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Unformatted text preview: #Part b. Obtain a 95% prediction interval for the muscle mass of a #woman whose age is 60. #Using R functions: new<data.frame(age=60) predict.lm(fit.1.27,new,interval="predict") # fit lwr upr # [1,] 84.94683 68.45067 101.443 #Interpretation: We can predict with 95% confidence that the muscle mass for a 60 year old woman not measured in the study will be between 68.45 and 101.44. #Compare the standard error for prediction to the standard error for Y.hat: Sheet1 Page 2 #Standard error for Y.hat: std.err.y60 < 8.173*sqrt(1/60 + (60mean(age))^2/sum((agemean(age))^2)) std.err.y60 #[1] 1.055131 #Standard error for Y.h(new): std.err.pred60 < 8.173*sqrt(1+1/60 + (60mean(age))^2/sum((agemean(age))^2)) std.err.pred60 # [1] 8.240827...
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 Spring '08
 Staff
 Linear Regression

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