Measure Variability of data points

#1
I'm struggling to find the appropriate solution to a project that I'm working on. I am trying measure the variability of lead time on products that we purchase I can't figure out what an appropriate calculation would be. Below are some of the details.

- I'm trying to compare vendors. Some tend to have short lead times (3-5 days) some are long (60-90 days). Whatever measure I use I need to be able to compare these vendors on the same scale.

- The data tends to have a lot of outliers.

My end goal is to be able to evaluate suppliers against each other. Are their lead times all over the place? Or do they tend to be relatively consistent.

Sample A - 32, 6, 9, 16, 3, 30, 34, 36, 33, 35, 42, 25, 38, 10

Sample B - 6, 7, 6, 6, 10, 8, 10, 8, 5, 8, 8, 7, 8, 6, 6, 6

Sample C - 7, 35, 6, 33, 31, 6, 32, 35, 30

Sample D - 28, 28, 20, 34, 16, 26, 27, 28, 15, 21, 23

Sample E - 2, 2, 4, 2, 1, 3, 4, 7, 11, 2, 1, 3, 2, 2, 2, 5

Any suggestions would be greatly appreciated!
 

Mean Joe

TS Contributor
#2
Coefficient of Variation?

Advantages
Whereas the standard deviation of data must always be understood in the context of the mean of the data. Instead, the actual value of the CV is independent of the unit in which the measurement has been taken, so it is a dimensionless number. For comparison between data sets with different units or widely different means, one should use the coefficient of variation instead of the standard deviation.

Disadvantages
When the mean value is close to zero, the coefficient of variation will approach infinity and is hence sensitive to small changes in the mean. This is often the case if the values do not originate from a ratio scale.
 
#3
Thanks Mean Joe. It seems that the CV is heavily influenced by outliers. Do you know of a good way to handle that?

Ex. Sample A - 75, 68, 61, 63,88, 84, 50, 77 - CV = .16929

Sample B - 5, 5, 71, 3, 6 - CV = 1.4732

Sample B is more consistent but because of the outlier they score very poorly.

Thanks
 

merik

New Member
#4
Here are a bunch of ideas:

1) Draw a histogram for each sample. If they are generally normally distributed and the outliers are visually separable (e.g. they all sit more than 2 Standard Deviations away from the mean), you can simply define a cutoff and exclude data beyond that cutoff (e.g. mean ± 2*SD). This is a simplified version of Chauvenet's criterion

2) You can also have a look at Grubbs' test for outliers. This also assumes that your data is normally distributed.

3) If your data is not normally distributed, there are also options. If you can get rid of outliers and the only concerns is regarding the statistical test that you are using for comparing the groups, then Central limit theorem is your friend; have a large number of data points in each group and you can assume noramlity. If you haven't gotten rid of outliers because of the weird distribution, you might look into the option converting the scales. Sometimes using log(x) instead of (x) gives you more felxibility.
 
#5
@mschweinzger.

Is this real data? Or have you or a teacher or someone made it up, as an example?

Are there any other explanatory factors, for example if some units were deliveries from a central inventory from far away? Like when the data goes from around 7 to suddenly 35.

How did you make the selection to choose units? Randomisation? I guess the vendors delivers thousands of units.
 
#6
@GretaGarbo

This is real data. There aren't any visible explanatory factors. The vendors delivery lead times simply vary. I'm trying to measure and evaluate vendors on how consistent (or inconsistent) the lead time is as this has significant implications on stock we keep on hand and the lead times we us when placing orders. I've attached a small sample of the data set that I'm working with. I'm trying to evaluate 12 months worth of deliveries. View attachment 2452
 
#7
What does this expression mean (from your first post):
“Are their lead times all over the place? “
From the xl-file: what is “rec_amout”? Is that not an explanatory variable?

What have you tried so far?
 
#8
What I meant by this was - Are their lead times highly variable?

rec_amount is the number of pieces that we received in the shipment (each row represents a receipt).

Currently, I take the median for each item number.

I then determine the acceptable lead time range for each item number based on the median. For items with a median lead time less than 20 days I've defined the acceptable range as +/- 2 days. For items with a median lead time greater than 20 days I've defined the acceptable range as +/- 10%.

For each receipt I then determine if the lead time fell inside or outside of the acceptable range. I then take the sum of all pieces received outside of the acceptable range / all pieces received total to determine the % of receipts out of range.

Other than that I've tried coefficient of variation but the significant outliers skew the results to heavily.