se's and CI's for fitted lines in multivariate regression analysis

Okay, I've now tried to the predict function and get the SE, although it seem
to calculate SE for each observation from the line (I assume), while I want
the CI-interval and SE for each line fitted line for the treatment. I do not
really understand what  parameter mean these SEs are calculated from when
there would be several means along the line...?. This is what I get with
predict:

predict(model, se.fit = TRUE, interval = "confidence")

Another way I can think of to show how well the lines fit the data is to
look at the intercepts and slopes instead. I can specify the line for each
level and would then get the estimate of slope and intercept, although I do
not know how I show the standard errors of the slope and intercept. 
lm(decrease[treatment=="A"]~colpos[treatment=="A"])

Call:
lm(formula = decrease[treatment == "A"] ~ colpos[treatment ==  "A"])

Coefficients:
            (Intercept)  colpos[treatment == "A"]  
                 2.5357                    0.4643  

Please let me know if you know how to find st. errors for (or st. error for
slope and intercept) of lines for each factor of a treatment.

The SE's for treatment "slope" will vary depending on the colpos values. Using `predict`, pick the mid-point of the colpos and rowpos variables (which is where the SE of the estimates will be minimized). This should be covered in any basic regression text.

model<-lm(decrease ~ rowpos  + colpos + treatment + treatment:colpos + 0, data=OrchardSprays)

# I do not think the use of the non-intercept version matters here and it's not in general a good practice, but it allows all the parameters to be labeled as you apparently expect.

predict(model, newdata=data.frame(colpos=4.5, treatment=unique(OrchardSprays$treatment), rowpos=mean(OrchardSprays$rowpos) ), se.fit = TRUE, interval = "confidence")
$fit
     fit        lwr       upr
1 35.000  20.331646  49.66835
2 63.125  48.456646  77.79335
3  7.625  -7.043354  22.29335
4 90.250  75.581646 104.91835
5 68.500  53.831646  83.16835
6 69.000  54.331646  83.66835
7 25.250  10.581646  39.91835
8  4.625 -10.043354  19.29335

$se.fit
       1        2        3        4        5        6        7        8 
7.291375 7.291375 7.291375 7.291375 7.291375 7.291375 7.291375 7.291375 

$df
[1] 47

$residual.scale
[1] 20.62312

unique(OrchardSprays$treatment)

[1] D E B H G F C A
Levels: A B C D E F G H

with(OrchardSprays, tapply(decrease, treatment, mean) )

A      B      C      D      E      F      G      H 
 4.625  7.625 25.250 35.000 63.125 69.000 68.500 90.250

David Winsemius, MD
Alameda, CA, USA