Dave

I have devlopped a spreadsheet that includes a macro that will claculate the Cochran value for any combination of sets and dgrees of freedom.

I used Cochran's original paper (1941) to test it and also tested it against published tables.

If you want a copy of the spreadshhet leet me know.

Lou



DaveCurti wrote:

Formula for Cochran's Critical Values
22-Jan-09

Hi,

I'm trying to do some data analysis using Cochran's test for outlying
variances.
I have 4 replicate numbers from each of 20 laboratories. I calculate the
variance of each set of data.
I can work out the Cochran's test value by dividing the maximum variance by
the sum of all the variances.
Then I need to compare this with the Cochran critical values, which are
available from tables, but these have gaps, so I'd like to be able to
calculate them.
Does anyone know of a formula to calculate these?

Dave

Previous Posts In This Thread:

On Thursday, January 22, 2009 11:08 AM
DaveCurti wrote:

Formula for Cochran's Critical Values
Hi,

I'm trying to do some data analysis using Cochran's test for outlying
variances.
I have 4 replicate numbers from each of 20 laboratories. I calculate the
variance of each set of data.
I can work out the Cochran's test value by dividing the maximum variance by
the sum of all the variances.
Then I need to compare this with the Cochran critical values, which are
available from tables, but these have gaps, so I'd like to be able to
calculate them.
Does anyone know of a formula to calculate these?

Dave

On Thursday, January 22, 2009 6:33 PM
LoriMille wrote:

it doesn't look like there's a simple formula for small samples, although it
it doesn't look like there's a simple formula for small samples, although it
approaches a Chi squared for larger ones (cf.
http://www.watpon.com/table/cochran.pdf).
Somewhat more accurate than a linear approximation would be to use cubic
interpolation around the neighbouring points eg for k=50 and v=1:

=TREND(B15:B18,A15:A18^{1,2,3},A25^{1,2,3})

gives 0.2599 as opposed to 0.2461 for the linear case. This formula can be
generalised by adapting Bernie's formula above (using offset(...-2,0,4,1) and
^{1,2,3}).


"Dave Curtis" wrote:

On Friday, January 23, 2009 4:01 AM
DaveCurti wrote:

Thanks for the info.
Thanks for the info.
I was hoping to be able to replicate Cochran's values with a formula, but
I've been unable to ascertain how they were derived.
Lori, your idea of a cubic interpolation seems a good one. I've only done
linear interpolations before. However, using your formula, I get a value of
0.2461, instead of the 0.2599 you obtain. Which bracketing points are best
for a cubic interpolation?
I'm not a statistician, so I'm groping in the dark a little here.

Thanks

Dave

"Lori Miller" wrote:

On Friday, January 23, 2009 9:12 PM
Lor wrote:

Dave, i think you're right - it was a typo.
Dave, i think you're right - it was a typo. It's best to use the neighbouring
points for this ie between the interval BC use the points ABCD, at the
endpoints you can use the two before or after.

"Dave Curtis" wrote:


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