unreliable to predict the brain weight of a blue whale. 13

14 6.4 Residuals Estimate the regression line (using JMP software or by calculating â and ˆb by hand). Then find the difference between each observed Y i and the predicted value Ŷi using the fitted line. These differences are called the residuals. Plot each difference against the corresponding X i value. This plot is called a residual plot. 14

15 If the assumptions for linear regression hold, what should one see in the residual plot? If the pattern of the residuals around the horizontal line at zero is: curved, then the assumption of linearity is violated. fan-shaped, then the assumption of constant standard deviation is violated (heteroscedasticity). filled with many outliers, then again the assumption of constant standard deviation is violated. shows a pattern (e.g., positive, negative, positive, negative,...) then the assumption of independent errors is violated. 15

16 When the residuals have a histogram that looks normal and when the residual plot shows no pattern, then we can use the normal distribution to make inferences about individuals. Suppose a residual plot had suggested that no transformation of the brainweight data were necessary. Then what percentage of 20-kilogram mammals have brains that weigh more than 180 grams? The regression equation says that the mean brainweight for 20 kilogram animals is * 20 = The sd of the residuals is Under the regression assumptions, the 20-kilogram mammals have brainweights that are normally distributed with mean and standard deviation The z-transform is ( )/ =.208. From the table, the area under the curve to the right of.208 is ( )/2 = %. 16