[R] Function for testing

R. Michael Weylandt michael.weylandt at gmail.com
Thu Jun 21 16:49:51 CEST 2012


Hi, this has the distinctive sound of homework, so I hesitate to give
too much help, but two things I note immediately.

In defining resx / resy (good preallocation!) you probably mean "n"
instead of "N".

A standard way to check how often a condition occurs is something like:

x <- rnorm(50)

mean(x < 1) # Will give percentage of x values less than 1.

Hope this helps,
Michael

On Thu, Jun 21, 2012 at 8:24 AM, a.b.carter <a.b.carter at hotmail.com> wrote:
> Hi all,
>
> I want to create a sample called x, with length 10 from a N(0,1)
> distribution. Next to that I want to create a sample called y, with length
> 10 from a N(0.5 ,1) distribution. Both samples are undergoing a t.test.
> The outcome must be that I can see how many times for x H0 was rejected. The
> same for y. I am testing under a confidence level of 0.05
> Down below is the function I must use:
>
> pv=function(n=10,N=100,m=1,...){
> resx=numeric(N)
> resy=numeric(N)
> for (i in 1:N){
> x=rnorm(n)
> y=rnorm(n,m)
> resx[i]=t.test(x,...)[[3]]
> resy[i]=t.test(y,...)[[3]]
> }
> z=list(resx,resy)
> names(z)=c("px","py")
> z}
>
> So I compute pv(10,100,0.5) which gives me p-values in px and py. But
> somewhere I must give a statement (I think at the ... but that gives me
> errors) that px<0.05 so I can see how many times H0 was rejected. Thus, the
> values that are computed must be compared with the statement <0.05.
>
> --
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> Sent from the R help mailing list archive at Nabble.com.
>
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