To scomurphy,
I am afraid the calculation of h(i) is based on the value of X, not the value of Y.

I can get SE(-5)=.866, but in order to get the E5* to match I believe I need to use h5=.6, which I'm unable to do. In the formula for Ei*, the only term that uses the (-i) model fit is the SE(-i) term, is this correct? For h5, I get (Xi-Xbar)^2 is 9, and the sum of the squares as 15. 1/n = .2, so h5 = .8. If we are supposed to use the (-i) model for hi as well, then I'm getting the the square as 12.25, and the sum of squares as either 5 (not including the 12.25) or 17.25, which neither give me h5=.6

Please help

I am still confused with the explanation, I have been reading the text so many times and it was stated there that "Suppose, however, that we refit the model deleting the i-th observation, obtaining an estimate of SE(-i) of sigma-E that is based on the REMAINING n-1 observations." My interpretation for this is that we will NOT include (5,0) in computing the regression, and thus the standard error of residuals is based on the first four observations only, that is, the standard error is 0.866. But based on this thread, it seems to be otherwise. Kindly enlighten me on this.

How did you come up with h5= 0.6?
I computed using the formula in page 245:
hi = 1/n + [(Xi-Xbar)^2/ summation from j=1 to n of(Xj - Xbar)^2]
shouldn't the answer for h5 = 1.2? However, if h5 = 1.2, I can't compute for sqrt(1-h5) since this will be a square root of a negative number. Please help.

for H5 this is my solution:
1/5 + (4/4) = 1.2 ???

If the residuals, Ei = Yi - Yi^ , then how could Ei and Yi not be correlated? Is this related to why we plot residual against the Yi^ values in order to detect nonconstant spread?

[NEAS: Correct. The residual is always correlated with the observed response values. We examine the correlation with the fitted values to test for heteroscedasticity.]

To calculate SE(-i), you delete the i-th observation and run a regression using the remaining n-1 observations. The standard error of this regression is SE(-i). For each studentized residual that you want to calculate, you would have to run a separate leave-one-out regression. Since the assignment only asks you to compute ONE studentized residual, you will only need to run ONE extra regression.

I'm not sure how to calculate SE(-i).  I'm coming up with what horshack calculated, but apparently that's not correct.

Oooh, thanks for posting that. I had the correct equation and everything was right until the last step. I had apparently made a mistake plugging it into my calculator and got a different answer (but close enough that I didn't notice it was wrong).

I also got SE(-5)=.866, which gives E5*=-2.556

[NEAS: Correct]

Horshack, I also got h5 = 0.6. However, I got SE(-5) = 0.866.