Gus Gassmann said on 9/25/02 11:45 AM:

>[EMAIL PROTECTED] wrote:
>
>> This is pretty standard stuff. For example, on page57 of Kleinbaum and
>> Kuppers book Applied Regressions Analysis...,  confidence bands are
>> graphically displayed for a regression model. The bands get wider towards
>> the ends of the regression slope, thus illustrating wider variation in the
>> extremes. The width of the confidence band is a function of the standard
>> error of the estimated y value at each level of the predictor variable. I
>> remember as a student asking Jamie Algina why this occured but did not get
>> an answer, and have not heard one since. Perhaps the band gets wider when
>> the predictor (x) is normally distributed but not when x is uniformly
>> distributed?
>
>It's pretty simple and has nothing to do with normal/not normal.
>If you have two lines, y = beta_0 + beta_1 x and y = b_0 + b_1 x,
>where b_i is close to beta_i, then the lines diverge and the distance between
>them increases as you move away from the center (where they typically
>intersect). So your confidence limits widen as you move out into the tails.
>You can couch this in fancy formulas, if you want, but that's all there is to
>it.

I've never really thought of it in these terms, so, to see if I 
understand correctly, let me state it differently. There is one point 
through which you are sure the regression line passe, X-bar,Y-bar The 
terms for slope and intercept are estimates based on the data and have 
associated error. The divergence of the 'confidence bands' is a 
consequence of the combination of extremes of slope and intercept. Is 
that what you were saying? Therefore the divergence has nothing to do 
with the relative paucity of data at the extremes, nor would 
'oversampling of the extremes' remove this divergence characteristic. 
Right?

Paul

Paul
.
.
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