Question

A manufacturer of computer memory chips produces chips in lots of 1000. If nothing has gone wrong in the manufacturing process, at most 7 chips each lot would be defective, but if something does go wrong, there could be far more defective chips. If something goes wrong with a given lot, they discard the entire lot. It would be prohibitively expensive to test every chip in every lot, so they want to make the decision of whether or not to discard a given lot on the basis of the number of defective chips in a simple random sample. They decide they can afford to test 100 chips from each lot. You are hired as their statistician.

There is a tradeoff between the cost of eroneously discarding a good lot, and the cost of warranty claims if a bad lot is sold. The next few problems refer to this scenario.

(Continues previous problem.) To have a chance of at most 3% of discarding a lot given that the lot is good, the test should reject if the number of defectives in the sample of size 100 is greater than or equal to (Q16)

The smallest number of defectives in the lot for which this test has at least a 97% chance of correctly detecting that the lot was bad is (Q19)

Answer 1

Sol:

Given that N=1000, n=100, k=5 and also given that sample size has a hypergeometric distribution

Let x be the number of defectives in the sample,then the pdf of hypergeometric distribution is

$P(X=x)=\frac{\binom{k}{x}\binom{N-k}{n-x}}{\binom{N}{n}}; x=0,1,2,..n$

16).

By using excel, we calculate the values

$P(X=0)=\frac{\binom{5}{0}\binom{995}{100}}{\binom{1000}{100}}=0.5898$

$P(X=1)=\frac{\binom{5}{1}\binom{995}{99}}{\binom{1000}{100}}=0.3291$

$P(X=2)=\frac{\binom{5}{2}\binom{995}{98}}{\binom{1000}{100}}=0.0726$

$P(X\geq 3)=1-P(X=0)-P(X=1)-P(X=2)$

                     $=1-0.5898-0.3291-0.0726=0.0093$

Since 0.0093 < 0.01, the test should reject if the number of defectives in the sample of size 100 is greater
than or equal to 3.

19).

The smallest number of defectives in the lot for which this test has at least a 97% chance of correctly detecting that the lot was bad is

$P(X=0)&plus;P(X=1)&plus;P(X=2)$

0.5898+0.3291+0.0726=0.9915

So, the smallest number of dedectives is 2

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