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Linear Models Assignment: VI

Subhadip Pal

Due date: 28th November 2009

1.

A) We are concentrating regression of BAC on NOB/(Weightp) for differentvalues of between 0 and 1. We have to estimate by obtaining, numerically, the

value of p that minimizes the residual squared error in the regression of BAC on

NOB/(Weightp). is a parameter. Based on the observed data we have to get an

estimate of the unknown constant.

To get an estimate we took a very crude method to calculate the Residual Sum Of

square (RSS) assuming the model for 10000 equidistant points

as a value of. Then we took the point as as estimate of optimal value of p for which

the RSS is minimum. And after implementing the process in R we get

=0.5793579

Fig1. Assuming the model for different

values of the p the Error Sum of Squares of the model

after fitting to the given data is plotted against the

different values of

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B)

Fig2. Scatter plot of ( , BAC) data points and 95%

prediction band for the future observation .The size of the datapoint is increasing function of NOB

C. Assume that in the regression of BAC on , we really do have

simple linear regression, with normal errors, and equal variances. The (point wise)coverage probability of the band is (c) greater than: 0.95 . Because to get the 95%band we used an estimate of the variance of the errors and the variance estimate is

. but so the band we have formed is wider than the

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actual 95% band that we could formed using the . Hence the probability of a

point would be inside the band is actually greater that 0 .95 .

2.

A) Amount of Sweat is related with that means Ln(S) is linearly

related with Ln(L) where S=Amount of Sweat and L=LoopLength . We are assuming

the model ln( . here is actually Ln(a) and =b . In R we fit the

linear model to get estimate of and and hence estimate of and and

we get and

B)

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Fig3. 95% prediction band for the amount of sweat as a

function of distance, as distance ranges continuously from 3 to 9.

Appendix:

R code used for problem 1:

######################data import part #########################

data_import

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library(lattice)p

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X=NOB/(Weight^opt[1,1])fit1=lm(BAC~X)library(lattice)trellis.device(pdf,file="C:\\My Computer\\edu\\Lm\\plot.pdf")par(cex.axis=1,cex.lab=1.2)par(mar=c(5,4+1,4,2))plot(X,BAC,type="p",cex=.3+NOB/mean(NOB),xlab="X : NOB/Weight^p_opt",ylab="Blood Alcohol content")abline(fit1)title("95% prediction band of BAC as a function of NOB/Weight^p")#legend(.6,0.0059,c("Optimal value of p"),col=c("red"),pch=c(21),cex=1.2)

grid=seq(0,max(X)+.1,length=1000)nd=data.frame(X=grid)#newData=grid#pred=predict(fit1,newdata=nd,interval="confidence")pred=predict(fit1,newdata=nd,interval="prediction")lines(grid,pred[,2],lty=1,col="blue")lines(grid,pred[,3],lty=1,col="blue")

dev.off()}

################################ to execute the functions########################################data_import()#ass6(10000,0) ###### l=10000 is precission parametre more l means moreprecission in estimation and more time to executeb=conf_int(10000)

R Code used for problem 2:

data_prep

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plot(LoopLength,Sweat)lines(fine.grid,exp(predY[,2]),col="red", lty=1)lines(fine.grid,exp(predY[,3]),col="red", lty=1)lines(fine.grid,exp(predY[,1]),col="brown", lty=1)#lines(fine.grid,b[1]*fine.grid^b[2],col="green" ,lty=1)return (z)}

## to execute the functionsdata_prep()z=ass6.2()