I used gls and it still does not provide me different estimates of variance
for each treatment group. Did I do anything wrong?
lm3<-gls(GSI~treatment,data=z,weights=varIdent(form=~treatment),method="ML")
summary(lm3)
Generalized least squares fit by maximum likelihood
Model: GSI ~ treatment
Data: z
AIC BIC logLik
9.174574 12.71483 0.4127128
Coefficients:
Value Std.Error t-value p-value
(Intercept) 1.0174923 0.1374462 7.402839 0.0000
treatmentHigh 0.7429293 0.1943783 3.822079 0.0028
treatmentLow -0.0146910 0.1943783 -0.075579 0.9411
treatmentMid 0.3869267 0.2099526 1.842924 0.0924
Correlation:
(Intr) trtmnH trtmnL
treatmentHigh -0.707
treatmentLow -0.707 0.500
treatmentMid -0.655 0.463 0.463
Standardized residuals:
Min Q1 Med Q3 Max
-1.9580366 -0.7219450 -0.1989222 0.9096203 1.8319336
Residual standard error: 0.2354039
Degrees of freedom: 15 total; 11 residual
Dieter Menne wrote:
>
>
> Feng, Jingyu wrote:
>>
>> I'am trying to develop some code if R, which would correspond to what I
>> did in SAS.
>> The data look like:
>>
>> Treatment Replicate group1 GSI
>>
>> ..
>> The SAS code is:
>> proc mixed data=data_name order=data method=ml; *scoring=10;
>> classes group1;
>> model GSI=group1/residual influence solution;
>> repeated /group=group1;
>>
>> run;
>>
>> Basically, I need different variance for each treatment group. I want to
>> do the similar thing in R.
>>
>> Here is what I get so far:
>> lm1<-lme(response~treatment,data=o,random=~1|as.factor(dummy),weights=varIdent(form=~1|treatment),method="ML")
>>
>> There should no random term in the model. However If I don't specify one,
>> lme won't work, so I made a dummy variable, which equals to 1 for every
>> observation.
>>
>>
>
> The much underused (quote Frank Harrell) gls in package nlme should do
> that. Quote PB (p250): It can be viewed as an lme function without the
> random argument.
>
> Dieter
>
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