420Hw07ans

# 420Hw07ans - STAT 420 (10 points) (due Friday, March 14, by...

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STAT 420 Spring 2008 Homework #7 (10 points) (due Friday, March 14, by 4:00 p.m.) 1. A forester seeking information on basic tree dimensions obtains the following measurements of the diameters 4.5 feet above the ground and the heights of 12 sugar maple trees. The forester wishes to determine if the diameter measurements can be used to predict the tree height. Diameter x (inches) 0.9 1.2 2.9 3.1 3.3 3.9 4.3 6.2 9.6 12.6 16.1 25.8 Height y (feet) 18 26 32 36 44.5 35.6 40.5 57.5 67.3 84 67 87.5 a) Make a scatterplot of x vs. y . Does a straight-line relationship seem appropriate? > x = c(0.9,1.2,2.9,3.1,3.3,3.9,4.3,6.2,9.6,12.6,16.1,25.8) > y = c(18,26,32,36,44.5,35.6,40.5,57.5,67.3,84,67,87.5) > plot(x,y) A straight-line relationship does NOT seem appropriate.

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b) Make a scatterplot of x ' = ln x vs. y ' = ln y . Does a straight-line relationship seem appropriate? > plot(x,y) * the least-squares regression line has been added to the scatterplot A straight-line relationship does seem appropriate. c) Fit a straight-line regression to the transformed data. Add the least-squares regression line to the scatterplot from part(b). Comment on the adequacy of the fit. > fit1 = lm(log(y) ~ log(x)) > abline(fit1\$coefficients) > summary(fit1) Call: lm(formula = log(y) ~ log(x)) Residuals: Min 1Q Median 3Q Max -0.15611 -0.11362 -0.02904 0.11127 0.18389
Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 3.06980 0.07467 41.11 1.74e-12 *** log(x) 0.46459 0.04039 11.50 4.35e-07 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 0.1348 on 10 degrees of freedom Multiple R-Squared: 0.9297, Adjusted R-squared: 0.9227 F-statistic: 132.3 on 1 and 10 DF, p-value: 4.347e-07 The least-squares regression line fits the data well. R 2 = 0.9297. d) Construct a 95% prediction interval for the height of the tree y with the diameter x = 10 inches. Hint: Construct a 95% prediction interval for ln y first. > predict.lm(fit1, data.frame(x=10), interval=c("prediction"), level=0.95) fit lwr upr [1,] 4.139553 3.820263 4.458843 OR > x2 = log(x) > y2 = log(y) > fit2 = lm(y2 ~ x2) > predict.lm(fit2, data.frame(x2=log(10)), interval=c("prediction"), level=0.95) fit lwr upr [1,] 4.139553 3.820263 4.458843 A 95% prediction interval for ln y is ( 3.820263 , 4.458843 ) e 3.820263 = 45.6162 e 4.458843 = 86.3875 A 95% prediction interval for ln y is ( 45.6162 , 86.3875 ) > exp(3.820263) [1] 45.6162 > exp(4.458843) [1] 86.3875

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The log rule : if the values of a variable range over more than one order of magnitude and the variable is strictly positive, then replacing the variable by its logarithm is likely to be helpful. 2.
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## 420Hw07ans - STAT 420 (10 points) (due Friday, March 14, by...

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