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View Full Version : Count by weight statistical analysis


smryan
27th May 2009, 11:40 AM
I do not have a strong statistics background and am hoping someone here can help.

For our smaller parts we stock and ship counts figured by weight. With some newly subcontracted parts there is quite the difference in count results. I've just taken 8 samples of 50 each - the low is 112.99g and the high is 116.52g. Depending on which sample I happened to grab I would get considerably different results. My 68kg box labeled 30,000 would either come up 91 pieces over or 821 pieces short.

How does one determine an appropriate sample size to ensure 99.7% accuracy? (+/- 3 sigma)

I know some of you quality geniuses will have this easily. :D

Why don't they ever ask me about Newton's Laws? :rolleyes::lmao:

Jennifer Kirley
27th May 2009, 01:14 PM
Based on your description, I can't think of any way to bring statistical confidence to your sampling except to count them out.

Bev D
27th May 2009, 01:34 PM
in order to help you I would need to know a few more things: to start off, what exactly do you need to know? the count? if so, what amount of allowable variation can you have in the count? (regardless of whether or not you use weight or actual counting. Even counting can be inaccurate particularly with large numbers of things.

smryan
27th May 2009, 01:57 PM
in order to help you I would need to know a few more things: to start off, what exactly do you need to know? the count? if so, what amount of allowable variation can you have in the count? (regardless of whether or not you use weight or actual counting. Even counting can be inaccurate particularly with large numbers of things.
Yes, the count is the goal. Customers order by quantity, not by weight. There is no machine to do the counting, so hand counting is out of the question. Shipping uses a conversion scale - counts out a sample, inputs the count and weighs it - then piles on till the count (based on weight) reads the target amount.

Most of our parts have little variation within a 'run'. The problem parts are running higher variation. (sub-contracted)

Acceptable margin of error.... well, we would rather err on the side of more (would not want a customer to think we were shorting them ever) but not too far, as we really can't afford to give product away. Not a great answer - sorry.:mg:

Statistical Steven
27th May 2009, 03:27 PM
This is how I would approach the problem. I would weigh out 30 individual parts, calculate a mean and standard deviation for the individuals. Then use simulation to simulate 1000 samples for a given sample size (n=50 for example). This will now give you distribution for the means (you can of course just use the the SD of individuals/sqrt(n) to get the estimate of the the variation if simulation is not your thing). Based on the 1000 samples I can set the specifications to control the percent of samples that will be on the low side and the percent on the high side. In other words, you do not want +/- 3S, what you really want is probably -0.5S/+3S to ensure you minimize under fulfillment while not rejecting too many that are overfilled.

I hope that makes sense.

smryan
27th May 2009, 04:46 PM
This is how I would approach the problem. I would weigh out 30 individual parts, calculate a mean and standard deviation for the individuals. Then use simulation to simulate 1000 samples for a given sample size (n=50 for example). This will now give you distribution for the means (you can of course just use the the SD of individuals/sqrt(n) to get the estimate of the the variation if simulation is not your thing). Based on the 1000 samples I can set the specifications to control the percent of samples that will be on the low side and the percent on the high side. In other words, you do not want +/- 3S, what you really want is probably -0.5S/+3S to ensure you minimize under fulfillment while not rejecting too many that are overfilled.

I hope that makes sense.
Nearly.
I had 8 groups of 50 each.
Of those I massed all 50 individuals of 3 groups (the hi, the low and a middle). The SD isn't the same on these 3 data sets - so how could I trust the info I was getting from just a single sample of 30?

How would one simulate 1000 samples from this? I'd like to try it.

At least the -0.5S/+3S makes perfect sense.:biglaugh:Thank you!

Miner
27th May 2009, 08:09 PM
Statistical Steven's approach makes a lot of sense.

To answer your question on the standard deviation, calculate the confidence interval for the standard deviations, then use the upper 95% (or greater) confidence limit for the standard deviation.

You could extend Statistical Steven's suggestion a bit further to calculate a sample size such that the 3 standard deviations of the average is less than the weight of a single part. This may result in a prohibitively large sample size, so you may have to compromise at 2 standard deviations or aim slightly high on count.

Jennifer Kirley
27th May 2009, 11:07 PM
Look at these great answers. There are times when the best I can do is make a post so as to bring the question back to the top of the list, where better expertise can take a crack at it. It appears to be working...:applause:

Statistical Steven
27th May 2009, 11:42 PM
Nearly.
I had 8 groups of 50 each.
Of those I massed all 50 individuals of 3 groups (the hi, the low and a middle). The SD isn't the same on these 3 data sets - so how could I trust the info I was getting from just a single sample of 30?

