## R-code (September 4, 2007)

## get the mathmarks data set
## note: if the first row in the txt file is one shorter than the rest, then R automatically sets header=T.
data <- read.table("mathmarks.txt", header=T)

## look at the data
data[1:3,]

## set the variable statistics to be the response (y) and algebra to be the covariate (x)
y <- data$statistics
x <- data$algebra

## plot x against y
plot(x, y)

## change the axis labels
plot(x, y, xlab="algebra", ylab="statistics", main="scatter plot")


##  let solve for the least-squares estimate of the best fitting line.
## remember that we are estimating a an intercept and a slope, so we have to
## create a vector of 1's and add it to the vextor x to make the matrix X.
X <- cbind(rep(1, length(x)), x)

X[1:4,]

beta.lse <- solve(t(X)%*%X)%*%t(X)%*%y

## add the least-squares line to the scatter plot
plot(x, y, xlab="algebra", ylab="statistics", main="scatter plot")
abline(beta.lse)

## put two plots on one page.
par(mfrow=c(2,1))
plot(x, y, xlab="algebra", ylab="statistics", main="scatter plot")

plot(x, y, xlab="algebra", ylab="statistics", main="scatter plot")
abline(beta.lse)


## suppose we couldn't determine the solution.  use optim() to optimize an expression.
## note the default is minimization.
ls.func <- function(beta){

out <- t((y - X%*%beta))%*%(y-X%*%beta)

return(out)
}

## test the function.  pick some beta to try.
beta.try <- c(2,5)
ls.func(beta.try)


## ok use optim() to minimize ls.func wrt b.
optim(par=c(0,0), fn=ls.func,  method="BFGS")