Hello,
If you want confidence intervals for the beta coefficients of the model,
try the following.
ci_lm <- function(object, level = 0.95){
sfit <- summary(object)
beta <- sfit$coefficients[, 1]
se <- sfit$coefficients[, 2]
df <- sfit$df[1]
alpha <- 1 - level
lower <- beta + qt(alpha/2, df = df)*se
upper <- beta + qt(1 - alpha/2, df = df)*se
data.frame(beta, lower, upper)
}
data(OrchardSprays)
model <- lm(decrease ~ rowpos + colpos * treatment, data = OrchardSprays)
ci_lm(model)
On the other hand, if you want to run regressions on each factor level
separately, use the argument 'subset' of lm().
model2 <- lm(decrease ~ colpos , subset = treatment == 'A', data =
OrchardSprays)
model2
I believe that you might be looking for this last one.
Rui Barradas
Em 16-10-2012 19:58, Sigrid escreveu:
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.
Thank you
~S
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