How would one simulate 1000 samples from this? I'd like to try it.

At least the -0.5S/+3S makes perfect sense.:biglaugh:Thank you!


I use software like SAS, though you can do it in Excel too. You will need to actually do the simulation twice, first to simulate 1000 groups of 50 parts, then another simulation to look at groups of the 8 means from the 1000 means. This is not a trivial problem, as the question you are asking is to set specifications for weights of 400 parts (8 groups of 50) that come from a population of individuals. The means of the 50 parts should be consistent, but as you saw, the variation wihtin the sample of 50 can vary considerable.

smryan
28th May 2009, 09:45 AM
Statistical Steven's approach makes a lot of sense.

To answer your question on the standard deviation, calculate the confidence interval for the standard deviations, then use the upper 95% (or greater) confidence limit for the standard deviation.


OK - lets give this a shot.... using an online tool (as I have zero experience of my own with this :o) I came up with the following table... could someone who understands this better than I please tell me if its anywhere near accurate? THANKS!

Jennifer Kirley
28th May 2009, 12:47 PM
Can anyone tell if this is a sound approach for the statistical problem?

Statistical Steven
28th May 2009, 01:15 PM
OK - lets give this a shot.... using an online tool (as I have zero experience of my own with this :o) I came up with the following table... could someone who understands this better than I please tell me if its anywhere near accurate? THANKS!

Looks like you used a power of 85%, is this true? In general, it looks right, but this is not the question you wanted answered. You are not taking a sample from the lot and weighing it to ensure the correct weight.

smryan
28th May 2009, 01:41 PM
Looks like you used a power of 85%, is this true? In general, it looks right, but this is not the question you wanted answered. You are not taking a sample from the lot and weighing it to ensure the correct weight.

No idea about the power of 85% - pic of tool used attached.

Correct, the weight is not the goal, an accurate count is. But the weight is what is used to determine the count, so if the weight is accurate then the count should be too.... right?:confused:

Miner
28th May 2009, 07:13 PM
No idea about the power of 85% - pic of tool used attached.

Correct, the weight is not the goal, an accurate count is. But the weight is what is used to determine the count, so if the weight is accurate then the count should be too.... right?:confused:
This calculator is specific to surveys. This is not appropriate for your situation.

smryan
29th May 2009, 09:17 AM
This calculator is specific to surveys. This is not appropriate for your situation.

:notme:I was hoping that wouldn't matter. When I say I have no statistical background I mean it very literally. Can someone help me find the approprite tool? I don't even know how to narrow down my search.:(

I use software like SAS, though you can do it in Excel too. You will need to actually do the simulation twice...
I'd like to try - I use Excel - but how? Or is this way beyond the scope of the forums?

Denis
29th May 2009, 10:54 AM
You have quite a bit of variation in the recorded weights.

Would it be possible to work with the supplier of these parts to reduce and understand the causes of the weight variation? I would look at trying to work with the supplier in the first instance.

Are they aware of the critical nature (to you) in this process?

Once you understand the root cause of the variation it might be easier to work on some solutions to eradicate the variation (never completely, but at least improve it).

Then repeat the study, or get the supplier to put some in process checks in place to detect the weight changes etc.

Miner
29th May 2009, 09:56 PM
:notme:I was hoping that wouldn't matter. When I say I have no statistical background I mean it very literally. Can someone help me find the approprite tool? I don't even know how to narrow down my search.:(


I'd like to try - I use Excel - but how? Or is this way beyond the scope of the forums?
Can you attach any actual, not summarized data for us to work with?

smryan
1st June 2009, 11:14 AM
OK - attached is the data I am currently working with. "Results" tab - the yellow & blue section I am pretty sure of. The green section is what I am unsure of. Which (if any) of the 3 would be the most true? (and why?):rolleyes:

The "data" tab has the masses of the 8 bags of 50 pieces. And the 150 pieces I had time to mass before they were shipped.

The parts are all within spec, so telling the sub to "make them better" would be poorly received. :whip: BUT the variance should be considerably less per run (it is here, anyway). So my one thought is to have the sub bag the pieces from each run separately instead of 30,000 or so pieces loose in the box. It would add a small step on both ends, but it could increase accuracy in shipping quantities.

So.... suppose the part variance becomes a much smaller issue. Given the possible non-normal distribution (based on this data) of what variance there still would be, can one statistically speculate that "A sample size ____% of the total count will give you ___% accuracy in your count"?

