Matt,

I've done an analysis below that I hope will be helpful.

Please don't hesitate to reply to the thread or to email me through the forum if you have any questions or comments.

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30 carts per day

25 orders per full cart

750 orders per stocker per day

(assuming 2.5 items per order)

1875 items per stocker per day

12,000 items per day

12,000 / 1875 = 6.4

so about 6 or 7 stockers in the warehouse.

The basic model is that when a stocker fills an order that is a trial which has some probability of success (we'll assume that this probability is either .95 or .99)

So each stocker fills about 750 orders in a day, so one could in principle have a sample size of 750 where the probability of success for each sample is (let's say .99).

The basic test to use is called in Devore's book Probability and Statistics for Engineering and The Sciences (5th edition) "tests concerning a population proportion" and is on page 335.

The formula is:

H_0 : p = p_0

H_a : p < p_0 (in this case we would probably want a one-tailed test)

(p_hat - p_0) / sqrt( (p_0) * (1 - p_0) / n )

p_hat -- estimated probability from data

p_0 -- probability under the null hypothesis

the constraints for this test are that (p_0 * n) >= 10 and

(1 - p_0)*n >= 10

(10 is what Devore chose, in another book it is said that other Statistics texts have a lesser constraint that (p_0 * n) >= 5 and (1 - p_0)*n >= 5 )

With a null hypothesis of p=.99 the more stringent constraint of

(1-p_0)*n >= 10 is not met with a sample size of 750:

750*.01 = 7.5

However, it is met with a null hypothesis of p=.95 :

750*.05 = 37.5

in fact, if one wanted only to get, say 50% about the limiting value of 10:

15 / .05 = 300

so a sample size of 300 would be sufficient to meet the constraints of the test (300*.05 = 15 >= 10) , or a sample size of 200 would just meet the published constraints for the test. (200*.05 = 10)

By increasing the sample size it is also possible to check for a higher accuracy (the 99% accuracy) (see note at the end)*

15 / .01 = 1500

10 / .01 = 1000

so similarly 1500 samples or 1000 samples would be required to check for the higher accuracy.

It is also possible under the model to compute the probability of a type-II error. In this case that would mean that the picking accuracy is some specified value lower than what is sought for, but the test does not find it.

David

*This might mean combining data from different stockers, and goes to a related question which is whether different stockers have different accuracy rates, and if they differ by how much do they differ, or could also mean combining data from different days.

p.s. Here are some other resources that might be helpful:

http://en.wikipedia.org/wiki/Hypothesis_testing#Common_test_statistics
http://www.stat.berkeley.edu/~stark/SticiGui/Text/zTest.htm#z_for_percentage