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

Dave, You could try to use linear interpolation: For a value in A2, with your table in D2:EXXX, and values in D are what should match A2: =TREND(OFFSET($E$2,MATCH(A2,$D$2:$D$XXX)-1,0,2,1),OFFSET($D$2,MATCH(A2,$D$2:$D$XXX)-1,0,2,1),A2) HTH, Bernie MS Excel MVP "Dave Curtis" <> wrote in message news:... > 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

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: > 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

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: > 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: > > > 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

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: > 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: > > > 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: > > > > > 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

Cochran's Values 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/tutorial...1-a2f309a021c2/biztalk-configure-and-sen.aspx

Re: Cochran's Values On Sun, 28 Mar 2010 06:37:43 -0700, Lou Janke wrote: >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 You could post it onto a free hosting site like 'mediafire' or the like, then post the link here. OR, you could post it as a template on the Microsoft template site. That would only be if it is macro free, or if you put all the macros into a worksheet as text, allowing the user to apply them into the VBeditor manually to get the workbook to function.

Maybe you find this article helpful: R.U.E. 't Lam, "Scrutiny of variance results for outliers: Cochrans test optimized", Analytica Chimica Acta 659 (2010) 68-84. Reprints available through In this article I develop an equation (Eq. 28) that will calculate accurate critical values Cc for the traditional Cochran's C test using Excel. The equation requires critical F values Fc as input parameter. Fc can be obtained from Excel function FINV: FINV[significance level, degrees of freedom 1, degrees of freedom 2]. Cc(alpha1,n,L) = 1 / {1 + (L-1) / FINV[alpha1/L, (n-1), (L-1)(n-1)]} Where alpha1 = one-sided significance level L = total number of data series n = total number of replicates in a single data series Equation 28 works for any significance level 0 <= alpha1 <= 1, any number of data series L >= 2, and any number of replicates per data series n >= 2.

Blog "Variance Outlier Test" I have started a blog with - Short introduction to the Variance Outlier Test (G test) - Additional explalantion - My updated manuscript - More extensives tables with critical values http://www.rtlam.blogspot.com/

On Thursday, January 22, 2009 5:08:11 PM UTC+1, Dave Curtis wrote: > 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 A formula to calculate critical values for Cochran’s C test can be found in: R.U.E. ’t Lam, "Scrutiny of variance results for outliers: Cochran’s test optimized", Analytica Chimica Acta 659 (2010) 68–84. Equation 28 calculates the exact critical values for the traditional Cochran's C test. Theformula works for any number of data sets, any number of replicates per data set, and at any confidence level. The equation requires critical F as input parameter. Critical F is obtained from Excel function FINV. To make the contents of the article readily available, and to give further directions on how to perform this variance outlier test, I maintain a blog: http://rtlam.blogspot.com/ Regards, Ruben 't Lam