Miner
1st June 2009, 08:16 PM
I looked at your data. Did you notice that the individual samples as well as the combined data set are bimodal? Furthermore, the two modes are the same on all three samples.

If you can resolve this, it would reduce your total variation significantly and reduce your count problem.

smryan
2nd June 2009, 09:12 AM
I looked at your data. Did you notice that the individual samples as well as the combined data set are bimodal? Furthermore, the two modes are the same on all three samples.

If you can resolve this, it would reduce your total variation significantly and reduce your count problem.

We need a smiley with a lightbulb going off =) :thanks:
No, I hand't noticed the bimodality in the data. My thought is the separate bagging of parts from each "run" would eliminate this - but if it doesn't is there another way to look at it?

Miner
2nd June 2009, 08:25 PM
You did not specify the type of product, so it is difficult to speculate.

Is it possible that the product comes from two different process streams (e.g., two cavity mold/die, two different manufacturing lines/facilities/sources)? If this is a possibility, can the supplier package the two streams separately?

smryan
3rd June 2009, 11:27 AM
You did not specify the type of product, so it is difficult to speculate.

Is it possible that the product comes from two different process streams (e.g., two cavity mold/die, two different manufacturing lines/facilities/sources)? If this is a possibility, can the supplier package the two streams separately?
Capacitors.
I presume their process mimics our pretty closely, which is why I am hoping to convince the powers that be to push for them to bag "runs" separately. They arrive in a box labeled with one date code and one lot number - which SHOULD indicate they are all of the same "run" on the same line. I'm having my doubts.

Jupitor
10th June 2009, 05:21 PM
Are we getting somewhere in this problem of count by weighing?

My client had a similar problem (and still has). They count a sample of 20 or 50 individual parts (depending on the size/weight of the parts) and use a ratio counter to 'count' 1000 nos. or whatever the quantity to be shipped. But occasionally they do get complaints of short supply by a few numbers.

The problem here is that the parts during manufacturing undergo a coating process and the amount of material that each piece picks up can and does vary even though by very small amount. And this contributes to the error. Cumulatively over say 200 lots amounting to about 200,000 pieces there may be a reported short supply of anything upto 50 pieces. Though this is only a very small percentage error and the company supplies free replacements it still constitutes a customer complaint and an embarrassment that nothing can yet be done to solve this problem effectively.

What I would therefore like to know is - do you have a similar source of non-uniformity in your parts

smryan
11th June 2009, 03:48 PM
Are we getting somewhere in this problem of count by weighing?

My client had a similar problem (and still has). They count a sample of 20 or 50 individual parts (depending on the size/weight of the parts) and use a ratio counter to 'count' 1000 nos. or whatever the quantity to be shipped. But occasionally they do get complaints of short supply by a few numbers.

The problem here is that the parts during manufacturing undergo a coating process and the amount of material that each piece picks up can and does vary even though by very small amount. And this contributes to the error. Cumulatively over say 200 lots amounting to about 200,000 pieces there may be a reported short supply of anything upto 50 pieces. Though this is only a very small percentage error and the company supplies free replacements it still constitutes a customer complaint and an embarrassment that nothing can yet be done to solve this problem effectively.

What I would therefore like to know is - do you have a similar source of non-uniformity in your parts

Yes, we too have a coating process that counts for some of the inconsistency. There is no concrete solution. The most I can ask for is that each "lot" is bagged separately, allowing us to take the sample from what should be a more consistent whole. We may then have to weigh out 3 or 4 samples to meet an order, but the count SHOULD be more accurate.

The resolution of the scale will be a % error that you have to "live with".

Do the same customers who complain when short 50/200,000 send back the overages when they get 200,050 instead of the 200,000? Never. They think, "Ah - bonus!". ;) Sigh.

bobdoering
11th June 2009, 04:25 PM
The problem here is that the parts during manufacturing undergo a coating process and the amount of material that each piece picks up can and does vary even though by very small amount. And this contributes to the error. Cumulatively over say 200 lots amounting to about 200,000 pieces there may be a reported short supply of anything up to 50 pieces. Though this is only a very small percentage error and the company supplies free replacements it still constitutes a customer complaint and an embarrassment that nothing can yet be done to solve this problem effectively.

What I would therefore like to know is - do you have a similar source of non-uniformity in your parts

I have had this kind of situation before, and it may require going to a mechanical counter rather than weigh counting. Sure, weigh counting is easy - hard to dispute that - but if there is too much variation between parts it can not be used accurately. The smaller the part, the more little discrepancies adds up in higher pack quantities. Plating is notorious for causing these problems.