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 = c(0L, 3L, 6L, 9L, 12L, 15L, 18L, 24L, 0L, 3L, 6L, 9L, 12L,
15L, 18L, 24L, 0L, 3L, 6L, 9L, 12L, 15L, 18L, 24L)), Names = c("y",
"lot","duration"), class = "data.frame", row.names = c(NA, -24L))
Data$lot <- as.factor(Data$lot)
summary(Data)
(ANCOVA <- lm(y ~ lot + duration, data=Data))
summary(ANCOVA)
anova(ANCOVA)
# diagram
color <- c("black", "red", "blue")
plot(y~duration, data=Data, pch=as.character(lot))
P <- predict(ANCOVA, int="c")
for(lev in levels(Data$lot)) {
index <- which(Data$lot==lev)
matlines(x=Data$duration[index], y=P[index,], col=color[as.integer(lev)])
}
Le 5 mai 2015 à 22:41, Bert Gunter <gunter.ber...@gene.com> a écrit :
> 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 statistical site or a local
> statistician, as these are not R issues.
>
> Cheers,
> Bert
>
>
> Bert Gunter
> Genentech Nonclinical Biostatistics
> (650) 467-7374
>
> "Data is not information. Information is not knowledge. And knowledge
> is certainly not wisdom."
> Clifford Stoll
>
>
>
>
> On Tue, May 5, 2015 at 9:53 AM, li li <hannah....@gmail.com> wrote:
>> 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 limit for the
>> mean of the each of
>> the regression lines. It doesnot seem straightforward to achieve this using
>> predict function. Can anyone give some suggestions?
>>
>> Here is my data. I only show the first 3 lots. Also I show the model I
>> used in the end. Thanks very much!
>> Hanna
>>
>> y lot time
>> [1,] 4.5 1 0
>> [2,] 4.5 1 3
>> [3,] 4.7 1 6
>> [4,] 6.7 1 9
>> [5,] 6.0 1 12
>> [6,] 4.4 1 15
>> [7,] 4.1 1 18
>> [8,] 5.3 1 24
>> [9,] 4.0 2 0
>> [10,] 4.2 2 3
>> [11,] 4.1 2 6
>> [12,] 6.4 2 9
>> [13,] 5.5 2 12
>> [14,] 3.5 2 15
>> [15,] 4.6 2 18
>> [16,] 4.1 2 24
>> [17,] 4.6 3 0
>> [18,] 5.0 3 3
>> [19,] 6.2 3 6
>> [20,] 5.9 3 9
>> [21,] 3.9 3 12
>> [22,] 5.3 3 15
>> [23,] 6.9 3 18
>> [24,] 5.7 3 24
>>
>>
>>> mod <- lm(y ~ lot+time)
>>> summary(mod)
>> Call:
>> lm(formula = y ~ lot + time)
>> Residuals:
>> Min 1Q Median 3Q Max
>> -1.5666 -0.3344 -0.1343 0.4479 1.8985
>> Coefficients:
>> Estimate Std. Error t value Pr(>|t|)
>> (Intercept) 4.74373 0.36617 12.955 2.84e-14 ***
>> lot2 -0.47500 0.41129 -1.155 0.2567
>> lot3 0.41250 0.41129 1.003 0.3234
>> lot4 0.96109 0.47943 2.005 0.0535 .
>> lot5 0.98109 0.47943 2.046 0.0490 *
>> lot6 -0.09891 0.47943 -0.206 0.8379
>> time 0.02586 0.02046 1.264 0.2153
>> ---
>>
>> [[alternative HTML version deleted]]
>>
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>
> ______________________________________________
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Stéphane Adamowicz
Chercheur / Scientist
stephane.adamow...@paca.inra.fr
Centre PACA - UR 1115 PSH
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