Wonderful! Thanks for the assistance.
Tom La Bone
-Original Message-
From: jiho [mailto:[EMAIL PROTECTED]
Sent: Friday, June 01, 2007 8:17 AM
To: [EMAIL PROTECTED]
Cc: 'R-help'
Subject: Re: [R] Problem with Weighted Variance in Hmisc
On 2007-June-01 , at 13:00 , Tom La
jiho wrote:
> On 2007-June-01 , at 01:03 , Tom La Bone wrote:
>> The function wtd.var(x,w) in Hmisc calculates the weighted variance
>> of x
>> where w are the weights. It appears to me that wtd.var(x,w) = var
>> (x) if all
>> of the weights are equal, but this does not appear to be the case.
On 2007-June-01 , at 13:00 , Tom La Bone wrote:
> The equation for weighted variance given in the NIST DataPlot
> documentation
> is the usual variance equation with the weights inserted. The
> weighted
> variance of the weighted mean is this weighted variance divided by N.
>
> There is anoth
that discusses the merits of the DataPlot
approach versus the Bevington approach?
Tom La Bone
-Original Message-
From: jiho [mailto:[EMAIL PROTECTED]
Sent: Friday, June 01, 2007 2:17 AM
To: [EMAIL PROTECTED]; R-help
Subject: Re: [R] Problem with Weighted Variance in Hmisc
On 2007-June-01 , at 0
On 2007-June-01 , at 01:03 , Tom La Bone wrote:
> The function wtd.var(x,w) in Hmisc calculates the weighted variance
> of x
> where w are the weights. It appears to me that wtd.var(x,w) = var
> (x) if all
> of the weights are equal, but this does not appear to be the case. Can
> someone point
The function wtd.var(x,w) in Hmisc calculates the weighted variance of x
where w are the weights. It appears to me that wtd.var(x,w) = var(x) if all
of the weights are equal, but this does not appear to be the case. Can
someone point out to me where I am going wrong here? Thanks.
Tom La Bo
