R help - Mar 2012 - How to test for the difference of means in population, please help

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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You should use mixed effects modeling to analyze data of this sort.
This is not a topic that has generally been covered by introductory
classes, so you should consult with a professional statistician on
your problem, or educate yourself well beyond the novice level (this
takes more than just reading 1 book, a few classes would be good to
get to this level, or intense study of several books).

Since everything is balanced nicely, you could average over the 4
repeats and use a 2 sample t test (assuming the assumptions hold, your
sample data would be fine) comparing the 2 sets of 400 means.  This
will test for a general difference in the overall means, but ignores
other information and hypotheses that may be important (which is why
the mixed effects model approach is much preferred).

On Tue, Mar 27, 2012 at 1:13 AM, ali_protocol
<mohammadianalimohammadian at 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)
>
> --
> View this message in context:
http://r.789695.n4.nabble.com/How-to-test-for-the-difference-of-means-in-population-please-help-tp4508089p4508089.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> 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.


-- 
Gregory (Greg) L. Snow Ph.D.
538280 at gmail.com

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... at 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)
>
> --
> View this message in
context:http://r.789695.n4.nabble.com/How-to-test-for-the-difference-of-means...
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-h... at r-project.org mailing
listhttps://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.