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

Thank

Dav

"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:

Submitted via EggHeadCafe - Software Developer Portal of Choice

BizTalk Configure and Send SMTP Mail Based on Message Within an Orchestration

http://www.eggheadcafe.com/tutorials...e-and-sen.aspx