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
---
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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.hlx at 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]] > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.
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 ? 22:41, Bert Gunter <gunter.berton at gene.com> a ?crit :
> 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.hlx at 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]]
>>
>> ______________________________________________
>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
St?phane Adamowicz
Chercheur / Scientist
stephane.adamowicz at paca.inra.fr
Centre PACA - UR 1115 PSH
Tel. : +33 1 (0)4 32 72 24 35
Fax : +33 1 (0)4 32 72 24 32
228, route de l'A?rodrome
84 914 Avignon Cedex 9
France
www.inra.fr