Dear Wu,

Thank you indeed very  much  for your response. Is it possible that I
use this method to compare the distribution of original population
that  this observations come from?

Thank you ideed very much.

On 3/28/12, alan wu [via R] <ml-node+s789695n4511310h33@n4.nabble.com> wrote:
>
>
> try anova if they have same observe points.
>
> conditions<-rep(c(1,2),each=1600)
> response<-sample (1:20,1600*2, replace= TRUE)
> points<-rep(rep(1:400, each=4),2)
> replicates<-rep(1:4,800)
> obs<-data.frame(response,conditions,points,replicates)
> fit <- lm(response~conditions+points,data=obs)
> anova(fit)
>
>
> On Mar 27, 3:13 am, ali_protocol <mohammadianalimohammad...@gmail.com>
> wrote:
>> Dear all,
>>
>> Novice in statistics.
>>
>> I have 2 experimental conditions. Each condition has ~400 points as its
>> response. Each condition is done in 4 repereats (so I have 2 x 400 x 4
>> points).
>>
>> I want to compare the means of two conditions and test whether they are
>> same
>> or not. Which test should I use?
>>
>> #populations
>> c = matrix (sample (1:20,1600, replace= TRUE), 400 ,4)
>> b = matrix (sample (1:20,1600, replace= TRUE), 400 ,4)
>>
>> #means of repeats
>> c.mean= apply (c,2, mean)
>> b.mean= apply (b,2,mean)
>>
>> #mean of experiment
>> c.mean.all= mean (c)
>> b.mean.all= mean (b)
>>
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>>
>> ______________________________________________
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>> PLEASE do read the posting
>> guidehttp://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
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