On Sat, Jul 16, 2011 at 12:05 AM, Rolf Turner wrote:
> (1) Why on earth should the n - 1 divisor be a problem? This divisor
> yields an unbiased estimator of sigma^2. Generally speaking,
> unbiasedness is a Good Thing.
Yes, it is an unbiased estimator of sigma^2 for a sample. My response
was p
On 16/07/11 17:49, Joshua Wiley wrote:
Hi,
The Z is basically:
(mean(x) - mean(y))/sqrt(var(x)/length(x) + var(y)/length(y))
and pnorm will give you a p-value, if you desire it.
If the n - 1 divisior used in var() is a problem for you, it is
trivial to work around:
(1) Why on earth should
Hi,
The Z is basically:
(mean(x) - mean(y))/sqrt(var(x)/length(x) + var(y)/length(y))
and pnorm will give you a p-value, if you desire it.
If the n - 1 divisior used in var() is a problem for you, it is
trivial to work around:
X <- cbind(x, y)
XX <- crossprod(X - tcrossprod(matrix(1, nrow(X)))
Hi Bogdan,
Look at ?pnorm
Josh
On Fri, Jul 15, 2011 at 9:10 PM, Bogdan Tanasa wrote:
> Hi,
>
> please could you recommend a R package that computes a 2 sample z-test ?
>
> thanks,
>
> Bogdan
>
> [[alternative HTML version deleted]]
>
> __
> R-h
Hi,
please could you recommend a R package that computes a 2 sample z-test ?
thanks,
Bogdan
[[alternative HTML version deleted]]
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PLEASE do read the posting guid
Dear all
I am learning R
I am doing Z-test with package 'BSDA'
here is my syntax and result:
> sdgr.ztest<-z.test(growth[type=='SD'& from_treeline=='above'], +
> growth[type=='SD'& from_treeline=='below'],alternative = "two.sided", + mu
> = 0, sigma.x =(sd(growth[type=='SD'& from_treeline==
Michael Pearmain wrote:
> Yes my mistake,
>
> I looked at the pwr.2p2n.test but i cannot place both n's and both p
> values to determine the sig value
> e,g *pwr.2p2n.test(h = , n1 = , n2 = , sig.level = , power = )
>
> or am i missing someting obvious?
>
> *
Not quite obvious, but h is the effect
Yes my mistake,
I looked at the pwr.2p2n.test but i cannot place both n's and both p values
to determine the sig value
e,g *pwr.2p2n.test(h = , n1 = , n2 = , sig.level = , power = )
or am i missing someting obvious?
i did the sam ein SPSS using a macro and the following code:
COMPUTE n1 = Contr
Michael Pearmain wrote:
> Hi All,
>
> I have a table based on ordial data and i want to compare proportions and
> i've seen in the pwr package i can use
> power.prop.test
>
> however i want to find out what the sig. value is based on n1,n2,p1,p2 and
> this package doesn't contain this..
> Does anyo
Hi All,
I have a table based on ordial data and i want to compare proportions and
i've seen in the pwr package i can use
power.prop.test
however i want to find out what the sig. value is based on n1,n2,p1,p2 and
this package doesn't contain this..
Does anyone know of a package that does or is it
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