Maybe this is what you wanted :
Data - structure(list(y = c(4.5, 4.5, 4.7, 6.7, 6, 4.4, 4.1, 5.3, 4, 4.2, 4.1,
6.4, 5.5, 3.5, 4.6, 4.1, 4.6, 5, 6.2, 5.9, 3.9, 5.3, 6.9, 5.7), lot = c(1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 3L), duration =
Hi all,
I have the following data in which there is one factor lot with six
levels and one continuous convariate time.
I want to fit an Ancova model with common slope and different intercept. So
the six lots will have seperate paralell
regression lines.I wanted to find the upper 95% confidence
I think you want CI's for intercepts, not means (what is a mean
for a line??). If so, the ?confint function will give this to you for
the lot effect estimates when applied to a model fitted without an
intercept:
myfit -lm(y~ lot-1+time)
confint(myfit)
Further discussion should be directed to a
3 matches
Mail list logo
