On Oct 16, 2012, at 11:58 AM, Sigrid wrote:
> 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
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