Read Myers section 2.7 and do problems 2.21 and 2.24 (due
next Tuesday)
Run a simple linear regression in both JMP and Splus.
Print the output. Circle everything that makes sense to
you (i.e. if the Durban-Watson statistic is printed and
you haven't any idea what it is, don't circle it!) Most
likely, more than 1/2 of the numbers will be mysterious.
SEs for coeficients (page 29-30)
b1 is Normal(beta1, sigma2/Sxx)
c, the centercept is Normal(beta0 + beta1 X-bar, sigma2/n)
b1 and c are independent (difficult)
b0 is Normal(beta0, sigma2(1/n + x-bar2/Sxx))
Confidence intervals for observations (section 2.9)
y-hat = c + b1 (x - x-bar)
So standard error is computed in the same ways as for intercept!
Alternative idea: shift data to make the x of interest be zero
CI for y is y-hat +/- 2 SEs
What does this interval forecast?
Prediction intervals for observations
Suppose we want to predict a future observation Y and we know x
use above to predict E(Y|x)
But Y - E(Y|x) has standard deviation sigma
So prediction - Y has variance = SE(prediction)2 + sigma2
What does this interval forecast?
Draw pictures of CI and PI
linear bounds
extrapolation not any where near wide enough
What if something doesn't lie in the prediction bounds?
Assumptions
linearity (y = a + bx)
zero mean errors (duh, how could they be anything else?)