Hi!
We will conduct a factorial experiment aiming to reduce variance in lathe process. How many replicates would be needed to be able to say that there has been a change in variance? Is there any way to calulate this?

Regards
Mike

No reply!!!!!.
Here is my 2 cents..
I am assuming you havent started a screening exp yet. If you are trying to do the screening exp. then i would say you might not need more than one replicate, if you ar tight on budget, you might not need a replicate in the screening phase,, it depends on many things.
Usually no one entertain to spend more than 1/4 of the total budget for screening.
If you can tell us more about the number of factors, levels etc etc.. we might able to help you
thanks

It took me a couple times through to spot a critical point that should have been obvious. To compare variances, you NEED to replicate (or repeat) the runs. With just a single measurement, then you can't estimate the variation in the process. :eek:

Minitab has a "2 Variance" test to compare two variances and a "Test For Equal Variances", the second of which sounds like it is the right idea here.

Unfortunately they don't give any way that I see to estimate the sample size needed.

So here is a rough way to estimate the sample size. Hopefully someone else will confirm this, because I am basically making this up as I go. There is an equation for the standard error of the standard deviation (I'm not making this part up):

0.707*s/N^0.5

So suppose you did 20 replicates and found s to be 1.0. The standard error of this would be 0.707*1/20^0.5 or 0.32. To be reasonably confident (95%) of a change, you would want to be about 2 standard errors from the original value.

So a variance of at least 1 + 0.32*2 = 1.64 or at most 1 - 0.32*2 = 0.36 would indicate a significant change. I do know that the distribution of standard deviations is skewed for small sample sizes, so the lower limit is probably conservative - it could be that 0.4 or 0.45 is actually significant at the 95% level.

Now this is for just 2 experimental setting. For a set of experiments from a factorial design, the results will be different (in ways that I'm not ready to even make a guess at), but this figure should put you in the ball park.

Overall, I have to say it sounds like bad news. Even with 20 parts, you would need a pretty large change in the st dev to be confident of a change. On top of that, you better be sure of the Gage R&R!

Tim F

It took me a couple times through to spot a critical point that should have been obvious. To compare variances, you NEED to replicate (or repeat) the runs. With just a single measurement, then you can't estimate the variation in the process. :eek:

Minitab has a "2 Variance" test to compare two variances and a "Test For Equal Variances", the second of which sounds like it is the right idea here.

Unfortunately they don't give any way that I see to estimate the sample size needed.

So here is a rough way to estimate the sample size. Hopefully someone else will confirm this, because I am basically making this up as I go. There is an equation for the standard error of the standard deviation (I'm not making this part up):


0.707*s/N^0.5

So suppose you did 20 replicates and found s to be 1.0. The standard error of this would be 0.707*1/20^0.5 or 0.32. To be reasonably confident (95%) of a change, you would want to be about 2 standard errors from the original value.

So a variance of at least 1 + 0.32*2 = 1.64 or at most 1 - 0.32*2 = 0.36 would indicate a significant change. I do know that the distribution of standard deviations is skewed for small sample sizes, so the lower limit is probably conservative - it could be that 0.4 or 0.45 is actually significant at the 95% level.

Now this is for just 2 experimental setting. For a set of experiments from a factorial design, the results will be different (in ways that I'm not ready to even make a guess at), but this figure should put you in the ball park.

Overall, I have to say it sounds like bad news. Even with 20 parts, you would need a pretty large change in the st dev to be confident of a change. On top of that, you better be sure of the Gage R&R!

Tim F


Tim:
there is way to calculate sample size in minitab.
you have to go here stat>power and sample size>
there are many options there, you can find the sample size for z test,t test, proportion,, one way anova, factorial exp .......

Peach,

I'm not sure that approach will work easily here. The powere tests - as far as I can tell, are all designed to compare means or proportions, not standard deviations.

As I read the original post, it is a question of reducing the variation, not changing the mean.

For example, suppose Mike wanted to know if the size of pieces coming from two different machines were the same or different - which would be a two sample t-test. We could go to Minitab and use the Stats/Power/Sample Size features to tells us how many pieces to run on the machines.

But does the same prediction work for deteremining if the standard deviation of the pieces coming off the two machines are different. I think that is a tougher question and would require more pieces to acheive the same degree of certainty.

Or here, think of this. In the calculations of sample size, you have to input an estimate of the st dev of your variable. Here the variable is the st dev, so we need to input the st dev of the st dev. However, the st dev of the st dev is itself a function of sample size.

This is just getting too confusing. We need a real statistician who has already thouhg through this question, or I would have to think harder than I am prepared to do ath the moment. I just asked Minitab support, so we'll see if this rates an answer from them.


Tim F