420Hw12ans - STAT 420 Homework #12 ppm 0 0 0 50 50 50 75 75...

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STAT 420 Fall 2008 Homework #12 ppm mV 0 1.72 0 1.68 0 1.74 50 2.04 50 2.11 50 2.17 75 2.40 75 2.32 75 2.33 100 2.91 100 3.00 100 2.89 150 4.47 150 4.51 150 4.43 200 6.67 200 6.66 1. Chemists often use ion-sensitive electrodes (ISEs)to measure the ion concentration of aqueous solutions. These devices measure the migration of the charge of these ions and give a reading in millivolts ( mV ). A standard curve is produced by measuring known concentrations ( in ppm ) and fitting a line to the millivolt data. The table on the right gives the concentrations in ppm and the voltage in mV for calcium ISE. a) Plot the points mV ( y ) versus ppm ( x ). Does linear model seem to be appropriate here? b) Use the Box-Cox method to determine the best transformation on the response variable mV. c) In part (a), a linear model does not seem appropriate. Fit a quadratic model. Does it seem to provide a better fit? 200 6.57 > Hw12_1 = read.table("http://www.stat.uiuc.edu/~stepanov/420Hw12_1.dat", header=T) > Hw12_1 x y 1 0 1.72 2 0 1.68 3 0 1.74 4 50 2.04 5 50 2.11 6 50 2.17 7 75 2.40 8 75 2.32 9 75 2.33 10 100 2.91 11 100 3.00 12 100 2.89 13 150 4.47 14 150 4.51 15 150 4.43 16 200 6.67 17 200 6.66 18 200 6.57
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> attach(Hw12_1) a) > plot(x,y) > abline(lm(y~x)$coefficients,lty=2) Linear model does NOT seem to be appropriate here. b) > library(MASS) > boxcox(fit,plotit=T) > boxcox(lm(y~x),plotit=T,lambda=seq(-1.0,-0.4,by=0.01)) λ – 0.7 seems to give the best transformation of the response variable.
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> fit1 = lm(y^(-0.7) ~ x) > plot(x,y^(-0.7)) > abline(fit1$coefficients) > xx = seq(0,200,by=0.1) > yy = (fit1$coefficients[1]+fit1$coefficients[2]*xx)^(1/(-0.7)) > plot(x,y) > lines(xx,yy) > par(mfrow=c(2,2)) > plot(fit1)
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c) > fit2 = lm(y ~ x + I(x^2)) > plot(fit2) > yy2 = fit2$coefficients[1] + fit2$coefficients[2]*xx + fit2$coefficients[3]*xx^2 > par(mfrow=c(1,1)) > plot(x,y) > lines(xx,yy2) > summary(fit2) Call: lm(formula = y ~ x + I(x^2)) Residuals: Min 1Q Median 3Q Max -0.085354 -0.047679 -0.004113 0.035984 0.143329 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.735e+00 3.442e-02 50.410 < 2e-16 *** x -3.772e-04 7.688e-04 -0.491 0.631 I(x^2) 1.242e-04 3.605e-06 34.452 1.07e-15 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.06341 on 15 degrees of freedom Multiple R-squared: 0.9988, Adjusted R-squared: 0.9987 F-statistic: 6500 on 2 and 15 DF, p-value: < 2.2e-16
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The linear term ( x ) is not significant. > fit3 = lm(y ~ I(x^2)) > yy3 = fit3$coefficients[1] + fit3$coefficients[2]*xx^2 > plot(x,y) > lines(xx,yy3) > lines(xx,yy2) > par(mfrow=c(2,2)) > plot(fit3) > summary(fit3) Call: lm(formula = y ~ I(x^2)) Residuals: Min 1Q Median 3Q Max -0.090685 -0.046398 -0.004792 0.036580 0.142152 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.722e+00 2.028e-02 84.89 <2e-16 ***
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420Hw12ans - STAT 420 Homework #12 ppm 0 0 0 50 50 50 75 75...

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